Class 12

MP Board Class 12th Maths Important Questions Chapter 8 समाकलनों के अनुप्रयोग

समाकलनों के अनुप्रयोग Important Questions

समाकलनों के अनुप्रयोग वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

(a) π
(b) \(\frac { \pi }{ 2 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -\pi }{ 4 } \)
उत्तर:
(c) \(\frac { \pi }{ 4 } \)

प्रश्न 2.
दीर्घवृत्त \(\frac { x^{ 2 } }{ y^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1 के प्रथम चतुर्थांश का क्षेत्रफल होगा –
(a) πab
(b) \(\frac{1}{2}\) πab
(c) \(\frac{1}{4}\) πab
(d) \(\frac{1}{8}\) πab
उत्तर:
(c) \(\frac{1}{4}\) πab

प्रश्न 3.
\(\int_{2}^{4} x^{3} d x\) का मान है –
(a) 60
(b) 50
(c) 70
(d) 256.
उत्तर:
(a) 60

प्रश्न 4.
\(\int_{0}^{2 a} f(x) d x=0\) होगा यदि –
(a) f (2a – x) = f (x)
(b) f (2a – x) = – f (x)
(c) f (x) सम फलन हो
(d) f (x) विषम फलन हो
उत्तर:
(b) f (2a – x) = – f (x)

प्रश्न 5.
\(\int_{0}^{2 \pi}|\sin x| d x\) बराबर है –
(a) 2
(b) \(\sqrt{3}\)
(c) 4
(d) 0.
उत्तर:
(c) 4

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

उत्तर:

1. 0
2. 0
3. 17
4. 0
5. \(\frac { \pi }{ 4 } \)
6. πa²

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

उत्तर:

1. सत्य
2. सत्य
3. असत्य
4. असत्य
5. सत्य
6. सत्य

प्रश्न 4.
सही जोड़ी बनाइए –

उत्तर:

1. (b)
2. (a)
3. (d)
4. (c)
5. (f)
6. (e)

प्रश्न 5.
एक शब्द/वाक्य में उत्तर दीजिए –

उत्तर:

1. 1
2. \(\frac { -\pi }{ 2 } \) log 2
3. 2
4. \(\frac { \pi }{ 2 } \)
5. -4
6. \(\frac { \pi }{ 12 } \)

समाकलनों के अनुप्रयोग दीर्घ उत्तरीय प्रश्न – II

प्रश्न 1.
दो वक्रों x² = 8y तथा y² = 8x के बीच घिरे क्षेत्र का क्षेत्रफल समाकल विधि से ज्ञात कीजिए।
हल:
दिये हुए परवलयों के समीकरण हैं:
x² = 8y ………… (1)
तथा y² = 8x ……………….. (2)
समी. (1) और (2) को हल करने पर,
( \(\frac { x^{ 2 } }{ 8 } \) )² = 8x
⇒ \(\frac { x^{ 4 } }{ 64 } \) = 8x
⇒ x⁴ = 64 × 8x = 8 × 8 × 8 × x
⇒ x³ = (8)³ या x = 8
समी. (1) से,

(8)² = 8y ⇒ y = 8
A के निर्देशांक (8, 8) हैं।
परवलय बिन्दु तथा A पर मिलते हैं।
∴ उभयनिष्ठ क्षेत्रफल = क्षेत्रफल OBALO – क्षेत्रफल OCALO

प्रश्न 2.
चक्र y = cosx, X – अक्ष और x = 0 तथा x = 2π से घिरे क्षेत्र का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
y = f (x) = cos x
जब x ∈ [0, \(\frac { \pi }{ 2 } \) ], cos x ≥ 0;
जब x ∈ [ \(\frac { \pi }{ 2 } \), \(\frac { 3\pi }{ 2 } \) ], cos x ≤ 0
जब x ∈ [ \(\frac { 3\pi }{ 2 } \), 2π], cos x ≥ 0

अभीष्ट क्षेत्रफल = OAB का क्षेत्रफल + BCD का क्षेत्रफल + DEF का क्षेत्रफल

प्रश्न 3.
वक्र y² = 4x तथा रेखा y = 2x के बीच का क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि से)
हल:
दिया गया वक्र और रेखा है –
y² = 4x और y = 2x
y² = 4x में y = 2x रखने पर,

(2x)² = 4x
⇒ 4x² = 4x
⇒ x² – x = 0
⇒ x(x – 1) = 0
∴ x = 0 या x = 1
तब y = 2x = 0 या y = 2x = 2 × 1 = 2
अत: वक्र y² = 4x तथा रेखा y = 2x के प्रतिच्छेद बिन्दु O(0, 0) एवं (1, 2) होंगे।
अत: अभीष्ट छायांकित क्षेत्र
= क्षेत्र (OPAMO) – क्षेत्र (OQAMO)

जहाँ y_c = वक्र y² = 4x तथा y_s = रेखा y = 2x के लिए,

= \(\frac{4}{3}\) [1 – 0] – \(\frac{2}{2}\) [1 – 0]
= \(\frac{4}{3}\) – 1 = \(\frac{1}{3}\) वर्ग इकाई।

प्रश्न 4.
समाकलन विधि द्वारा वक्रों x² = 4y तथा x = 4y – 2 के अंतर्गत क्षेत्रफल ज्ञात कीजिए।
हल:
प्रश्न क्रमांक 3 की भाँति हल करें।

प्रश्न 5.
समाकलन विधि से रेखाओं |x| + |y| = a से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए।
हल:
दी गयी रेखाओं |x| + |y| = a द्वारा निरूपित रेखायें निम्न होंगी –
x + y = a ……………… (1)
– x – y = a ………………. (2)
x – y = a …………….. (3)
– x + y = a …………….. (4)
⇒ \(\frac{x}{a}\) + \(\frac{y}{a}\) = 1, \(\frac{x}{-a}\) + \(\frac{y}{-a}\) = 1
⇒ \(\frac{x}{a}\) + \(\frac{y}{-a}\) = 1, \(\frac{x}{-a}\) + \(\frac{y}{a}\) = 1

उक्त रेखाओं के ग्राफ से स्पष्ट है, कि उक्त रेखाएँ क्रमशः PQ, RS, PS और QR रेखाओं द्वारा निरूपित हैं।
अतः उक्त रेखाओं द्वारा घिरा क्षेत्रफल
= 4 × ∆OPQ का क्षेत्रफल

= 4 × \(\frac { a^{ 2 } }{ 2 } \) = 2a² वर्ग इकाई।

प्रश्न 6.
वृत्त x² + y² का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
दिया है:
वृत्त x² + y² = a²
⇒ y² = a² – x²
⇒ y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)
वृत्त X – अक्ष तथा Y – अक्ष के सापेक्ष सममित आकृति है अतः
वृत्त का क्षेत्रफल = 4 × ∆ABC का क्षेत्रफल

प्रश्न 7.
वृत्त x² + y² = 25 का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
प्रश्न क्रमांक 6 की भाँति हल करें। [संकेत – a = 5 रखें।]

प्रश्न 8.
वक्रों y² = 4ax तथा x² = 4ay, a > 0 के अन्तर्गत क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि द्वारा)
हल:
दिये हुए परवलयों के समीकरण हैं:
y² = 4ax ………….. (1)
तथा x² = 4ay ……………. (2)
समी. (1) और (2) को हल करने पर,

\(\frac { x^{ 4 } }{ 16a^{ 2 } } \) = 4ax
⇒ x⁴ – 64a³x = 0
⇒ x(x³ – 64a³) = 0
अतः या तो x = 0 अथवा x³ – 64a³ = 0
जब x³ – 64a³ = 0
जब x³ – 64a³ = 0 तो x = 4a
जब x = 0 तो y = 0
और जब x = 4a तो y = 4a
अत: दोनों परवलय बिन्दुओं (0, 0), (4a, 4a) पर मिलते हैं।
∴ उभयनिष्ठ क्षेत्रफल = क्षेत्रफल OBALO – क्षेत्रफल OCALO

प्रश्न 9.
वक्रों y² = 4x और x² = 4y के बीच का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
दिए हुए वक्रों के समीकरण हैं:
y² = 4x ……………… (1)
xc = 4y ………………. (2)
समी. (2) से y का मान समी. (1) में रखने पर,

⇒ \(\frac { x^{ 4 } }{ 16 } \) = 4x
⇒ x⁴ – 64x = 0
⇒ x(x³ – 64) = 0
⇒ x = 0 या x³ = 64
⇒ x = 0 या x = 4
∴ बिन्दु 0 पर x = 0 तथा बिन्दु M पर x = 4
अतः अभीष्ट क्षेत्रफल = क्षेत्र. OAPBO
= क्षेत्र. OAPMO – क्षेत्र. OBPMO

प्रश्न 10.
वक्र y² = x और x² = y के बीच का क्षेत्रफल समाकलन विधि से ज्ञात कीजिए।
हल:
प्रश्न क्रमांक 9 की भाँति हल करें।

प्रश्न 11.
दीर्घवृत्त \(\frac { x^{ 2 } }{ a^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1 से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि द्वारा)
हल:
दिया हुआ वक्र X – अक्ष और Y – अक्ष के सापेक्ष सममित है। अतएव सम्पूर्ण क्षेत्रफल क्षेत्र OAB का 4 गुना होगा।
यहाँ y = \(\frac{b}{a}\) \(\sqrt { a^{ 2 }-x^{ 2 } } \)
वक्र पर बिन्दु P लिया। तब पट्टी PM का क्षेत्रफल ydx होगा।
x का मान O पर शून्य तथा A पर x = a है।

प्रश्न 12.
परवलय y² = 4ax तथा इसकी नाभिलम्ब जीवा से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (समाकलन विधि द्वारा)
हल:
परवलय का समीकरण है:
y² = 4ax
⇒ y = ± 2\(\sqrt{ax}\)
LSL’ नाभिलम्ब है तथा S(a, 0) और 0(0, 0) है।
अतः अभीष्ट क्षेत्रफल = 2 × क्षेत्रफल OSL

प्रश्न 13.
परवलय y² = 4ax और रेखा y = mx के बीच घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए।
हल:
परवलय का समी. y² = 4ax ……………. (1)
रेखा का समी. y = mx …………….. (2)
मूलबिन्दु O(0, 0) है। P परवलय और रेखा का प्रतिच्छेद बिन्दु है। समी. (1) और (2) को हल करने पर,
y² = 4ax
⇒ (mx)² = 4ax
⇒ m²x² – 4ax = 0
⇒ x(m²x – 4a) = 0
∴ x = 0, x = \(\frac { 4a }{ m^{ 2 } } \)

प्रश्न 14.
प्रथम चतुर्थांश में x² = 4y, y = 2, y = 4 एवं Y – अक्ष से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (NCERT)
हल:
दिए गए वक्र का समीकरण है:
x² = 4y
⇒ x = 2\(\sqrt{y}\)
∴ अभीष्ट क्षेत्रफल = ABCD का क्षेत्रफल

प्रश्न 15.
छेदक रेखा x = \(\frac { a }{ \sqrt { 2 } } \) द्वारा वृत्त x² + y² = a² के छोटे भाग का क्षेत्रफल ज्ञात कीजिए। (NCERT)
हल:
वृत्त का समीकरण है:
x² + y² = a²
y² = a² – x²
y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)
रेखा का समीकरण है –
x = \(\frac { a }{ \sqrt { 2 } } \)
समी. (1) और (2) को हल करने पर,

( \(\frac { a }{ \sqrt { 2 } } \) )² + y² = a²
y² = a² – \(\frac { a^{ 2 } }{ 2 } \) = \(\frac { a^{ 2 } }{ 2 } \)
y = \(\frac { a }{ \sqrt { 2 } } \)
वृत्त और रेखा का प्रतिच्छेद बिन्दु A( \(\frac { a }{ \sqrt { 2 } } \), \(\frac { a }{ \sqrt { 2 } } \) ) है।
D के निर्देशांक (a, 0) होंगे।
∴ अभीष्ट क्षेत्रफल = ABD का क्षेत्रफल = 2 × ACD का क्षेत्रफल

प्रश्न 16.
यदि वक्र x = y² एवं रेखा x = 4 से घिरा हुआ क्षेत्रफल x = a द्वारा दो बराबर भागों में विभाजित होता है, तो a का मान ज्ञात कीजिए। (NCERT)
हल:
परवलय का समीकरण है:
x = y²
⇒ y = \(\sqrt{x}\) ………… (1)
परवलय तथा रेखा x = 4 से घिरे क्षेत्रफल को रेखा x = a दो बराबर भागों में विभाजित करती है।
क्षेत्रफल OEC = क्षेत्रफल EFCB

प्रश्न 17.
प्रथम चतुर्थांश में वृत्त x² + y² = 4 एवं रेखाओं x = 0, x = 2 से घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (NCERT)
हल:
वृत्त का समीकरण है:
x² + y² = 4
⇒ y² = 4 – x²
⇒ y = \(\sqrt { (2)^{ 2 }-x^{ 2 } } \)

प्रश्न 18.
प्रथम चतुर्थांश में वृत्त x² + y² = 32 रेखा y = x एवं X – अक्ष द्वारा घिरे क्षेत्र का क्षेत्रफल ज्ञात कीजिए। (NCERT; CBSE 2018)
हल:
दिए गए वृत्त का समीकरण है:
x² + y² = 32
y² = 32 – x²
वृत्त का समीकरण y = \(\sqrt { 32-x^{ 2 } } \)
रेखा का समीकरण है:
समी. (1) और (2) को हल करने पर,
x² + x² = 32
2x² = 32
x² = 16
x = 4

x का मान समी. (2) में रखने पर,
y = 4
रेखा और वृत्त का प्रतिच्छेद बिन्दु O(0,0) तथा A(4, 4) है। B के निर्देशांक (4, 0) तथा C के निर्देशांक (4\(\sqrt{2}\),0) हैं।
अभीष्ट क्षेत्रफल = क्षेत्रफल OACBO
= क्षेत्रफल OAB + क्षेत्रफल ABC

प्रश्न 19.
उस त्रिभुज का क्षेत्रफल समाकलन द्वारा ज्ञात कीजिए जिसकी भुजाएँ y = 2x + 1, y = 3x + 1 तथा x = 4 हैं। (NCERT)
हल:
माना कि ∆ABC की भुजाओं AB, AC तथा BC के समीकरण क्रमशः हैं –
y = 2x + 1 ……….. (1)
y = 3x + 1 ……….. (2)
और x = 4 ……….. (3)
समी. (1) और (2) को हल करने पर, बिन्दु A (0, 1)
समी. (1) और (3) को हल करने पर, बिन्दु B (4, 9) और
समी. (2) और (3) को हल करने पर, बिन्दु C (4, 13) हैं।
अत: अभीष्ट क्षेत्रफल

MP Board Class 12 Maths Important Questions

MP Board Class 12th Hindi Swati Solutions पद्य Chapter 6 शौर्य और देशप्रेम

शौर्य और देशप्रेम अभ्यास

शौर्य और देशप्रेम अति लघु उत्तरीय प्रश्न

प्रश्न 1.
पं. बालकृष्ण शर्मा ‘नवीन’ ने कविता में किसका आह्वान किया है? (2009)
उत्तर:
पं. बालकृष्ण शर्मा ‘नवीन’ ने अपनी कविता में अपने देश के युवकों का आह्वान किया है।

प्रश्न 2.
“क्षणिक आतंक” से क्या तात्पर्य है? (2015)
उत्तर:
क्षणिक आतंक से तात्पर्य तनिक देर के लिए आए हुए भय से है। वह भय (बाधा) एक क्षण में ही नष्ट हो जाएगा, वह तुम्हारा कुछ भी न बिगाड़ सकेगा।

प्रश्न 3.
“नींव के पत्थर” से कवि का क्या तात्पर्य है? (2009, 14)
उत्तर:
“नींव के पत्थर” से कवि का तात्पर्य उन वीर पुरुषों से है जिन्होंने देश की स्वाधीनता के लिए अपने प्राणों का बलिदान किया है।

प्रश्न 4.
किस धरोहर को हाथ से न जाने देने की बात कही गई है?
उत्तर:
देश की स्वाधीनता रूपी धरोहर को हाथ से न जाने देने की बात कही गई है।

प्रश्न 5.
“देश की समृद्धि की प्रबल आशा” से क्या तात्पर्य है? (2009)
उत्तर:
“देश की समृद्धि की प्रबल आशा” से तात्पर्य है भारत की उत्तरोत्तर बढ़ती हुई प्रगति की प्रबल अभिलाषा जो इस देश के मौजवानों से अपेक्षित है।

शौर्य और देशप्रेम लघु उत्तरीय प्रश्न

प्रश्न 1.
कवि ने भारतीय स्वर्णिम इतिहास के किन अंशों का उल्लेख कर युवाओं को प्रेरित किया है
उत्तर:
भारत के पूर्वजों का इतिहास बड़ा गौरवपूर्ण और वीरतापूर्ण रहा है। हमेशा उन्होंने विजय श्री का ही वरण किया है। फूंक से पर्वत उड़ाने की कहावत भारतीयों पर ही सटीक बैठती है। उनके क्रोध से पर्वत भी उन्हें मार्ग दे देते थे। तुम उन्हीं वीर भारतीयों की सन्तान हो, तुम्हारे अन्दर भी वही रक्त, वही वीरता और वही देशप्रेम है। तुम भी उन गुणों से प्रेरणा लेकर अपने मार्ग की बाधाओं को हटाते हुए प्रगति के पथ पर निरन्तर अग्रसर होते रहो।

प्रश्न 2.
भारतीय युवकों का क्रोध किस प्रकार का बताया गया है?
उत्तर:
भारतीय युवाओं का क्रोध इतना भयंकर है कि इसे देखकर भूकम्प आ जाएगा और पृथ्वी और आकाश काँप उठेंगे। आकाश में आग की लपटें उठेगी जो शत्रुओं को जलाकर भस्म कर देंगी। इसी क्रोध को देखकर समाज में फैली हुई बुरी प्रथाएँ भी समाप्त हो जाएँगी। भारतीय युवाओं का क्रोध पृथ्वी को भी इधर-उधर डोलने पर मजबूर कर सकता है। सीता स्वयंवर के समय जनक जी की अनुचित वाणी सुनकर लक्ष्मण जी ने जो गर्जना की थी तो आकाश और पृथ्वी दोनों काँप गए थे। इसी संदर्भ में यह चौपाई देखिए-
“लखन सकोप बचन जे बोले,डगमगानि महि दिग्गज डोले ॥”

प्रश्न 3.
“तुम चलाते हो सदा चिर चेतना के तीर” का आशय स्पष्ट कीजिए। (2012)
उत्तर:
देश का युवा वर्ग दिशा और काल रूपी धनुष को धारण करने वाला धनुर्धर है जो चिर चेतना (उद्बोधन) रूपी तीर चलाकर देश की रक्षा करने वाले पहरुओं को सावधान करता है। युवा अपनी वीरता से देश की रक्षा तो करता ही है साथ-ही-साथ में अन्य लोगों में देश की स्वाधीनता के प्रति जागृति प्रदान करता है। देश के नौजवानों पर ही देश की प्रगति की सुनहरी आशा टिकी हुई है।

प्रश्न 4.
वतन के लाड़लों से क्या अपेक्षा की गई है? (2016)
उत्तर:
वतन के लाड़लों से यह अपेक्षा की गई है कि वे इस देश की स्वाधीनता पर आँच आने न दें अर्थात् शहीदों के बलिदानों से प्राप्त स्वाधीनता की रक्षा करें। देश के युवा ही देश के शृंगार हैं और देश के उत्थान की अभिलाषा की साकार मूर्ति हैं। देश का सुनहरा भविष्य उसके लाड़ले युवाओं पर ही निर्भर है। वे ही अपने देश की स्वतन्त्रता की रक्षा करते हुए उसके अभ्युदय के द्वार खोल देंगे। देश का युवा ही इस धरोहर (स्वाधीनता) की रक्षा करने में समर्थ है। वह कभी भी देश के शत्रुओं को उनके इरादों में सफल नहीं होने देगा।

प्रश्न 5.
“दाँव पर गौरव लगे तो शीश दे देना विहँस कर”-पंक्ति से कवि ने क्या भाव व्यक्त किया है? (2011)
उत्तर:
देश की संस्कृति और देश की रक्षा का भार हमारे ऊपर ही है। इसलिए हमें इस हेतु सदैव तत्पर और सजग रहना चाहिए। कवि देश के युवाओं का आह्वान करता है कि यदि देश के गौरव (सम्मान) पर आँच आए तो तुम अपने प्राणों का उत्सर्ग करने से भी न चूकना देश के युवाओं के उत्सर्ग पर ही देश की सम्पूर्ण समृद्धि का भवन खड़ा है। भारत देश का युवा अपने प्राण रहते हुए कभी भी अपने देश की स्वतन्त्रता पर आँच न आने देगा। धरोहर रूप में प्राप्त इस स्वाधीनता को इस देश के निवासी अपने प्राणों से भी बहुमूल्य मानते हैं। वे अपने प्राण रहते कभी भी देश के गौरव को क्षति नहीं पहुँचने देंगे।

प्रश्न 6.
“समय की शिला पर चिह्न छोड़ने” यह पंक्ति किस संदर्भ में कही गई है?
उत्तर:
देश के युवा जब भी अपने वीरतापूर्ण कार्यों से देश की प्रगति को आगे बढ़ाते हैं या देश की स्वाधीनता की रक्षा करते हैं तो इतिहास उनके शौर्य की गाथा को स्वर्णाक्षरों में अंकित कर लेता है। देश के वीरों के ये चिह्न ऐसे ही नहीं मिटते जैसे पत्थर की शिला पर खुदे हुए चिह्न। वीर पुरुषों के उत्सर्ग और वीरतायुक्त कार्यों को इतिहास स्वर्णाक्षरों में लिख लेता है जिसे समय का परिवर्तन मिटा नहीं पाता। ऐसे अनेक भारतीय वीर हुए हैं जिन्होंने समय की शिला पर चिह्न छोड़े हैं और वे इतिहास में अमर हो गए हैं।

शौर्य और देशप्रेम दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
कवि नवीन ने भारतीय युवाओं को “काल के भी काल” क्यों कहा है? (2013)
उत्तर:
भावना और कर्तव्य के संपुट में ही देश प्रेम पलता है। देश की रक्षा और उसकी प्रतिष्ठा के लिए जिस पराक्रम की आवश्यकता होती है,वही पराक्रम देश प्रेम के अन्तर्गत शौर्य रूप में प्रकट होता है। यहाँ के वीर काल के भी काल हैं। अर्थात् काल से भी ये निडर रहते हैं और आवश्यकता पड़ने पर उससे भिड़ जाते हैं। भारतीय वीर मृत्यु से अभीत रहते हैं। वे प्राणों को हथेली पर रखकर देश की रक्षा के लिए रणक्षेत्र में प्रवेश करते हैं और विजय प्राप्त करके ही लौटते हैं। उन वीरों से मृत्यु भी भयभीत रहती है। मृत्यु उन्हीं लोगों को भय से त्रस्त करती है जो कायर पुरुष हैं। भारत के युवा काल की परवाह किए बिना देश की रक्षा के लिए अपने प्राणों की बाजी लगा देते हैं। अतः कवि ने उन्हें काल के भी काल की उपमा दी है।

प्रश्न 2.
“औ, पधारेगा सृजन कर अग्नि में सुस्नान” के द्वारा कवि के किस दृष्टिकोण का पता चलता है?
उत्तर:
उक्त पंक्ति से कवि का तात्पर्य यह है कि युद्ध की अग्नि में स्नान करके ही सृजन पदार्पण करता है या यों कहें कि युद्ध के बाद ही शान्ति स्थापित होती है। देश की स्वाधीनता की रक्षा के लिए युवकों को युद्ध रूपी अग्नि प्रज्ज्वलित करनी पड़ेगी तब उसमें से गुजरकर स्वाधीनता प्राप्त होगी। देश की स्वतन्त्रता में ही देश का निर्माण निहित है। देश के युवाओं का कवि ने आह्वान किया है और उन्हें युद्ध करके भी देश की रक्षा करनी है। देश की रक्षा के लिए वीर युवक अपने प्राणों की चिन्ता नहीं करते। जो जाति अपने प्राणों को हथेली पर रखकर चलती है,वही जाति स्वतन्त्रता को भोगने की अधिकारिणी है। जवानी ने अपना शीश देकर ही स्वतन्त्रता प्राप्त की है।

प्रश्न 3.
कवि सरल ने युवाओं का आह्वान किस हेतु किया है? (2017)
उत्तर-कवि श्रीकृष्ण सरल ने देश के युवाओं का आह्वान देश की स्वतन्त्रता की रक्षा के लिए किया है। देश का युवा ही अपने अपूर्व बलिदान के द्वारा देश की स्वाधीनता की रक्षा कर सकता है। यह स्वतन्त्रता सस्ते में नहीं प्राप्त हुई है,बल्कि असंख्य वीरों के बलिदान के बाद ही इसकी प्राप्ति हुई है। इस स्वतन्त्रता रूपी महल को स्थायित्व प्रदान करने के लिए भारत के अनेक वीर युवा नींव के पत्थर बने हैं। उन्हें हमें स्मरण रखना है। भारत के युवाओं को देशद्रोहियों से अपने देश को बचाना है। जो लहू तुम्हें विरासत में मिला है वह खून तुमसे कह रहा है कि तुम वीर पुत्र हो और सिंह की खेती को स्यारों को खाने मत देना। अर्थात् भारत की पवित्र भूमि को कहीं कायर देशद्रोही कोई नुकसान न पहुँचा दें।

प्रश्न 4.
“सिंह” और “स्यार” किसे और किस संदर्भ में कहा है?
उत्तर:
“सिंह” का तात्पर्य भारत के वीर पुरुषों से है और “स्यार” कायर एवं देश के गद्दार लोगों से है। भारत वीरों का देश है जो सदा से वीर पुत्रों का निवास स्थान रहा है। कुछ लोग ऐसे होते हैं जो कायरता से बद्ध होकर चोरी छुपे देशद्रोह का कार्य करते हैं। उन लोगों को कवि ने बड़ा उपयुक्त स्यार’ नाम दिया है। इनमें कुछ करने का साहस तो होता नहीं, लेकिन सिंह की खेती चरना चाहते हैं। यह कार्य उन्हें रास तो नहीं आता। कभी-कभी जान के लाले भी पड़ जाते हैं। अत: देश के युवा वीरों को ध्यान रखना है कि देश के अन्दर रहने वाले और देश के बाहर रहने वाले शत्रु अपने देश को नुकसान न पहुँचाने पाएँ।

प्रश्न 5.
निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए
(i) काल का तब धनुष ………. सृजन-नाश-हिलोर।
(ii) वह निपट आतंक ……….. अभयता के स्रोत।
(iii) तुम न समझो ……….. पीकर ही खिली है।
(iv) जो विरासत में मिला ………. खाने न देना।
उत्तर:
(i) संदर्भ :
प्रस्तुत पंक्तियाँ ‘अरे तुम हो काल के भी काल’ नामक कविता से उद्धृत की गई हैं। इसके रचयिता ओजस्वी कवि बालकृष्ण शर्मा ‘नवीन’ हैं।

प्रसंग :
इस कविता में कवि ने यह स्पष्ट किया है कि ध्वंस से ही सृजन की संभावना प्रकट होती है। कवि ने यहाँ व्यक्ति को उसके साहस और दृढ़ निश्चय से परिचित कराते हुए स्पष्ट किया है कि उसका लक्ष्य विजय प्राप्त करना ही है।

व्याख्या :
कवि देश के गौरवशाली निवासियों को प्रोत्साहित करते हुए कहता है कि तुम्हारा काल भी कुछ नहीं बिगाड़ सकता,क्योंकि तुम काल के भी महाकाल और विकराल हो। काल का तुम्हारा धनुष है और दिशाओं की प्रत्यंचा है। जब उस धनुष की प्रत्यंचा में सिहरन होगी तो नाश होगा और फिर सृजन की हिलोर उठती हुई दिखाई देगी अर्थात् तुम्हारे युद्ध करने से ही पुनः सृजन का चक्र चलेगा। तुम दिशा और काल रूपी धनुष को धारण करने वाले धनुर्धर हो। तुम सदा चेतना के तीर चलाकर सुप्त वीरों को जाग्रत करते हो और स्वदेश-रक्षा में तत्पर रहते हो।

(ii) संदर्भ :
प्रस्तुत पंक्तियाँ ‘अरे तुम हो काल के भी काल’ नामक कविता से उद्धृत की गई हैं। इसके रचयिता ओजस्वी कवि बालकृष्ण शर्मा ‘नवीन’ हैं।

प्रसंग :
इस कविता में कवि ने यह स्पष्ट किया है कि ध्वंस से ही सृजन की संभावना प्रकट होती है। कवि ने यहाँ व्यक्ति को उसके साहस और दृढ़ निश्चय से परिचित कराते हुए स्पष्ट किया है कि उसका लक्ष्य विजय प्राप्त करना ही है।

व्याख्या :
क्षणिक आतंक तुम्हारा कुछ भी नहीं बिगाड़ सकता। कोई भी आतंक या शत्रु तुम्हें नष्ट नहीं कर सकता। तुम निर्दोष हो और अपने देश की रक्षा के लिए सदा तत्पर हो। तुम्हारे सामने जो भयावना आतंक है वह क्षणिक है और तुम चिरन्तन अभयता के स्रोत हो। अर्थात् तुम्हारे अन्दर अभयता भरी हुई है जिसके सामने आतंक भी भययुक्त परिलक्षित होता है। तुम कभी भी यह मत सोचो कि तुम कभी नष्ट हो जाओगे। तुम अनश्वर (अमर) हो और तुम्हारा भाग्य तुम्हारी कर्मठता के कारण स्पष्ट तौर पर शुभ दिखाई दे रहा है।

(iii) संदर्भ :
प्रस्तुत पंक्तियाँ ‘राष्ट्र के शृंगार’ नामक कविता से उद्धृत की गई हैं। इसके रचयिता ओजस्वी कवि श्रीकृष्ण सरल हैं।

प्रसंग :
इन पंक्तियों में कवि ने देश के वीर नागरिकों और बलशाली युवाओं का आह्वान किया है कि वे अपने देश की रक्षा करें। अपने देश की स्वतन्त्रता पर वे आँच न आने दें।

व्याख्या :
कवि भारत के देशवासियों को सम्बोधित करते हुए कहता है कि हे मेरे राष्ट्र के श्रृंगार और उसकी अभिलाषाओं को पूर्ण करने वाले वीरो ! तुम अपने देश की स्वतन्त्रता पर आँच न आने देना। जिन शहीदों ने अपना बलिदान देकर इस स्वाधीनता रूपी बगीचे को लहलहाया है,उन वतन के प्रेमियों की याद तुम बिसरा न देना। अर्थात् उन स्वतन्त्रता सेनानियों को हमें याद रखना है जिनके उत्सर्ग के परिणामस्वरूप ही हमें स्वाधीनता प्राप्त हुई। तुम इस स्वाधीनता को सस्ता समझने की भूल मत करना, यह हमें अनेक बलिदानों के बाद प्राप्त हुई है। इस स्वतन्त्रता रूपी चमन की हर कली, मस्त सुगन्ध और फूलों का रूप-रंग शहीदों के रक्त से सींचा गया है।

जिस प्रकार नींव का पत्थर स्वयं धरती के नीचे दफन होकर अपने ऊपर सुन्दर महल को आकार और स्थायित्व देता है, उसी प्रकार उन शहीदों ने अपने को देश की स्वतन्त्रता के लिए बलिदान कर, नींव का पत्थर बनाया है। वे ही शहीद जो नींव के पत्थर बने हैं, अपनी सौगन्ध देकर उनके द्वारा दी गई अमानत (देश की स्वाधीनता) की रक्षा करने के लिए कहते हैं। तुम अपने देश की स्वतन्त्रता पर आँच न आने देना।

(iv) शब्दार्थ :
भेड़िए = गद्दार; देशद्रोही = देश के शत्रु; कल्याण = समाज की प्रगति; सिसकियाँ भरना = रोना; ध्वंस = विनाश; आग-यौवन के धनी = वीर युवक जिनके हृदय में क्रोध की ज्वाला धधक रही है; विरासत = उत्तराधिकार; स्यार = गीदड़ (कायर पुरुष)।

संदर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि देश की भूमि को हड़पने के लिए भेड़िए (देश के शत्रु) ललचा रहे हैं और देश के अन्दर रहने वाले देशद्रोही देश के इतिहास से छेड़छाड़ कर उसे विनाश की ओर ले जा रहे हैं। देश की प्रगति देशद्रोहियों के कारण रो रही है अर्थात् प्रगति अपने को पतन की ओर अग्रसित होते हुए महसूस कर रही है। देश के निर्माण को उसका बैरी विनाश खाए जा रहा है। विनाश ने अपने विकराल हाथ चारों ओर फैला दिये हैं। ऐसे विकराल समय में कवि उन युवकों का आह्वान कर रहा है जिनके हृदय में क्रोध की अग्नि जल रही है।

उनसे कवि कहता है कि तुम व्यर्थ में आपस में मत लड़ो और न ही देश की सम्पत्ति को नुकसान पहुँचा कर अपने बल का ह्रास मत करो। तुम वीर पूर्वजों की सन्तान हो, तुम उन भेड़ियों (देशद्रोहियों) के दाँत ही नहीं तोड़ो बल्कि उनकी गर्दन मरोड़कर मृत्यु की शैया पर सुला दो,सिंह की खेती को कहीं गीदड़ न चर जाएँ ! अर्थात् सिंह के समान वीर पुत्रों के देश को कायर पुरुष दीमक की तरह चट न कर जाएँ। हे वीर पुरुष ! तुम जागो और अपने देश की स्वतन्त्रता की रक्षा करो।

प्रश्न 6.
“श्रीकृष्ण सरल राष्ट्रीय विचारधारा के कवि हैं।” संकलित कविता के आधार पर इस कथन की विवेचना कीजिए।
उत्तर:
श्रीकृष्ण सरल राष्ट्रीय विचारधारा के कवि हैं। उनकी देशभक्ति की उत्कृष्ट रचनाएँ इस कथन की साक्षी हैं। इनका लेखन इतिहास जैसा प्रामाणिक और शोधपूर्ण है। देशभक्ति से परिपूर्ण और क्रान्तिकारी लेखन के लिए उन्होंने देश-विदेश की अनेक यात्राएँ कीं। उन्होंने स्वयं निम्न पंक्तियाँ लिखकर यह स्पष्ट कर दिया कि उनका सम्पूर्ण जीवन राष्ट्र के लिए समर्पित था-
“कर्तव्य राष्ट्र के लिए समर्पित हों अपने, हो इसी दिशा में उत्प्रेरित चिन्तन-धारा,
हर धड़कन में हो राष्ट्र, राष्ट्र ही साँसों में, हो राष्ट्र समर्पित मरण और जीवन सारा।”

इनके मन में क्रान्तिकारियों के प्रति पहले से ही अनुराग था। उन्होंने देश के वीर पुत्रों का आह्वान किया है कि वे अपने देश की स्वाधीनता पर आँच न आने दें। उन्होंने लिखा है-
“राष्ट्र के शृंगार ! मेरे देश के साकार सपनो !
देश की स्वाधीनता पर आँच तुम आने न देना।
जिन शहीदों के लहू से लहलहाया चमन अपना
उन वतन के लाड़लों की
याद मुझाने न देना।”

आगे उन्होंने देश प्रेम की अनुभूति इस प्रकार व्यक्त की है-
“जो विरासत में मिला, वह खून तुमसे कह रहा है
सिंह की खेती किसी भी स्यार को खाने न देना।
देश की स्वाधीनता पर आँच तुम आने न देना।”

इन उदाहरणों से यह तथ्य स्पष्ट हो जाता है कि सरल जी राष्ट्रीय विचारधारा के कवि हैं।

शौर्य और देशप्रेम काव्य सौन्दर्य

प्रश्न 1.
निम्नलिखित शब्दों के विलोम शब्द लिखिए
विजय, स्वाधीनता, सम्मान, निर्माण, देशद्रोही।
उत्तर:

प्रश्न 2.
दिये हुए शब्दों के दो-दो पर्यायवाची लिखिए
दाँत, पर्वत,खून, आग, बाग।
उत्तर:
(i) दाँत – दशन,दंत,रदन।
(ii) पर्वत – गिरि, अचल, भूधर।
(iii) खून – लहू, रक्त,रुधिर।।
(iv) आग – अग्नि, अनल,पावक।
(v) बाग – उद्यान,वाटिका, उपवन।

प्रश्न 3.
अलंकार छाँटिए
(क) क्या करेगा यह विचारा तनिक-सा अवरोध?
(ख) अरे ! तुम हो काल के भी काल।
(ग) देश का कल्याण गहरी सिसकियाँ जब भर रहा हो।
उत्तर:
(क) अनुप्रास।
(ख) अनुप्रास।
(ग) अनुप्रास,मानवीकरण।

प्रश्न 4.
बालकृष्ण शर्मा ‘नवीन’ के संकलित अंश में से ओज गुण की अन्य पंक्तियों को अर्थ के साथ उद्धृत कीजिए।
उत्तर:
“क्या करेगा यह विचारा तनिक-सा अवरोध?
जानता है जग, तुम्हारा है भयंकर,क्रोध !”

अर्थ :
तुम्हारे सामने आया हुआ थोड़ा-सा अवरोध (रुकावट) तुम्हारा कुछ नहीं बिगाड़ सकता,क्योंकि तुम वीर हो। संसार तुम्हारे भयानक क्रोध से परिचित है। तुम्हारा क्रोध संसार में प्रलय ला देता है।

प्रश्न 5.
निम्नलिखित पंक्तियों में कौन-सा रस है? पहचान कर लिखिए-
तुम प्रबल दिक्-काल-धनु-धारी सुधन्वा वीर
तुम चलाते हो सदा चिर चेतना के तीर !
उत्तर:
उक्त पंक्तियों में वीर रस है।

अरे तुम हो काल के भी काल भाव सारांश

प्रस्तुत कविता ‘अरे तुम हो काल के भी काल’ वीर रस के ओजपूर्ण कवि ‘बालकृष्ण शर्मा ‘नवीन’ द्वारा लिखित है। इस कविता में कवि ने वीर पुरुषों को प्रोत्साहित किया है।

बालकृष्ण शर्मा ‘नवीन’ आधुनिक युग के ऊर्जावान कवि हैं। उनके काव्य में विप्लव की हुंकार दृष्टिगोचर होती है। देश प्रेम की उत्ताल तरंगें उनके काव्य सरोवर में निरन्तर उठती हैं। वे इस समाज से सडी-गली रूढियों और निर्बलताओं को समाप्त करना चाहते हैं। संकलित कविता में उन्होंने स्पष्ट किया है कि ध्वंस से ही सृजन की संभावना प्रकट होती है। असामाजिक प्रथाओं को समाप्त करना नवीन जी का लक्ष्य है। इस कविता में वे व्यक्ति को उसके साहस और दृढ़ निश्चय से परिचित कराते हुए स्पष्ट करते हैं कि उसका लक्ष्य विजय प्राप्त करना ही है। उसके मन में कहीं से भी निर्बलता एवं निराशा का भाव जाग्रत नहीं होना चाहिए। वह अनश्वर है और काल को भी चुनौती देने में समर्थ है। जब व्यक्ति वीरता के साथ खड़ा होगा तभी आतंक का दौर समाप्त होगा। कवि वीर पुरुषों को प्रोत्साहित करता है कि उनके क्रोध के सामने कोई भी विरोध टिक नहीं पाएगा। अन्त में वीर पुरुष की ही विजय होती है।

अरे तुम हो काल के भी काल संदर्भ-प्रसंग सहित व्याख्या

1. कौन कहता है कि तुमको खा सकेगा काल?
अरे? तुम हो काल के भी काल अति विकराल?
काल का तव धनुष, दिक् की धनुष की डोर,
धनु विकम्पन से सिहरती सृजन-नाश-हिलोर।
तुम प्रबल दिक्-काल-धनु-धारी सुधन्वा वीर
तुम चलाते हो सदा चिर चेतना के तीर !
क्या बिगाड़ेगा तुम्हारा, यह क्षणिक आतंक?
क्या समझते हो कि होगे नष्ट तुम अकलंक
यह निपट आतंक भी है भीति-ओत-प्रोत !
और तुम? तुम हो चिरन्तन अभयता के स्रोत॥
एक क्षण को भी न सोचो कि तुम होगे नष्ट
तुम अनश्वर हो ! तुम्हारा भाग्य है सुस्पष्ट !

शब्दार्थ :
काल = मृत्यु; विकराल = भयंकर; विकम्पन = काँपना; सृजन = निर्माण; प्रबल = शक्तिशाली; दिक्-काल-धनु-धारी-सुधन्वा = दिशा और काल रूपी धनुष को धारण करने वाला धनुर्धर; चिर = हमेशा रहने वाला; चेतना = चेतनता; क्षणिक = थोड़ी देर रहने वाला; आतंक = भय; अकलंक = निर्दोष; चिरन्तन = हमेशा रहने वाली; अनश्वर = नष्ट न होने वाला; सुस्पष्ट = साफ।

संदर्भ :
प्रस्तुत पंक्तियाँ ‘अरे तुम हो काल के भी काल’ नामक कविता से उद्धृत की गई हैं। इसके रचयिता ओजस्वी कवि बालकृष्ण शर्मा ‘नवीन’ हैं।

प्रसंग :
इस कविता में कवि ने यह स्पष्ट किया है कि ध्वंस से ही सृजन की संभावना प्रकट होती है। कवि ने यहाँ व्यक्ति को उसके साहस और दृढ़ निश्चय से परिचित कराते हुए स्पष्ट किया है कि उसका लक्ष्य विजय प्राप्त करना ही है।

व्याख्या :
कवि देश के गौरवशाली निवासियों को प्रोत्साहित करते हुए कहता है कि तुम्हारा काल भी कुछ नहीं बिगाड़ सकता,क्योंकि तुम काल के भी महाकाल और विकराल हो। काल का तुम्हारा धनुष है और दिशाओं की प्रत्यंचा है। जब उस धनुष की प्रत्यंचा में सिहरन होगी तो नाश होगा और फिर सृजन की हिलोर उठती हुई दिखाई देगी अर्थात् तुम्हारे युद्ध करने से ही पुनः सृजन का चक्र चलेगा। तुम दिशा और काल रूपी धनुष को धारण करने वाले धनुर्धर हो। तुम सदा चेतना के तीर चलाकर सुप्त वीरों को जाग्रत करते हो और स्वदेश-रक्षा में तत्पर रहते हो।

क्षणिक आतंक तुम्हारा कुछ भी नहीं बिगाड़ सकता। कोई भी आतंक या शत्रु तुम्हें नष्ट नहीं कर सकता। तुम निर्दोष हो और अपने देश की रक्षा के लिए सदा तत्पर हो। तुम्हारे सामने जो भयावना आतंक है वह क्षणिक है और तुम चिरन्तन अभयता के स्रोत हो। अर्थात् तुम्हारे अन्दर अभयता भरी हुई है जिसके सामने आतंक भी भययुक्त परिलक्षित होता है। तुम कभी भी यह मत सोचो कि तुम कभी नष्ट हो जाओगे। तुम अनश्वर (अमर) हो और तुम्हारा भाग्य तुम्हारी कर्मठता के कारण स्पष्ट तौर पर शुभ दिखाई दे रहा है।

काव्य सौन्दर्य :

1. शौर्य और ओजपूर्ण कविता है।
2. वीर रस का अनूठा सामंजस्य है।
3. अनुप्रास,रूपक अलंकारों का प्रयोग किया गया है।
4. कविता उत्साहवर्द्धक है।

(2) चिर विजय दासी तुम्हारी, तुम जयी उबुद्ध
क्यों बनो हतआश तुम, लख मार्ग निज अवरुद्ध?
फॅक से तुमने उड़ायी भूधरों की पाँत,
और तुमने खींच फेंके काल के भी दाँत
क्या करेगा यह विचारा तनिक-सा अवरोध?
जानता है जग, तुम्हारा है भयंकर क्रोध !
जब करोगे क्रोध तुम, तब आयेगा भूडोल
काँप उठेंगे सभी भूगोल और खगोल,
नाश की लपटें उठेगी गगन-मण्डल बीच
भस्म होंगी ये असामाजिक प्रथाएँ नीच !
औ पधारेगा सृजन कर, अग्नि में स्नान,
मत बनो गत आश ! तुम हो चिर अनन्त महान !

शब्दार्थ :
उबुद्ध = प्रबुद्ध; हतआश = निराश; अवरुद्ध = रुका हुआ; भूधर = पर्वत; पाँत = पंक्ति; तनिक-सा = थोड़ा-सा; अवरोध = रुकावट; भूडोल = भूकम्प; प्रथाएँ = रीतियाँ; असामाजिक = बुरी; सृजन = निर्माण; अग्नि में सुस्नान = युद्ध के उपरान्त; गतआश == निराश; चिर = हमेशा रहने वाला; अनन्त = जिसका अन्त न हो।

संदर्भ :
एवं

प्रसंग :
पूर्ववत्।

व्याख्या :
कवि भारतवासियों से उद्बोधन करते हुए कहता है कि विजय हमेशा से तुम्हारी चेरी रही है। तुम प्रबुद्ध जयी हो। फिर अपने मार्ग की तनिक-सी बाधा को देखकर तुम निराश क्यों हो जाते हो? तुम उन पूर्वजों की सन्तान हो, जिन्होंने फूंक से पहाड़ों को उड़ाया है और काल के भी दाँत तोड़कर उसके हाथ में दे दिए हैं। कहने का तात्पर्य है कि कठिन-से-कठिन कार्य तुमने पूर्व में किए हैं और सफलता प्राप्त की है और मृत्यु का भी अभयता से वरण किया है। यानी मृत्यु भी उनके साहस को तोड़ नहीं पाई है। फिर तुम्हारे रास्ते में आयी थोड़ी-सी बाधा तुम्हारा क्या बिगाड़ सकती है? संसार तुम्हारे क्रोध से परिचित है। जब भी तुमने क्रोध किया है संसार में प्रलय जैसा वातावरण उत्पन्न हो गया है।

जब भी तुम क्रोध करोगे भूकम्प आ जाएगा यानी धरा और अम्बर काँप उठेगे। (सीता स्वयंवर के समय जब लक्ष्मण जी ने जनकजी के अनुचित वाक्यों को सुनकर क्रोध किया था, तो धरा और गगन काँप उठे थे। उदाहरण देखिए-“लखन सकोप वचन जे बोले। डगमगानि महि दिग्गज डोले।” जब तुम युद्ध के लिए तत्पर हो जाओगे,तो गगन-मण्डल के बीच नाश की लपटें उठने लगेंगी और उसके साथ ही समाज में फैली बुराइयाँ भी नष्ट हो जाएंगी और फिर तुम देखोगे कि युद्ध की ज्वाला में स्नान करके ही नव निर्माण विकसित होगा अर्थात् युद्ध के बाद ही शान्ति और सृजन का उद्भव होता है। इसलिए तुम कभी भी अपने मन में निराशा का भाव मत लाओ। तुम चिर वीर, अनन्त और महान् हो। व्यक्ति के उत्सर्ग में ही उसकी वीरता और महानता आत्मसात है।

काव्य सौन्दर्य :

1. ओजपूर्ण कविता है।
2. वीर रस का प्रयोग किया गया है।
3. देशभक्ति की प्रेरणा दी गई है।
4. अतिशयोक्ति अलंकार का प्रयोग प्रेरणाप्रद है।

राष्ट्र के शृंगार भाव सारांश

प्रस्तुत कविता ‘राष्ट्र के श्रृंगार’ वीर रस के उल्लेखनीय कवि ‘श्रीकृष्ण सरल’ द्वारा रचित है। इस कविता में देश प्रेम की भावना दृष्टिगत होती है। कवि ने स्वाधीनता की रक्षा के लिए युवाओं का आह्वान किया है।

सरल जी ने युवाओं का आह्वान करते हुए उन्हें प्रेरणा दी है कि शहीदों के बलिदान से अर्जित स्वाधीनता की हर हाल में रक्षा करनी है। इस देश के शहीद अपने देश के लिए नींव के पत्थर बने। देशद्रोहियों और देश के शत्रुओं से अपने देश की रक्षा करनी है। तुम देश के गौरव हो,वीर हो और तुम नाश की धरती पर सृजन की कथा लिखते हो। देश तुमसे अपने उत्थान की अपेक्षा करता है। अपने अन्दर धधकती ज्वाला को कजलाने न देना।

राष्ट्र के शृंगार संदर्भ-प्रसंग सहित व्याख्या

(1) राष्ट्र के श्रृंगार ! मेरे देश के साकार सपनों !
देश की स्वाधीनता पर आँच तुम आने न देना।
जिन शहीदों के लहू से लहलहाया चमन अपना
उन वतन के लाड़लों की
याद मुझाने न देना।
देश की स्वाधीनता पर आँच तुम आने न देना।
तुम न समझो, देश की स्वाधीनता यों ही मिली है
हर कली इस बाग की, कुछ खून पीकर ही खिली है।
मस्त सौरभ, रूप या जो रंग फूलों को मिला है,
यह शहीदों के उबलते खून का ही सिलसिला है।
बिछ गए वे नींव में, दीवार के नीचे गड़े हैं,
महल अपने, शहीदों की छातियों पर ही खड़े हैं।
नींव के पत्थर तुम्हें सौगन्ध अपनी दे रहे हैं
जो धरोहर दी तुम्हें,
वह हाथ से जाने न देना।
देश की स्वाधीनता पर आँच तुम आने न देना।

शब्दार्थ :
श्रृंगार = सजावट; साकार = साक्षात्; सपनों = अभिलाषाएँ; आँच न आने देना = नुकसान न होने देना; चमन = बगीचा; वतन = देश; सौरभ = सुगन्ध; धरोहर = अमानत,थाती।

संदर्भ :
प्रस्तुत पंक्तियाँ ‘राष्ट्र के शृंगार’ नामक कविता से उद्धृत की गई हैं। इसके रचयिता ओजस्वी कवि श्रीकृष्ण सरल हैं।

प्रसंग :
इन पंक्तियों में कवि ने देश के वीर नागरिकों और बलशाली युवाओं का आह्वान किया है कि वे अपने देश की रक्षा करें। अपने देश की स्वतन्त्रता पर वे आँच न आने दें।

व्याख्या :
कवि भारत के देशवासियों को सम्बोधित करते हुए कहता है कि हे मेरे राष्ट्र के श्रृंगार और उसकी अभिलाषाओं को पूर्ण करने वाले वीरो ! तुम अपने देश की स्वतन्त्रता पर आँच न आने देना। जिन शहीदों ने अपना बलिदान देकर इस स्वाधीनता रूपी बगीचे को लहलहाया है,उन वतन के प्रेमियों की याद तुम बिसरा न देना। अर्थात् उन स्वतन्त्रता सेनानियों को हमें याद रखना है जिनके उत्सर्ग के परिणामस्वरूप ही हमें स्वाधीनता प्राप्त हुई। तुम इस स्वाधीनता को सस्ता समझने की भूल मत करना, यह हमें अनेक बलिदानों के बाद प्राप्त हुई है। इस स्वतन्त्रता रूपी चमन की हर कली, मस्त सुगन्ध और फूलों का रूप-रंग शहीदों के रक्त से सींचा गया है।

जिस प्रकार नींव का पत्थर स्वयं धरती के नीचे दफन होकर अपने ऊपर सुन्दर महल को आकार और स्थायित्व देता है, उसी प्रकार उन शहीदों ने अपने को देश की स्वतन्त्रता के लिए बलिदान कर, नींव का पत्थर बनाया है। वे ही शहीद जो नींव के पत्थर बने हैं, अपनी सौगन्ध देकर उनके द्वारा दी गई अमानत (देश की स्वाधीनता) की रक्षा करने के लिए कहते हैं। तुम अपने देश की स्वतन्त्रता पर आँच न आने देना।

काव्य सौन्दर्य :

1. ओजस्वी कविता है।
2. देशप्रेमियों को उत्सर्ग के लिए प्रेरित किया गया है।
3. वीर रस की कविता होते हुए भी यह सरल और सुबोध है।
4. रूपक अलंकार और प्रतीकों का सुन्दर समायोजन है।

(2) देश के भूगोल पर जब भेड़िए ललचा रहे हों
देश के इतिहास को जब देशद्रोही खा रहे हों
देश का कल्याण गहरी सिसकियाँ जब भर रहा हो
देश के निर्माण को, जब ध्वंस डटकर चर रहा हो।
आग-यौवन के धनी ! तुम खिड़कियाँ शीशे न तोड़ो,
भेड़ियों के दाँत तोड़ो, गर्दनें उनकी मरोड़ो।
जो विरासत में मिला, वह खून तुमसे कह रहा है
सिंह की खेती
किसी भी स्यार को खाने न देना
देश की स्वाधीनता पर आँच तुम आने न देना।

शब्दार्थ :
भेड़िए = गद्दार; देशद्रोही = देश के शत्रु; कल्याण = समाज की प्रगति; सिसकियाँ भरना = रोना; ध्वंस = विनाश; आग-यौवन के धनी = वीर युवक जिनके हृदय में क्रोध की ज्वाला धधक रही है; विरासत = उत्तराधिकार; स्यार = गीदड़ (कायर पुरुष)।

संदर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि देश की भूमि को हड़पने के लिए भेड़िए (देश के शत्रु) ललचा रहे हैं और देश के अन्दर रहने वाले देशद्रोही देश के इतिहास से छेड़छाड़ कर उसे विनाश की ओर ले जा रहे हैं। देश की प्रगति देशद्रोहियों के कारण रो रही है अर्थात् प्रगति अपने को पतन की ओर अग्रसित होते हुए महसूस कर रही है। देश के निर्माण को उसका बैरी विनाश खाए जा रहा है। विनाश ने अपने विकराल हाथ चारों ओर फैला दिये हैं। ऐसे विकराल समय में कवि उन युवकों का आह्वान कर रहा है जिनके हृदय में क्रोध की अग्नि जल रही है।

उनसे कवि कहता है कि तुम व्यर्थ में आपस में मत लड़ो और न ही देश की सम्पत्ति को नुकसान पहुँचा कर अपने बल का ह्रास मत करो। तुम वीर पूर्वजों की सन्तान हो, तुम उन भेड़ियों (देशद्रोहियों) के दाँत ही नहीं तोड़ो बल्कि उनकी गर्दन मरोड़कर मृत्यु की शैया पर सुला दो,सिंह की खेती को कहीं गीदड़ न चर जाएँ ! अर्थात् सिंह के समान वीर पुत्रों के देश को कायर पुरुष दीमक की तरह चट न कर जाएँ। हे वीर पुरुष ! तुम जागो और अपने देश की स्वतन्त्रता की रक्षा करो।

काव्य सौन्दर्य :

1. रूपक अलंकार व मुहावरों का प्रयोग सुन्दर बन पड़ा।
2. वीर रस की कविता है।
3. भाषा सरल और ओजस्वी है।

(3) तुम युवक हो, काल को भी काल से दिखते रहे हो,
देश का सौभाग्य, अपने खून से लिखते रहे हो।
ज्वाल की, भूचाल की साकार परिभाषा तुम्ही हो,
देश की समृद्धि की सबसे प्रबल आशा तुम्ही हो।
ठान लोगे तुम अगर, युग को नई तस्वीर दोगे,
गर्जना से शत्रुओं के, तुम कलेजे चीर दोगे।
दाँव पर गौरव लगे तो शीश दे देना विहँस कर
देश के सम्मान पर
काली घटा छाने न देना।
देश की स्वाधीनता पर आँच तुम आने न देना। (2009)

शब्दार्थ :
काल के भी काल = विकराल; ज्वाल = अग्नि; भूचाल = भूकम्प; साकार = साक्षात्; समृद्धि = उन्नति,वैभव; प्रबल = शक्तिशाली; ठान लेना = पक्का निश्चय कर लेना; गर्जना = ललकार; कलेजे चीर देना = भयाक्रान्त कर देना; गौरव = सम्मान,स्वतन्त्रता।

संदर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि देश के युवकों को अपने शौर्य से देश की तस्वीर बदलने के लिए आह्वान करते हुए कहता है कि हे युवक ! तुम काल के भी काल हो अर्थात् काल भी तुमसे भयभीत रहता है। तुम देश का सौभाग्य (उत्कर्ष) अपने रक्त से लिखते हो अर्थात् अपने बलिदान से देश के भाग्य को सौभाग्य में बदल देते हो। तुम अपनी वीरता से धरा पर अग्नि वर्षा कर सकते हो और भूकम्प ला सकते हो। कहने का तात्पर्य है कि तुम्हारे विरुद्ध कोई भी शत्रु टिक नहीं सकता। देश को तुमसे ही अपनी प्रगति की फलदायी आशा है। यदि तुम अपने मन में दृढ़ निश्चय कर लो तो इस युग को भी नये युग के शुभारम्भ में बदल सकते हो। तुम्हारी ललकार शत्रुओं के हृदय को भेद देती है। यदि तुम्हारे गौरव का प्रश्न उठे तो अपने प्राण भी हँसते-हँसते दे देना। तुम्हें ध्यान रखना है कि देश के सम्मान पर काली घटा न छाने पाये अर्थात् देश की स्वतन्त्रता पर काले बादल न छा जाएँ। तुम देश की स्वतन्त्रता को कोई हानि न होने देना।

काव्य सौन्दर्य :

1. वीर रस का सुन्दर समायोजन है।
2. कहावतों का समुचित प्रयोग किया गया है।
3. राष्ट्रवादी कविता है।

(4) वह जवानी, जो कि जीना और मरना जानती है,
गर्भ में ज्वालामुखी के जो उतरना जानती है।
बाहुओं के जोर से पर्वत जवानी ठेलती है,
मौत के हैं खेल जितने भी, जवानी खेलती है।
नाश को निर्माण के पथ पर जवानी मोड़ती है,
वह समय की हर शिला पर चिह्न अपने छोड़ती है।
देश का उत्थान तुमसे माँगता है नौजवानो !
दहकते बलिदान के अंगार
कजलानेन देना।
देश की स्वाधीनता पर आँच तुम आने न देना। (2009)

शब्दार्थ :
गर्भ = अन्दर; बाहुओं = भुजाओं; ठेलती = सरका देती है; शिला = चट्टान; उत्थान = प्रगति; दहकते = जलते हुए; कजलाने = बुझने जैसे।

संदर्भ :
पूर्ववत्।

प्रसंग :
यहाँ कवि ने जवानी का आह्वान कर देश को निर्माण के पथ पर ले जाने का आग्रह किया है।

व्याख्या :
कवि कहता है कि जवानी अत्यन्त जोशीली और मूल्यवान है,वह जिस प्रकार वीरता से जीना जानती है उसी प्रकार वीरता से मरना भी जानती है। वह ज्वालामुखी के अन्दर जाने में भी भयभीत नहीं होती। यानी विकराल और मृत्यु-सम परिस्थिति में भी अपनी वीरता प्रदर्शित करती है। अपनी भुजाओं के बल से जवानी पर्वतों को भी खिसका देती है, जितने भी खतरनाक खेल हैं, उनको जवानी हँसते-हँसते खेलती है। यह अपने देश के विनाश को भी निर्माण के पथ पर ले जाती है अर्थात् देश की विषम परिस्थितियों में भी वीरता का परिचय देकर सुखद परिस्थिति में बदल देती है।

यह अपने वीरतापूर्ण कार्यों के द्वारा चट्टान पर अपने सुकार्यों के चिह्न छोड़ देती है अर्थात् जवानी के वीरतापूर्ण कार्य इतिहास में स्वर्णाक्षरों में लिखे जाते हैं। कवि का यह उद्बोधन है कि हे नौजवानो ! तुम्हारा देश तुमसे अपने उत्थान की आशा लगाए बैठा है। अतः तुम अपने बलिदान रूपी अंगारों को बुझने न देना अर्थात् अपने उत्सर्ग से देश की स्वाधीनता को बनाए रखना। हे जवानो ! अपने देश की स्वाधीनता पर आँच मत आने देना।

काव्य सौन्दर्य :

1. देश-प्रेम का अपूर्व उदाहरण इस कविता में प्रस्तुत किया गया है।
2. ओजस्वी उद्बोधन प्रशंसनीय है।
3. भाषा सरल और बोधगम्य है।

MP Board Class 12th Hindi Solutions

MP Board Class 12th Physics Important Questions Chapter 10 Wave Optics

Wave Optics Important Questions

Wave Optics Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
A light ray appears to travel in a straight path because :
(a) It is made up of tiny particles
(b) Its velocity is high
(c) Its wavelength is very small
(d) None of the above.
Answer:
(c) Its wavelength is very small

Question 2.
Which of the following phenomenon does not give evidence of wave nature of light:
(a) Refraction
(b) Reflection
(c) Interference
(d) Polarisation.
Answer:
(b) Reflection

Question 3.
Two coherent sources or light must have the same :
(a) Amplitude
(b) Phase difference
(c) Frequency
(d) Both (b) and (c).
Answer:
(c) Frequency

Question 4.
Which of the following wave can not be polarised :
(a) X – rays
(b) Radio waves
(c) Longitudinal waves
(d) Transverse waves.
Answer:
(c) Longitudinal waves

Question 5.
The intensity of central fringe in Young’s experiment is I. One slit is closed then intensity becomes I₀. Which of the following is true :
(a) I = I₀
(b) I = 2I₀
(c) I = 4I₀
(d) No relation between I and I₀.
Answer:
(c) I = 4I₀

Question 6.
A colourless soap solution in white light appears to be coloured because of :
(a) Reflection
(b) Refraction
(c) Interference
(d) Diffraction.
Answer:
(d) Diffraction.

Question 7.
The ratio of amplitudes of light waves is 2 : 3. Due to their interference, ratio of maximum and minimum intensity of light will be :
(a) 2 : 3
(b) 4 : 9
(c) 5 : 1
(d) 25 : 1.
Answer:
(d) 25 : 1.

Question 8.
Young’s double – slit experiment proves that:
(a) Light rays consist of particles
(b) Light rays are waves
(c) Light rays are neither particles nor waves
(d) None of these.
Answer:
(b) Light rays are waves

Question 9.
Which of the following cannot be explained by wave nature of light : (MP 2009 Set B)
(a) Polarisation
(b) Interference
(c) Diffraction
(d) Photoelectric effect.
Answer:
(d) Photoelectric effect.

Question 10.
Waves are carrier of :
(a) Energy
(b) Amplitude
(c) Wavelength
(d) Matter.
Answer:
(a) Energy

Question 2.
Fill in the blanks :

1. The light is transverse wave, it is proved by ………………. of light.
2. In Young’s double slit experiment, the distance between two successive bright fringes is known as ……………….
3. The phase difference between two coherent sources of light is ……………….
4. In constructive interference of light path difference is always ………………. multiple of \(\frac {λ}{2}\)
5. Young’s double – slit experiment proves ………………. theory of light.
6. Bending of waves around the comers of an obstacle is known as ……………….
7. The phase difference between two points situated in the same wave front is ……………….
8. At polarising angle, the angle between reflected and refracted ray is ………………. (MP 2012)
9. The surface of medium at an instant of time, in which all the particles of medium vibrates with same phase is called ……………….
10. To take clear photographs of clouds is used before the lens of camera. (MP 2009 Set A)
11. ………………. rays are used as germs killer. (MP 2009 Set B)
12. At polarising angle, the angle between the reflected ray and the refracted ray from a surface is ………………. (MP 2011,13)
13. i_p+ r = ………………. (MP 2013)

Answer:

1. Polarisation
2. Width of dark fringes
3. Zero or constant
4. Even
5. Wave
6. Diffraction
7. Zero
8. 90°
9. Wavefront
10. Polaroids
11. Ultraviolet rays
12. 90°
13. 90°

Question 3.
Match the Columns :
I. (MP 2012)

Answer:

1. (c)
2. (e)
3. (d)
4. (b)
5. (a)

II. (MP 2010)

Answer:

1. (e)
2. (d)
3. (b)
4. (c)
5. (a)

Question 4.
Write the answer in one word/sentence :

1. What type of wave front will emerge from a point source?
2. Which phenomenon explains the colours on soap bubble?
3. What is the effect on fringe width if the distance between the coherent sources is decreases?
4. How is the wave front related to the direction of corresponding rays.
5. What is the phase difference corresponding to a path difference of λ?

Answer:

1. Spherical
2. Interference
3. Increases
4. Perpendicular
5. 2p.

Wave Optics Very Short Answer Type Questions

Question 1.
What is the phase difference between two points on a wavefront?
Answer:
Zero.

Question 2.
The line perpendicular to wavefront express the direction of what?
Answer:
Direction of propagation of wave.

Question 3.
What is the cause of colours on the thin film of liquid surface?
Answer:
Interference.

Question 4.
What should be the phase difference between two waves producing destructive interference?
Answer:
ϕ = π, 3π, 5π … (2n + 1) π.

Question 5.
Which type of light sources are required for interference of light?
Answer:
Coherent sources.

Question 6.
What should be the order of aperture of slit to obtain diffraction pattern?
Answer:
Of the order of wavelength of light.

Question 7.
Write formula of fringe width in Young’s double slit experiment.
Answer:
β = \(\frac {Dλ}{d}\)
Where
D → Distance between slit and screen
d → Separation between the slits
λ → Wavelength of light.

Question 8.
Two bulbs each of 60 W are kept very closed to each other. Will they produce interference?
Answer:
No, they will not produce interference because the bulbs are two different sources, they never be coherent.

Question 9.
What is angle between reflected ray and refracted ray at the angle of polarization?
Answer:
90°.

Question 10.
What do you mean by crossed polaroids?
Answer:
If the transmission axis of the two polaroids are perpendicular then these polaroids are called as crossed polaroids.

Question 11.
When a small circular obstacle is placed in the path of light coming from a far distance then a bright spot is observed at the centre of shadow of obstacle. Why?
Answer:
When a small circular obstacle is placed in the path of light then the diffraction takes place at its comers (edges). The diffracted wave produce the constructive interferemce at the centre of shadow. So a bright spot is observed.

Question 12.
The longitudinal waves do not show the phenomenon of polarization why?
Answer:
In the longitudinal waves the vibration of particles of medium takes place along the direction of propagation of waves. Hence they
do not show the polarization.

Question 13.
Define crossed polaroids.
Answer:
The two polaroids whose transmission axis are perpendicular to each other are called crossed polaroids.

Question 14.
For which arrangement of polarizer and analyzer in the analysis of polarized light, the intensity of light is minimum?
Answer:
When the axis of polarizer and analyser are mutually perpendicular.

Wave Optics Short Answer Type Questions

Question 1.
What is the principle of superposition of waves?
Answer:
When two or more waves reach to a point of a medium simultaneously, the particle vibrates with the resultant amplitude equal to the vector sum of the individual amplitudes.
Let n waves are reaching to a point and displacement of the particle for each wave is \(\vec { { y }_{ 1 } } \) , \(\vec { { y }_{ 2 } } \) , \(\vec { { y }_{ 3 } } \) , ………….
∴ The resultant displacement, y = \(\vec { { y }_{ 1 } } \) + \(\vec { { y }_{ 2 } } \) + \(\vec { { y }_{ 3 } } \) + …….. + \(\vec { { y }_{ n } } \)

Question 2.
When a white light is incident on the soap bubble or on a thin film of oil on f water surface, the beautiful colours are seen why? The soap bubble appears colourful why?
Answer:
When the sunlight is made to incident on a thin film whose thickness is of order of wavelength of light then some of its part gets reflected from its upper surface and remaining part of light is reflected from its lower surface.The interference of these reflected surface takes place. Since the white light is consist of seven colours, so due to interference of these reflected rays of different colours the film or soap bubble appears to be colourful.

Question 3.
In a Young’s double slit experiment the two slits are illuminated by two different lamps which are emitting the light of same wavelength. Will you observe interference pattern? Give reason.
Answer:
The sustained interference pattern will not be observed because two different source may never be coherent for sustained interference the sources should be coherent.

Question 4.
Explain, interference follows the law of conservation of energy.
Or
When two waves interfere, the energy of few particles become zero, is the energy lost?
Answer:
In the phenomenon of interference the total energy of waves remains constant, only the energy is redistributed. The energy which is lost at destructive interference is gained at the constructive interference. Hence, the energy is conserved.

Question 5.
How would you know that two sources are coherent?
Or
What are coherent sources of light? What are the conditions of two sources being coherent?
Answer:
Coherent sources:
The sources which emit the waves of same frequency with constant phase difference (zero), are called coherent sources.

Conditions:

1. The frequency of both the sources should be equal.
2. The phase difference should be constant or zero.

Question 6.
What do you mean by interference of light?
Or
What is interference light? Write its one example.
Answer:
When two light waves of same frequency and approximately same amplitude travel in a medium in the same direction then due to the superimposition the intensity of light changes. This phenomenon is called interference.

Example:
The soap bubble and thin film of oil on the water appear to be coloured.

Question 7.
Can the interference be obtained from white light?
Answer:
Yes, in the white light there are seven colours. The position of central bright fringe for all colours remains same. Hence the central bright fringe is white. For the different colours the positions of bright fringes are different.

Question 8.
In Young’s double slit experiment if the wavelength of monochromatic light is increased, then what will be the effect on fringe width?
Answer:
Now, we know that fringe width, β = \(\frac {λD}{d}\)
If D and d are constants, then β ∝ A.
Therefore, if wavelength is increased, fringe w’idth will also be increased.

Question 9.
In Young’s double slit experiment, what happens if screen is moved away from the slit?
Answer:
From the formula, β = \(\frac {λD}{d}\)
If λ and dare constants, then β ∝D. i.e., Fringe width will be increased.

Question 10.
If the apparatus of Young’s double slit experiment is immersed into water then what will be the effect on fringewidth ?
Answer:
From the formula β = \(\frac {λD}{d}\) it is clear that if D and d are constant then β ∝ λ .
Since the value of λ in the water decreases so the fringe width will decrease.

Question 11.
If the distance between two coherent sources is decreased then what will be the effect on fringe width ?
Answer:
Fringe width will increase because β ∝\(\frac {1}{d}\)

Question 12.
On what factors does the fringe width in Young’s double slit experiment depends? Write only name of factors.
Answer:
According to the formula β = \(\frac {λD}{d}\) the fringe width depends upon

1. Wavelength of light
2. Distance of slits from the screen and
3. Distance between the slits.

Question 13.
What do you mean by diffraction?
Answer:
When an obstacle comes in the path of waves then waves get bent at the obstacle. The bending of the waves around the comers is called diffraction.

Question 14.
How many types of diffraction are there? Write their examples.
Answer:
Diffraction is of two types :

1. Fresnel class diffraction
2. Fraunhofer class diffraction.example of Fresnel class diffraction is diffraction from straight edge and example of Fraunhofer diffraction is diffraction from a single slit.

Question 15.
What do you mean by polarization?
Answer:
Polarization of light is the phenomenon in which vibrations of electric vectors are restricted in a certain direction on a plane perpendicular to the direction of propagation

Question 16.
Differentiate plane polarized light and unpolarized light.
Answer:
Difference between plane polarized and unpolarized light:
Plane polarized light:

• In plane polarized light, the electric vector vibrates in one direction, normal to the direction of propagation of light.
• When polarized light is passed through a polaroid, which is rotated about the ray, then intensity becomes maximum and minimum.

Unpolarized light:

• The electric vector vibrates in all the directions normal to the direction of propagation of light.
• When this light is passed through a polaroid and polaroid is rotated then intensity of light does not change.

Question 17.
Light waves can be polarized but sound waves can’t. Why?
Answer:
Polarization is possible only in transverse waves not in longitudinal waves. Sound waves are longitudinal waves. Since, light waves are transverse waves, hence they can be polarized.

Question 18.
How the phenomenon of polarization is different from interference and diffraction.
Answer:
Interference and diffraction can occur in transverse as well as in longitudinal wave, but the polarization can take place only in transverse waves.

Question 19.
When an aircraft is travelling at low height, then the pictures on the T.V. screen undergo vibration. Why?
Answer:
When the aircraft is travelling at a lower height, then it reflects the T.V. signals. Interference occurs between the incoming signals and the reflected signals. Hence, vibrations are seen on the pictures in the T.V. screen.

Question 20.
Write down why the sun glasses made of polaroids are much useful than the goggles.
Answer:
Light reflected from the objects are partially plane polarized. The polaroid stops the horizontal component of partially plane polarized light decreasing the dazzling effect. But goggles cannot do so.

Question 21.
What will be the effect on interference pattern in a Young’s double slit experiment if one slit is closed?
Answer:
The interference pattern will not be obtained.

Question 22.
In which of the following waves the polarization is possible why?

1. X – rays
2. Sound waves
3. Radio waves.

Answer:
In X – rays and Radio waves polarization is possible because these are transverse waves. In sound waves polarization is not possible because they are longitudinal.

Question 23.
Write two limitations of Huygen’s wave theory.
Answer:

1. With the help of this theory the photoelectric effect could not be explained.
2. According to this theory there is a mediam called ‘ether’ but experimentally it could not be verified.

Question 24.
What is the shape of interference fringes in Young’s double slit experiment?
Answer:
The interference fringes in double slit experiment are like the parallel lines. These lines are parallel to the slit and are bright and dark alternately.

Question 25.
Define the following :

1. Spherical wave front
2. Plane wavefront

Answer:
1. Spherical wavefront : When the light source is of point type then at any instant at finite distance the locus of all the particles vibrating with same phase is a sphere. It is called spherical wavefront.

2. Plane wavefront : When the source of light is at infinite distance then the wave front formed is called plane wavefront.

Question 26.
Which type of wave do not show the polarization?
Answer:
Longitudinal waves do not show the polarization.

Question 27.
In Young’s double slit experiment if white light is used then only white and dark fringes are obtained. Is this statement true? Why?
Answer:
This statement is not true, because in this condition only central bright fringe is white while the other fringes are coloured.

Question 28.
Two persons separated by a high obstacle like a wall can hear their voice but cannot see each other. Why?
Answer:
For the diffraction of waves it is necessary that the size of obstacle should be of order of the wavelength of wave. The wavelength of sound waves is of order of size of obstacle so the sound waves are diffracted and the person can hear their voice.
The wavelength of light waves is very small while the aperture of obstacle is large so light waves are not diffrected and the persons can not see each other.

Question 29.
On what factors does the diffraction depend?
Answer:
The diffraction depends upon two factors :

1. Size of aperture and
2. Wavelength.

Question 30.
What is interference of light? Write the necessary conditions for it.
Or
Write any four necessary conditions for interference of light.
Answer:
Interference:
When two light waves of same frequency and approximately same amplitude travel in a medium in the same direction then due to the superimposition the intensity of light changes. This phenomenon is called interference.

Example:
The soap bubble and thin film of oil on the water appear to be coloured.

Conditions for interference:

1. Frequencies of light emitted by both sources should be equal.
2. Amplitudes of both waves should be almost equal.
3. Both waves should move in the same direction.
4. Both the sources should be placed close to each other.
5. Both the light sources should be narrow.
6. Phase difference of the two waves should be zero or should be constant.
7. Light sources should be monochromatic.
8. If light rays are polarized, then their plane of polarization should be parallel.
9. Both the sources should be coherent (i.e phase difference between the two waves should be zero or constant).

Question 31.
State Huygen’s secondary wavelets theory.
Answer:
Huygen’s secondary wavelets theory : The main postulates of this theory are given below:

1. When a disturbance is created in a medium the particles of the medium starts vibrating. The surface in the medium, on which each particle vibrate in same phase is called the wavefront. If the source is like a point, at a finite distance, the wavefront is always spherical. If the source is in form of a slit at a finite distance the wavefront is cylindrical.

2. Every point of the medium on a given wavefront is a source of originating new
disturbances. These new disturbances are called secondary wavelets.

3. These secondary wavelets advances in the medium with same velocity as of original wave.

4. The common tangential surface drawn on these secondary wavelets at any instant, shows the position of new wavefront at that instant.

Question 32.
Write the path difference for constructive and destructive interference in terms of wavelength l.
Answer:
We know that
Phase difference = \(\frac {2π}{λ}\) x path difference
or ϕ = \(\frac {2π}{λ}\) x y … (1)
For constructive interference ϕ = 2nπ … (2)
From eqns. (1) and (2)
\(\frac {2π}{λ}\) = 2nπ
or
For n^th bright fringe (constructive interference)
y = y_n
or y_n = 2n\(\frac {λ}{2}\)
Hence for constructive interference the path difference should be even multiple of \(\frac {λ}{2}\)
For minimum intensity ϕ = (2n +1 )π
or \(\frac {2π}{λ}\) y = (2n +1 )π
or y = (2n +1) \(\frac {λ}{2}\)
or y = (2n + l) \(\frac {λ}{2}\)
For n^th dark fringe y = y_n
or y_n = (2n + 1)\(\frac {λ}{2}\)

Question 33.
What should be the path difference between the two waves
y₁ = a₁ sin ωt and y₂ = a₂ cos(ωt + ϕ)?
Answer:
For first wave y₁ = a₁ sin ωt
and for second wave y₂ = a₂ cos(ωt + ϕ)
ory₂ = a₂ sin (ωt + ϕ + \(\frac {π}{2}\))
The phase difference between first and second wave.

Question 34.
If the distance between the slits is doubled and distance between slit and screen is halved then what will be the effect on fringe width in Young’s double slit experiment?
Answer:
Formula : Fringe width β = \(\frac {Dλ}{2}\)
Let d’, D’ and β’ are changed values of slit separation
Distance between slits and fringe width respectively then

∴ Fringe width will be one fourth of its previous value.

Question 35.
If angle of polarization is 30° then find out refractive index of medium.
Solution:
Given : Angle of polarization i_p = 30°
µ = tani_p ⇒ µ = tan30°
µ = \(\frac { 1 }{ \sqrt { 3 } }\)

Question 36.
Define the following:

1. Plane of vibration
2. Plane of polarization.

Answer:
1. Plane of vibration : Plane of vibration is that plane which contains the vibra-tions of electric vector and direction of propagation of wave.

2. Plane of polarization : A plane perpendicular to the plane of vibration and which contains the direction of propagation of waves is called plane of polarization.

ABCD → Plane of vibration, EFGH → Plane of polarization.

Question 37.
How many types of interference are there? Explain.
Answer:
There are two types of interference.

1. Constructive interference
2. Destructive interference.

Interference of light:
When two light waves of same frequency and approximately same amplitude travel in a medium in the same direction then due to the superimposition the intensity of light changes. This phenomenon is called interference.
Example:
The soap bubble and thin fiLm of oil on the water appear to be coloured.

Constructive interference:
When two wave super imposed together at a point with same phase i.e. the fall on crest and trough on trough then at that point the intensity of light becomes maximum. This interference is called constructive interference.

Destructive interference:
When two waves superimposed together at a point in such a way that there is a phase difference of 180° [or (2n – 1) π] then at those points the intensity of light becomes minimum or zero. This interference is called as destructive interference

Conditions:
For constructive interference the phase difference should zero or even multiple of π i.e. ϕ = 2nπ. For destructive interference the phase difference should be 180° or odd multiple of π. i.e. ϕ = (2n + 1) π, where n = 0, 1, 2, 3, ……

Question 38.
What do you mean by coherent sources? Can two independent sources of light produce interference ? How can the coherent waves be obtained?
Answer:
Coherent Sources:
The sources which emit the waves of same frequency with constant phase difference (zero), are called coherent sources.

The phase difference between the waves coming from two different sources varies with time. Hence the two different sources cannot produce interference. The coherent waves can be obtained by different methods such as Fresnel’s biprism method, Amplitude division method etc.

Question 39.
What do you mean by diffraction of light? What is essential condition to obtain diffraction? The diffraction of sound w aves is easier than that of light waves. Why?
Answer:
Diffraction of light:
When an obstacle comes in the path of waves then waves get bent at the obstacle. The bending of the waves around the comers is called diffraction.

Condition:
To obtain diffraction the size of obstacle should be of order of wave length of waves.

In case of sound waves the size of obstacle is of order of wavelength while in case of light the obstacle is usually not of order of wavelength of light so the diffraction of sound waves is easier than that of light waves.

Question 40.
What is the difference between interference and diffraction?
Answer:
Difference between interference and diffraction :
Interference:

• Interference fringes are formed when the two light rays from the coherent sources are superimposed.
• All bright fringes are equally illuminated.
• Width of all fringes are always equal.
• Minimum intensity is zero, so fringes can be distinguished very well.

Diffraction:

• Fringes are formed due to superimposition of light rays from the same source.
• Central fringe is brightest and all other fringes have intensity’ in decreasing order.
• Fringe width decreases when we move away from the centre.
• Minimum intensity fringes are not perfectly darker so it is relatively difficult to distinguish bright and dark fringes.

Question 41.
What is polaroid? State its applications in daily life.
Answer:
Polaroids are large sized manufactured polarizing films capable of producing plane polarized beam of light

Uses:

1. These are used in laboratory to produce and analyse plane polarized light.
2. These are used in sun glasses.
3. These are used in headlights and windscreen of automobile to cut off the dazzling light.
4. These are used for viewing three dimensional pictures.

Question 42.
What are polaroids? How will you obtain the plane polarized light by polaroids? How will you distinguish between unpolarized light and plane polarized light?
Answer:
Polariods:
Polaroids are large sized manufactured polarizing films capable of producing plane polarized beam of light. The light obtained from an ordinary source is made to fall on polariod. The light emerging out from the polaroides is plane polarized light. To identify (or distinguish) between polarized and non – polarized light. The beam is made to incident on another polariod. If there is no change in intesity of light on rotating the Polaroid then the light is unpolarized. But if there is variation in the intensity of light then the light is plane polarized.

Question 43.
What is polaroid? Write its construction and working.
Answer:
Polaroid:
Polaroids are large sized manufactured polarizing films capable of producing plane polarized beam of light.

Construction:
It is a thin film whice is sandwiched between two glass plates. To form this film, the crystal of an organic compound ‘herapathite’ or Iodo – quinine sulphate are kept on a thin film of Nitro cellulose in such a way that the axis of all crystals are parallel to each other.

Working:
The polaroid has a characteristic plane called transmission plane. When an unpolarized light falls on it only the vibrations of electric vector, parallel to the plane of transmission get transmitted. Thus the transmitted light obtained is plane polarized.

Wave Optics Long Answer Type Questions

Question 1.
Explain Huygen’s theory of secondary wavelets.
Answer:
Huygen’s secondary wavelets theory:

1. Each point of wavefront behaves like a new source of light which emits the new waves. These are called secondary wavelets.

2. The secondary wavelets advances in the medium with the velocity same as that of the original waves.

3. The common tangential surface drawn on these secondary wave lets at any in stant shows the position of new wave front at that instant.

Let S is point type light source AB is a part of spherical wavefront. After time ‘t’ what will be the position of this wavefront it is to be determined. According to Huygen’s secondary wave. let theory every point of wavefront behaves like a source of light. Take points P₁ P₂, P₃, P₄, on this wavefront.

The distance travelled by secondary wavelet in time ‘t’ with velocity V be vt. Now by taking P₁ P₂, P₃, P₄, …… as centre draw circles of radius vt. Now draw a comman tangential surface A₁B₁ It will show the new wavefront.

Question 2.
What do you mean by wavefront? How many types of wavefront are there?
Answer:
Wavefront:
An imaginaiy plane made of medium perpendicular to the direction of propagation of waves in which each particle of medium vibrates with same phase is called wavefront. There are three types of wavefronts :

1. Spherical wavefront : When the light source is point type and at finite distance then the wavefront is called spherical wavefront

2. Cylindrical wavefront : When light source is slit type and at a finite distance then the wavefront is called as cylindrical wavefront.

3. Plane wavefront : When light source is at infinite distance then the wavefront obtained is called plane wavefront.

Question 3.
What is interference of light? Write its essential conditions.
Or
State essential condition for interference of light.
Answer:
Interference:
When two light waves of same frequency and approximately same amplitude travel in a medium in the same direction then due to the superimposition the intensity of light changes. This phenomenon is called interference.

Example:
The soap bubble and thin film of oil on the water appear to be coloured.

Conditions for interference:

1. Frequencies of light emitted by both sources should be equal.
2. Amplitudes of both waves should be almost equal.
3. Both waves should move in the same direction.
4. Both the sources should be placed close to each other.
5. Both the light sources should be narrow.
6. Phase difference of the two waves should be zero or should be constant.
7. Light sources should be monochromatic.
8. If light rays are polarized, then their plane of polarization should be parallel.
9. Both the sources should be coherent (i.e phase difference between the two waves should be zero or constant).

Question 4.
What do you mean by interference of light? Define constructive and de – structive interference and write essential conditions for it
Answer:
Interference of light:
When two light waves of same frequency and approximately same amplitude travel in a medium in the same direction then due to the superimposition the intensity of light changes. This phenomenon is called interference.

Example:
The soap bubble and thin fiLm of oil on the water appear to be coloured.

Constructive interference:
When two wave super imposed together at a point with same phase i.e. the fall on crest and trough on trough then at that point the intensity of light becomes maximum. This interference is called constructive interference.

Destructive interference:
When two waves superimposed together at a point in such a way that there is a phase difference of 180° [or (2n – 1) π] then at those points the intensity of light becomes minimum or zero. This interference is called as destructive interference

Conditions:
For constructive interference the phase difference should zero or even multiple of π i.e. ϕ = 2nπ. For destructive interference the phase difference should be 180° or odd multiple of π. i.e. ϕ = (2n + 1) π, where n = 0, 1, 2, 3, ……

Question 5.
How does the diffraction of light take place from a straight edge? Draw energy distribution curve also.
Or
Describe the diffraction of light from a straight edge under the following heads :

1. Experimental arrangement
2. Observation
3. Energy distribution curve.

Answer:
1. Experimental arrangement:
In the fig. S is a narrow slit which is illuminated by a monochromatic light source. A straight edge is placed infront of slit S in such a way that its edge is parallel to the slit. At some distance from the obstacle AB a screen XY is placed. WW’ is a cylindrical wavefront.

2. Observation:
According to geometrical optics above the point O on the screen XY the intensity should be uniform because in this region light rays can reach without any obstruction.

Below the point O there should be complete darkness because the light rays are obstructed by obstacle AB. But actually if not happens. Actually we get the following pattern:

(a) In the region of geometrical shadow some light is reached. The intensity of light rapidly decreases and at point P, the intensity becomes zero. Thus after point P₁ complete darkness is obtained.

(b) In the illuminated region OX the dark and bright of decreasing intensity and width are obtained. Ultimately above point P the fringes vanishes and light of uniform intensity is obtained.

3. Energy distribution curve : It is shown in adjacent fig.

Question 6.
Write any four differences between interference and diffraction.
Answer:
Interference:

• Interference fringes are formed when the two light rays from the coherent sources are superimposed.
• All bright fringes are equally illuminated.
• Width of all fringes are always equal.
• Minimum intensity is zero, so fringes can be distinguished very well.

Diffraction:

• Fringes are formed due to superimposition of light rays from the same source.
• Central fringe is brightest and all other fringes have intensity’ in decreasing order.
• Fringe width decreases when we move away from the centre.
• Minimum intensity fringes are not perfectly darker so it is relatively difficult to distinguish bright and dark fringes.

Question 7.
What do you mean by polarization of light? Describe on experiment to prove that light waves are transverse in nature.
Answer:
The lack of symmetry about the direction of travel of a wave is called polarization. To demonstrate that light waves are transverse we use two polaroids A and B. These polaroids are placed in such a way that their planes parallel to their vertical axis.

Now beam of light is incident normally on A and the light is seen through B from the other side. In this condition the intensity of light is maximum. Now keeping ‘A’ fixed, the B is rotated by doing this the intensity of light decreases gradually. When the axis of B becomes perpendicular to that of A then the intensity of light becomes minimum. If the ‘B’ is further rotated then the intensity again increases gradually. The intensity becomes maximum again when the axis of A and B again becomes parallel.

By this experiment it is proved that the light waves are transverse in nature

Question 8.
Explain polarization of light by reflection.
Answer:
When a beam of ordinary light is reflected from a transparent medium such as glass etc. then it get partially polarized. With increase in the value of angle of incident, at a particular value of angle of incidence the reflected light is obtained completely plane polarized. This value of angle of incidence is called as angle of polarization.

In the fig. AB is incident ray, BC is reflected ray and BD is refracted ray. MN is the interface of air and medium when the light is incident at angle i_p (angle of polarization) then the reflected ray along BC is completely plane polarized. The vibrations of plane polarized light are perpendicular to the plane of paper. Hence they are expressed by dots The light obtained in the direction BD is partially polarized.

Explanation:
In a non – polarized light the electric vectors can be dissociated in two rectangular components

1. Parallel to the plane of incidence
2. Perpendicular to the plane of incidence

These are expressed in the fig. by dots and arrows. When the light falls on refracting surface then its electrons vibrates in the two directions. At the angle of polarization the reflected ray and refracted rays are mutually perpendicular. The vibration parallel to the plane of incidence are along the wave. Hence they can not produce transverse waves along the reflected light. As a result the electric vectors vibrates in a plane perpendicular to the plane of incidence and the completely plane polarized light is obtained.

Question 9.
State Brewster’s law. Prove that at angle of polarization, reflected rays and refracted rays are mutually perpendicular.
Or
What is Brewster’s law? Prove that for light incident on a transparent surface at Brewster’s angle the reflected and refracted rays are normal to each other.
Answer:
Brewster’s law:
The refractive index of a medium is equal to the tangent of angle of polarization, i.e.,
μ = tan i_p
Let a light ray is incident at the angle of polarization ip, to the medium of refractive index μ.

By Snell’s law, μ = \(\frac { sin{ i }_{ p } }{ sinr }\)
Where, r is the angle of refraction. But, μ = tan i_p
From eqns. (1) and (2), we get
tan i_p = \(\frac { sin{ i }_{ p } }{ sinr }\)
or \(\frac { sin{ i }_{ p } }{ cos{ i }_{ p } }\) = \(\frac { sin{ i }_{ p } }{ sinr }\)
∴ cos i_p = sinr
or sin(90°- i_p) = sinr
or 90° – i_p = r
or i_p + r = 90°
But we have, i_p + r + ∠CBD = 180°
or 90°+∠CBD = 180°
or ∠CBD = 180°
Thus, reflected ray and refracted ray are mutually perpendicular.

Question 10.
What do you mean by polarization of light? Why the light get polarized by scattering? Explain.
Answer:
Polarization of light:
In plane polarized light, the electric vector vibrates in one direction, normal to the direction of propagation of light. When polarized light is passed through a polaroid, which is rotated about the ray, then intensity becomes maximum and minimurp.

Explanation of polarization by scattering:
It is observed that the sun light is scattered by the molecules of gases in the atmosphere. Due to scattering light rays gets polarized in the adjacent fig. the polarization of light by scattering due to nitrogen molecule is shown.

Light incident from sun is unpolarized in which electric field vector can vibrate in all possible directions perpendicular to the direction of propagation of incident light. All these vibrations can be resolved into two components :

1. Paralled to plane of paper shown by double arrowed lines.
2. Perpendicular to plane of papers shown by dots.

When light waves falls on nitrogen molecules then electrons of molecule start vibrating in their perpendicular directions. The vibration parallel to the plane of paper are along the observer. Therefore these vibrations cannot produce transverse waves in the direction of observer. The vibrations perpendicular to the plane of paper can only produce transverse wave in the direction of the observer. Thus the light entering the eye of observer is plane polarized light.

Question 11.
What are polaroids? Write its construction, working and uses?
Answer:
Polaroids are large sized manufactured polarizing films capable of producing plane polarized beam of light

Uses:

1. These are used in laboratory to produce and analyse plane polarized light.
2. These are used in sun glasses.
3. These are used in headlights and windscreen of automobile to cut off the dazzling light.
4. These are used for viewing three dimensional pictures.

Construction:
It is a thin film whice is sandwiched between two glass plates. To form this film, the crystal of an organic compound ‘herapathite’ or Iodo – quinine sulphate are kept on a thin film of Nitro cellulose in such a way that the axis of all crystals are parallel to each other.

Working:
The polaroid has a characteristic plane called transmission plane. When an unpolarized light falls on it only the vibrations of electric vector, parallel to the plane of transmission get transmitted. Thus the transmitted light obtained is plane polarized.

Question 12.
Obtain the conditions for maximum and minimum intensity of light in Young’s double slit experiment
Or
Find out the conditions for constructive and destructive interference of two waves theoriticaily.
Or
What do you mean by interference of light? Two waves having amplitudes a₁, a₂ and angular frequencies co are superimposed together. Find out intensity in the constructive and destructive interference.
Answer:
Interference:
When two light waves of same frequency and approximately same amplitude travel in a medium in the same direction then due to the superimposition the intensity of light changes. This phenomenon is called interference.

Example:
The soap bubble and thin film of oil on the water appear to be coloured.

Let two waves having same frequency are propagating in same direction in same medium. Their equation are
y₁ = a₁Sinωt … (1)
and y₂ = a₂ sin (ωt + ϕ) … (2)
Where, a₁ and a₂ are their amplitudes and ϕ is phase difference between them.
If both the wave are superimposed together then by the principle of superposition of waves,
\(\vec { y }\) = \(\vec { { y }_{ 1 } } \) + \(\vec { { y }_{ 1 } } \)
or \(\vec { y }\) = a₁ sinωt + a₂ sinωt (ωt + ϕ)
or y = a₁ sinωt + a₂ sinωt cosϕ + a₂ cosωt sinϕ
or y = (a₁ + a₂ cosϕ)sinωt + a₂ϕ cosωt … (3)
Now let
a₁ = a₂ cosϕ = R cosϕ … (4)
and a₂ sinϕ = R sinθ … (5)
Where, R and θ are constants
Then by eqn. (3),
y = R cosθ sinωt + R sinθcosωt
or y = R sin(ωt + θ) … (6)
Equation (6) shows the resultant wave. Its amplitude is R
Squaring and adding equations (4) and (5)
R² cos²θ + R² sin²θ = (a₁ + a₂ cosϕ)² + (a₂ sinϕ)²
or R² (cos²θ + sin²) = a₁ + 2a₁a₂ cosϕ + a₂² cos²ϕ + a₂² sin²ϕ
or R² = a₁² + 2a₁a₂ cosϕ + a₂²
or R² = a₁² + a₁²+ 2a₁a₂ cosϕ
∵ Intensity is proportional to the R²
I = K(a₁² + a₁² + 2a₁a₂ cosϕ) … (7)
Where K is a proportionality constant.
From equation (7) it is clear that the resultant intensity depends upon the phase difference between the two waves.
Condition for constructive interference : from equation (7) it is clear that for maximum intensity
cosϕ= 1
or ϕ = 2nπ, (where, n = 0, 1,2, 3, )
At the points where the waves with same phase are superimposed, the intensity becomes maximum.
Thus I_max = K(a₁² + a₁² + 2a₁a₂)
= K(a₁ + a₂)²
If the path difference for constructive interference is ∆x, then
∆x = \(\frac {λ}{2π}\) x ϕ
∆x = \(\frac {λ}{2π}\) x 2nπ
∆x = 2n x \(\frac {λ}{2}\) = even mul of \(\frac {λ}{2}\)

Condition for destructive interference : From equation (7) it is clear that for minimum intensity
cosϕ = -1
ϕ = (2n -1)π (where, n = 1,2,3, ……… )
or
Hence if the waves are opposite in phase then the destructive interference takes place. If the path difference for destructive interference is ∆x, then
∆x = \(\frac {λ}{2π}\) x ϕ
∆x = \(\frac {λ}{2π}\)(2n – 1)π
∆x = (2n – 1)\(\frac {λ}{2}\) = odd multiple of \(\frac {λ}{2}\)

Expression for intensity in the constructive interference In equation (7) putting cosϕ = 1
I = K(a₁² + a₁² + 2a₁a₂ ) = K(a₁ + a₂)²
If a₁ = a₁ = a, then
I = K(a + a)² or I = 4Ka².

Expression for intensity in destructive interference
Putting cosϕ= -1 in equation (7)
I = K(a₁² + a₂² + 2a₁a₂) = K(a₁-a₂)²
If a₁ = a₂ = a, then
If a₁ = a₁ = a, then
I = K(a – a)² = 0

Question 13.
Derive formula for the width of fringe in Young’s double – slit experiment.
Or
Prove that β = \(\frac { λD}{d}\), where symbols have their usual meanings.
Answer:
Let S is a source which is illuminated with monochromatic light. S₁ and S₂ are the two slits placed close to each other which are parallel and equidistant from S. Therefore, Si and S2 behave as coherent sources.
Suppose that S₁ S₂ = d
Screen is placed at distance D from the sources S₁ S₂ on which interference fringes are formed.
Path difference at point O on screen is
S₁O – S₂O = 0
∴Intensity at point O will be maximum and will be central fringe.
Let a point P on the screen XY at distance x from the centre.
From S₁, drop perpendicular S₁K on screen and from S₂L, drop perpendicular S₂L on screen.
In right angled ∆S₁KP,

Now, for the bright fringe,

If distances of n^th and (n + 1)^th fringes from the centre are x_n and x_n+x, therefore from equation (1),
x_n = \(\frac {nDλ}{d}\)
and x_n+1 = \(\frac {(n+1)Dλ}{d}\)
Distance between the two consecutive bright fringes or width of the dark fringe,
x_n+1 – x_n = \(\frac {(n+1)Dλ}{d}\) – \(\frac {nDλ}{d}\)
and x_n+1 – x_n = \(\frac {Dλ}{d}\) … (3)
Similarly, distance of n^th and (n + l)^th dark fringes from the point O are x_n and x_n+1, therefore from equation (2),
x_n = \(\frac {(2n – 1)Dλ}{2d}\)
and x_n+1 = \(\frac {(2n + 1)Dλ}{2d}\)
Distance between the two consecutive dark fringes or width of the bright fringe,

Hence, it is clear from equations (3) and (4), that the width of the fringes are always equal and is given by
β = \(\frac {Dλ}{d}\)

Question 14.
State law of malus and explain it.
Answer:
Malus law:
If a beam of completely plane polarized light is incident on an analyser, then the resultant intensity (I) of light transmitted from the analyser varies directly as the square of the cosine of the angle between the transmission direction of polariser and analyser. i.e., I ∝ cos²θ

Derivation:
Let OP and OA are the pass axes of polarizer and analyser respectively. Let the angle between OP and OA be θ and a be the B amplitude of emergent light from polarizer along axis OP. Amplitude a can be resolved in two components :

• acosθ along pass axis OA of analyzer
• asinθ along pass axis OB of polarizer

Component acos0 only emerges out of analyzer. Therefore intensity of light emerging out from analyzer
I ∝ (acosθ)²
or I = Ka² cos²θ
Where, K is proportionality constant
or I = I₀ cos²θ
Where I₀ = ka² … (1)
∴ I ∝ cos²θ
This is law of malus.

Wave Optics Numerical Questions

Question 1.
What is the Brewster’s angle for transmission from air to glass? (µ = 1.5) (NCERT)
Solution:
Given µ = 1.5
∵µ = tan i_p
∴ 1.5 = tan i_p
0r i_p = tan^-1(1.5) = 56.4°.

Question 2.
In a Young’s double slit experiment of the fringe width is 4 mm. Obtain distance of third dark fringe from central bright fringe.
Solution:
Given: β = 4mm = 4 x 10^-3 m
For third dark fringe n = 3

⇒ x_n =10 x 10^-3 m
= 10^-2 m
= 10 mm.

Question 3.
The angle of polarization for a transparent medium is 60°. Find the angle of refraction and refractive index of medium.
Solution:
Given : Angle of polarization i_p = 60°
∴ i_p + r = 90°
∴ 60° + r = 90°
r = 90° – 60°
∴ r = 30°.
Again by Brewster’s Law
µ = tan i_p = tan 60°
µ = 1.73.

Question 4.
In a Young’s experiment the distance of slits from the screen is 2m. A light of 6000 A is incident on the slits. (i) If the fringe width is 2mm then find distance between the slits. (ii) If the light of wavelenght 4800 Å is incident then find fringe width.
Solution:
Given D = 2m, λ= 6000Å = 6000 x 10^-10m = 6 x 10^-7m

MP Board Class 12th Physics Important Questions

MP Board Class 12th Physics Important Questions Chapter 7 Alternating Current

Alternating Current Important Questions

Alternating Current Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The mean value of an A.C. over one complete cycyle is :
(a) \(\frac { { 2I }_{ 0 } }{ \pi }\)
(b) –\(\frac { { 2I }_{ 0 } }{ \pi }\)
(c) Infinite
(d) Zero
Answer:
(d) Zero

Question 2.
The relation between peak value V₀ and rms value Vrms of alternating voltage is:
(a) V₀ = 0.707 V_ms
(b) V_ms = 0.707 V₀
(c) V_ms = 0.637 V₀
(d) V₀= 0.637 V_ms
Answer:
(c) V_ms = 0.637 V₀

Question 3.
A capacitor allows through it:
(a) Only d.c
(b) Only a.c
(c) a.c and d.c both
(d) Neither a.c nor d.c.
Answer:
(b) Only a.c

Question 4.
In a circuit wattless current flows if it contains :
(a) Only R
(b) R – L
(c) R – C
(d) Only L.
Answer:
(d) Only L.

Question 5.
A device which convert mechanical energy into electrical energy is :
(a) d.c moter
(b) a.c generator
(c) Transformer
(d) Choke coil.
Answer:
(b) a.c generator

Question 6.
The relation between peak value on a.c I₀ and virtual value Irms is :
(a) I_{0 =} \(\frac { 1 }{ \sqrt { 3 } }\) = I_ms
(b) I_ms = \(\sqrt { 3{ I }_{ 0 } }\)
(c) I₀ = I_ms
(d) I₀ = \(\sqrt { 3{ I }_{ ms } }\) I_ms
Answer:
(d) I₀ = \(\sqrt { 3{ I }_{ ms } }\) I_ms

Question 2.
Fill in the blanks :

1. The cause of production of induced e.m.f. is change in
2. The dimensional formula of inductance is
3. In an ideal coupling the value of coupling coefficient K is
4. The ohmic resistance of an ideal inductor is
5. Due to eddy currents the decay of electrical energy is in the form of

Answer:

1. Magnetic flux
2. [ML²T^-2A^-2]
3. One
4. Zero
5. Heat energy.

Question 3.
Match the Column :
I.

Answer:

1. (c)
2. (d)
3. (b)
4. (e)
5. (a)

II.

Answer:

1. (e)
2. (c)
3. (b)
4. (d)
5. (a)

Question 4.
Write the answer ¡n one word/sentence:

1. What is the frequency of direct current?
2. Which effect is exhibited by alternating current?
3. What is the magnitude of inductive reactance for direct current?
4. What ¡s the magnitude of capacitive reactance for direct current?
5. What is the unit of \(\sqrt { LC }\)?
6. What ¡s average value of alternating current for one complete cycle?
7. What will be the magnitude of power coefficient for a L – C – R circuit in resonance state?
8. How many times the direction of alternating current charges for one complete cycle?
9. What will be the magnitude of power coefficient for pure inductor or capacitor?
10. What is the phase difference between e.m.f. and a.c. pure resistance, pure inductor and pure capacitor?

Answer:

1. Zero
2. Heating effect
3. Zero
4. Infinity
5. Second
6. Zero
7. One
8. Two times
9. Zero
10. Phase difference ϕ = 0

Alternating Current Very Short Answer Type Questions

Question 1.
In direct current circuit contaning pure inductance, what will be the value of inductive reactance?
Answer:
Since X_L= ωL for d.c. ω = 0, therefore X_L will be zero.

Question 2.
Can battery be charged with alternating current?
Answer:
No, it can’t be charged.

Question 3.
Which is more dangerous a.c. or d.c. at same voltage?
Answer:
Alternating current.

Question 4.
How will you recognize whether the current flowing through the coil of bulb in a.c. or d.c.?
Answer:
If on bringing magnet near the coil, if it vibrate then it will be alternating current.

Alternating Current Short Answer Type Questions

Question 1.
Define rms and peak value of alternating current and state the relation between them.
Answer:
Root mean square value:
Root mean square value of an a.c. is defined as that value of steady current which would generate the same amount of heat in a given resistance in a given time, as is done by the a.c. when passed through the same resistance for the same time. It is also called virtual value.

Peak value:
The maximum value of ax. is called its peak value or amplitude i.e.,
Peak value = \(\sqrt {2}\) x rms value
or I₀ = \(\sqrt {2}\) I_rms

Question 2.
a.c. does not show magnetic and chemical effects, why?
Answer:
a.c. flows in one direction during the first half cycle and in reverse direction in the second half cycle. Therefore, the average value of current for a complete cycle is zero. Thus, it does not show magnetic and chemical effects.

Question 3.
Why a.c. is more dangerous than d.c. of the same voltage?
Answer:
The maximum value of alternating e.m.f. is \(\sqrt {2}\) times of its virtual e.m.f. If virtual value is 220 V, then the peak value of e.m.f. = 220 x \(\sqrt {2}\) = 311.13 V. Hence, the voltage of ax. fluctuates between +311.13 V to – 311.13 V. Therefore, ax. is more dangerous.

Question 4.
a. c. cannot be measured by moving coil galvanometer, why?
Answer:
Moving coil galvanometer works on the principle of magnetic effect of current. a.c. does riot show magnetic effect because when ax. is passed through the coil, the direction and magnitude of current will frequently change and so the deflecting couple produces. But, in each complete cycle, the average deflecting couple is zero. So, there will be no deflection in the coil.

Question 5.
A bulb and a solenoid are connected in series to an a.c. source. If a soft – iron core is inserted inside the solenoid, what will be its effect on the intensity of light coming out of the bulb?
Answer:
The bulb will glow dimmer i.e., intensity decreases. This is because on introducing soft – iron core in the solenoid, the inductance increases. So, inductive reactance increases (by X_L = ωL), hence the current through the bulb decreases.

Question 6.
Write the expression for pure inductive reactance in an a.c. circuit and explain how it depends on the frequency of a.c.? Write its unit.
Answer:
Reactance of pure inductive component,
X_L = ωL = 2 πf L
Where, f= Frequency of a.c and L = Inductance of the coil
Clearly, X_L ∝ f
i.e., the reactance of pure inductive component is proportional to the frequency of a.c. Unit of reactance is ohm.

Question 7.
Write the expression for reactance of capacitor in an a.c. circuit. Explain how does it depend on frequency of a.c.? Write its unit.
Answer:
Capacitive reactance

Where, f = Frequency of a.c. and C = Capacitance of the capacitor. i.e., the capacitive impedance is inversely proportional to the frequency of a.c. Unit of capacitive reactance is ohm.

Question 8.
A capacitor stops d.c., why?
Answer:
In an a.c. circuit, reactance offered by capacitor i.e., capacitive reactance is

Hence, capacitor stops d.c.

Question 9.
What is the resistance of a coil of self – inductance L in a d.c. circuit?
Answer:
Resistance of coil in a.c. circuit i.e., inductive reactance is
X_L = ωL = 2πf L
For d.c., f = 0
Hence, resistance offered by the coil in d.c. circuit
X_L = 2π.0.L = 0.

Question 10.
In order to reduce the current in a.c. circuit, an inductor is more suitable than a resistance. Why?
Answer:
If a resistance is used in an a.c. circuit, a part of electrical energy is lost in the form of heat. Instead, if we use an inductance, there is no energy loss in the form of heat because power factor for inductive component is zero.

Question 11.
An electric heater is operated by direct current and alternating current respectively on the same potential difference. Will the heat liberated, in these two cases be the same? Give reasons for your answer.
Answer:
An electric heater containing a coil has resistance as well as inductance. So, its resistance is large for a.c. than for d.c.
From the formula for produced heat,
H = \(\frac { { V }^{ 2 } }{ R }\)
We conclude, less heat will therefore be produced by a.c.

Question 12.

1. What will be the power loss in a pure inductive or pure capacitive circuit when it is joined to a.c. source?
2. When will voltage and current in an LCR a.c. circuit be in same phase?

Answer:
1. When an inductor is joined to an a.c. source, then voltage is ahead of current by π/2, hence ϕ = π/2.
The power factor is P = V_rmsI_rms cosϕ
= V_rmsI_rms cos\(\frac {π}{2}\) = 0
Hence, there is no consumption of power. When a capacitor is joined, then voltage is lagging behind the current by \(\frac {π}{2}\) . Hence ϕ = \(\frac {π}{2}\) = 0
The power factor is P = V_rmsl_rms cosϕ
= V_rmsl_rms cos(- \(\frac {π}{2}\)) = 0
= 0
Hence, there is no consumption of power.

2. Voltage and current in an LCR a.c. circuit will be in the same phase when frequency is f = \(\frac { 1 }{ 2\pi \sqrt { LC } }\)

Question 13.
What is Wattless current?
Answer:
In a.c. circuit, when the average consumed power is zero, then that current is called wattless current.
If a pure inductance or pure capacitance is used in a.c. circuit, then the phase difference between current and e.m.f. becomes\(\frac {π}{2}\).
P_av = V_rms x I_rms x cosϕ
= V_rms x I_rms x cos\(\frac {π}{2}\)
= V_rms x I_rms x o = 0
Choke coil is made up of thick copper wire. Its ohmic resistance is negligible. Hence, the phase difference between current and e.m.f. is \(\frac {π}{2}\). Thus, the average consumed power becomes zero. Therefore, the current is called wattless.
By using a choke coil in an alternating circuit, we can obtain a wattless current.

Question 14.
For a circuit used for transporting electric power, a low power factor implies large power loss. Why?
Answer:
The power factor is given by cosϕ = \(\frac {R}{Z}\). Therefore, for low power factor, the resistance R will be less. Hence a large current will flow through the secondary coil of the transformer, which will produce more heat as H x I². Thus, a large power will be lost.

Question 15.
A transformer works with a.c. supply but not with d.c. Why?
Answer:
It is because transformer works on the principle of mutual induction. The input a.c. changes the flux (through the soft – iron core) linked with the secondary coil. This produces output current in the secondary coil. This is not possible with d.c. supply input.

Question 16.
In houses, electricity is supplied through a.c. but not through d.c. Why?
Answer:
This is because the strength of current in the form of a.c. can be reduced and supplied from one place to another by using a transformer and thus without power loss. Transformer cannot work with d.c.

Question 17.
Why laminated core is used in transformer?
Answer:
Effect of eddy currents is minimized in a transformer by using laminated core. Lamination causes greater resistance to eddy currents produced. Thus, loss of electrical energy in the form of heat is reduced.

Question 18.
Give difference between resistance and reactance.
Answer:
The obstruction to the flow of alternating current or direct current by the conductor is called resistance. The obstruction to the flow of alternating current offered by inductor or capacity is called reactance.

Question 19.
In series resonance L – C – R circuit what is the value of impendence.
Answer:
Since Z =\(\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } }\)
But in resonance stage X_L = X_C
∴ Z = R
That is impendence is equal to resistance.

Question 20.
What is the value of peak to peak value of alternating e.m.f.?
Answer:
It will be equal to sum of peak value of e.m.f. in positive half cycle and negative half cycle, i.e., peak to peak value of alternating e.m.f. will be 2E_m = 2\(\sqrt { 2 }\) E_rms.

Question 21.
What are the reasons of energy losses in transformers? How these losses are renuced?
Answer:
Energy losses and their removal:
1.Copper loss:
Some part of the energy is wasted in the form of heat due to the heating effect of current in primary and secondary coil because the coils have some resistance. The amount of heat produced is I² Rt. To reduce it, thick coils are used in the primary coil of step – up transformer and in the secondary coil of a step – down transformer.

2. Iron loss:
Eddy current is produced in the iron core of the transformer, causes heating. This loss is called iron loss. To minimize this loss, the core is laminated.

3. Magnetic flux leakage:
All the magnetic flux produced by primary coil may not be transferred to the secondary coil. Therefore, some energy is wasted. To minimize this loss, soft – iron core is used.

4. Hysteresis loss:
Some amount of energy is wasted because the iron core becomes magnetized during the first half and then gets demagnetized during the other half. This wastes loss of magnetic energy is called hysteresis loss. It is minimized by taking soft- iron core which has thin hysteresis loop.

Question 22.
Give differences between step – up and step – down transformer.
Ans.
Differences between a step – up and step – down transformer :
Step – up transformer

• It increases the voltage.
• It decreases the current strength.
• The number of turns in secondary coil is greater than that of primary coil.
• Its transformation ratio is more than one.

Step – down transformer

• It decreases the voltage.
• It increases the current strength.
• The number of turns in primary coil is greater than that of secondary coil.
• Its transformation ratio is less than one.

Question 23.
What do you mean by root mean square value of a.c. Obtain expression for I_rms.
Answer:
Root mean square value of a.c.:
Root mean square value of an a.c. is defined as that value of steady current which would generate the same amount of heat in a given resistance in a given time, as is done by the a.c. when passed through the same resistance for the same time. It is also called virtual value.

Expression for I_rms:
Let the alternating current is given by the equation
I = I₀sin ωt
This current flows through a resistance R, for a small interval of time dt, then the small amount of heat dH is produced in the resistance R.
Then dH = I² Rdt … (2)
For a complete cycle, the total heat produced will be obtained by integrating equation. (2)

If root mean square value of current is reprsented by Ims then the amount of heat produced in the same resistance R in the same interval of time will be
H = I²_rms RT … (5)
Now from equation (4) and (5), we get

This is required expression.

Question 24.
Prove for an a.c. circuit:
P_av = V_rms x I_rms x cosϕ
Where, the symbols have their usual meaning.
Answer:
Let the potential difference and current of a.c. circuit is given by
V=V₀sm ωt … (1)
and I = I₀sm(ωt – ϕ) … (2)
Where, ϕ is the phase difference between voltage and current.
Here the current lags behind the voltage
Now, instantaneous power, P = V.I
= V₀sin ωt.I₀sin{ωt – ϕ), [from eqns. (1) and (2)]
= V₀I₀sinωt.sin(ωt – ϕ) = \(\frac {1}{2}\)V₀I₀.2sinωt.sin(ωt – ϕ)
= \(\frac {1}{2}\)V₀I₀[cosϕ – cos(2ωt – ϕ)]
= \(\frac {1}{2}\)V₀I₀cosϕ – = \(\frac {1}{2}\)V₀I₀cosϕ(2ωt – ϕ)
But, the value of cos(2ωt – ϕ) for a complete cycle is zero. Therefore, average power for a complete cycle is
P_av = \(\frac {1}{2}\)V₀I₀cosϕ or P_av = \(\frac { { V }_{ 0 } }{ \sqrt { 2 } }\).\(\frac { { I }_{ 0 } }{ \sqrt { 2 } }\).cosϕ
P_av = V_rms x I_rms x c0sϕ
Where, V_rms and I_rmsare average alternating e.m.f. and current and cosϕ is called power factor, its value depends upon the nature of the circuit.

Question 25.
What is meant by wattless current? When the current becomes wattless? Why the current flowing through choke coil is wattless?
Answer:
In a.c. circuit, when the average consumed power is zero, then that current is called wattless current.
If a pure inductance or pure capacitance is used in a.c. circuit, then the phase difference between current and e.m.f. becomes \(\frac {π}{2}\)
P_av = V_rms x I_rms x c0sϕ
= V_rms x I_rms x cos\(\frac {π}{2}\)
= V_rms x I_rms x 0= 0
Choke coil is made up of thick copper wire. Its ohmic resistance is negligible. Hence, the phase difference between current and e.m.f. is \(\frac {π}{2}\). Thus, the average consumed power becomes zero. Therefore, the current is called wattless. By using a choke coil in an alternating circuit, we can obtain a wattless current.

Question 26.
What is the principle of choke coil? Explain its use for controlling current. Why is current flowing in choke coil known as wattless current?
Answer:
Principle of choke coil:
A coil of very low resistance and high inductance is used to control current in an a.c. circuit. This coil is known as choke coil. It causes very little loss of energy. If an a.c. circuit contains zero resistance and a pure inductance, the phase difference between e.m.f. and current is 90°. Under this condition, average power

P_av = V_rms.I_rms cosϕ
= V_rms.I_rms x cos90°
=0, ( ∵ cos90° = 0)
So, there is no loss of energy. But in actual practice, the resistance in the circuit cannot be zero. Therefore, a small part of electrical energy is lost as heat. A choke coil is made up of a thick copper wire wound around a soft – iron core. Due to soft – iron core and so many turns of wire, the inductance is increased and the thick wire reduces its resistance.

Use:
Current in a.c. circuit can be controlled by using a choke coil in the circuit. Since, impedance of the choke coil is XL = caL. So, the current is reduced if inductance is high and is increased if the inductance is low.

Wattless current:
Current in a choke coil is called wattless current because average power consumed is zero (due to 90° phase difference between e.m.f. and current in the coil). The loss of energy due to small resistance of the coil is negligible.

Question 27.
Give differences between a.c. and d.c.
Answer:
Differences between a.c. and d.c. :
a.c.:

• Its direction and magnitude both changes.
• Transformers are used in a.c.
• It cannot be used for electroplating.
• It is generally not used for electro – magnet.
• The measuring instruments are based on its heating effect.
• It is dangerous.

d.c.:

• The magnitude may change but direction does not change.
• Transformers cannot be used.
• It can be used in electroplating.
• It is used for electromagnet.
• Its measuring instruments are based on its magnetic effect.
• It is less dangerous than a.c.

Question 28.
For a circuit used for transporting electric power, a low power factor implies large power loss. Why?
Answer:
The power factor is given by cosϕ =\(\frac {R}{Z}\). Therefore, for low power factor, the resistance R will be less. Hence a large current will flow through the secondary coil of the transformer, which will produce more heat as H x I². Thus, a large power will be lost.

Question 29.
An a.c. circuit contains a pure resistive component. Obtain an expression for current. What is the phase difference between current and voltage? Draw current voltage graph.
Answer:
Let an a.c. source be connected with a resistance R as shown in figure. The alternating e.m.f. is given by the equation.
V = V₀ sinωt … (1)

The corresponding value of current at any instant is
I = \(\frac {V}{R}\) = \(\frac { { V }_{ 0 }sin\omega t }{ R }\)
or I = I₀ sinωt, (∵ I₀ = \(\frac { { V }_{ 0 } }{ R }\) … (2)
Here, V = IR is valid because Ohm’s law is applicable for both a.c. and d.c.

It is. clear from equation (1) and (2) that the phase of both E and I is cot. Hence, current and voltage are in the same phase. Fig. shows variation of current and voltage in an a.c. circuit with respect to time.

Question 30.
An alternating e.m.f. V = V₀ sinωt is applied across a purely inductive a.c. circuit. Calculate the expression for the following :

1. Current flow
2. Inductive reactance and
3. Phase relation between current and voltage and its graph.

Answer:
Suppose an alternating e.m.f. V = V₀ sinωt … (1)
is applied across an a.c. circuit of inductance L. At the instant, when the current is maximum, the rate of change of current is zero. Hence, potential difference is zero, so induced e.m.f. = -L \(\frac {dI}{dt}\)

To maintain the flow of current, the applied voltage must be equal and opposite to the induced e.m.f.

This is the required expression for current flow.

Where, aL is called inductive reactance and is denoted by XL.
Thus, X_L = ωL = 2πfL
Where, f = Frequency of current.
Phase relation between current and voltage:
From eqns. (1) and (3), we find that the phase of voltage is at, whereas that of current is at ωt – \(\frac {π}{2}\). Hence, current lags behind the voltage by \(\frac {π}{2}\). This is shown in the phaser diagram below.

Question 31.
An alternating voltage V = V₀ sinωt is applied across a capacitor C. Obtain an expression for

1. Circuit current
2. Capacitive reactance and
3. Phase relation between current and voltage.

Answer:
1. Suppose the instantaneous voltage at any instant of time t is
V = V₀ sinωt … (1)
If Q is the charge acquired by the capacitor at this instant, c
then
c = \(\frac {Q}{V}\)
or Q = CV or Q = C V₀ sinωt

But the current at any instant is given by

2. From Equation (3), \(\frac { { V }_{ 0 } }{ { I }_{ 0 } }\) =  \(\frac {1}{ωC}\)
Comparing with Ohm’s law, \(\frac {1}{ωC}\)is called capacitive reactance denoted by X_c.
X_c = \(\frac {1}{ωC}\) =\(\frac {1}{2πfC}\)

3. Phase relation between current and voltage:
From eqns. (1) and (2), we find that the phase of voltage is ωt, whereas of current is ωt + \(\frac {π}{2}\) hence current is ahead of voltage by \(\frac {π}{2}\) or we can say that voltage is lagging behind the current by n! 2. This is shown in the figure below:

Question 32.
Explain the behaviour of inductor and capacitor towards a.c. and d.c.
Answer:
The reactance offered by an inductor is given by the equation
X_L = ωL = 2πfL.
Where f is the frequency of the source and L is the inductance of the inductor
For a d.c. f = 0 thus X_L = 0.
For an a.c. f ≠ 0 and X_L ∝L.
Flence, an inductor provides no reactance to the flow of d.c. and provide reactance of the a.c. i.e. an inductor by passes d.c. and blocks a.c.
The reactance offered by a capacitor is given by the equation,
X_c = \(\frac {1}{ωC}\) = \(\frac {1}{2πfC}\)
Where C is the capacitance of the capacitor.
For d.c. f = 0 thus, X_c = ∞
For an a.c. f ≠ 0 and X_c ∝\(\frac {1}{f}\)
Hence, a capacitor provides infinite reactance to d.c. and with increasing frequency of an a.c. the reactance decreases, i.e. a capacitor by passes a.c. and blocks d.c.

Question 33.
What is Q – factor? Write the expression for it.
Answer:
In an L – C – R circuit when a capacitor gets charged, electrical energy is stored in it, and when current flows through the conductor, magnetic energy gets stored in it. During the half cycle of a.c. magnetic energy is maximum and electrical energy is zero and in the next half cycle magnetic energy becomes zero and electrical energy is maximum.

Moreover some amount of energy gets dissipated in the form of heat in the ohmic resistor. Less is the dissipation of energy in the circuit, more sharp will be the resonance. The measurement of the sharpness is done by a dimensionless quantity called ‘Quality factor’ or ‘Q – factor’.

Definition:
In an L – C – R circuit, the ratio of resonant frequency to the difference of its neighbouring frequencies so that their corresponding current is \(\frac { 1 }{ \sqrt { 2 } }\) times of the peak value is called Q factor of the circuit. Its expression is Q factor = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)

Q factor to be maximum :

1. Value of R should be less.
2. Value of \(\frac {L}{C}\) should be more.

Alternating Current Long Answer Type Questions

Question 1.
Describe the transformer on the basis of follow ing points :

1. Principle
2. Types of transformer
3. Any three energy losses.

Or
Explain construction, principle and working method of transformer.
Answer:
Labelled diagram :

Construction:
A transformer consists of laminated core, primary coil and secondary coil. The laminated core is obtained by piling a number of laminated rectangular strips of soft iron. Two insulated copper wires are wound on the opposite arms, in the form of coils. The coil connected to the input of a.c. is called primary coil and the other through which output is taken is called secondary coil.

Method of working and principle:
Let the number of turns in primary and secondary coils are N_P and N_s respectively. If the magnetic flux linked with the primary coil at any instant is ϕ, then the e.m.f. induced in the primary coil will be
E_P = – N_P\(\frac {dϕ}{dt}\) … (1)
If there is no loss of flux, the secondary coil will also be linked with the same flux ϕ. So, the e.m.f. induced in the secondary coil will be
E_s = – N_s\(\frac {dϕ}{dt}\) … (1)
Dividing equation (2) by equation (1), we get

If there is no loss of energy in the primary coil, then the induced e.m.f. produced in the primary coil will be nearly equal to the applied potential difference (V_P) between its ends. Similarly, because the secondary coil is open, hence the potential difference across its ends will be equal to the e.m.f. induced in it i.e., under ideal conditions

Let I_p and I_s be the current through primary and secondary coils respectively. Then under ideal conditions,
Instantaneous input power = Instantaneous output power
i.e., Power in primary coil = Power in secondary coil

The quantity r in equation (5) and (7) is called transformation ratio.
For step – up transformer:

So, a step-up transformer increases the voltage, but decreases the strength of current. Also, the number of turns in the secondary coil is more than that of primary coil.

For step – down transformer:

So, a step – down transformer decreases the voltage but increases the strength of current. The number of turns in secondary coil is less than that of primary coil.

Energy losses:
1.Copper loss:
Some part of the energy is wasted in the form of heat due to the heating effect of current in primary and secondary coil because the coils have some resistance. The amount of heat produced is I²Rt. To reduce it, thick coils are used in the primary coil of step – up transformer and in the secondary coil of a step – down transformer.

2. Iron loss:
Eddy current is produced in the iron core of the transformer, causes heating. This loss is called iron loss. To minimize this loss, the core is laminated.

3. Magnetic flux leakage:
All the magnetic flux produced by primary coil may not be transferred to the secondary coil. Therefore, some energy is wasted. To minimize this loss, soft – iron core is used.

Effect of eddy currents is minimized in a transformer by using laminated core. Lamination cause greater resistance to eddy currents produced. Thus, loss of electrical energy in the form of heat is reduced.

Question 2.
Describe a transformer on the basis of following points :

1. Principle
2. Its types and labelled diagram
3. Energy losses in transformer.

Or

1. Working principle
2. Types and labelled diagram
3. Energy losses
4. Equation of transformation ratio.

Answer:
1. Principle :
Let the number of turns in primary and secondary coils are N_P and N_s respectively. If the magnetic flux linked with the primary coil at any instant is ϕ, then the e.m.f. induced in the primary coil will be
E_P = – N_P\(\frac {dϕ}{dt}\) … (1)
If there is no loss of flux, the secondary coil will also be linked with the same flux ϕ. So, the e.m.f. induced in the secondary coil will be
E_s = – N_s\(\frac {dϕ}{dt}\) … (1)
Dividing equation (2) by equation (1), we get

If there is no loss of energy in the primary coil, then the induced e.m.f. produced in the primary coil will be nearly equal to the applied potential difference (V_P) between its ends. Similarly, because the secondary coil is open, hence the potential difference across its ends will be equal to the e.m.f. induced in it i.e., under ideal conditions

Let I_p and I_s be the current through primary and secondary coils respectively. Then under ideal conditions,
Instantaneous input power = Instantaneous output power
i.e., Power in primary coil = Power in secondary coil

The quantity r in equation (5) and (7) is called transformation ratio.

2. Types and labelled diagram :

3. Energy losses:
(i) Copper loss:
Some part of the energy is wasted in the form of heat due to the heating effect of current in primary and secondary coil because the coils have some resistance. The amount of heat produced is I²Rt. To reduce it, thick coils are used in the primary coil of step – up transformer and in the secondary coil of a step – down transformer.

(ii) Iron loss:
Eddy current is produced in the iron core of the transformer, causes heating. This loss is called iron loss. To minimize this loss, the core is laminated.

(ii) Magnetic flux leakage:
All the magnetic flux produced by primary coil may not be transferred to the secondary coil. Therefore, some energy is wasted. To minimize this loss, soft – iron core is used.

(iv) Hysteresis loss:
Some amount of energy is wasted because the iron core becomes magnetized during the first half and then gets demagnetized during the other half. This wastes loss of magnetic energy is called hysteresis loss. It is minimized by taking soft – iron core which has thin hysteresis loop.

4. Equation of transformation ratio:
Let the number of turns in primary and secondary coils are N_P and N_s respectively. If the magnetic flux linked with the primary coil at any instant is ϕ, then the e.m.f. induced in the primary coil will be
E_P = – N_P\(\frac {dϕ}{dt}\) … (1)
If there is no loss of flux, the secondary coil will also be linked with the same flux ϕ. So, the e.m.f. induced in the secondary coil will be
E_s = – N_s\(\frac {dϕ}{dt}\) … (1)
Dividing equation (2) by equation (1), we get

Question 3.
Obtain an expression for impedance in L – C – R circuit. Under what condition, the current will be maximum?
Or
What is resonant circuit ? Find the resonant frequency of L – C – R circuit.
Or
Obtain the condition of resonant frequency in the L – C – R circuit. Where this circuit is used ?
Or
Explain the resonance of a series L – C – R circuit under the following heads :

1. Resultant voltage
2. Impedance of circuit
3. Frequency of resonance.

Answer:
In L – C circuit or in L – C – R circuit, when the value of impedance is least, the current becomes maximum. This phenomenon is called resonance. In this state, the frequency of the applied e.m.f. is equal to the natural frequency of the L – C circuit called resonant circuit.

Expression for the impedance:
Let inductance L, capacitor C and resistance R are joined in series to an a.c. circuit as shown in the Fig. (a).
Let the instantaneous current is I.
Potential difference across L will be, V_L=I.X_L … (1)
Potential difference across C will be, V_C = I.X_C … (2)
and Potential difference across R will be, V_R= I. R … (3)
Now, V_R and f are in the same phase and V_L leads I by 90° and V_C lags I by 90°. Hence, the angle between V_L and V_C will be 180°.

Now, resultant of V_L and V_C is V_L – V_C, then angle between V_L – V_C and V_R is 90°.
Resultant,
V² = V²_R + (V_L – V_C )²
V² = I²R² + I² (X_L – X_C)²
or R² + (X_L – X_C)²

By comparing equation (4) with Ohm’s law, we get the resultant resistance also

∴ Impedance of the circuit,

This is the equation for impedance from Fig. (a).
If ϕ be the phase difference between V and I, then from Fig. (b). V,

Now, the current flowing through the circuit will be given by
I = I₀ sin(ωt – ϕ)
With respect to voltage, the current is behind ϕ.

Since, impedance is minimum and current is maximum, therefore this state is called resonance.
Again, under resonance condition,

This is the required expression.
This circuit is used for tuning the radio receiver.

Alternating Current Numerical Questions

Question 1.
For a L – R circuit, the reactance of inductor is 3Ω and ohmic resistance is 4Ω . Find out impedance of the circuit.
Solution:

Question 2.
In a transformer, the ratio of number of turns in primary coil to that of secondary coil is 20:1. If it is operated at 200 volt then find out voltage obtained across secondary coil.
Solution:

Question 3.
Alternating current of \(\sqrt { 2A }\) flows through a circuit, what is the maximum value of current ?
Solution:
Maximum value I₀ = \(\sqrt {2}\) × I_rms
I₀ = \(\sqrt {2}\) × \(\sqrt {2}\) = 2A.

Question 4.
The peak value of alternating current is \(\sqrt { 2A }\) . Find out r. m. s. value of the current.
Solution:
Formula:
I_rms = \(\frac { { I }_{ 0 } }{ \sqrt { 2 } }\) = \(\frac { 2\sqrt { 2 } }{ \sqrt { 2 } }\)

Question 5.
In domestic circuit mains voltage is 220 V, find out its maximum value.
Solution:

Question 6.
The r.m.s. value of current in an a.c. circuit is 10A. Find the peak value of current.
Solution:
Given, I_rms = 10A
Formula: I_rms = \(\frac { { I }_{ 0 } }{ \sqrt { 2 } }\) or I₀ = I_rms x \(\sqrt {2}\)
Putting the value in the formula, we get
I₀ = 10 x \(\sqrt {2}\) = 10 x 1.414 = 14.1A.

Question 7.
The equation of a alternating voltage is V = 141 sin50πt. Determine r.m.s. value of voltage.
Solution:
Given, V = 141 sin 50πt.
Comparing with the equation V=V₀ sinωt, we get

Question 8.
An e.m.f. of a.c. source is given by E = 300 sin 314t. Obtain the peak voltage and the frequency of the source.
Solution:
Comparing with E = E₀ sinωt
∴ Peak voltage E₀ = 300 V.
Here, ω = 314
or 2πf = 314
or f = \(\frac {314}{2π}\)
= \(\frac {314}{2×3.14 }\)
= 50 Hz

Question 9.
Self – inductance of a coil is 0.025 henry, calculate the inductive impedance for 50 Hz a.c.
Solution:
Formula: X_L= ωL = 2πfL.
Given: f = 50 Hz and L = 0.025 H.
X_L = 2 x 3.14 x 50 x 0.025
= 78.5Ω

Question 10.
A 1.5µF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the current in the circuit. (NCERT)
Solution:

Question 11.
The potential drops across a resistor, capacitor and inductor in an a.c. circuit are 80 V, 100 V and 40 V respectively. Calculate the e.m.f. of the source of given a.c. circuit.
Solution:
Formula: V² = V²_R + (V_L – V_C)²
Given : V_R = 80 V, V_L = 40V and V_C = 100V.
Substituting these values in the formula,
V² =80²+(40 – 100)²
= 6400 + 3600 = 10000
∴ V = 100V.

Question 12.
The ratio of number of turns in primary to that of secondary coil of a step up transformer is 1:10. If it is connected to the mains of 220 V, a current of 5 A flows. Calculate the induced e.m.f. and induced current in the secondary coil.
Solution:

Question 13.
A step – down transformer changes 2200 V to 220 V. The number of turns in primary coil is 5000. If its efficiency is 90% and output power is 8 kW, calculate :

1. Number of turns in secondary coil and
2. Input power.

Solution:

MP Board Class 12th Physics Important Questions

MP Board Class 12th Physics Important Questions Chapter 9 Ray Optics and Optical Instruments

Ray Optics and Optical Instruments Important Questions

Ray Optics and Optical Instruments Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The virtual and smaller image can be obtained by :
(a) Plane mirror
(b) Convex mirror
(c) Concave mirror
(d) None of these
Answer:
(b) Convex mirror

Question 2.
Keeping the incident ray unchanged if the mirror is rotated by 30° angle then the reflected ray will be deviated through :
(a) 30°
(b) 60°
(c) 15°
(d) 45°.
Answer:
(b) 60°

Question 3.
A light ray falls normally on a mirror. The value of angle of reflection will be :
(a) 0°
(b) 90°
(c) 135°
(d) 45°.
Answer:
(a) 0°

Question 4.
The relation between focal length and radius of corvature of a spherical mirror is:
(a) f = 2R
(b) f = R
(c) R = 2f
(d) None of these.
Answer:
(c) R = 2f

Question 5.
Magnified and virtual image can be obtained by :
(a) Convex mirror
(b) Concave mirror
(c) Plane mirror
(d) All the above.
Answer:
(b) Concave mirror

Question 6.
The image formed by a concave mirror cannot be obtained on the screen if the object is placed between :
(a) Pole and focus
(b) Focus and centre of curvature
(c) Centre and infinity
(d) At centre of curvature.
Answer:
(a) Pole and focus

Question 7.
Which of the following statement is false to form a real image by concave mirror:
(a) Object is at focus
(b) Object is at centre of curvature
(c) Object is at infinity
(d) Object is between pole and focus.
Answer:
(d) Object is between pole and focus.

Question 8.
The image formed by a convex mirror is :
(a) Virtual
(b) Real
(c) Magnified
(d) Inverted.
Answer:
(a) Virtual

Question 2.
(I) Fill in the blanks :

1. The phenomenon of deviation in the path of a light ray while entering from one optical medium to the other optical medium is called ……………..
2. The ratio of sine of angle of incidence to the sine of angle of refraction is called as ……………..
3. If the path of a light ray after multiple refractions is reversed then it retraces the path from which it came. This phenomenon is called ……………..
4. The angle of incidence corresponding to which the angle of refraction becomes 90° is called as ……………..
5. The illusion of water in desert is called ……………..
6. In cold countries at the port trust the ship appears to be inverted in air. This illusion is called as ……………..
7. The prism in which the inverted image can be obtained in same direction in erected position is called ……………..
8. The is a device which is based on the total internal reflection through which the optial signal can be transmitted from one place to other place by a zig – zag path.
9. The distance of pole from first principle focus is called ……………..
10. The limit of resolution of healthy eye is …………….. (MP 2016)
11. The instruments by which the far distant or nearby object can be seen are called ……………..
12. The angle subtended by an object at the eye is called ……………..
13. On increasing the aperture of objective lens of astronomical telescope, its …………….. increases.
14. The ratio of angle of vision subtended by image to that subtended by object when object is kept at least distance of distinct vision is called …………….. of microscope.
15. To increase the magnifying power of a microscope, two different lenses are used. This microscope is called ……………..
16. In a compound microscope the lens towards the object is called and the lens towards the eye is called ……………..
17. …………….. is an optical instrument with the help of which the far distant object are seen.
18. With the help of …………….. the astronomical bodies are seen.
19. With the help of …………….. the far distant objects situated on earth are seen.

Answer:

1. Refraction of light
2. Refractive index
3. Principle of reversibility
4. Critical angle
5. Desert mirage
6. Cold mirage
7. Inverting prism
8. Optical fibre,
9. Angle of vision
10. Resolving power
11. Magnifying power
12. Compound microscope
13. Objective lens, eyepiece
14. Telescope
15. Astronomical telescope
16. Terrestrial telescope.

Question 3.
Match the Column :
I.

Answer:

1. (e)
2. (d)
3. (b)
4. (a)
5. (c)

II.

Answer:

1. (c)
2. (e)
3. (d)
4. (b)
5. (a)

Question 4.
Write the answer in one word / sentence:

1. If a ray of light incident normally on a mirror, then what will be angle of reflection?
2. From which spherical mirror we get magnified image?
3. From which spherical mirror we get always a virtual and smaller than object?
4. Write the relation between radius of curvature and focal length for a spherical mirror?
5. Conjugate foci is true for which type of images?
6. When a spherical mirror is immersed in water, what changes we get in its focal length?
7. What will be the magnitude of magnification of a convex mirror for its real image and virtual image?

Answer:

1. Zero
2. Concave mirror
3. Convex mirror
4. f = \(\frac {R}{2}\).
5. For real images
6. There is no changes in focal length
7. For real image it is negative and for virtual image it is positive.

Question 4.
Write the answer in one word / sentence :

1. If a ray of light incident normally on a mirror, then what will be angle of reflection?
2. From which spherical mirror we get magnified image?
3. From which spherical mirror we get always a virtual and smaller than object?
4. Write the relation between radius of curvature and focal length for a spherical mirror?
5. Conjugate foci is true for which type of images?
6. When a spherical mirror is immersed in water, what changes we get in its focal length?
7. What will be the magnitude of magnification of a convex mirror for its real image and virtual image?

Answer:

1. Zero
2. Concave mirror
3. Convex mirror
4. f = \(\frac {R}{2}\)
5. For real images
6. There is no changes in focal length
7. For real image it is negative and for virtual image it is positive.

Ray Optics and Optical Instruments Very Short Answer Type Questions

Question 1.
Which property remain unchanged in the phenomenon of refractions of light?
Answer:
Frequency.

Question 2.
Which property get changed in the phenomenon of refraction of light?
Answer:
Velocity, wavelength and amplitudes.

Question 3.
Write one characteristics of image formed by convex mirror.
Answer:
Image is formed behind the mirror.

Question 4.
If a water tank is seen inclinedly what will its apparent depth will be seen?
Answer:
Its apparent depth will decrease when seen inclinedly.

Question 5.
What are the conditions for total internal reflection?
Answer:

1. Angle of incidence should be more than critical angles.
2. Light rays must enter from denser medium to rarer medium.

Question 6.
On which phenomenon optical fibre is depend?
Answer:
Total internal reflection.

Question 7.
When the value of relative refractive index is less than one and more than one?
Answer:
Refractive index of rarer medium with respect to denser medium is less than one and refractive index of denser medium with respect to rarer medium is more than one.

Question 8.
If a concave mirror and convex lens is dipped in water, what will be the changes in their focal length?
Answer:
The focal length of concave mirror will remain unchanged and that of convex lens will increase.

Question 9.
Can a lens be used in that medium from which it is made of?
Answer:
No.

Question 10.
If an object is kept at focus of a concave lens where its image will formed?
Answer:
Image will be formed between focus and optical centre of the lens.

Question 11.
The power of which lens be positive?
Answer:
Convex lens.

Question 12.
The power of which lens is negative?
Answer:
Concave lens.

Question 13.
The power of a lens is more in water or air?
Answer:
Air.

Question 14.
What is focal length and power of a rectangular slab of glass?
Answer:
P = \(\frac {1}{f}\) = \(\frac {1}{∞}\) = 0. Power is zero.

Question 15.
Write the unit of Dispersive power.
Answer:
It has no unit.

Question 16.
Write the formula for Rayleigh’s law.
Answer:
I ∝\(\frac { 1 }{ { \lambda }^{ 2 } }\)

Question 17.
Why sky is seen blue?
Answer:
Sky is seen blue due to scattering of light.

Question 18.
Write the formula for angular dispersion.
Answer:
θ = A(µ_v – µ_r)
or θ = A(δ_v – δ_r).

Question 19.
When the light rays is parallel to the base of prism?
Answer:
At angle of minimum deviation.

Question 20.
Write the relation between angular dispersion and dispersive power.
Answer:
ω = \(\frac {θ}{δ}\) = θ = ω x δ.

Question 21.
Which microscope is used as reading lens?
Answer:
Simple microscope.

Question 22.
How resolving power of microscope depends on wavelength of light?
Answer:
When wavelength of light decreases, its resolving power increases.

Question 23.
Write the formula of magnification for terrestrial telescope.
Answer:
m = f₁ – f_e.

Question 24.
The length of which telescope is more among reflective or refractive telescope.
Answer:
The length of refractive telescope is more.

Question 25.
How image is turned of simple microscope?
Answer:
Erect and larger than object.

Question 26.
How final image is formed for compound microscope?
Answer:
Virtual, inverted and larger than object.

Question 27.
If the focal length of eye lens of a microscope is decreased then what will be the effect in its magnifications?
Answer:
Its magnification will increase.

Question 28.
If the focal length of objective lens of microscope and telescope is increased, then what will be the effect in their magnifications?
Answer:
Magnifications of microscope will decrease and that of telescope will get increased.

Ray Optics and Optical Instruments Short Answer Type Questions

Question 1.
What is mirror? What are the types of it?
Answer:
The highly polished reflective surface is called mirror, it is of three types :

1. Plane mirror
2. Concave mirror
3. Convex mirror.

Question 2.
Write the definition of convex mirror and concave mirror.
Answer:
1. Convex mirror:
If the internal surface of a spherical mirror is polished and reflection of light occur from external surface, it is called convex mirrors.

2. Concave mirror:
If the external surface of a spherical mirror is polished and reflection of light occur from internal surface, it is called concave mirror.

Question 3.
Write the formula for focal length of spherical mirror and explain it.
Answer:
\(\frac {1}{f}\) = \(\frac {1}{v}\) + \(\frac {1}{u}\) where f is focal length, v is image distance and u is object distance for a spherical mirror.

Question 4.
Define magnification of a spherical mirror.
Answer:
The ratio of size of image to size of object is known as magnification.
m = \(\frac {h’}{h}\) = \(\frac {v}{u}\)

Question 5.
Concave mirror is used for shaving, why?
Answer:
When the object is within the focal length of the concave mirror then virtual, erect and magnified image is formed. This is the reason that concave mirror is used for shaving.

Question 6.
If a glass slab is kept in alphabet of different colour. Will all the alphabet be seen to be raise uniformly?
Answer:
No, red colour alphabet will be seen to be raised more as its refractive index is less.

Question 7.
What is the cause of brilliance of diamond?
Answer:
Diamond shines due to total internal reflection. Refractive index of diamond is 2.42 and critical angle is 24.41°, which is very less. Diamond is cut in such a way that light is incident on its faces at an angle more than 24.41°. The incident light undergoes total internal reflection repeatedly and hence gets collected inside, but it comes out only through few faces. Hence diamond sparkles when seen in the direction of emergent light.

Question 8.
What will be the effect on the focal length of a concave mirror and convex lens when they are immersed in water?
Answer:
The focal length of concave mirror will remain same but the focal length of convex lens will be increased.

Question 9.
Define refractive Index.
Answer:
When the light rays enter from one conjugate medium to another conjugate medium, the ratio of sine of angle of incidence to sine of angle of refraction is known as refractive index of medium.

Question 10.
Write the reason for twinkling of stars.
Answer:
As we go on higher and higher altitude the density of air goes on decreasing and hence the refractive index. Also, the atmosphere consists of several moving masses of air and the temperature is also changing, because of all these reasons the rays coming from the stars continuously changing the path.

Therefore, the quantity of rays entering the eye of an observer, changes. When these rays enter the eye the star appears brighter and so. Thus, the stars appear twinkling. The moon is very near to the surface of earth. The light coming out of the moon does not undergo much changes when it crosses the earth’s atmosphere. So, moon does not appear twinkling.

Question 11.
Write any two conditions for total internal reflection.
Answer:

1. The ray of light must travel from denser medium to rarer medium.
2. The angle of incidence (i.e., critical angle) for which angle of refraction should be 90°.

Question 12.
Write the formula for refraction for a lens.
Answer:
\(\frac {1}{f}\) = (µ – 1)(\(\frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } }\))

Question 13.
Define power of a lens.
Answer:
The reciprocal of focal length of a lens in metre is called its power.
p = \(\frac {1}{f}\)(m)

Question 14.
A concave lens and a convex lens of same focal length is kept in contact. What will be the focal length and power of the combination?
Answer:
\(\frac {1}{F}\) = \(\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } }\)
f₁ = – f and f₂ = f
∴ \(\frac {1}{F}\) = –\(\frac {1}{f}\)+\(\frac {1}{f}\) = 0
or F = 0
and power p = \(\frac {100}{f}\) = \(\frac {100}{∞}\)

Question 15.
What do you mean by magnification? Write its formula.
Answer:
The ratio of sine of image to sine of object is defined as magnifications, its formula is
m = \(\frac {I}{O}\) = \(\frac {v}{u}\)

Question 16.
Draw ray diagram of image formation when an object is placed at focus of a convex lens.
Answer:
The image formed at infinity. Image is real and inverted.

Question 17.
A glass prism causes dispersion while a glass plate does not. Why?
Answer:
When white light passes through a prism, it splits up into its constituent colours because refractive index of glass is different for different colours. The angular dispersion is (µ_V – µ_R) A and angular deviation is (µ – 1) A. The refracting faces of glass plate are parallel, so, A = 0. Hence, there is neither dispersion nor deviation. All the emergent rays are parallel to incident ray.

Question 18.
Define dispersive power.
Answer:
The ratio of the angular dispersion between the violet and red emergent rays and the deviation suffered by the mean ray (yellow colour) is called dispersive power (ω) of the material of the prism.

Question 19.
Define angular dispersion. On what factors it depends and how?
Answer:
When light rays get dispersed into constituent colours, angle between any two colours (in general violet and red) is called angular dispersion.
If δ_V is the deviation for violet light and SR for the red light, then angular dispersion (θ) is,
θ = δ_V – δ_R.

Question 20.
Danger signals are always red, why?
Answer:
When light falls on the signal, all colours are scattered much more than that of red colour. So it suffers least scattering and reachs directly to the eye of the observer. Hence it can be seen from far – off distances.

Question 21.
Sun appears red during the sunrise and sunset, why?
Answer:
Sunlight has to travel through the longer distances at the time of sunset and sunrise, meanwhile almost all the coloured light waves get scattered except red as its wavelength is more, it is unable to scatter due to atmospheric molecules. Therefore, only red light rays reach our eyes and sun appears red.

Question 22.
What do you mean by dispersion and spectrum?
Answer:
When white light passes through the prism, it gets splitted into its constituent colours. This process is called dispersion. Colours in group is called spectrum. Order of colours in spectrum is violet, indigo, blue, green, yellow, orange, red.

Question 23.
Why does the light rays disperse through prism?
Answer:
White light consists of continuous range of wavelength. Refractive index of prism material is different for different colours and is given by Cauchy relation :
µ = a + \(\frac { b }{ { \lambda }^{ 2 } } +\frac { c }{ { \lambda }^{ 2 } }\)
Where a. b and c are constants. For small angle of prism, the angle of deviation is given by
δ = (µ – 1)A ⇒ δ ∝µ
Now,
λ_red > λ_violet
∴ µ_red < µ_violet
∴ δ_red < δ_violet
i.e., the red colour is deviated less than the violet colour. Other colours are deviated by intermediate angles. So, different colours of white light get dispersed on refraction through prism.

Question 24.
What do you mean by angular dispersion? Derive an expression for it. On “what factors does it depends?
Answer:
The angular dispersion is defined as difference between angle of deviations for any two colours. If angle of deviations for red and violet colours are δ_r and δ_v respectively then,
Angular dispersion θ = δ_v – δ_r
∴ δ_v = A(µ_v – 1), δ_r = A (µ_r – 1)
∴ θ = A(µ_v – 1) – A(µ_r – 1)
or = θ = A(µ_v – µ_r)
This is the required expression for angular dispersion.
θ depends on :

1. Angle of prism (A) : With increase in A, θ increases.
2. Material of prism : With increase in ‘µ’ θ also increases.

Question 25.
If the aperture of the objective lens of telescope is increased, then what would be the effect on :

1. Magnifying
2. Resolving power

Answer:
1. The magnifying power remains unchanged.

2. Resolving power is increased as
Resolving power = \(\frac {a}{1.22λ}\)
Where, a = aperture.

Question 26.
How the magnifying power of an astronomical telescope is increased?
Answer:
The magnifying power of an astronomical telescope is given by :
m = – \(\frac { { f }_{ 0 } }{ { f }_{ e } }\)
Hence, for greater magnifying power of the value f₀, i.e., focal length of objective lens should be large and f_e, i. e. focal length of eye lens should be small.

Question 27.
If the telescope is inverted, then the size of the image becomes very small, but it does not happen in microscope. Why?
Answer:
The magnifying power of a telescope is m = – \(\frac { { f }_{ 0 } }{ { f }_{ e } }\). If the telescope is inverted, then the magnification will become m = – \(\frac { { f }_{ 0 } }{ { f }_{ e } }\). As f_e << f₀ hence, the size of the image will be very less.

In compound microscope the magnification is m = – \(\frac { { v }_{ 0 } }{ { u }_{ 0 } } \frac { D }{ { f }_{ e } }\) The value of u₀ is little more than f₀. Hence, the formula for magnification can be written as m = – \(\frac { { v }_{ 0 } }{ { u }_{ e } } \frac { D }{ { f }_{ e } }\) reversing the microscope, the angle subtended by v₀ will be the same. So, the magnification remains unchanged.

Question 28.
The moon is several times smaller than sun. Even then at the time of eclipse, the entire sun is covered by moon. How?
Answer:
The distance of sun from the earth is 1000 times much larger than the distance of moon from earth. During solar eclipse, the moon comes in between sun and earth. But the sun and moon subtend equal angle at our eyes. Hence, the moon covers the sun completely.

Question 29.
What do you mean by the resolving power of a microscope?
Answer:
Resolving power:
The resolving power of a microscope is determined by the least distance between two point objects which can be distinguished by the microscope.
Resolving power = \(\frac {2µsinθ }{λ}\) Where, µ is refractive index of the medium between the object and objective, λ is the wavelength of light used to illuminate the object, θ is the semivertical angle of the cone of light from the point object.

Question 30.
What do you mean by the resolving power of a telescope?
Answer:
Resolving power:
The resolving power of a telescope is defined as the reciprocal of the smallest angular separation between two distant objects whose images are separated in the telescope.
i. e., Resolving power = \(\frac {a }{1.22λ}\)
Where, X is wavelength of light and a is the diameter of objective lens.

Question 31.
In compound microscope, why the focal length of objective should be less than that of eyepiece?
Or
Why the focal length of the objective of a compound microscope should be small?
Answer:
The magnifying power of a compound microscope is given by :
m = – \(\frac { { v }_{ 0 } }{ { u }_{ 0 } }\) (1 + \(\frac { D }{ { f }_{ e } }\))
For greater magnifying power u₀ should be smaller, but the object should be placed out of the focus of the objective lens. Therefore, the focal length of the objective lens should be small.

Question 32.
What is astronomical telescope?
Answer:
The telescope by which we can observe the objects at space is known as astronomical telescope.

Question 33.
Is it possible that, when a microscope is inverted, it becomes telescope? Give reason.
Answer:
No, because of the difference of focal lengths of objective and eyepiece in microscope is less while in telescope it is greater.

Question 34.
Microscope and telescope must have high magnifying power and resolving power. Why?
Answer:
When magnifying power is high The object is seen big and clear whereas for low resolving power. There is no clearity of objects.

Question 35.
Why the aperture of photographic lens is less?
Answer:
When the aperture of photographic lens is less, the depth of focus of it increases.

Question 36.
What is reflection? Write the laws of reflection.
Answer:
Reflection:
The phenomenon of bouncing back of the part of light in the same medium when light is incident on the polished surface (minor surface) is called reflection of tight.

Laws of reflection :

1. Angle of incidence is always equal to the angle of reflection (i.e., i = r).
2. The incident ray, reflected ray and normal at the point of incidence all lie on the same plane.

Question 37.
Show the position, nature and size of image when an object is placed between pole and focus infront of concave mirror.
Answer:
Let a ray parallel to principal axis incident from point A on the mirror and reflected through F. Another ray is incident normally and reflects back tracing the same path. These rays after reflection appears to come from point A’. Hence A ‘B’ be the virtual image of object AB.

Thus image is virtual, erect and larger in size than object.

Question 38.
What are conjugate foci?
Answer:
The pair of points on the principal axis of a spherical mirror, such that the object placed at one of the point forms the image on the another point, are called conjugate foci. It is applicable only for real image.

Question 39.
What is parallax?
Answer:
When two objects are seen on a straight line, they seem to be coincident. If the objects are at different places and the eye is moved sideways then there is a relative displacement between them. Thus, when two objects are seen in a straight line and the eye is moved side ways, then relative displacement is called parallax.

Question 40.
What type of mirror is used in vehicle to observe the traffic at the back?
Answer:
Convex mirror, as the image formed by a convex mirror is virtual, erect and smaller, therefore wider field of view of the traffic is given by the convex mirror.

Question 41.
What are the factor upon which refractive index of the medium depends?
Answer:

1. On the pair of medium : Refractive index of denser medium with respect to rarer medium is always greater than 1 but refractive index of rarer medium with respect to denser medium is less than 1.
2. Density of the medium : If optical density of any medium is more than its refractive index will also be more.
3. Wavelength of light : If the wavelength is less more is the refractive index and more the wavelength less is the refractive index.
4. Temperature of medium : On increasing temperature, it becomes optically rarer hence refractive index decreases.

Question 42.
What is principle of reversibility? Explain.
Answer:
Principle of reversibility:
If the path of a r’of light after number of reflections or refractions, is reversed, then the ray retraces the same path in opposite direction i.e., it will follow the same path by which it came.

In fig. PQ is a refractive surface (air – glass interface) made of glass.
AO is incident ray and OB is refracted ray. If the path of ray is reversed, then it return back in OA direction.
Obviously, aµ_g= \(\frac {sin i}{sin r}\) … (1)
Where, i is angle of incidence and r is angle of refraction.
Now, if light ray refracts from glass to air, then
gµ_a = \(\frac {sin r}{sin i}\) … (2)
From equation. (1) and (2),
aµ_g × gµ_a = \(\frac {sin i}{sin r}\) × \(\frac {sin r}{sin i}\)
or aµ_g × gµ_a = 1
This is principle of reversibility.

Question 43.
Refractive index of diamond is much greater than normal glass. Why?
Answer:
Refractive index of diamond (µ = 2.42) is much greater than normal glass (µ = 1.5) so the critical angle of diamond is approximately 24°. Due to this for diamond. The light rays which incidences between 24° to 90° suffer’s total internal reflection. Number of turns due to which the brilliance of diamond increases.

Question 44.
What do you understand by refraction of light? Write its law.
Answer:
The phenomenon of change in the path of light as it passes obliquely from one transparent medium to another is called refraction of light.

Laws :
1. The incident ray, refracted ray and normal to the surface at the point of incidence all lie on the same plane.

2. For a monochromatic light the ratio of sine of the angle of incident to the sine of angle of refraction is constant for given pair of medium i.e., \(\frac {sin i}{sin r}\) = µ
Where µ is known as refractive index of the medium.

Question 45.
Define the following for lens :

1. Optical centre
2. Conjugate foci
3. First and second principal focus.

Answer:
1. Optical centre:
If the emergent ray from the lens is parallel to incident ray, then the point where the refracted ray intersect principal axis is called optical centre.

2. Conjugate foci:
It is a pair of points on the principal axis such that when an object is placed at one point its image forms at the another and when the object is placed at the second point the image is formed at the first point. Such pair of points are called conjugate foci.

3. First principal focus F₁:
It is a point on principal axis of a lens such that the rays of light starting from it (convex lens) or appearing to meet at it (concave lens), after refraction from the lens become parallel to principal axis.

Second principal focus F₂:
It is the point on the principal axis of lens such that the rays of light incident parallel to principal axis after refraction from the lens pass through this point (convex lens) or appear to diverge from this point (concave lens).

Question 46.
When an object is kept in between optical centre and focus in front of convex lens draw a ray diagram of image formation.
Answer:

Question 47.
Write Cartesian sign convention for a lens. According to it which lens will have positive and negative focal length?
Answer:

1. The incident ray is taken from left to right for drawing ray diagram.
2. All the distance measured in the direction of incident ray is taken as positive and opposite to it is taken as negative.
3. Height measured perpendicular to principal axis taken as positive and downwards taken as negative.

The focal length of convex lens is taken as positive whereas for concave lens it is negatives.

Question 48.
What do you mean by linear magnification? For a convex lens show that:
\(\frac {I}{O}\) = \(\frac {v}{u}\), where symbols have their usual meanings.
Answer:
The ratio of the size of the image to that of the size of the object measured perpendicular to principal axis is called linear magnification.

Expression for magnification :

This is the required formula.

Question 49.
On which factor the focal length of any lens depends?
Answer:
1. Material of the lens:
\(\frac {1}{f}\) = (µ – 1)(\(\frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } }\)) or \(\frac {1}{f}\) ∝ µ or f ∝\(\frac {1}{µ}\) i.e., focal length will be greater for smaller refractive index.

2. Colour of light or wavelength:
The focal length for violet colour is lesser than that of red colour for any lens.

3. Radius of curvature:
For thick lens (R is more) focal length is less whereas for thin lens (R is less) focal length is more.

Question 50.
Explain the formation of image in case of concave lens.
Answer:
Let AB be an object kept perpendicular to principal axis. An incident ray parallel to principal axis appear to passing through principal focus (F). Second rays which passes through optical centre O after refraction moves in straight line. When both refracted ray are traced beyond they meet at point A’ where a vertical image is formed. Hence A’ B’ is the final image of the object.

Question 51.
When a lens is dipped in liquid its focal length increases, why?
Answer:

When a lens is dipped in liquid its focal length increases.

Question 52.
Why does sky appear blue?
Answer:
According to the Rayleigh, light rays with smaller wavelengths are scattered more and we know that blue colour light has smaller wavelength than the red light. There fore, when light falls on atmospheric molecules, blue light waves scatter more and spread into the sky. That is why, sky appears blue. As moon has no atmosphere hence there will be no scattering of light. Hence, sky will appear dark.

Question 53.
What is scattering of light? Explain scattering with the help of theory of electromagnetic waves.
Answer:
Scattering of light:
When light falls on particles of size smaller than wavelengths of light much of the light is picked up by particle and given out in some other direction. This is known as the scattering of light.

Rayleigh’s law:
According to Rayleigh, the intensity of scattered light is inversely proportional to the fourth power of wavelength of light.
i.e I ∝\(\frac { 1 }{ { \lambda }^{ 4 } }\) As the wavelength ofBlue light is less than that of red light, therefore blue light is scattered more than red light.

Question 54.
How refraction of light occur in prism? Explain.
Answer:
As shown in fig., let ABC is a prism, AB and AC are its refracting faces. Suppose a light ray PQ is being incident at point Q of the surface AB. MO is normal drawn at point Q.

As at point Q the light ray is entering from air (rarer medium) to glass (denser medium), so it bends towards normal following the path QR and falls at point R of the surface AC, NO is normal at point R. At point R the refraction of light ray takes place from glass (denser medium) to air (rarer medium), so it goes away from normal and emerges out in RS direction. Obviously PQ be the incident ray, QR be the refracted ray and RS be the emergent ray.

∠PQM = angle of incidence (i), ∠RQO = angle of refraction (r) and SRN = angle of emergence (e). So, it is clear that when light ray passes through prism n gets deviated from us path. The deviation is measured by angle between direction of incident ray and emergent ray. The angle between incident ray and emergent ray is called angle of deviation. It is represented by S. In fig, UTS is an angle of deviation.

Question 55.
What are the factors on which angle of minimum deviation depends?
Answer:
1. Angle of prism:
It is clear from eqn. (6) that δ_m ∝A , so if the value of angle of prism is greater than the value of minimum angle of deviation will also be greater.

2. Refractive index of material of prism:
δ_m ∝ (µ – 1), so greater the value of refractive index of material of prism greater will be the value of angle of minimum deviation.

3. Colour of light:
The value of angle of minimum deviation also depends upon colour of light. Refractive index of prism is greater of violet colour and is less for red colour. So, violet colour is deviated most and minimum deviation will be for red.

Question 56.
What is angle of minimum deviation. Draw its graphs.
It has been found by the experiments that the value of angle of deviation depends on the angle of incidence. As the value of angle of incidence is increased, the value of angle of deviation decreases. For a particular value of angle of incidence the angle of deviation becomes minimum, further if the value of angle of incidence is increased, then the value of angle of deviation increases.

The minimum value of angle of deviation is called angle of minimum deviation. It is denoted by δm. In this position the prism is said to be under the position of minimum

Question 57.
Write similarities between Astronomical telescope and Galilean telescope.
Answer:
Similarities:
Astronomical telescope :

• The focal length of objective lens is greater than of eye lens.
• When final image is formed at infinity m = \(\frac { { f }_{ 0 } }{ { f }_{ e } }\)
• The aperture of objective is greater than that of eye lens.

Galilean telescope

• The focal length of objective lens of this telescope is also greater than that of eye lens.
• When final image is formed at infinity m = \(\frac { { f }_{ 0 } }{ { f }_{ e } }\)
• The aperture of objective is greater than that of eye lens.

Question 58.
Write dissimilarities between Astronomical telescope and Galilean telescope.
Answer:

Question 59.
Differentiate between magnifying power and resolving power of the telescope.
Answer:
Difference between magnifying power and resolving power:
Magnifying power:

• It is defined by the ratio of visual angle of final image to visual angle of object.
• For greater magnifying power the focal length of objective should be greater and that of eye piece should be small.

Resolving power:

• It is defined as the reciprocal of the smallest angular separation between two distant objects whose images are separated in the telescope.
• For greater resolving power the diameter of objective should be greater.

Ray Optics and Optical Instruments Long Answer Type Questions

Question 1.
Establish the relation between focal length and radius of curvature for spherical mirror.
Or
Prove for spherical mirror:
f = \(\frac {R}{2}\)
Establish the relation between focal length and radius of curvature concave mirror.
Answer:
Let MPN be a concave mirror and its pole is P, focus F, centre of curvature is C and PC is principal axis.
Now, a ray AB is incident on the mirror parallel to principal axis. Join CB, which will be the normal to the mirror.
Now, by laws of reflection,

Incident ∠ABC = Reflected ∠CBF = θ (let)
AB ∥ PC and BC is a transversal line
∠ABC = ∠PCB = θ, (alternate angles)
∠CBF = ∠PCB = θ
∴ In ∆BCF, the opposite sides will be equal.
∴ FC = BF
As the aperture of the mirror is small.
∴ BF = PF (approx)
∴ FC = PF
or PC = PF
or R = 2f
or f = \(\frac {R}{2}\)

Question 2.
Define focal length of a spherical mirror. Find out the relation between u, v and f for spherical mirror.
Or
What are conjugate foci? For any spherical mirror prove that:
\(\frac {1}{f}\) = \(\frac {1}{u}\) + \(\frac {1}{v}\)

Or
Establish the relation amongst u, v and f for concave mirror.
Or
What is mirror formula? Derive mirror formula for spherical mirror.
Answer:
Focal length:
The distance between the principal focus and pole of mirror is called focal length of spherical mirror.

Conjugate foci:
The pair of points on the principal axis of a spherical mirror, such that the object placed at one of the point forms the image on the another point, are called conjugate foci. It is applicable only for real image.
Let MPN is a concave mirror, P is pole, F is a focus and C is its radius of curvature.

If A’B’ is an image of any object AB placed on principal axis, before the mirror.
Now, join AP and AP. AP is an incident ray and PA’ is a reflected ray at point P.
Draw a normal on principal axis from Q.
Now, ∆ A’B’F and ∆ QRF are similar.

Question 3.
Prove that _aµ_w x _wµ_g x _gµ_a = 1
Answer:
When a ray of light passes through several parallel media and if the first and last medium is same, then the incident and emergent beam will be parallel to each other but they will be laterally displaced.
Referring to Fig., let AB be an incident beam falling on the air – water interface at an angle of incidence i. It undergoes refraction and bends towards the normal as BC with angle of refraction as r₁.

By Snell’s law,
_aµ_w = \(\frac { sini }{ sin{ r }_{ 1 } }\) … (1)
At C, the beam BC undergoes refraction at the water – glass interface and moves as CD.
∠N₂BC = ∠BCN₃ = r₁ (alternate angles)
So, at the water – glass interface, r₁ is the incident beam and r₂ is the refracted beam. Hence, by Snell’s law,
_wµ_g = \(\frac { sin{ r }_{ 1 } }{ sin{ r }_{ 2 } }\) … (2)
At D, the beam CD strikes the glass-air interface. As the beam moves from denser to rarer medium, so it will bend away from the normal. But all the interfaces are parallel to each other and the first and last medium are the same, hence the angle of emergence is equal to angle of incidence.

Question 4.
Establish relation between real and apparent depth, tns. Let an object P is placed at the bottom of a water tank (denser medium). An observer is observing that object. A light ray PQ coming from object incident at Q and get refracted. Another ray of light falls normally on the water air interface and refracts without deviation. These refracted rays appears to meet at point P’.
∴ ∠OPQ = ∠PQN = i (alternating angles)
∠OPQ = ∠MQR = r (corresponding angles)
Let PO = h(alternating angles)
and P’O = h'(corresponding angles)

Question 5.
What is total internal reflection? Write its conditions. Establish the relation refractive index of the medium and critical angle. On what factors does the critical angle depend?
Or
Establish the relation between critical angle and refractive index of medium.
Answer:
Critical angle:
When ray of light travel from denser medium to rarer medium. It is the angle of incidence for which the angle of refraction is 90°.
Total internal reflection:
When light ray moves towards rarer medium from denser medium and if angle of incidence is greater than the critical angle, light ray cannot pass through the plane joining the two media and gets reflected back to the same medium. This phenomenon is called total internal reflection.

Conditions:

1. Light ray must travel from the denser medium to rarer medium.
2. Angle of incidence should exceed the critical angle value.

Relation between critical angle and refractive index:
Let PQ is an incident ray travelling through denser medium to rarer medium and refracted along QR.
∴ Angle of incidence = i_c
and angle of refraction = 90°
∴ _gµ_a = \(\frac { sin{ i }_{ c } }{ sin90° }\) = sin i_c
or _gµ_a = \(\frac { 1 }{ { _{ a }{ \mu }_{ g } } }\) = sin i_c, (by principle of reversibility of light)
∴ _gµ_a = \(\frac { 1 }{ sin{ i }_{ c } }\)
Which is the required relation.

Examples:

1. Formation of mirage in desert
2. Brilliance of diamond.

Factors affecting the critical angle :
1. Colour of light ray used : The refractive index for violet ray is greater therefore critical angle will be least for violet colour (because c ∝\(\frac {1}{µ}\)).

2. The nature of pair of media:
The critical angle also depends upon the pair of media used. For glass-air pair its value is 42°. For diamond air pair its value is 24°.

3. Temperature:
With the increase of temperature the refractive index of the material of medium decreases therefore the critical angle increases.

Question 6.
What are optical fibres? Write its construction and working method with diagram. Writfe any two uses of optical fibres.
Or
What afe optical fibres? Write its working principle. How does it transmit light signal? Write its main uses.
Answer:
An optical fibre is a thin strand of highly pure glass or quartz core with outside coating of glass or slightly different chemical composition having different refractive index, as shown in fig.

The main parts of optical fibre are as follows :
1. Central core:
It is the innermost core of the fibre made of thin and fine quality transparent glass. The refractive index of material of core is about 1.7.

2. Cladding:
It is a layer made of glass or plastic which surrounds the central core of optical fibre. Its refractive index is less than that of central core about µ = 1.5.

3. Plastic jacket:
The central core and cladding are enclosed in a protective jacket made of plastic for providing safety and strength to optical fibre.

Working:
When light falls on the interface separating the core and coating at an angle of incidence greater than the critical angle, then the phenomena of total internal reflection takes place because the refractive index of coating is less than the core. This phenomena take place continuously and light travels the entire length of the fibre even if it is curved or twisted.

Uses:

1. For transmission of light without any loss in its intensity.
2. In endoscopy, to examine the internal parts of body such as stomach and intestines.

Question 7.
What will be the effect on focal length and nature of a lens when dipped in liquid?
Answer:
Media of both the sides of the lens : If the focal length of a lens in air is f_a, then

and if the focal length of a lens in a liquid is f₁, than

Dividing equation (1) by equation (2), we get

Hence, in the liquid, focal length of lens increases.

Case I.
If the lens is dipped in a liquid whose refractive index is less than that of lens, i.e.,

Hence, the nature of lens will remain the same.

Case II.
If the lens is dipped in a liquid whose refractive index is more than that of lens, i.e.,

So, the nature of the lens will change i.e., the lens which behaves as convex in air will behave as concave in liquid. This is the reason why an air bubble appears as concave lens in water.

Case III.
If the refractive index of lens is equal to that of liquid, then _aµ₁ = _aµ_g

Hence, the lens will behave as glass plate.

Question 8.
Prove for any spherical refracting surface :
\(\frac { \mu -1 }{ R }\) = \(\frac { \mu }{ v } -\frac { 1 }{ u }\)
Where symbols have their usual meanings.
Or
Derive an expression \(\frac { \mu }{ v } -\frac { 1 }{ u }\) = \(\frac { \mu -1 }{ R }\) showing relation among p,u, v and R for aconcave spherical refracting surface.
Answer:
Let APB is a cross – section of any spherical refracting surface. P is its pole and C is its centre of curvature. Left of the surface has air as a medium and in right, medium has refractive index µ.

Any point object is placed at point O on the principal axis of which a virtual image is formed at I due to refracting surface APB. According to the figure, angle of incidence,
∠OMC = i
Angle of refraction,
∠LMN = ∠IMC = r
Let ∠MOP= α, ∠MIP = β and ∠MCP = θ.
From Snell’s law,
µ = \(\frac {sin i}{sin r}\)
or µ = \(\frac {i}{r}\), (as i and r are very small)
or i = µr … (1)
From ∆OMC, θ = i + a, (∵exterior angle = sum of opposite interior angles)
or i = θ – α … (2)
and in ∆ IMC, θ = r + β
or r = θ – β … (3)
From eqns. (2) and (3), putting the values of i and r in eqn. (1),
θ – α = µ(θ – β)
or µβ – α = (µ – 1)θ … (4)

This is the required formula.

Question 9.
Two lenses of focal lengths f₁ and f₂ are kept in contact, if the focal length of the combination is F, then prove that :
\(\frac {1}{F}\) = \(\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } }\)
Or
Establish the formula for the focal length of the combination of two thin lenses.
Or
Two thin convex lenses having focal lengths f₁ and f₂ respectively are kept in contact. Describe the method to determine the focal length of the system on the basis of following points:

1. Labelled ray diagram
2. Derivation of formula used
3. Two important precautions.

Answer:
L₁ and L₂ are two convex lenses in contact with each other. P is the object pin whose final image is formed at Q after refracting through lenses L₁ and L₂. The image of P is formed at R due to lens L₁ with optical centre O₁ . The image of P at R acts as image for the lens L₂ which forms its final image at Q.

The optical centre of L₂ is O₂ , u is the object distance of the pin P for lens L₁ and v’ is the image distance for the lens L₁ which acts as the object distance for the lens L₂ forming the final image at Q at a distance v from L₂.

Determination of formula:
Two lenses of focal length f₁ and f₂ are in contact with each other. The focal length of the combination is ‘f’ P is the object on the common principal axis of the two lenses. The image formed by L₁ is at R. The image by L₂ acts as object for the lens L₂ which in turn forms the final image at Q.
∴ PO₁= u, O₁R = v’, O₂Q = v
From the relation,
\(\frac {1}{f}\) = \(\frac {1}{v’}\) – \(\frac {1}{u}\)
For L₁ of focal length f₁,
\(\frac { 1 }{ { f }_{ 1 } }\) = \(\frac {1}{v’}\) – \(\frac {1}{u}\) … (1)
For L₂ of focal length f₂,
\(\frac { 1 }{ { f }_{ 2 } }\) = \(\frac {1}{v}\) – \(\frac {1}{v’}\) … (2)
Adding eqns. (1) and (2), we get
\(\frac { 1 }{ { f }_{ 1 } }\)+\(\frac { 1 }{ { f }_{ 2 } }\) = \(\frac {1}{v}\) – \(\frac {1}{u}\) … (3)
If the focal length of the combination is F, then
\(\frac {1}{F}\) = \(\frac {1}{v}\) – \(\frac {1}{u}\) … (4)
From eqns. (3) and (4),

By knowing u and v, the focal length of combination of lenses can be determined.

Precautions:

1. The tips of the pins and optical centres of the lenses must be in one line and also this line must be parallel to the optical bench.
2. The eye should be placed nearly 25 cm away while removing the parallax.
3. The lenses must be neat and clean.

Question 10.
Draw the ray diagram of refraction through a lens and derive the refraction formula.
Or
prove for a lens : \(\frac {1}{f}\) = (µ – 1)(\(\frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } }\))
Or
Where, the symbols have their usual meanings.
Or
What is lens maker formula? Derive it for a convex lens.
Or
Derive refraction formula for a thin lens.
Answer:
Let any lens is made up of two spherical surfaces AB and CD of radius of curvatures R₁ and R₂ . Any point object is placed on principal axis at O. Image of an object O is formed at 7, due to AB and this image acts as an object for the second surface CD and image is produced at I₁. In this way object’s image through the lens is formed at 7.
Let distance of object O from the lens is u, Image I₁ is at distance v₁ , Image I is at distance v.
∴ Through the first refracting surface AB,

\(\frac { (\mu -1) }{ { R }_{ 1 } }\) = \(\frac { \mu }{ { v }_{ 1 } }\) – \(\frac {1}{u}\) … (1)

Now, if u = ∞, then v = f
Substituting these values in equation (3), we have

Which is refraction formula.

Question 11.
Draw ray diagram for refraction through a prism and establish a relation between refractive index of material of prism, angle of prism and angle of minimum deviation.
Answer:
Let ABC be a prism, AB and AC are refracting surfaces. A is the angle of prism, PQ is incident ray, QR is refracted ray, RS is emergent ray, MO and NO are normals, i₁, is angle of incidence, r₁ is angle of refraction, e is angle of emergence, r₂ is angle of incidence of AC.

Now, from figure,
Angle of deviation, δ = ∠UTS
or δ = ∠TQR + ∠TRQ,
(exterior angle is sum of interior opposite angles)
= (i₁ – r₁) + (i₂ – r₂)
In the position of minimum deviation,
i₁ = i₂ = i, (say)
and r₁ = ₂
∴ δ_m = (i – r) + (i – r) = 2i – 2r … (1)
In quadrilateral AQOR,
∠AQO + ∠ARO= 180°, (each angle is equal to 90°)
∠A + ∠QOR = 180° … (2)
In ∠QOR,
∠QOR + r₁ +r₂ = 180° … (3)
From eqns. (2) and (3),
∠A + ∠QOR = ∠QOR + r₁+ r₂
or ∠A = r₁ + r₂
By condition of minimum deviation,
∠A = r + r = 2r or r = \(\frac {λ}{2}\)
Putting it in equation (1),

This is the required formula

Question 12.
What is angle of minimum deviation? On what factor does it depend?
Answer:
It has been found by die experiments that the value of angle of deviation depends on the angle of incidence. As the value of angle of incidence is increased, the value of angle of deviation decreases. For a particular value of angle of incidence die angle of deviation becomes minimum, further if the value of angle of incidence is increased, then the value of angle of deviation increases.

The minimum value of angle of deviation is called angle of minimum deviation. It is denoted by δ_m. In this position the prism is said to be under the position of minimum deviation.

Angle of deviation depends upon following factors :
1.Angle of prism:
It is clear that δ_m∝A , so if the value of angle of prism is greater then the value of minimum angle of deviation will also be greater.

2. Refractive index of material of prism:
δ_m ∝ (µ – 1), so greater the value of refractive index of material of prism greater will be the value of angle of minimum deviation.

3. Colour of light:
The value of angle of minimum deviation also depends upon colour of light. Refractive index of prism is greater of violet colour and is less for red colour.

So, violet colour is deviated most and minimum deviation will be for red.

Question 13.
What is rainbow? Explain the formation of primary and secondary rainbow with the help of ray diagram.
Answer:
Sometimes, at the time of rain or after a shower, an observer standing with his back towards the sun looks a circular arc of spectrum colours in the sky, called rainbow.

Sometimes two rainbows are seen:

1. Primary rainbow and
2. Secondary rainbow.

In primary rainbow, red is seen outside and violet inside, while in secondary rainbow, the violet colour is at outside and red colour inside. In primary rainbow, the red colour emerges out at an angle of 42° and violet rays at an angle of 40°, hence the angular width of the rainbow is 2°. It is comparatively bright. In secondary rainbow, the violet rays emerges out at an angle of 55° and red rays at an angle of 52°. Hence, the angular width of the rainbow is 3°. It is comparatively less bright.

1. Primary rainbow : The primary rainbow is produced by two refractions and one total internal reflection of sun rays by the rain drops. It is shown in the Fig. (a).

2. Secondary rainbow: The secondary rainbow is formed by two refractions along with two internal reflections by the rain drops as shown in Fig. (b).

Question 14.
What are differences between primary and secondary rainbow?
Answer:
Differences between primary and secondary rainbow :
Primary rainbow:

• In this rainbow, red colour is outside and violet is inside.
• Its brightness is more.
• Its angular width is 2°.
• It lies below the secondary rainbow.
• It is formed by two refractions and one internal reflection.

Secondary rainbow

• In this rainbow, violet colour is outside and red is inside.
• It is less bright.
• Its angular width is 3°.
• It lies above the primary rainbow.
• It is formed by two refractions and two internal reflections.

Question 15.
What is simple microscope? Draw the ray diagram and explain its principle. Write its two uses.
Or
Draw the ray diagram of simple microscope and obtain an expression for its magnifying power.
Or
Describe the working of simple microscope under following points:

1. Ray diagram for the formation of image when formed at D.
2. Derivation for the magnifying power when final image formed at D and at infinity.

Answer:
Simple microscope is a device with the help of which magnified image of small objects are seen. It is a convex lens of less focal length.

Principle:
When an object is placed before a convex lens between its focus and optical centre then a virtual, erect and magnified image is formed.

Use:

1. As reading lens.
2. To see the magnified image of small objects, example to see biological slides.

Ray diagram:
Ray diagram for a simple microscope is given below.

Expression for magnifying power :
The ratio of the visual angle of image to the visual angle of object, placed at the least distance of distinct vision, is called magnifying power.

Now OB’ = D, (least distance of distinct vision)
By sign convention OB’ = D(-ve) and OB = u(-ve)
∴ m = \(\frac {-D}{-u}\)
m = \(\frac {D}{u}\)
Where, u is the distance of object from the lens.
When the image is formed at the least distance of distinct vision :

When the image is formed at infinity: If the image is formed at infinity, then u = f.
∴ m = \(\frac {D}{u}\) = \(\frac {D}{f}\)
For greater magnification the focal length should be small.

Question 16.
Describe compound microscope on following heads :

1. Ray diagram of formation of image.
2. Derivation of formula for magnifying power when the image is formed (a) At least distance of distinct vision, (b) At infinity, (c) Condition for greater magnification.

Or
Write the nature and kind of final image formed by compound microscope. How the magnifying power can be increased?
Answer:
1. Ray diagram :

2. Formula derivation for magnifying power:
The ratio of visual angle of image to the visual angle of object, placed at a distance of least distance of distinct vision is called magnifying power of compound microscope.

If it is assumed that the object AB is kept at least distance of distinct vision D and the visual angle formed at eye is α, then
tan α = \(\frac {AB}{D}\)
The angle formed by final image A”B” is β.
∴ tan β = \(\frac {A”B”}{D}\)
If the angles α and β are very small, then tan α = α and tanβ = β

In Δ ABO₁ and ΔA’B’O₁

(a) When the final image is formed at least distance of distinct vision :

This is the required formula, – ve sign shows that the image is inverted.

(b) When the final image is formed at infinity :
The eye – lens is so adjusted that the image A’B’ formed by the objective is on the focus of eyepiece, then u_e = f_e.
From equation (4),

(c) Condition for better magnification :

1. u₀ should be small, i.e., the object should be placed near the objective lens. But the object should be away from the focus, hence the focal length of the objective should be small.
2. f_e should be small, i.e., focal length of eyepiece should be small.
3. v₀ should be greater, i.e., the distance between objective lens and eyepiece should be greater.

Question 17.
Describe terrestrial telescope on following points :

1. Ray diagram
2. Calculation of magnifying power when the image is formed at (a) Least distance of distinct vision, (b) Infinity.

Answer:
Terrestrial telescope:
These telescopes are used to observe the far object on the earth. In astronomical telescope image is formed inverted but in terrestrial telescope the image formed is erect. Therefore, these telescopes are used to observe the objects on earth.

1. Ray diagram :

2. Expression for magnifying power:
Magnification is defined as the ratio of the visual angle of final image to that of the visual angle of object. Visual angle of final image to that of the visual angle of object.

From fig. (b),
Angle made by the final image P”Q” on eyepiece L_e is
β = ∠P”O”Q” = ∠P’OQ’
Angle formed by object lens is
α = ∠P’OQ’
As the angles α and β are very small, hence

(a) When the image is formed at least distance of distinct vision :
The image formed by the erecting lens lies within the focal length of eyepiece i.e., in between O’ and f, for eyepiece f_e = v_e and v_e = -D.

(b) When the image is formed at infinity : In this case u_e = f_e

Question 18.
Describe the astronomical telescope on following points :
(i) Ray diagram for the formation of image.
(ii) Formula derivation for magnifying power when the image is formed at (a) Least distance of distinct vision and (b) Infinity.
Length of telescope.
Or
Draw a labelled ray diagram of an astronomical telescope and going only formula for magnification. What are the conditions for maximum magnification?
Answer:
1. Ray diagram for the formation of image:

2. Formula derivation for magnifying power:
Magnification is defined as the ratio of the visual angle of final image to that of the object.

From fig. (b),
Angle made by the final image P”Q” on eyepiece L_e is
β = ∠P”O’Q’ = ∠P’OQ’
Angle formed by object on objective lens is
α = ∠P’OQ’
As the angles a and P are very small, hence

(a) When the final image is formed at least distance of distinct vision:
The distance of eye – lens (L_e) from final image P”Q” is O’Q” = v_e = D(- ve). The object is P’Q’, situated at distance O’Q’ from eyepiece, i.e., ue = O’Q'(- v_e).

(b) When the final image is formed at infinity:
The image p’Q’ formed by the objective lens (L₀) lies on the first focus of eye – lens (Le) i.e., u_e = f_e
Putting in eqn. (iv), we get m = \(\frac { { f }_{ 0 } }{ { f }_{ e } }\).

(iii) Tube length is f₀ + f_e.

Question 19.
Describe Galilean telescope under the following heads :

1. Ray diagram when final image is formed at least distance of distinct vision.
2. Expression for magnifying power when final image is formed at least distance of distinct vision D.

Answer:
1. Ray diagram :

2. Formula derivation for magnifying power:
Magnification is defined as the ratio of the visual angle of final image to that of the object.

Angle formed by final image β = ∠P’O₂Q” = ∠P’O₂Q’
Angle formed by object on objective lens α = ∠P’O₁Q’
As angles and are very less, tanα ≈ α and tan β ≈ β

If the final image is formed at least distance of distinct vision : For eye – lens v = – D, u = u_e and f = – f_e, (By cartesian sign convention)

In this case, the tube length is f₀ – f_e.

Ray Optics and Optical Instruments Numerical Questions

Question 1.
The focal length of a convex lens is 10 cm. If it is immersed into a liquid of refractive index 1.75, then find out its nature and focal length.
Solution:
Given, f_a = 10 cm, _aµ₁ = 1.75, _aµ_g = 1.5

i.e., The lens behaves like a concave lens.

Question 2.
A lens forms an image of a needle placed at a distance of 45 cm from it, on a screen situated at distance 90 cm on other side of it. Name the kinds of lens. Calculate its focal length. If the size of needle is 5 cm, what will be the size of the image?
Solution:
Given, u = – 45 cm, v = 90 cm, size of object O = 5 cm

∵ Focal length is positive therefore the lens would be a convex lens.
Again, magnification m = \(\frac {I}{O}\) = \(\frac {v}{u}\)
or \(\frac {I}{5}\) = \(\frac {90}{-45}\)
I = – 5 x 2 or I = – 10
(-) sign shows here that the image formed be inverted and real.

Question 3.
Two thin lenses of power + 12D and – 2D respectively are placed in contact. Find the power focal length and nature of the combination.
Solution:
Given,
P₁ = + 12D, P₂ = – 2D
P = P₁ + P₂ = + 12D – 2D = 10D
But power = \(\frac {1}{f}\)
∴ f = \(\frac {1}{P}\) = \(\frac {1}{10}\)m = 0.1 m
∵ Focal length is positive so the lens is of converging nature.

Question 4.
Velocity of light in glass (µ = 1.5) is 2 x 10⁸m/s. Find out the velocity of light in air.
Solution:
Formula µ = \(\frac { { V }_{ a } }{ V_{ g } }\) where _aµ_g = 1.5, V_g =2 x 10 ⁸m/s
V_a =V_g x _aµ_g = 1.5 x 2 x 10⁸
or  V_a = 3.0 x 10⁸ m/s

Question 5.
The refractive index of glass and water are 3/2 and 4/3. Find the ratio of velocity of light in glass and water.
Solution:

Question 6.
The refractive index of material of prism and its refracting angle are and 60° respectively. Find angle of incidence for the position of minimum deviation.
Solution:

Question 7.
Refractive index of water w.r.t. glass is \(\frac {9}{8}\) if the velocity of light in water is 2.25 x 10⁸m/s then find the velocity of light in glass.
Solution:

Question 8.
The refractive indices for red and violet colour of a crown glass prism are 1.513 and 1.523 respectively. Find out dispersive power of crown glass prism.
Solution:

Question 9.
The refractive angle of a prism is 60°. If angle of minimum deviation is 38° then obtain refractive index of material of prism, (sin 49° = 0.7547)
Solution:

Question 10.
If length of astronomical telescope is 80 cm and its magnification is 15, then find focal length of objective lens and eyepiece.
Solution:
Given : f₀ + f_e = 80 cm … (1)
m = – 15
Formula: m = – \(\frac { { f }_{ 0 } }{ { f }_{ e } }\)
– 15 = – \(\frac { { f }_{ 0 } }{ { f }_{ e } }\)
or f₀ = 15 f_e
put this value in equation (1),
15 f_e + f_e = 80
or 16 f_e = 80
f_e = 5cm
Now f₀ + 5 = 80
or f₀ = 75 cm.

Question 11.
The magnifying power of an objective lens of a compound microscope is 8, if the magnifying power of the compound microscope is 32. Then calculate magnifying power of an eyepiece lens.
Solution:
Given: m₀ = 8, m = 32, me = ?
Formula : m = m₀ x m_e
32 = 8 x m_e
or m_e = \(\frac {32}{8}\) = 4
or m_e = 4

Question 12.
Diameter of an objective lens of a telescopes is 01 m and the wavelength of light is 6000 A. What will be its resolving limit?
Answer:

MP Board Class 12th Physics Important Questions

MP Board Class 12th Economics Important Questions Unit 7 Determination of Income and Employment

Micro Economics Determination of Income and Employment Important Questions

Micro Economics Determination of Income and Employment Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Income and employment are determined by:
(a) Total demand
(b) Total supply
(c) Total demand and total supply both
(d) By market demand.
Answer:
(c) Total demand and total supply both

Question 2.
The relation between consumption and savings are:
(a) Inverse
(b) Direct
(c) Inverse and direct both
(d) Neither inverse nor direct.
Answer:
(a) Inverse

Question 3.
When an economy their to save all its extra income then investment calculation will be:
(a) 1
(b) Uncertain
(c) 0
(d) Infinite.
Answer:
(a) 1

Question 4.
“Supply creates it own demand”. This statement was given by the economist:
(a) Keynes
(b) Pigou
(c) J. B.Say
(d) Adam Smith.
Answer:
(c) J. B.Say

Question 5.
Classical theory is based on the assumption of:
(a) Say’s law of market
(b) Flexibility in wage rates
(c) Flexibility in interest rate
(d) All of the above.
Answer:
(d) All of the above.

Questions 2.
Fill in the blanks:

1. ……………………. refers to highest rate of return over cost expected from marginal or additional unit of a capital asset.
2. Deflationary gap is the measurement of ……………………. demand.
3. Deficit demand indicates ……………………. gap.
4. In case of excess demand bank rate …………………….
5. Multiplier can also ……………………. in opposite direction.
6. The point at which aggregate demand and aggregate supply are equal is known as …………………….
7. Unemployment is the result of …………………….
8. Propensity to consume shows relation between ……………………. and distributed income.

Answer:

1. Marginal efficiency of capital
2. Deficit
3. Deflationary gap
4. Increase
5. Work
6. Effective demand
7. Deficit
8. Consumption

Question 3.
State true or false:

1. Full employment does not mean zero unemployment.
2. Increase in interest rate in future, will reduce the savings.
3. Consumption expenditure does not increase in the same proportion as income increases.
4. Theories of employment was propounded by Marshall.
5. Unemployment is the result of deficit demand.
6. Keynes theory also applies to underdeveloped countries.
7. Keynes theory is based on the concept of full employment.

Answer:

1. True
2. False
3. True
4. False
5. False
6. False
7. True.

Question 4.
Match the following:

Answer:

1. (d)
2. (a)
3. (c)
4. (f)
5. (b)
6. (e).

Question 5.
Answer the following in one word / sentence:

1. What will be the effect of multiplier, if multiplier leakage is high?
2. Excess demand gives birth to?
3. Which is the most appropriate measure to check deficit demand?
4. Who propounded the deficit demand theory of unemployment?
5. Name the interest theory propounded by Prof. Keynes.
6. Which is the most effective measure to check it deficit demand?

Answer:

1. Less
2. Money inflation
3. Increase in public expenditure
4. Keynes
5. Liquidity preference
6. Increase in exports.

Determination of Income and Employment  Very Short Answer Type Questions

Question 1.
What do you mean by marginal propensity to consume?
Answer:
Marginal Propensity to Consume (M.P.C.):
It refers to the ratio of change in the consumption to the change in income. It may be algebraically expressed as under:

or
M.P.C = \(\frac { ∆C }{ ∆Y } \)

Question 2.
What is meant by marginal propensity to save?
Answer:
Marginal Propensity to Save (M.P.S.):
It refers to the ratio of change is saving to a change in income. It can be obtained by dividing change in savings by change in income. Algebraically it can be expressed as:

M.P.S = \(\frac { ∆S }{ ∆Y } \)

Question 3.
What is the difference between actual investment and induced investment?
Answer:
Actual investment:
Actual investment refers to realizing investment. It is what the investors actually make at different levels of income and employment in the economy.

Induce investment:
Induce investment in that investment which depends on the expected profitability of level of income.

Question 4.
What is effective demand?
Answer:
Effective demand means that amount of money which people of a country spend on consumption and investment in a specific time. That level of demand is effective which is equal to aggregate supply. According to Keynes, “By point of aggregate demand at which the aggregate curve intersect the price at that point is called effective demand.”

Question 5.
Explain voluntary and involuntary unemployment.
Answer:
Voluntary:
It signifies the situation where in all the person willing to work at existing rate of remuneration are getting employment.

Involuntary:
It refers to that situation where in all the persons are willing to work at existing rate of remuneration do not get employment. It signifies the state where the available resources are not fully utilized.

Question 6.
Explain any two measures which can be taken to remove the excess demand (Write any two).
Answer:
Followings are the two measures to be taken to check the excess demand:
1. Bank rate:
The rate at which the central bank lends money to its member banks is called bank rate. In a situation of excess demand, the central bank should raise the bank rate.

2. Open market operations:
The buying and selling of government securities and bonds in the open market by the central bank are called open market operations. In the time of excess demand, central bank sells government securities to commercial banks or to their customers.

Determination of Income and Employment Short Answer Type Questions

Question 1.
Write the characteristics of traditional theory of income and employment.
Answer:
Followings are the characteristics of theory of income and employment:
1. Automatic balance between demand and supply:
According to J.B. Say, the supply creates its own demand. When goods are produced and supplied, it automatically creates demand for some other goods, the classical economists always believed in automatic full employment through the free play of market mechanism, the forces of demand and supply.

2. Idea of unemployment is baseless:
The classical economist were not scared of the problem of unemployment. There cannot be any involuntary unemployment in classical and neoclassical world. According to them excess production is impossible so there is no question of normal unemployment.

3. Non – Interference society:
According to traditional economists, if there is no interference of the government in the economy of a nation then the situation of unemployment will never rise. If the government interferes in the matters then unemployment will always arise.

4. Supply creates its own demand:
According to classical theory supply, creates its own demand. Supply leads to production, expenditure income and demand. Therefore aggregate supply becomes equal to aggregate demand.

Question 2.
Write assumptions of traditional theory or classical theory of income and employment
Answer:
Assumption of Classical theory: Followings are the assumptions of classical theory of economics:

1. The economy enjoys the state of full employment as the aggregate demand is equal to aggregate supply.
2. There is no saving in the economy as all that is earned is invested.
3. There is free competition among the products and factors.
4. There is always possibility of extension of market.
5. There is end of interference of government and non – government and no protection is given.
6. There is no accumulation of wealth.
7. This theory works only when all the goods produced are sold in market.

Question 3.
What is unemployment according to Keynes?
Answer:
According to Keynes, there cannot be an automatic equality between aggregate demand and full employment. He holds the state of full employment rarely exits. It is possible that aggregate demand may full short of aggregate supply and state of full employment is disturbed. According to him there is a state of involuntary unemployment where able persons work at current rates of remuneration face unemployment. Unemployment exits due to lack of effective demand and lack of expenditure on consumption and investment.

Question 4.
What are the components of aggregate demand? Explain.
Answer:
Followings are the components of aggregate demand:

1. Household consumption demand:
In this category comes, the goods and services to satisfy their consumption needs. The aggregate consumption expenditure constitutes consumption demand. Household consumption depends on level of income.

2. Investment demand:
This component of aggregate demand comes from the business firms. Investment goods like machines, factory buildings are not demanded by household consumer but by business or firms.

3. Government demand for goods and services:
Like consumers and firms, government may also demand goods and services either for consumption or for investment purposes.

4. Net export demand:
To find out aggregate demand of a country items of net export demand is included in it. For this, import is subtracted from exports.

Question 5.
Write short note on:

• Full employment
• Under employment.

Or
What do you mean by full employment and under employment?
Answer:
Full employment:
The term “full employment” signifies the situation wherein all the persons willing to work at existing rate of remuneration are getting employment. Full employment never means zero unemployment.

Under employment:
It refers to that situation wherein all the persons are willing to work at existing rate of remuneration do not get employment. It signifies the state where the available resources are not fully utilized the aggregate supply of goods and services.

Question 6.
State measures to correct excess demand.
Answer:
1. Bank rate:
The rate at which the central bank lends money to its member banks is called bank rate. In a situation of excess demand, the central bank should raise the bank rate. Increase in bank rate will automatically lead to increase in banks lending rates to their borrowers. This, in turn, will reduce the volume of banks loans and advances. Consequently, expenditure on both consumption and investment is reduced.

2. Increasing the reserve ratio:
Reserve ratio refers to a minimum percentage of deposits of a commercial bank in the Reserve Bank. If the reserve ratio is increased it will reduce the credit creation capacity of the banks. When there is a situation of excess demand, central bank can increase this ratio and thereby cash resources of the banks can be reduced. This will force the commercial banks to contract credit.

3. Open market operations:
The buying and selling of government securities and bonds in the open market by the central bank are called open market operations. In the time of excess demand, central bank sells government securities to commercial banks or to their customers. This will reduce the cash reserves with the commercial banks directly and indirectly thereby affecting their capacity to make loans. It will reduce the cash reserves of the commercial banks.

4. Reducing credit inflation:
To reduce the excess demand or to reduce inflation it is essential to reduce credit inflation. By selling securities, by increasing bank rate by central bank credit inflation can be reduce. ‘

5. By taking old currency and circulating new currency:
In situation of excess demand the government cancel the old currency and in place of it new currency was introduced for circulation.

Question 7.
Distinguish between Aggregate Demand and Aggregate Supply.
Answer:
Differences between Aggregate Demand and Aggregate Supply:
Aggregate Demand:

1. Aggregate demand is refers to the total volume of goods and services demanded in an economy in a year.
2. It implies the total amount of expenditure incurred by the community on purchase of all goods and services.
3. Aggregate demand and aggregate expenditure is the same thing.

Aggregate Supply:

1. Aggregate supply refers to the total supply of goods and services in the economy in a year.
2. According to classical concept aggregate supply is perfectly in elastic with respect to the price.
3. According to Keynesian concept the aggregate supply is perfectly elastic with respect to price level.

Question 8.
Differentiate between Induced Investment and Autonomous Investment
Answer:
Differences between Induced Investment and Autonomous Investment:
Induced Investment:

1. Induced investment is that investment which depends on the expected profitability of level of income.
2. It is income elastic.
3. This investment is generally made in private sector.
4. Induced investment curve tends to move upwards indicating increased investment of the increased level of income.

Autonomous Investment:

1. Autonomous investment is that investment which is not influenced by the expected profitability.
2. It is not income elastic.
3. This investment is generally made in public sector.
4. Autonomous investment curve is a horizontal straight line parallel to X-axis. It indicating that autonomous investment is independent of the level of income.

Question 9.
Explain paradox of thrift.
Answer:
Paradox of thrift:
The paradox of thrift, in brief, states that the higher ratio of saving in an economy will create condition that result in a fall in aggregate savings. This can be explained as follows If all the people in the economy start saving more, the aggregate demand for consumption will decrease which will lead to low investment, low production and low income. Consequently the savings will reduce. This is known as paradox of thrift.

Question 10.
Are the following entered:

• On the credit side or the debit side and
• In the current accour or cap Hal account in the Balance of Payment account. You must give reasons for your

(a) Investments from abroad
(b) Transfer of funds to relatives abroad.

Answer:
(a) Investment from abroad:
Investment from abroad is a component of capital account. It comes on the credit side of capital account of balance of payment. It implies investment from abroad in shares of Indian companies, in Indian branches of foreign companies, in real estate, etc. It also includes repatriation of Indian investment abroad etc. These bring in foreign exchange.

(b) Transfer of funds to relatives abroad:
This is shown on the credit side of current account of balance of payment under the heading “invisibles” .Transfer means receipts/ payments between residents and non-residents for which the provider does not receive a quid proquo in return.

Question 11.
Measure the level of ex-ante aggregate demand when autonomous investment and consumption expenditure (A) is Rs 50 crores and MPS = 0 – 2 and the level of income (Y) is Rs 4000.00 crores. State whether the economy is in equilibrium.
Answer:
MPC = 1 – MPS = 1 – 0 – 2 = 0-8
AD = (C +1) + CY
= 50 + 0 – 8 x 4000
= 50 + 3200
= ₹ 3250
crores AS (Y) = ₹ 4000 crores
Here,
AD < AS. So, the economy is not in equilibrium.

Question 12.
What do you mean by marginal propensity to consume? How is related to marginal propensity to save?
Answer:
Marginal Propensity to Consume (MPC):
Marginal propensity to consume can be defined as,“The ratio of change in consumption (∆C) to change in income (∆Y). It indicates that part of additional income which is not spent on consumption”.
Symbolically, MPC=\(\frac { ∆C }{ ∆Y } \)
Here, DC indicates change in consumption whereas DY means change in income. For example, if national income of a country increases from 11,000 crore to ₹ 1,200 crore and, as a result, consumption expenditure goes up from ₹800 crore to ₹ 1900 crore, then
MPC will be 0-5 = \(\frac { 100 }{ 200 } \)
Relation between M.P. save and M.P.S.
Marginal propensity to save: It is the relation between change in saving and change in income.
Equation,
MPS = \(\frac { ∆S }{ ∆Y } \)
The sum of M.P.C. and M.P.S. is equal to 1
Y = C + S
Or
∆Y = ∆C +∆S.
Dividing both sides by ∆Y,we get the following:
= \(\frac { ∆Y }{ ∆Y } \) \(\frac { ∆C }{ ∆Y } \) + \(\frac { ∆S }{ ∆Y } \)
Or
1 = MPC + MPS.
So, it is proved that sum of MPC and MPS = 1.

Determination of Income and Employment Long Answer Type Questions

Question 1.
Distinguish between Classical and Keynesian Theory of Income.
Answer:
Differences between Classical and Keynesian Theory of Income:
Classical Theory:

1. The classical economist always believed in automatic full employment through the free play of market mechanism (The force of demand and supply).
2. According to it, full employment is the most normal situation of capitalistic economy.
3. This theory is based on the market.
4. According to it, “the supply creates its own demand”. Supply leads to the production, expenditure income and demand. So aggregate supply becomes equal to aggregate demand.
5. This principle is based on long term assumptions.
6. According to this principle in economy there is no possibility of more production and normal unemployment.
7. According to this principle, similarity between savings and investment is established by change in rate of interest.

Keynesian Theory:

1. According to Keynes, the state of full employment rare exists. He says that income and employment level is fixed on that point where aggregate demand fixed on that point where aggregate demand is equal to aggregate supply. So there is no need of full employment.
2. According to him, there is the state of involuntary unemployment.
3. It is based on psychological principle of consumption.
4. In the opinion of Keynes tustiy there cannot be an automatic equality between aggregate demand and full employment. It is possible that aggregate demand may fall short of aggregate supply and the state of full employment is disturb.
5. This principle is based on short term assumption.
6. According to this principle normal more production and normal unemployment is possible.
7. According to this principle similarity between saving and investment is established by change in income.

Question 2.
Explain the concept of multiplier.
Answer:
Multiplier:
According to J. M. Keynes, of all the variables involved in the determination of income and employment, investment is most important. Change in investment leads to change in national income. According to him, change in national income is a multiple of change in investment. This multiple is termed as the multiplier or investment multiplier. Thus, investment multiplier is a measure of change in national income as a result of change in investment.

Put symbols:
K = ∆Y/∆I
Where,
K = Multiplier
∆Y = Change in income
∆I = Change in investment

For example, if national income of a country increases by ₹ 300 crore as a result of increase in investment of ₹100 crore, the value of multiplier here will be 3. Because change in income is three times more than change in investment.

“Investment multiplier tells us that when there is an increment of aggregate investment income will increase by an amount which is K times the increment of investment. ” —Keynes

“Investment multiplier is the ratio of an increase in income to given increase in investment. ” —Dillard

Question 3.
Explain the factors affecting marginal efficiency of capital.
Answer:
Followings are the factors affecting marginal efficiency of capital:
1. Demand tendency:
If the demand for a product is expected to increase in future then marginal efficiency of capital goes up. This result in increases in investment. On the contrary if the demand for a product is expected to fall the marginal efficiency of the capital also goes down.

2. Cost and Prices:
If the cost of production and price of a product is expected to fall in future then the marginal efficiency of capital also falls. This reduces investment also. On the contrary if the cost of production and prices is expected to increases in futures then marginal efficiency of capital increases and investment also increases.

3. Liquidity of resources:
If an entrepreneur has a large amount of resources in terms of cash he then tries to earn profit from different types of investment. This increases the marginal efficiency of capital. On the contrary if the entrepreneur has less amount of resources as cash he has problem in making new investment and marginal efficiency of capital also reduces.

4. Changes in income:
Sudden changes in income due to unexpected profits and loss and heavy taxes or concession in taxes also affects marginal efficiency of capital.

5. Consumption tendency:
If consumption tendency is high in short-term then the marginal efficiency of capital also increases. On the contrary if consumption tendency is less in the short-term then marginal efficiency of capital also decreases.

Question 4.
Explain income and employment equilibrium with aggregate demand and aggregate supply with the help of diagram.
Answer:
According to Keynes, equilibrium level of income and employment is deter-mined where aggregate demand and aggregate supply are equal. This can be explained with the help of imaginary figures.

AD is smaller than AS, AD > AS
In the above aggregate table, the aggregate demand and aggregate supply are equal at Rs. 400 crores of National, Income. At this level saving is also equal to investment and any change on this level cannot be stable and remain for a long period of time.

Explanation with the help of curve:
This equilibrium between income and employment can be further explained with the help of given curves:

In the above figure of part-A aggregate demand (AD) cuts aggregate supply (AS) at the point E. A1 this point the equilibrium level of income and employment is determined at Rs. 400 crores. In figure of part-B show the saving and investment at the same income level.

At the equilibrium point income, saving and investment is also equal. Keynes has defined it as “Effective demand”. According to Keynes, the point at which the aggregate demand intersects with the aggregate supply the price of that point is called effective demand. According to Keynes, for equilibrium full employment is necessary.

According to him determination of equilibrium is possible only at full employment level or before and after full employment also. If equilibrium is established before full employment then it is called “imperfect employment equilibrium”. The equilibrium at full employment is called “full employment equilibrium”, which is a special situation of economy of a country.

Question 5.
Write any four factors determining propensity to consume.
Answer:
The following factors determines, propensity to consume:

1. Money income:
If there is increase in money income, the consumption expenditure will also increase.

2. Distribution of income:
If the income reaches the hands of poor people in stead of rich people, the consumption expenditure will rise. The consumption expenditure will reduce if the distribution of income is unequal and it will rise if the distribution of income is equal.

3. Fiscal policy:
The propensity to consume will decrease if more indirect taxes are imposed and it will rise if more direct taxes all imposed indirect taxes adversely affect the poor while direct taxes adversely affect the rich.

4. Future expectation of rise or fall in prices:
If the prices are expected to rise in future, present consumption will be more. If the prices expected to fall down in future present consumption will be less.

5. Effect of exposure:
The countries where goods are exposed or advertise more, there propensity to consume is more.

Question 6.
Throw light on the concept of full employment and involuntary unemployment.
Answer:
(1) Full employment: The term “Full employment” signifies the situation. Where in all the persons willing to work at existing rate of remuneration are getting employment. Full employment never means zero unemployment. According to Ward “Full employment refers to level of employment associated with a normal level of unemployment.” The full employment equilibrium refers to that equilibrium which is arrived at by equality of aggregate demand and aggregate supply.

The point of full employment will be where, the aggregate demand curve and the aggregate supply curve both intersect each other as is shown in the following diagram. Here, at O.P. (National income level) aggregate demand curve and aggregate supply curve both meet at point E. Hence, EP is represents full employment and E is called the point of full employment equilibrium.

(2) Involuntary unemployment or Under employment:
The term “unemployment” refers to that situation where in all the persons are willing to work at existing rate of remuneration do not get employment. It signifies Y the state where the available resources are not fully utilized. The aggregate supply of goods and services is perfectly elastic.

In other words the supply of can change when there is a change in demand. In this situation, the equilibrium level of output and employment will depend on the aggregate demand. It is a situation below full employment. The aggregate anticipated expenditure is less than the available aggregate output.

This can be explained by diagram:
In diagram point ‘E’ is of full employment. At this point the demand of labours is equal to the supply of labours. On OW₁ . wages ON₁ labours are willing to work. Enterprises requires only ON₂ labour demand. Thus N₁ N₂ will be known as involuntary employment.

Question 7.
What do you mean by deficit demand? What is the impact of deficient demand on production output, employment on prices.
Answer:
Deficit demand is a situation where the equilibrium between aggregate demand and aggregate supply takes place at a point below the full employment point. It is also known as under employment equilibrium. In other words deficient demand refers to the situation where available aggregate output exceeds the anticipated aggregate expenditure.

In such situation the available resources are not fully utilized. This shortage of aggregate demand does not enable us to reach the level of full employment so it is called deficient demand.

The impact of deficient demand can be studied under the following heads:

I. Impact of deficit demand on output:
if the aggregate demand is less in compare to the aggregate supply is an economy it means that what ever goods and services are available in the country are not fully demanded by people. This situation has impact on total output.
The impact are as follows:

• Due to less demand the producers will be forced to reduce the production.
• In such situation producers will not be able to utilize their present production capacity fully.
• Thus factors of production will not be fully utilized by the producers.
• Due to the situation of deficient demand the present existing firms will stop their work and new firms will not enter in the market.
• The cost of production will be high in this situation.

II. Impact of deficit demand on employment:
In an economy when the situation of ‘ deficit demand arises then it affect the employment situation also. The impact is seen in the following way:

1. Due to shortage of demand the producers will reduce the production and thus less no. of workers will be required.
2. If in the country there is a situation of partial employment then some people will have to lose their jobs.
3. If in a country already the situation of under employment prevails then the condition will became from bad to worse.

III. Impact of deficit demand on prices:
When the situation of deficit demand arises there is impact on prices of goods also. The impact is as follows:

1. If aggregate demand is less as compared to aggregate supply then in the country price level will be less.
2. Due to reduction in price level the difference of profit to the producers will be less.
3. The consumers will get the profit of this situation because they will get the goods and services at less price.

Question 8.
Explain the determination of effective demand.
Answer:
The principle of effective demand is the foundation stone of income and employment theory of Prof. Keynes. According to Keynes in an economy, employment depend on effective demand. Due to reduction in effective demand unemployment spreads in a country and only by increasing effective demand unemployment can be removed. By effective demand we mean the amount of money which people of a country spent on consumption and investment in a specific time. That level of demand is effective which is equal to aggregate supply.

According to Keynes, “The point of aggregate demand at which the curve of aggregate curve intersect the price at that point is called the effective demand. Here, AD curve represent aggregate demand function and A₁S curve represent aggregate supply function, AD and A₁S curves meet each other at N point which is a point of effective demand. At this point OP is the employment available and here OQ = receipts = costs. At any other point two curves cannot be in equilibrium. Here AD and AS are exactly equal.

Question 9.
Write the importance of effective demand.
Answer:
Following are the importance of effective demand:
(1) Determination of employment:
According to Keynes, the employment depend on effective demand and aggregate demand is equal to aggregate supply. Thus effective demand determines the quantity of demand on one hand and on the other hand the amount of income produced by it.

(2) Receiving income:
Effective demand also determines income. Labour get employment due to effective demand. Employment gives wages to labours. This is the effect of effective demand only.

(3) Importance of investment:
Effective demand also determines investment. Due to it producers are motivated to produce more. To increase production the investors and producers are encouraged to invest in new-new capital resources.

(4) The concept of full employment is rejected:
The concept of full employment is fully rejected with the help of effective demand. In practical life this concept is not accepted because employment is created by effective demand. When effective demand increases the employment is also created.

By getting employment the situation of employment is improved. By decrease in effective demand the level of employment reduces and unemployment spread in the society. Thus retrenchment of people is done even if he is capable. i > o the concept of full employment is not accepted.

Question 10.
What do you mean by excess demand? Write it causes.
Answer:
Meaning: When planned aggregate expenditure (i.e., aggregate demand) is greater than the aggregate supply (i.e., available output) at full employment level, the situation is termed as excess demand. When the anticipated expenditure in the economy is more than the value of full employment output, there exists a gap between aggregate demand and aggregate supply.

The gap is technically termed as inflationary gap. (Excess demand or Inflationary gap is a situation of escalating expenditure in the economy without a rise in the flow of goods and services.
Excess demand = Inflationary gap = ADE – ADF = FC.

Causes of Excess demand:
Excess demand may caused by several factors. Important among them are mentioned below:

• Excess demand may be caused by an increase in money supply due to deficit financing.
• It may also be caused by increase in consumption demand due to rise in propensity to consume.
• An increase in autonomous investment without corresponding increase in savings may cause excess demand.
• A fall in the supply of consumer goods and services due to diversion of resources from the production of consumer goods to the production of defense goods, will result in excess demand. The available output for the households will not be sufficient to meet their aggregate demand.
• Excess demand may also arise when demand for exports increases.

Question 11.
Explain the employment principle by Keynes.
Answer:
Keynes, in his book ‘National Principle had analyzed income and employment principle. He propounded this principle during the depression period of 1930. According to his theory the level OL. Income and employment is maintained by effective demand. In the absence of effective demand unemployment increases. Effective demand is determined by

1. Aggregate demand,
2. Aggregate supply.

The point where AD = AS, at this point Effective demand is determined. At this point the amount of production, and level of employment is determined. This can be explained by chart:
Effective Demand = Total production
Total income = Employment

Question 12.
What do you mean by Deficit demand? What are its causes?
Or
State any five causes of Deficit demand.
Answer:
Deficit demand is a situation where the equilibrium between aggregate demand and aggregate supply takes place at a point below the full employment point. It is also known as under employment equilibrium. In other words deficient demand refers to the situation where available aggregate output exceeds the anticipated aggregate expenditure.

In such situation the available resources are not fully utilized. This shortage of aggregate demand does not enable us to reach the level of full employment so it is called deficient demand. Cause of Deficient demand: In any economy the situation of Deficient demand arises due to following reasons:

• Fall in public expenditure,
• Reduction in export demand,
• Fall in investment demand.
• Fall in disposable income,
• Fall in consumption demand,
• Fall in supply of money,
• Fall in creation of credit.

Question 13.
State measures to correct excess demand.
Answer:
1. Bank rate:
The rate at which the central bank lends money to its member banks is called bank rate. In a situation of excess demand, the central bank should raise the bank rate. Increase in bank rate will automatically lead to increase in banks lending rates to their borrowers. This, in turn, will reduce the volume of banks loans and advances. Consequently, expenditure on both consumption and investment is reduced.

2. Increasing the reserve ratio:
Reserve ratio refers to a minimum percentage of deposits of a commercial bank in the Reserve Bank. If the reserve ratio is increased it will reduce the credit creation capacity of the banks. When there is a situation of excess demand, central bank can increase this ratio and thereby cash resources of the banks can be reduced. This will force the commercial banks to contract credit.

3. Open market operations:
The buying and selling of government securities and bonds in the open market by the central bank are called open market operations. In the time of excess demand, central bank sells government securities to commercial banks or to their customers. This will reduce the cash reserves with the commercial banks directly and indirectly thereby affecting their capacity to make loans. It will reduce the cash reserves of the commercial banks.

4. Reducing credit Inflation:
To reduce the excess demand or to reduce inflation it is essential to reduce credit inflation. By selling securities, by increasing bank rate by central bank credit inflation can be reduce.

5. By taking old currency and circulating new currency:
In situation of excess demand the government cancel the old currency and in place of it new currency was intro duced for circulation.

MP Board Class 12th Economics Important Questions

MP Board 12th Class Physics Important Questions Chapter 8 Electromagnetic Waves

Electromagnetic Waves Important Questions

Electromagnetic Waves Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The electromagnetic waves are produced due to :
(a) Accelecrated charge
(b) Charge at rest
(c) A charge moving with uniform speed
(d) Current carrying conduction.
Answer:
(b) Charge at rest

Question 2.
Which of the following is not an electromagnetic wave :
(a) X – ray
(b) Ultraviolet ray
(c) Beta rays
(d) Visible light.
Answer:
(c) Beta rays

Question 3.
Electromagnetic waves were first verified experimentally by :
(a) Maxwell
(b) Hertz
(c) Bose
(d) Marconi.
Answer:
(b) Hertz

Question 4.
The formula for the velocity of electromagnetic wave in vacuum is given by :

Answer:

Question 5.
The ozonosphere absorbs :
(a) Visible light
(b) Microwaves
(c) Infrared rays
(d) Ultraviolet radiations.
Answer:
(d) Ultraviolet radiations.

Question 6.
If in an electromagnetic wave the electric vector is along X – direction magnetic vector is along Y – direction then the direction of propagation will be :
(a) X – direction
(b) Y – direction
(c) Z – direction
(d) None of these.
Answer:
(c) Z – direction

Question 2.
Fill in the blanks :

1. The sound waves are the longitudinal waves. The electromagnetic wave is …………….
2. The electromagnetic waves are sinusoideal oscillations of ……………. and ……………. which are mutually perpendicular.
3. The colour of visible light depends upon its …………….
4. The ……………. rays are used as germ killer.
5. The wave length range of visible light is …………….
6. The electromagnetic wave having maximum frequency is …………….
7. The waves having frequency of order of mega hertz are called as …………….

Answer:

1. Transverse waves
2. Electric and magnetic fields
3. Frequency
4. Ultraviolet
5. 400 nm to 800 nm
6. Gamma
7. Radio waves.

Question 3.
Match the Column :

Answer:

1. (b)
2. (d)
3. (e)
4. (c)
5. (a)

Question 4.
Write the answer in one word/sentence :

1. Name the scientist who first predicted the existence of electromagnetic waves.
2. Write the SI unit of displacement current.
3. What is displacement current due to?
4. Is displacement current like conduction current also a source of magnetic field?
5. Write down Maxwell’s equations for steady magnetic field.
6. Which of the following has the lowest frequency : microwaves, ultra – violet rays and X – rays?

Answer:

1. James Clerk Maxwell
2. The SI unit of displacement current is ampere (A)
3. Displacement current exists due to a time – varying electric field or changing electric flux
4. Yes, displacement current also produces a magnetic field
5. (i) \(\int { \vec { B } .\vec { dS } }\) = 0
   (ii) \(\int { \vec { B } .\vec { dl } }\) = 0.6. Microwaves.

Electromagnetic Waves Very Short Answer Type Questions

Question 1.
Answer the following questions in one word/one sentence :

1. Who discovered electromagnetic waves?
2. Are electromagnetic waves deflected by electric or magnetic field?

Answer:

1. Maxwell
2. No.

Question 2.
Write the formula by which the speed of electromagnetic waves can be determined in vacuum.
Answer:
Speed of electromagnetic wave is given by the formula :
c = \(\frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } }\)
Where, µ₀ = Permeability of free space and ε₀ = Permittivity of free space.

Question 3.
How do the electric field and magnetic field propagate in space?
Answer:
In the space the electric field and magnetic field travel perpendicular to each othe also perpendicular to the propagation of wave.

Question 4.
What type of oscillation takes place in electromagnetic waves? Are they transverse or longitudinal?
Answer:
In electromagnetic waves, both electric field and magnetic field oscillate perpendicular to each other and also perpendicular to the direction of propagation of waves. They are transverse in nature.

Question 5.

1. State the relation between the amplitudes of electric field and magnetic field of electromagnetic waves travelling through vacuum.
2. The electromagnetic waves are transverse in nature, how?

Answer:

1. c = E/B, where c is the speed of electromagnetic wave, E is amplitude of electric field and B is amplitude of magnetic field.
2. The electric field and magnetic field are perpendicular to each other and also perpendicular to the direction of propagation of wave. Hence, electromagnetic waves are transverse in nature.

Question 6.

1. What was the limit of wavelength of electromagnetic waves produced by Hertz?
2. What was the limit of wavelength of electromagnetic waves produced by J.C. Bose ?

Answer:

1. About 5 m,
2. 5 to 25 mm.

Question 7.
Write the frequency of electromagnetic spectrum in ascending order.
Answer:
Radio waves, Microwaves, Infrared rays, Visible light, Ultraviolet rays, X – rays and γ – rays.

Question 8.
Arrange the following electromagnetic waves in ascending order of their wavelength.
Microwaves, γ – rays, radio waves, violet rays.
Answer:
γ – rays, violet rays, microwaves and radio waves.

Question 9.
Answer the following questions :

1. What is the range of wavelength of visible rays?
2. What is the wavelength of X – rays?

Answer:

1. Wavelength of visible rays ranges from 4000 Å to 7800 Å.
2. 1 Å to 100 Å.

Question 10.
Which waves are used as signals through fog and why?
Answer:
Infrared rays are used for signals through fog because its wavelength is greater and hence the scattering is least.

Question 11.
What is displacement current?
Answer:
When electric flux changes with time in vacuum or in medium, then it is equivalent to an electric current. This current is called displacement current.

Question 12.
What are electromagnetic waves? What is its nature?
Answer:
Electromagnetic waves:
Electromagnetic waves are those waves which are produced by the oscillating magnetic and electric fields perpendicular to one another and also perpendicular to the direction of propagation of waves. The nature of electromagnetic waves are transverse.

Question 13.
Write four properties of electromagnetic waves.
Answer:

1. The electromagnetic waves are produced by oscillatory or accelerated charge.
2. They are transverse in nature.
3. They can propagate through vacuum.
4. The speed of electromagnetic waves in vacuum is same as the speed of light (= 3 x 10⁸m/s).

Question 14.
Dim red light is used in film developing room, why?
Or
Why red light is used in photographic room?
Answer:
The energy of photons of red light is less therefore, it hardly affects the films.

Question 15.
The bulbs of ultraviolet lamps are made up of quartz, why?
Answer:
The ultraviolet rays are absorbed by glass but not by quartz, therefore ultraviolet rays pass through quartz but not through the glass.

Question 16.
Optical telescope and radio telescopes are used in earth but X – ray telescopes are used in space, why?
Answer:
Visible rays and radio waves are not absorbed by atmosphere of earth while X – rays are absorbed, therefore X – ray telescopes can be used only in space or in satellite.

Question 17.
Had there been no atmosphere near to the surface of the earth, what would have been the surface temperature ? Answer with reasons.
Or
How would the absence of atmosphere affect the temperature of earth surface?
Answer:
The temperature of the earth would be lower because the green house effect of the atmosphere would be absent.

Question 18.
Where is the ozone layer situated in the atmosphere? What role it plays in the survival of human beings?
Or
How, the thin ozone layer is crucial for animal survival on earth?
Answer:
Ozone layer absorbs ultraviolet radiation coming from sun and other heavenly bodies and hence prevents it from reaching to the earth. Ultraviolet radiations cause genetic effect and damage the life of animals. Ozone layer is situated in between 30 to 50 km above the earth.

Question 19.
On what factors does the ionization of atmosphere depend?
Answer:
The state of ionization of atmosphere depends upon the ultraviolet rays coming from the sun and other heavenly bodies and also the height of atmosphere from the surface of earth.

Question 20.
What are the various ways by which radio waves can propagate?
Answer:
Radio waves can propagate from one place to the other on the earth by the following three modes in accordance with their frequency :

1. By ground wave
2. By space wave
3. By sky wave.

Question 21.
What are ground waves?
Answer:
The radio waves transmitted from radio transmission travelling along the earth surface are called ground waves. The wavelength of 200 m and more than that are transmitted as ground waves. This is called as medium wave band.

Question 22.
On what factors does the transmission of ground waves depend?
Answer:
The transmission of ground waves depends upon :
Wavelength:
The wavelength of waves will be 200 m and more are transmitted as ground waves. The wavelengths less than 200 m are absorbed by earth.

The curvature of earth:
The earth surface is curved hence ground waves cannot be transmitted for far distance.

Question 23.
What are sky waves? Explain.
Answer:
Those radio waves which are transmitted through lower atmosphere but reflected by ionosphere are called sky waves. The radio waves in the frequency range from 1500 Hz to 40 MHz are propagated by this mode. This is called short – wave band.

Question 24.
Why artificial satellites are used for long distance T.V. communication?
Answer:
T.V. signals are not reflected by ionosphere, therefore these waves are reflected by satellite.

Question 25.
Write one use of each :

1. Microwaves
2. Infrared ray
3. Ultraviolet rays
4. Gama rays.

Answer:

1. Microwaves : In radar.
2. Infrared ray : Radiation purpose in some diseases.
3. Ultraviolet rays : For killing germs.
4. Gama rays : For treatment of cancer.

Question 26.
Write the type of radio wave and write its one use :
Answer:
Waves:

1. Short radio wave
2. Wireless wave

Uses:

1. For propagation of radio and television signal
2. In navigation and police radio transmission

Electromagnetic Waves Short Answer Type Questions

Question 1.
What are electromagnetic waves? Write their four properties. Draw suitable figure for showing direction of electric field, magnetic field and propagation of wave.
Or
What are electromagnetic waves? Write their main properties.
Answer:
Electromagnetic waves are those waves which are produced by oscillating magnetic and electric field perpendicular to each other and also perpendicular to the direction of propagation.

The properties of electromagnetic waves are :

1. The electromagnetic waves are transverse in nature.
2. They can travel through vacuum also.
3. They are not affected by magnetic field and electric field.
4. The velocity of electromagnetic waves in vacuum is constant i.e., 3 x10⁸ m/s.
5. These waves show the properties of reflection, refraction, interference, diffraction, polarization etc.
6. These waves are produced by oscillating or accelerated charge.
7. At every point in the space c = E/B, where c = Velocity of wave, E = Electric field and B = Intensity of magnetic field.

Question 2.
Write the differences between sound waves and electromagnetic waves.
Answer:
Differences between sound waves and electromagnetic waves :
Sound waves:

• They are mechanical waves.
• They are longitudinal waves.
• The sound waves do not show the the phenomenon of polarization.
• For the propagation of sound waves, medium is required.
• The velocity of sound waves is near about 332 m/s at 0°C in air.

Electromagnetic waves:

• They are not mechanical waves.
• They are transverse in nature.
• Electromagnetic waves show the pheno – menon of polarization.
• Electromagnetic waves can travel in vacuum. For propagation of electromagnetic waves, medium is not required.
• Its velocity is very high; about 3 x 10⁸ m/s.

Question 3.
What is electromagnetic spectrum? Write down the name of different type of spectrum and their uses. (One use for each constituent of spectrum)
Answer:
The array obtained on arranging all the electromagnetic waves in order, on the basis of their wavelength is called electromagnetic spectrum. In order of increasing wavelength they are :

1. Gamma rays
2. X – rays
3. Ultraviolet radiation
4. Visible light
5. Infrared radiation
6. Microwaves
7. Radio waves.

Uses of electromagnetic spectrum :

1. γ – rays : To kill unwanted cell in the treatment of cancer.
2. X – rays : To detect broken bones and to study the structure of crystals.
3. Ultraviolet rays : To kill the germs, to detect the purity of ghee, diamond and jewels.
4. Visible rays : To see the objects.
5. Infrared rays : To use as signals during war, in photography at night or in mist or fog, for therapeutic purpose by the doctors.
6. Microwaves (Hertzian waves) : In Radar, television, microwave oven.
7. Radio waves (Wireless waves): In television, navigation and polic radio transmission.

Question 4.
Write the uses of different region of electromagnetic spectrum.
Answer:
Uses of electromagnetic spectrum :

1. γ – rays : To kill unwanted cell in the treatment of cancer.
2. X – rays : To detect broken bones and to study the structure of crystals.
3. Ultraviolet rays: To kill the germs, to detect the purity of ghee, diamond and jewels.
4. Visible rays : To see the objects.
5. Infrared rays : To use as signals during war, in photography at night or in mist or fog, for therapeutic purpose by the doctors.
6. Hertzian waves : Radio and T. V. transmission.
7. Wireless waves : In navigation and police radio transmission.

Question 5.
Why electromagnetic waves when strike a surface exerts pressure on it?
Answer:
Electromagnetic waves carry momentum, so it exerts pressure on surface when strike on it.

Question 6.
What are micro waves? State one of its important uses.
Answer:
The waves whose order of wavelength in few milimetre are called microwaves. It is used in radar system.

Question 7.
Write the names of the scientists who discovered the different constituent of electromagnetic spectrum.
Answer:
The scientists who discovered the different constituents are as follows :

1. γ – rays : Becquerel
2. X – rays : Roentgen
3. Ultraviolet rays : Ritter
4. Visible rays : Newton

Question 8.
Describe the Hertz’s experiment of the production of electromagnetic waves in following points:

1. Labelled diagram of apparatus
2. Experimental arrangement.

Answer:
It consists of two parallel plates P₁, and P₂ connected to two metal spheres S₁ and S₂. The distance between P₁ and P₂ is 60 cm. The gap between S₁ and S₁ can change by sliding along the rods R₁ and R₂.R₁ and R₂ are connected to the terminals of the secondary coil of an induction coil, so that a high voltage can be establi shed between S₁ and S₂ and spark between the gap is produced, P₁ and P₂ form a capacitor of low capacitance.

D is a detector. When the charge starts oscillating, between the gap, series of sparks are produced. This results in the production of electromagnetic waves. These radiations are received by detector D.

Question 9.
What is displacement current? Why was this concept introduced?
Answer:
Displacement current:
It is that current which comes into existence (in addition to conduction current) whenever the electric field and hence the electric flux changes with time. It is equal to s0 times the rate of change of electric flux through a given surface
I_d = ε₀\(\frac { d{ \phi }_{ E } }{ dt }\) = ε₀A\(\frac {dE}{ dt }\)
Maxwell introduced the concept of displacement current for the following reasons :

1. To make the Ampere’s circuital law logically consistent.
2. To satisfy the property of continuity of electric current along any closed path.

Question 10.
Prove that I_D = ε₀\(\frac {d}{ dt }\)ϕ_E, where the symbols have usual meaning.
Answer:

Question 11.
What conclusion can be drawn between displacement current and conduction current?
Answer:
Like conduction currents displacement current is also a source of magnetic field. Both the current are equal to each other. Displacement current is originated due to charge in electric field between two plates of a condensor. I_D current is in between the two plates of condensor whereas conduction current is in the wire connecting both plates and is a source of electromotive force.

Electromagnetic Waves Long Answer Type Questions

Question 1.
Describe different components of electromagnetic spectrum.
Answer:
Electromagnetic waves have a large range of wavelength and frequency. When Maxwell predicted about the existence of electromagnetic waves at that time he was aware of visible light only. Gamma rays, X – rays, UV – rays and infrared rays were discovered at the end of 19^th century. Later on radio waves and microwaves came into existence. So, electromagnetic waves have a large frequency or wavelength range.

All these electromagnetic wayes have general properties. But they do have some individual features. So, the classification on the basis of frequency or wavelength is called “Electromagnetic spectrum”. “An orderly arrangement of radiations according to wavelength (or frequency) is called electromagnetic spectrum”. Names of rays in decreasing frequency and increasing wave length are as follows :

1. Gamma rays (y rays)
2. X – rays (X – rays)
3. UV rays
4. Visible rays
5. IR rays
6. Microwaves
7. Radio waves

Following facts related to electromagnetic spectrum are to be remembered.

1. Gamma rays:
These rays were discovered by Henery Becqurel in 1896. These rays have frequency from 10¹⁸ Hz to 10²² Hz (i.e., from wavelength 10^-14 m to 10^-10 m).

Production / generation:
The rays are produced due to nuclear reactions, radioactive elements like uranium and radium emit these rays by themselves.
Properties:

1. These rays affect photographic plates.
2. These rays ionize gases.
3. These rays produce luminiscence.
4. These rays have high penetration power.

Uses:

1. Making polythene from Enelene.
2. To detect the nuclear structure.
3. In destroying unwanted celis.
4. In treatment of cancer.

2. X – rays:
These rays were discovered by Rontegen in 1895. These rays have frequency from 10¹⁶ Hz to 10²⁰ Hz (or wavelength from 10^-13 m to 10² m).

Production:
These rays are produced by bombarding of electrons on heavy metals.

Properties:

1. These rays affect photographic plate.
2. These rays ionize gases.
3. These rays produce luminiscence.
4. These rays have less penetration power as compared to Gamma rays.

X – rays can penetrate human body, but can be obstructed by bones and metallic materials.

Uses:
In medical investigation, like in the detection of fractured bone, the presence of metallic things in the body such as coin, bullet etc.

1. In the study of crystalline structure.
2. In the treatment of cancer and tumour.
3. In the engineering field, in detecting cracks and defects etc.
4. They are used in metal detector in restricted are as to find the prohibited things.

3. Ultraviolet rays:
These rays were discovered by J. W. Ritter in year 1801. These waves have frequency from 7.5 x 10¹⁴ Hz to 5 x 10¹⁷Hz (or wavelength from 6 x 10^-10m to 4 x 10^-7m.)

Production:
Ultraviolet rays are generated from discharge tube or carbon arc etc. Sun is an important source of ultraviolet rays. These rays are absorbed by ozone layers before reaching to atmosphere.

Properties:

1. These rays affect photographic plate and are more active in visible region for chemical changes.
2. These rays also produce fluorescence.
3. These rays show photoelectric effect.
4. These rays ionize the gases.

Uses:

1. In killing micro-organisms of diseases.
2. Producing photoelectric effect.
3. In detection of jewels, gems, fresh eggs and of pure Ghee.
4. In detection of finger print.
5. In preservation of food materials.

4. Visible light:
It has frequency from 4 x 10¹⁴ Hz to 7 x 10¹⁹ Hz (or wavelength from 4 x 10^-7 m to 7.8 x 10^-77).

Production:
Sun is main source of visible light. The light obtained from electric bulb, candle and eye lamp is visible light. It is emitted when atoms of excited state reach to the ground state. It was first studied by Newton in 1966. The visible spectrum has many colour, the prominent being violet, indigo, blue, green, yellow, orange and red.

Properties:

1. The visible light produces sight sensitivity.
2. It shows chemical effect. As a consequence, it affects the photographic plate.
3. The visible light shows photoelectric effect.

5. Infrared rays:
These rays were discovered by William Herschell in 1800. These rays have frequency from 3 x 10¹¹ Hz to 4.0 x 10¹⁴Hz (or wavelength from 7.8 x 10^-7 to 10^-3m).

Production:
Infrared rays are produced by hot bodies and molecules. So, these are also called thermal rays.

Properties:

1. These rays show thermal effect.
2. Due to greater wavelength, these rays get less scattered, so these rays can be propagated up to long distances in fog also.
3. These rays do not affect photographic plate.

Uses:

1. Infrared lamps are used in treatment of diseases.
2. These rays are also used in photography in dark because these rays do not affect photographic plate.
3. To send signals from one place to another in fog and mist, because they have longer wavelength so their scattering is least and penetrating power is high.
4. These rays are used to produce greenhouse effect.

6. Microwaves:
These waves were discovered in 1888 by Hertz. These rays have frequency from 109 Hz to 10¹² Hz (i.e., wavelength from 10^-3 m to 0.3 m.). These are short wavelength radio waves.

Production:
These rays are produced in special type of vacuum tube like magnetron, klystron etc.

Properties:
These rays have very short length. So, they do not get diffracted by the obstruction of large size. These rays move forward in the form of a narrow beam of rays. So, these rays can be sent in a certain direction without any expansion.

Uses:

1. These waves are mainly used in Radar system and Remote communication.
2. These waves are detected by crystal detectors or semiconductors diode.

7. Radio waves:
These rays were discovered in 1899 by Marconi. Frequency of these waves are generally measured in kHz or MHz. These rays have frequency from 500 kHz to 1000 MHz or wavelength from 0.1 m to 6 x 10² m. These waves are also called wireless rays.

Production:
These waves can be produced by oscillator circuit and can also be produced by passing high frequency alternating current through any conductor.

Properties:
Due to higher wavelength, they get diffracted.

Uses:

1. These waves are used in radio or TV communication as follows :
2. 530 kHz to 1710 kHz radio waves are used in amplitude modulation.
3. 54 MHz high frequency short waves are used in radio bands.
4. From 50 MHz to 890 MHz waves are used as television rays.
5. From 88 MHz to 108 MHz waves are used in frequency modulation.

Electromagnetic Waves Numerical Questions

Question 1.
A parallel plate capacitor has circular plates, each of radius 5.0 cm it is being charged so that electric Field in the gap between its plates rises steadily at the rate of 10¹² Vm^-1 s^-1. What is the displacement current? (NCERT)
Solution:
Here r = 5 cm = 5 x 10^-2 m,
\(\frac {d}{ dt }\) = 10¹² Vm^-1 s^-1 dt
Displacement current,
I_d = ε₀\(\frac { d{ \phi }_{ E } }{ dt }\) = ε₀ A \(\frac {dE}{ dt }\)
= ε₀.πr² \(\frac {dE}{ dt }\)
= 8.85 x 10^-12 x π x(5 x 10^-2) x 10¹² A
= 0.07 A.

Question 2.
The voltage between the plates of a parallel – plate capacitor of capacitance 1.0µF is changing at the rate of 5 Vs^-1. What is the displacement current in the capacitor? (NCERT)
Solution:
Here C = 1.0µF = 1.0 x 10^-6F,
\(\frac {dV}{ dt }\) = 5Vs^-1
Displacement current,

Question 3.
Electromagnetic waves travel in a medium at a speed of 2.0 x 10⁸ms^-1. The relative permeability of the medium is 1.0. Find the relative permittivity.
Solution:
Speed of an electromagnetic wave in a medium is given by

Question 4.
A plane electromagnetic wave of frequency 25 MHz travels in free space along the X – direction. At a particular point in space and time, \(\vec { E }\) = 6.3\(\hat { j }\) Vm^-1. What is \(\vec { B }\) at this point?
Solution:
Magnitude of \(\vec { E }\),E = 6.3 Vm^-1.
Magnitude of \(\vec { B }\) ,

We note that E is along Indirection and the wave propagates along X – direction. Now an electromagnetic wave propagates in the direction of the vector \(\vec { E }\) x \(\vec { B }\). So this vector must point along X – direction.

MP Board Class 12th Physics Important Questions

MP Board Class 12th Economics Important Questions Unit 8 Money and Banking

Micro Economics Money and Banking Important Questions

Micro Economics Money and Banking Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
“Money is what money does”. Who said this:
(a) Hartley Withers
(b) Harte
(c) Prof. Thomas
(d) Keynes.
Answer:
(a) Hartley Withers

Question 2.
Function of money is:
(a) Medium of exchange
(b) Measure of value
(c) Store of value
(d) All of the above.
Answer:
(a) Medium of exchange

Question 3.
Meaning of money supply is:
(a) Money deposits in the bank
(b) Cash available with public
(c) Savings in the post office
(d) All of the above.
Answer:
(d) All of the above.

Question 4.
What is the Central Bank of India:
(a) Commercial Banks
(b) Central Bank
(c) Private Bank
(d) None of these.
Answer:
(a) Commercial Banks

Question 5.
Through which method we can withdraw money from the bank:
(a) Drawing letter
(b) Cheque
(c) A.T.M.
(d) All of the above.
Answer:
(d) All of the above.

Question 6.
Who is the guardian of Indian Banking System:
(a) Reserve Bank of India
(b) State Bank of India
(c) Unit Trust of India
(d) Life Insurance Company of India.
Answer:
(a) Reserve Bank of India

Question 7.
Narasimham Committee is related to what:
(a) Improvement in Taxation
(b) Improvement in Banking
(c) Improvement in Agriculture
(d) Improvement in Infrastructure.
Answer:
(b) Improvement in Banking

Question 2.
Fill in the blanks:

1. Central Bank of India is………………..
2. Bank rate is also known as………………..
3. The bank generates……………….. deposits in credit creation.
4. When CRR decreases, credit creation………………..
5. Measure of deferred payment is ……………….. function of money.
6. Medium of exchange is ……………….. function of money.
7. The static and dynamic function of money is divided by………………..

Answer

1. Reserve Bank of India
2. Redemption
3. Derivative
4. Increases
5. Secondary
6. Primary
7. Paul Einzig.

Question 3.
State true or false:

1. Money is needed for day-to-day transactions.
2. The precautionary demand for money increases with the proportionate increase in income.
3. Reserve Bank of India provides loan to public.
4. Along with the Reserve Bank of India, Commercial banks are also authorized to issue currency.
5. Reliable money also include cheques.
6. Reserve Bank of India cannot become the owner of any real estate.

Answer:

1. True
2. True
3. False
4. False
5. False
6. True.

Question 4.
Match the following:

Answer:
1 – c
2 – a
3 – d
4 – e
5 – b

Question 5.
Answer the following in one word / sentence:

1. Name the bank that provides long term loans to the farmers.
2. Trade cycle is which evil of money?
3. By increasing liquid fund ratio, which will be the effect on money supply?
4. “The Banks are not only generators of money but are also creators of money”.
5. Who has the right to issues paper money?
6. When was NABARD established?

Answer:

1. Agricultural or Cooperative banks
2. Economic
3. Decreases
4. Sayer’s
5. Central Bank
6. 1982.

Micro Economics Money and Banking Very Short Answer Type Questions

Question 1.
Write two taboo works of Reserve Bank of India.
Answer:
Two taboo works of RBI are:

1. Reserve Bank cannot accept deposits on interest from the public.
2. Reserve Bank cannot provide loans for a fixed term.

Question 2.
What do you mean by overdraft facility?
Answer:
Clients who have current account with the bank are granted” the facility of withdrawing more money than actually lying in their accounts. It is called overdraft. This facility is available to reliable person for a short term.

Question 3.
What is commercial bank?
Answer:
Commercial banks are those banks who are established under the Indian Company Act, and they perform all the functions of banks.

Question 4.
Write the meaning of money.
Answer:
According to Prof. J.M. Keynes, Money is that by the delivery of which debt contracts and price contracts are discharged and in the shape of which a store of general purchasing power is held.

Question 5.
Write any two economic defects of Money.
Answer:
Two economic defects of money are:

1. Money gives birth to trade cycle.
2. Currency is the trust worthy custodian of money.

Question 6.
When was Reserve Bank of India nationalised?
Answer:
Reserve Bank of India was nationalised on 1st January 1949.

Question 7.
Which is the Central Bank of India? When was it established?
Answer:
Central Bank of India is Reserve Bank of India. It was established on 1st April, 1935.

Question 8.
What do you mean by Bank?
Answer:
Nowadays, the word ‘Bank’ is very common and popular so, general public is acquainted with it. Generally, bank means that institution which deals with transactions of money.

According to Prof. Wicksell:
“Bank is the heart and central point of modem currency system”.

Question 9.
Write definitions of money.
Answer:
According to Prof. Hartley Withers, “Money is what money does”. According to Prof. Seligman, “Money is one thing that possesses general acceptability”.

Question 10.
What is the problem of double coincidence of wants in barter exchange?
Answer:
1. Problem of double coincidence:
In barter system, goods are to be exchanged for goods, hence, it is essential that both the parties should need the goods which other has in exchange of their goods, then only the exchange is possible.

2. For example, ‘A’ possesses a chair and ‘B’ possesses a table. Now if ‘A’ wants to exchange his change for a table then he has to search a person, who needs chair in exchange of table. Thus, becomes a complex process for him to find such man.

Question 11.
‘Money is the medium of exchange,’ Explain.
Answer:
Medium of exchange:
This is the most important function of money. It acts as a medium of exchange. All the exchanges of goods and services are taken place in terms of money. (MPBoardSolutions.com) By paying predetermined price, money can be exchanged with the desired goods and services. A money has the general acceptability, therefore, all the exchanges in an economy take place in terms of money. It is because of this reason that money has been defined as generally acceptable purchasing power.

Question 12.
“Measure of value is the main function of money”. Explain.
Answer:
“Measure of value is the main function of money”:
The main function of money is that it measures the value of goods and services. In other words, the prices of all goods and services are expressed in terms of money. In ancient times it was not possible to measure the value of clothes and wheat. With the introduction of money, this difficulty of measurement disappeared and it became very convinient to measure the value of money. Money acts as a unit of account for all goods, wages salaries, interests etc. The national income, capital formation and other are measured in terms of money.

Question 13.
Write the importance of the money as the store of exchange value.
Answer:
Money acts also as a store of value because:

1. Money can be stored very easily. Money is a liquid form of capital. It require less place to store.
2. Money has the merit of general acceptibility.
3. It is convinient to store money.
4. Value of money remains relatively stable. Due to all these reasons money is important.

Question 14.
What is meant by money supply? What are its measures?
Answer:
Money supply refers to the total volume of money held by the public at a particular point of time in an economy.
Measures of money supply are:
M₁ = Currency and coins with public + Demand deposits of Commercial banks + Other Deposites with RBI.
M₂ = M₁ + Savings deposits with post office Saving bank.
M₃ = M₂ + Net time deposits with banks.
M₄ = M₃ + Total deposits with post office saving bank.

Question 15.
Write difference between Scheduled Banks and Non – Scheduled Banks.
Answer:
Differences between Scheduled Banks and Non – Scheduled Banks:

Scheduled Banks:

• Scheduled banks are those banks whose names appear in the second schedule of RBI.
• The paid up share capital together with reserve fund must not be less than Rs. 5 lake.

Non-Scheduled Banks:

• Non – scheduled banks are those banks whose names do not appear in the second schedule of RBI.
• The paid up capital of such banks together with cash reserve is less than Rs. 5 lake.

Question 16.
If in an economy, all the customers of the bank deposit their money in the banks and if all of them withdraw their money from the bank then what will be its effect on the economy?
Answer:
If in any economy, bank customers deposit their money in the banks then there will be shortage of money for production and investment activity. As a result, the process of development will come to a stand still. In the same way, if all the customers withdraw their money from the banks, the banking would collapse and economy will face a problem. If in any bank financial crisis arises then central bank is the only bank to settle this problem and all banking systems will collapse.

Question 17,
Write the meaning of Commercial banks?
Answer:
A commercial bank is a financial institution which accepts deposits from the public and gives loans for purposes of consumptions and investment.

Companies Act, 1949:
A banking company is one which transacts the business of banking which means the accepting for the purpose of lending or investment of deposits of money from the public repayable on demand or otherwise and withdrawal by cheque or otherwise.

Question 18.
Write the types of Commercial bank.
Answer:
There are two types of Commercial banks:

1. Scheduled banks:
Scheduled banks are those banks whose names appear in the second schedule of RBI.

2. Non – scheduled banks:
Non – Scheduled banks are those, banks whose names do not appear in the second schedule of RBI. In other words, those banks which do not fulfill the requisite conditions as explained above are called non-scheduled banks.

Question 19.
What is paper money?
Answer:
Paper money is a country’s official paper currency that is circulated for transaction related purposes of goods and services. The printing of paper money is typically regulated by a country’s Central bank / treasury in order to keep the flow of funds in line with monetary policy.

Paper money is that money which is issued by the order of the government. Paper money is in the form of paper notes or currency notes. In India, the Reserve Bank of India enjoys the sole monopoly of issuing currency notes (Paper money).

Question 20.
What do you mean by money supply?
Answer:
Supply of money is a stock concept. It refers to total stock of money (of all types) held by the people of a country at a point of time.

Question 21.
What do you mean by Fiat money or Legal tender money?
Answer:
Fiat money which serves as money on the basis of fiat (order) of the government It is issued by authority of the government. It includes notes, coins.

Question 22.
What do you mean by C.R.R.?
Answer:
Commercial banks are required under law to keep a certain percentage of their total deposit in the Central bank in the form of cash reserves. This is called Cash Reserve Ratio (C.R.R.).

Question 23.
What do you mean by credit money?
Answer:
It refers to that money of which money value is more than commodity value.

Question 24.
Why is paper money called the Legal tender money?
Answer:
Paper money or Paper notes are called as legal tender money because nobody can refuse its acceptance as medium of exchange. It is legal tender. It means people have to accept it legally for different payments.

Question 25.
What do you mean by high powered money?
Answer:
It consist of currency (Notes and Coins in circulation with public and valt cash of commercial banks) and deposits held by the government of India and commercial banks with R.B.I.

Micro Economics Money and Banking Short Answer Type Questions

Question 1.
“Money is a good servant but a bad master”. Explain.
Answer:
Money is becoming essential part of life, it is controlling our activities in such a way that we are completely under it. It should be means but is has become end of life. Till money acts as servant it is useful for human being but as money goes out of control it becomes harmful for the entire economy. This situation arises when supply of money exceeds its demand. Money becomes master instead of servant. It has become the only means of satisfying our wants. We cannot think of our life without money that is why, it is said that “money is good servant but a bad master.”

Question 2.
Define Central bank.
Answer:
Central bank of any country is the highest financial institution of that country. All other banks of the country work under its guidance. According to bank of international settlements. “The central bank in any country is that which has been entrusted the duty of regulating the volume of currency and credit in the country”.

Question 3.
What do you mean by clearing house of Central bank?
or
How Central bank does the work of clearing house?
Answer:
Clearing house:
Representatives of different banks in a city meet at the clearing house of the Central bank. Payments from one bank to other bank are settled through a simple book of adjustments without involving transfers of funds. Daily differences in the clearing between the banks are adjusted by means of debit and credit entries in their respective account with the Central bank.

Question 4.
Write two advantages and defects of Commercial banks.
Answer:
Following are the two advantages of commercial banks:

1. Use of savings for production purposes:
Banks collects small and big savings of the country and use them for various purposes. Banks provide timely financial assistance to traders and industries. Banks help in the distribution of surplus capital in regions where it is not wanted to those regions where it can be used. Large scale transactions would be quite impossible without banks.

2. Facility of capital transfer:
Banks spend money from one place to another easily in less time. They send it through draft, cheques, etc. For the development economy money can be transferred from one place to another.

Defects of commercial banks:

1. Less banking facilities:
Compare to other countries there are less banking facilities available in India.
Some banks do not get proper guidance and instructions from reserve bank of India. It has unfavourable effect on banking system. So, Indian Commercial banking system has not succeeded in attracking the Indian public.

2. Inefficiency:
Most of the commercial banking systems are unable to provide efficient and quick services to their customers. Moreover in India the distribution of commercial banks are in imbalance way.

Question 5.
Differentiate between Reserve Bank and Commercial Bank.
Answer:
Differences between Reserve Bank and Commercial Bank:

Reserve Bank:

• It is the highest bank of India. It is a controlling bank.
• There is only one central bank (Reserve bank) in India.
• It enjoys the monopoly right for note issue.
• It has no dealing and direct relation with public.
• The ownership of this bank is in the hands of central government.

Commercial Bank:

• It is a part of banking system. It is con trolled by Reserve Bank of India.
• There are many commercial banks in India.
• A commercial bank does not have such right.
• It has direct relation and dealing with public.
• The ownership of this bank is in the hands of government or public.

Question 6.
Why is speculative demand for money inversely related to the rate of interest?
Answer:
Speculative demand for money is made for return in the form of interest. Hence, rate of interest determiners the speculative demand for money of the rate of interest higher the speculative demand for money will be low and vice-versa. The reason is that people will convert their money in bonds when the interest on bond is higher.

In that case, demand for speculative purpose will be low. On the other hand, the rate of interest is lower than the expected interest, people will not convert their money in bonds and keep money in hand for speculative purpose. In this way, a specular demand for money in inversely related to the of interest.

Question 7.
What role of RBI is known as lender of last resort?
Answer:
Lender of last resort:
The Central Bank aslo plays the role of last resort. Lender of last resort means that a Commercial Bank fails to meet his financial requirements from other sources, it can as a last resort, approaches to the Central Bank for loans and advances. The Central Bank consists such bank through discounting of approved securities and bills of exchange. As a last resort, Central Bank exercise control over the entire banking system of the conuntry.

Question 8.
What is liquidity trap?
Answer:
Liquidity Trap:
Everyone in the economy holds his wealth in money balance and if additional money is injected within the economy, it will be used to satiate people’s craving for money balances without increasing the demand for bond and without further lowering the rate of interest below the flow min. Such a situation is called liquidity trap.

Question 9.
What is barter system? What are its defects?
Answer:
Barter system:
Barter system means the exchange of goods and services directly for goods and services. In other words economic exchange without the medium of money is referred to as Barter system.

Drawbacks of barter system : Following are main drawbacks of the barter system:

1. Lack of double coincidence:
The lack of double coincidence of wants is the major drawback. It is very rare when the owner of same goods or service could find some are who wants his good or service and possessed that good service that the first person wanted. No exchange is possible if the double coincidence is not there.

2. Absence of common unit of measurement:
The second main drawback of barter is the absence of a common unit of measurement in which the value of goods and services can be measured. In the absence of a common unit proper accounting system is not possible.

3. Lacks of standard deferred payments:
Thirdly, the barter system lacks any satisfactory unit to engage in contracts involving future payments. In a barter economy future payments would have to be stated in specific goods of services which may involve disagreement over the quality of goods or even on the commodity used for repayment.

4. Lack of storing wealth:
Fourthly, the barter system does not provide for any method of storing purchasing power.

Question 10.
Suppose a bond promises Rs 500 at the end of two years with no intermediate return. If the rate of interest is 5% per annum, what is the price of the bond?
Answer:
= 500 x \(\frac { 1 }{ \left\lceil 1+\frac { 5 }{ 100 } \right\rceil \left\lceil 1+\frac { 5 }{ 100 } \right\rceil } \)
= 500 x \(\frac { 20 }{ 21 }\) x \(\frac { 20 }{ 21 }\)
=\(\frac { 10,000 x 20 }{ 21 x 21 }\)
=\(\frac { 2,00,000}{ 441 }\)
= Rs 453.51.

Micro Economics Money and Banking Long Answer Type Questions

Question 1
What do you mean by stock of money and flow of money?
or
Write the differences between the stock of money and flow of money.
Answer:
Differences between Stock of money and Flow of money:

Stock:

• Stock is related to a point of time.
• A stock is a quantity measurable at a particular point of time such as 10:00 am etc.
• It has no time dimension.
• National capital is stock.
• It is a static concept.

Flow:

• Flow is related to a period of time.
• A flow is a quantity which is measured over a specific period of time such as an hour, a day, a week etc.
• It has time dimension.
• National income is a flow.
• It is a dynamic concept.

Question 2.
Explain five functions of RBI in detail.
Answer:
Followings are the functions of Reserve Bank of India:

1. Issue of currency notes:
The Reserve Bank of India has been given the monopoly of issuing currency notes. At present, it is authorised to issue currency notes in the denomination of Rs. 5,10,20,50,100,500 and 2,000. One rupee note and coins are issued by government of India. The notes issued by R.B.I. are legal currency.

2. Credit control:
The main functions of R.B.I. is to control and regulate volume of credit and currency in India. It controls the credit granted by commercial banks. It controls the credit inflationary impact of the ever increasing development financing in five year plans. R.B.I. controls the credit by increasing or decreasing bank rates buying and selling government securities in the open market, by increasing the cash reserves, by issuing specific directions to banks etc.

3. Banker agent and advisor to government:
The reserve bank of India acts as banker, agent and advisor to the central and state governments. As a banker to government R.B.I. performs many functions. It manages the public debt, makes all payments, receives all the revenues in government account managing foreign exchange, etc. R.B.I. has appointed experts of all fields of specialization who gives proper advice to central and state governments in various matters from time-to-time.

4. Banker’s bank:
R.B.I. is banker to banks in India. It accepts deposits of commercial banks and lends of them from time – to – time. It extends short term loans and advances against eligible securities and promissory notes. It makes purchases and sales of foreign currencies also. The commercial and other banks have to deposit with R.B.I. a certain percentage of their total liabilities to enjoy various facilities from it.

5. Foreign exchange management:
R.B.I. controls all receipts and payments in foreign exchange under F.E.M. A. All payments to be made in foreign exchange exceeding the limit prescribed under the act must seek RBI’s permission.

Question 3.
Explain the methods to control the credit taken by the reserve bank.
Or
Explain how the Reserve Bank of India controls credit.
Answer:
Followings are the methods to control the credit:

1. Bank rate:
The rate at which Reserve Bank of India gives loan to other banks on the basis of securities is called bank rate. R.B.I. increases or decrease the bank rate for reducing or expanding the credit from time-to-time.

2. Activities of open market:
R.B.I. buys and sells the government securities in the open market for expanding and reducing the volume of credit. This is called open market operation. By this activities R.B.I. increases or decreases the amount of money.

3. Changing the ratio of bank’s reserves:
R.B. I. controls the credit by increasing the percentage of cash reserves which is kept by scheduled commercial banks compulsorily with R.B.I.

4. Liquidity ratio-system:
Reserve Bank of India increases or decreases the liquidity ratio for reducing or expanding credit.

5. Selective credit control policy:
Reserve bank adopts the selective credit control policy in respect of certain commodities which have been sensitive.

6. By issuing specific directions to banks:
R.B.I. issues specific directions to banks in general regarding lending rates.

7. Increase or Decrease refinancing facility:
R.B.I. increases or decreases refinancing facility to commercial banks for credit control. Rigid attitude for refinancing reduces credit and liberal attitude for refinancing expands credit.

Question 4.
Suggest some measures to remove the defects of commercial bank.
Answer:
Following steps should be taken to improve the conditions of Indian commercial banks:

1. For the balance development of commercial banks their new branches should be open in villages and backward areas.
2. To solve the problem of shortage of capital the money deposits schemes of the banks should be made attractive.
3. Knowledge and education regarding the banks should be given clearly.
4. For increasing the banking systems the trained and efficient officers should be appointed.
5. Reserve bank should make such rules which will control the corruption in the banking system.
6. Banks should follow the policy of cooperation with other banks instead of following the policy of competition with each other.

Question 5.
Write any five functions of Commercial Banks.
Answer:
Following are the functions of Commercial banks:

I. Primary Functions

1. Acceptance of deposits:
All commercial banks accept money on deposits. By taking money on deposits a bank provides safe keeping for people’s money. For deposition of money a bank provides facility of five types of customer accounts:

• Current account
• Fixed deposit account
• Savings bank account
• Home savings account and
• Recurring deposit account.

2. Lending and Investment:
The second primary function of a bank is to lend money to borrowers bank keeps a part of the total deposits with itself as cash reserves and lends the balance. Bank charges interest on such lending of money, which is also the main source of profit to most of the banks. Money lending may be in the form of overdrafts, cash credits or loans.

II. Subsidiary Functions

In order to provide several kinds of facilities to their customers commercial banks perform many functions. They are as follows:

1. Issuing of credit instruments:
All commercial banks issue various instruments of credit such as bill of exchange, hundi, draft and cheque etc.

2. Arrangement of foreign exchange:
For foreign trade banks convert the currency of one country into the currency of another country. This work is done by exchange Banks.

3. Keeping valuables safely:
Banks provide the facility of lockers for the customers to keep their valuable things gold, diamonds, Jewellary etc. are kept in these lockers.

4. Discounting of bills of exchange and hundies:
Banks discount the bill and hundies of their customers before the date of maturity of such instruments.

5. Collecting payments:
Commercial banks collect payments of their customers bills, cheques and hundies etc. Banks charge commissions for these services.

6. Making payments:
Banks make payments on behalf of their customers. The make payments of installments of loans, interest, donations, insurance etc.They charge nominal commission for their services.

7. Selling and Purchasing of securities:
Commercial banks sell and purchase shares, debentures Govt.bonds etc.

Question 6.
What are the main functions of money? How does money remove the defect of barter system?
Answer:
Followings are the functions of money:

Primary Functions:

1. Medium of exchange:
This is the most important function of money. It acts as a medium of exchange. All the exchanges of goods and services are taken place in terms of money. By paying predetermined price, money can be exchanged with the desired goods and services.

2. Measure of value:
The second basic function of money is that it measures the value of goods and services. In other words, the prices of all goods and services are expressed in terms of money. In ancient times it was not possible of measure the value of clothes and wheat. With the introduction of money this difficulty of measurement disappeared and it became very convinient to measure the value of money. Money acts as a unit of account for all goods, wages salaries, interests etc. The national income, capital formation and other are measured in terms of money.

3. Store of exchange value:
Money can be stored very easily. Money is a liquid form of capital. It require less place to store. There is very less fluctuation in the value of money in comparison to goods. That is why mentry to save a part of their income for future. Money provides a base for store value.

Store of value means store of wealth.People can now keep their wealth in the form of money. Under barter system, storing of wealth was possible only in terms of commodities which had its defects like perishable nature of some goods, cost of storage etc. But storing of value in the form of money has solved all these difficulties. It was Keynes who first realised the store value function of money. He regarded money as link between the present and the future. Money allows us to store purchasing power which can be used at any time in future to purchase goods and services.

4. Transfer of value:
With the economic development the trade and commerce also increased rapidly. This causes the need of transfer of money from one place to another. Sometimes it crosses the national boundary too. Money is a liquid assets so it can be transferred from one place to another.

Through money, value can be easily and quickly transferred from one place to another because money is acceptable everywhere. For example a person can transfer money from Katakana to Delhi through bank draft, bill of exchange etc. Money thus, facilitates movement of wealth and capital. Under barter system it was difficult to transfer value in the from commodities.

Question 7.
Explain the functions of commercial banks. (Any two)
Answer:
Functions of a Commercial Bank:
Following are main functions of a commercial Bank:

1. Acceptance of deposits:
This is an important primary function of the Commercial Banks. The Commercial Banks accept deposits from individuals, business firms and other institutions. This is economically useful function in the sense that it helps in the mobilization of savings for production purposes. The commercial banks accept deposits in several forms according to the requirements of different sections of the society.

2. Advancing of Loans:
Extending loans is another important primary function of the commercial banks. It is also the main source of their income. Traditionally, bankers charged a service charge from the depositors, as they did not use the deposits for lending purposes. Gradually, they realized that.there is no point in keeping all the money which they received from depositors as revenue. All the depositors never approached bankers to withdraw their money at one point of time. In the beginning, their lending out of deposits were confined to short term loans to provide working funds for current business operations.

Now, banks have extended their lending activities to investment in long term bonds. Furthermore, today, commercial banks lend to consumers and government units besides financing trade and industry to meet the divergent needs of their customers. In the manner, they could find safe and lucrative outlets for their funds. The normal performance for banks is for secured loans, but they often give loans to business firms of high credit standing without security. .

Question 8.
What is money multiplier? How will you determine its value?
Answer:
Money multiplier:
Money multiplier may be defined as “the ratio of the stock of money to the stock of high powered money in an economy.” In equation:
Money multiplier = \({ M }{ H }\)
M = Stock of money
H = Stock of high powered money in an economy
The value of money multiplier is greater than 1. The value of money multiplier is determined by applying the following formula:
Money multiplier = \({ 1- cdr }{ cdr – rdr }\)
Method of finding out multiplier
Supply of money = Money + Deposit
M = Cu + DD
= (1 + Cdr) DD
Cdr = Cu / DD
H = Cu + R
= CdrDD + rdr DD
= (Cdr + rdr) DD
Money multiplier = M/H
\(\frac { (1+Cdr)DD }{ (Cdr+rdr)DD }\) = \(\frac { 1+Cdr }{ Cdr+rdr }\)
1 + Cdr > Cdr + rdr.

Question 9.
What are the methods adopted by Reserve Bank of India to regulate the credit?
Answer:
There are those instruments of monetary policy which after overall supply of money / credit in the economy. They are as follows:

1. Bank Rate:
The bank rate is the minimum rate at which the Central Bank of a country (as a lender of last season) is prepared to give credit to the Commercial Bank. The increase in bank rate increases the rate of interest and credit becomes dear. Accordingly, the demand for credit is reduced. On the other hand, decrease in the bank rate lowers the market rate of interest charged by the commercial banks from their borrowers. Credit becomes cheap. The Reserve Bank adopts dear money policy when the supply of credit needs to be reduced during periods of inflation. It adopts cheap money policy when credit needs to be expanded during deflation.

2. Open Market Operations:
Open Market Operations refer to the sale and purchase of securities in the open market by the Central Bank. By selling the securities (like, National Saving Certificates-NSCs), the Central Bank withdraws cash balances from within the economy. And, by buying the securities, the Central Bank contributes to cash balances in the economy.

3. Cash Reserve Ratio (CRR):
It refers to the minimum percentage of a bank’s total deposits required to be keep with the Central Banks. Commercial Banks have to keep with toe Central Bank a certain percentage of their deposits in the form of cash reserves as a matter of law. When the cash flow or credit is to be increased, minimum reserve ratio is reduced and when the cash flow or credit is to be reduced, minimum cash reserve ratio is increased.

4. Statutory Liquidity Ratio (SLR):
Every bank is required to maintain a fixed percentage of its assets in the form of cash or other liquid assets, called SLR. With a view to reducing the flow of credit in the market, the Central Bank increases this liquidity ratio. However, in case of expansion of credit, the liquidity ratio is reduced.

Qualitative Method of Credit Control
Or
Qualitative Instruments of Monetary Policy

1. Margin Requirement:
The margin requirement of loan refers to the difference between the current value of the security offered for loans and the value of loans granted. Suppose, a person mortgages an article worth Rs 100 with the bank and the bank gives him loan of Rs 80. The margin requirement in this case would be 20 percent. In case, the flow of credit is to be restricted for certain specific business activities in the economy, the margin requirement of loan is raised for those very activities. The margin requirement is lowered in case the expansion of credit is desired.

2, Rationing of Credit:
Rationing of credit refers to fixation of credit quotas for different business activities. Rationing of credit is introduced when the flow of credit is to be checked particularly for speculative activities in the economy. The Central Bank fixes credit quota for different business activities. The Commercial Banks cannot exceed the quota limits while granting loans.

MP Board Class 12th Economics Important Questions

MP Board Class 12th Maths Important Questions Chapter 5B अवकलन

अवकलन Important Questions

अवकलन लघु उत्तरीय प्रश्न

प्रश्न 1.
x के सापेक्ष sin [cos (x²) ] का अवकलन गुणांक ज्ञात कीजिए। (NCERT)
हल:
माना y = sin [cos (x²) ]
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) sin [cos (x²) ]
= \(\frac{d}{dx}\) sin t, [cos² = t, रखने पर]
= \(\frac{d}{dt}\) sin t \(\frac{dt}{dx}\) sin t
= cost \(\frac{d}{dx}\) cos x²
= cos (cos x²) \(\frac{d}{dx}\) cos u, [x² = u रखने पर]
= cos (cos x²) \(\frac{d}{du}\) cos u \(\frac{du}{dx}\)
= – cos(cos x²) sin u \(\frac{d}{dx}\) x²
= -2x cos(cos x²).sin x²

प्रश्न 2.
यदि y = sec [tan\(\sqrt{x}\) ] हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
हल:
दिया है:
y = sec [tan\(\sqrt{x}\) ]

प्रश्न 3.
यदि y = log [cos e^x] हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए। (NCERT)
हल:
दिया है:
y = log [cos e^x]
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) [log(cos e^x)]
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) log t, [cos e^x = t रखने पर]
= \(\frac{d}{dt}\) log t \(\frac{dt}{dx}\)
= \(\frac{1}{t}\). \(\frac{d}{dx}\) cos e^x

प्रश्न 4.
यदि y = cos [log x + e^x] हो, तो 4 का मान ज्ञात कीजिए। (NCERT)
हल:
दिया है:
y = cos [log x + e^x]

प्रश्न 5.
यदि y = cos^-1(e^x) हो, तो \(\frac{dy}{dx}\) ज्ञात कीजिए।
हल:
दिया है:
y = cos^-1(e^x)
∴ \(\frac{dy}{dx}\) = \(\frac{d}{dx}\) cos^-1(e^x)
e^x = t रखने पर,

प्रश्न 6.
यदि y + sin y = cos x हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए। (NCERT)
हल:
दिया गया फलन y + sin y = cos x
x के सापेक्ष अवकलन करने पर, का मान ज्ञात कीजिए।
\(\frac{d}{dx}\) (y + sin y) = \(\frac{d}{dx}\) cos x
⇒ \(\frac{dy}{dx}\) + cos y \(\frac{dy}{dx}\) = – sin x
⇒ \(\frac{dy}{dx}\) (1 + cos y) = – sin x
⇒ \(\frac{dy}{dx}\) = \(\frac { -sinx }{ 1+cosy } \)

प्रश्न 7.
यदि 2x + 3y = sinx हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए। (NCERT)
हल:
दिया है:
2x +3y = sin x
x के सापेक्ष अवकलन करने पर,
\(\frac{d}{dx}\) (2x + 3y) = \(\frac{d}{dx}\) sin x
2 \(\frac{d}{dx}\) x + 3 \(\frac{dy}{dx}\) = cos x
⇒ 2 + 3 \(\frac{dy}{dx}\) = cos x
⇒ 3 \(\frac{dy}{dx}\) = cos x – 2
∴ \(\frac{dy}{dx}\) = \(\frac { cosx-2 }{ 3 } \)

प्रश्न 8.
\(\frac{dy}{dx}\) का मान ज्ञात कीजिए यदि
x = a cos θ, y = a sin θ (NCERT)
हल:
दिया है:
x = a cos θ
y = a sin θ
अब समी. (1) का के सापेक्ष अवकलन करने पर,
\(\frac { dx }{ d\theta } \) = – a sin θ
पुनः समी. (2) का θ के सापेक्ष अवकलन करने पर,

प्रश्न 9.
\(\frac{dy}{dx}\) का मान ज्ञात कीजिए यदि (NCERT)
x = at², y = 2 at
हल:
दिया है:
x = at²
∴ \(\frac{dx}{dt}\) = 2 at
पुनः y = 2 at
\(\frac{dy}{dt}\) = 2a

प्रश्न 10.
यदि y = x² + 3x + 2 हो, तो \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) का मान ज्ञात कीजिए। (NCERT)
हल:
दिया है –
y = x² + 3x + 2

x के सापेक्ष पुनः अवकलन करने पर,

प्रश्न 11.
यदि y = x³ + tan x हो, तो का मान ज्ञात कीजिए। (NCERT)
हल:
दिया है:
y = x² + tan x
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) [x³ + tan x]
= \(\frac{d}{dx}\) x³ + \(\frac{d}{dx}\) tan x
⇒ \(\frac{dy}{dx}\) = 3x² + sec² x
x के सापेक्ष पुनः अवकलन करने पर,

अवकलन दीर्घ उत्तरीय प्रश्न – I

प्रश्न 1.
y = tan^-1 \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) का x के सापेक्ष अवकलन कीजिए।
हल:
दिया है:
y = tan^-1 \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \)
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) tan^-1 \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \)

पुनः 1 + x² = u रखने पर,

प्रश्न 2.
\(\frac{dy}{dx}\) का मान ज्ञात कीजिए यदि
x = a (t + sint), y = a (1 – cost)
हल:
दिया है:
x = a (t + sint)
∴ \(\frac{dx}{dt}\) = a (1 + cos t)
पुनः y = a (1 – cos t)
∴ \(\frac{dy}{dt}\) = a (0 + sint) = a sint

प्रश्न 3.
यदि x = a (2θ – sin 2θ) तथा y = a (1 – cos 2θ), तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिये जबकि θ = \(\frac { \pi }{ 3 } \) (CBSE 2018) .
हल:
दिया है:
समी. (2) को θ के सापेक्ष अवकलन करने पर,
x = a (2θ – sin 2θ) ………………………. (1)
y = a (1 – cos 2θ) ………………………… (2)
\(\frac { dx }{ d\theta } \) = a (2.1 – cos 2θ.2)
= 2a (1- cos 2θ)
= 2a.2 sin²θ
= 4a sin²θ …………………….. (3) |
समी. (2) को θ के सापेक्ष अवकलन करने पर,
\(\frac { dy }{ d\theta } \) = a(0 + sin 2θ.2)
= 2a sin 2θ
= 2a. 2 sinθ cosθ
= 4a sinθcosθ ……………………….. (4)
समी. (4) ÷ (3) से,

जब θ = \(\frac { \pi }{ 3 } \), तब
\(\frac{dy}{dx}\) = cot \(\frac { \pi }{ 3 } \) = \(\frac { 1 }{ \sqrt { 3 } } \)

प्रश्न 4.
यदि y = a sin mx + b cos mx हो, तो सिद्ध कीजिए कि
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + m² y = 0
हल:
दिया है:
y = a sin mx + b cos mx …………………….. (1)
समी. (1) के दोनों पक्षों में x के सापेक्ष अवकलन करने पर,
\(\frac{dy}{dx}\) = am cos mx – bm sin mx ……………… (2)
पुनः (2) का x के सापेक्ष अवकलन करने पर,

प्रश्न 5.
(A) यदि y = esin^{-1 x} हो, तो सिद्ध कीजिए कि (1 – x²), y₂ – xy₁ – m² y = 0
हल:
दिया है:

पुनः अवकलन करने पर,

(B)
यदि y = e^mtan-1x हो, तो सिद्ध कीजिए कि (1 + x²)y₂ + (2x – m) y₁ = 0
हल:
प्रश्न क्र. 5 (A) की भाँति हल करें।

(C)
यदि y = e^mcos-1x हो, तो सिद्ध कीजिए कि (1 – x²)y₂ + (2x – m)y₁ = 0
हल:
प्रश्न क्र. 5 (A) की भाँति हल करें।

प्रश्न 6.
sin^-1 \(\frac { 2x }{ 1+x^{ 2 } } \) का x के सापेक्ष अवकलन कीजिए।
हल:

प्रश्न 7.
यदि y = cot^-1 \(\sqrt { \frac { 1+x }{ 1-x } } \) हो, तो \(\frac{dy}{dx}\) ज्ञात कीजिए।
हल:
दिया है,
y = cot^-1 \(\sqrt { \frac { 1+x }{ 1-x } } \)
माना x = cos θ

समी. (1) में इसका मान रखने पर,

अब x के सापेक्ष दोनों पक्षों का अवकलन करने पर,
\(\frac{dy}{dx}\) = – \(\frac { 1 }{ 2\sqrt { 1-x^{ 2 } } } \)

प्रश्न 8.
यदि y = tan^-1 \(\sqrt { \frac { 1+x }{ 1-x } } \) हो, तो ज्ञात कीजिए।
हल: प्रश्न क्र. 7 की भाँति हल करें।
[उत्तर – \(\frac{dy}{dx}\) = \(\frac { 1 }{ 2\sqrt { 1-x^{ 2 } } } \)

प्रश्न 9.
यदि y = cot^-1 \(\frac { cosx+sinx }{ cosx-sinx } \) हो, तो \(\frac{dy}{dx}\) ज्ञात कीजिए।
हल:
दिया है:

प्रश्न 10.
y = tan^-1 \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) का x के सापेक्ष अवकलन कीजिए।
हल:
y = tan^-1 \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \)
समी. (1) x = tan θ में रखने पर, तब θ = tan^-1 x

प्रश्न 11.
यदि y = cot^-1 [ \(\frac { \sqrt { 1+x^{ 2 }+1 } }{ x } \) ] हो, तो \(\frac{dy}{dx}\) ज्ञात कीजिए।
हल:
y = cot^-1 [ \(\frac { \sqrt { 1+x^{ 2 }+1 } }{ x } \) ]
x = tan θ रखने पर,

प्रश्न 12.
यदि y = x^sinx हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
हल:
दिया है:
y = x^sinx
समी. (1) के दोनों पक्षों का log लेने पर, …………………………….. (1)
दोनों पक्षों का अवकलन करने पर,

प्रश्न 13.
यदि y = \(\sqrt { \frac { 1-x }{ 1+x } } \) हो, तो सिद्ध कीजिए \(\frac{dy}{dx}\) = \(\frac { y }{ x^{ 2 }-1 } \)
हल:
दिया है:

प्रश्न 14.
यदि y = (sin x)^{sinxsinx…………… ∞} हो, तो 4 का मान ज्ञात कीजिए।
हल:
दिया है:
y = (sin x)^{sinxsinx…………… ∞}
⇒ y = (sin x)^y
⇒ log y = y log sinx

x के सापेक्ष अवकलन करने पर,

प्रश्न 15.
(A) यदि y = \(\sqrt { sinx+\sqrt { sinx+……+\infty } } \) हो, तो सिद्ध कीजिए कि
\(\frac{dy}{dx}\) = \(\frac { cosx }{ 2y-1 } \)
हल:
दिया है:

(B)
यदि y = \(\sqrt { cotx\sqrt { cotx+\sqrt { cotx+…….+\infty } } } \) हो, तो सिद्ध कीजिए कि
\(\frac{dy}{dx}\) = \(\frac { cosec^{ 2 }x }{ 1-2y } \)
हल:
प्रश्न क्र. 15 (A) की भाँति हल करें।

(C) यदि y = \(\sqrt { tanx+\sqrt { tanx+\sqrt { tanx+…..\infty } } } \) हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
हल:
प्रश्न क्र. 15 (A) की भाँति हल करें।
[उत्तर: \(\frac { sec^{ 2 }x }{ (2y-1) } \) ]

प्रश्न 16.
यदि y = e^{x+ex+ex+e…………… ∞} हो, तो सिद्ध कीजिए कि
\(\frac{dy}{dx}\) = \(\frac{y}{1-y}\)
हल:
दिया है:
y = e^{x+ex+ex+e…………… ∞}
⇒ y = e^x+y
दोनों पक्षों में log लेने पर,
log y = loge^x+y
⇒ log y = x + y

प्रश्न 17.
फलन \(\frac { 1 }{ (x+a)(x+b)(x+c) } \) का x के सापेक्ष अवकलन कीजिये।
हल:
माना y = \(\frac { 1 }{ (x+a)(x+b)(x+c) } \)
दोनों पक्षों में log लेने पर
log y = log1 – log(x + a) – log(x + b) – log(x + c)
x के सापेक्ष अवकलन करने पर

प्रश्न 18.
log ( \(\sqrt{x}\) + \(\frac { 1 }{ \sqrt { x } } \) का x के सापेक्ष अवकल गुणांक ज्ञात कीजिये।
हल:

x के सापेक्ष अवकलन करने पर,

प्रश्न 19.
यदि y = tan^-1 ( \(\frac { sinx }{ 1+cosx } \) ) हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिये।
हल:
दिया है:
y = tan^-1 ( \(\frac { sinx }{ 1+cosx } \) )

दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
अत:
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) ( \(\frac{x}{2}\) ) = \(\frac{1}{2}\)

प्रश्न 20.
फलन f (x) = x² के लिए अंतराल [-1, 1] में रोले प्रमेय का सत्यापन कीजिए। (NCERT)
हल:
यहाँ f (x) = x², a = -1, b = 1

1. चूँकि f (x) = x² एक बहुपद फलन है इसलिए f (x) संवृत अंतराल [-1, 1] में संतत है।
2. f’ (x) = 2x विवृत्त अंतराल (-1, 1) में परिभाषित है इसलिए f (x) विवृत अंतराल (-1, 1) में अवकलनीय है।
3. f (-1) = (-1)² = 1, f(1) = (1)² = 1

∴ f (-1) = f (1)
अतः फलन f (x) रोले प्रमेय के सभी प्रतिबंधों को संतुष्ट करता है।
∴ f'(c) = 0
⇒ 2c = 0 [∵ f'(x) = 2x]
⇒ (c) = 0 ∈ (-1, 1)
अतः रोले प्रमेय सत्यापित होता है।

प्रश्न 21.
फलन f (x) = x² + 2x – 8 के लिए अंतराल [-4, 2) में रोले प्रमेय का सत्यापन कीजिए। (NCERT)
हल:
यहाँ f (x) = x² + 2x – 8, a = – 4, b = 2.

1. f (x) = x² + 2x – 8 एक बहुपद फलन है इसलिए f (x) संवृत अंतराल [-4, 2] में संतत है।
2. f’ (x) = 2x + 2 विवृत्त अंतराल (-4, 2) में परिभाषित है इसलिए f (x) विवृत अंतराल (-4, 2) में अवकलनीय है।
3. f (-4) = (-4)² + 2 (-4) – 8

= 16 – 8 – 8 = 0
f (2) = (2)² + 2 × 2 – 8 = 0
f (- 4) = f (2)
अतः f(x) रोले प्रमेय की तीनों शर्तों को संतुष्ट करता है।
∴ f'(c) = 0
⇒ 2c + 2 = 0
⇒ c = -1 ∈ (-4, 2)
अतः रोले प्रमेय सत्यापित होता है।

प्रश्न 22.
फलन f (x) = 2x² + x² – 4x – 2 के लिए रोले प्रमेय का सत्यापन कीजिए।
हल:
चूँकि बहुपद फलन सभी वास्तविक संख्याओं के लिए संतत और अवकलनीय होता है इसलिए f (x) प्रत्येक अंतराल में संतत और अवकलनीय है।
अब f (x) = 0
⇒ 2x³ + x² – 4x – 2 = 0
⇒ x²(2x + 1) – 2 (2x + 1) = 0
⇒ (x² – 2) (2x + 1) = 0
⇒ x² = 2, 2x + 1 = 0
⇒ x = ± \(\sqrt{2}\), x = – \(\frac{1}{2}\)
⇒ x = – \(\sqrt{2}\), \(\sqrt{2}\), \(\frac{-1}{2}\)
अब अंतराल [-\(\sqrt{2}\), \(\sqrt{2}\) ] लेते हैं।

1. फलन f (x) संवृत अंतराल [-\(\sqrt{2}\), \(\sqrt{2}\) ] में संतत है।
2. f'(x) = 6x² + 2x – 4 का अस्तित्व विवृत अंतराल [-\(\sqrt{2}\), \(\sqrt{2}\) ] में है इसलिए f (x) विवृत अंतराल [-\(\sqrt{2}\), \(\sqrt{2}\) ] में अवकलनीय है।
3. f ( –\(\sqrt{2}\) ) = f ( \(\sqrt{2}\) ) = 0

अतः f (x) रोले प्रमेय के सभी प्रतिबंधों को संतुष्ट करता है।
∴ f'(c) = 0
⇒ 6c² + 2c – 4 = 0, [∵f'(x) = 6x² + 2x – 4]


अतः रोले प्रमेय सत्यापित होता है।

प्रश्न 23.
अंतराल [1,3] में फलन f (x) = x + \(\frac{1}{x}\) के लिए मध्यमान प्रमेय का सत्यापन कीजिए।
उत्तर
हल:
f (x) = x + \(\frac{1}{x}\) = \(\frac { x^{ 2 }+1 }{ x } \), x ∈ [1, 3]

1. चूँकि f(x), x ≠ 0 सहित एक परिमेय फलन है और प्रत्येक परिमेय फलन संतत होता है, इसलिए f (x), [1, 3] में संतत फलन है।
2. f'(x) = 1 – \(\frac { 1 }{ x^{ 2 } } \) विवृत अंतराल (1, 3) में परिभाषित है और प्रतेक परिमेय फलन संतत होता है इसलिए f (x), [1, 3] में संतत फलन है।
3. f (1) = 2, f (3) = \(\frac{10}{3}\)

अतः मध्यमान प्रमेय के दोनों प्रतिबंध संतुष्ट होते हैं।

अत: लैग्रांज का मध्यमान प्रमेय सत्यापित होता है।

प्रश्न 24.
फलन f (x) = log x का अंतराल [1, e] में मध्यमान प्रमेय का सत्यापन कीजिए।
हल:
f(x) = log x, x ∈ [1, e], x > 0.
a = 1, b = e

1. चूँकि f (x) = log x, x > 0 एक संतत फलन होता है इसलिए f (x), [1, e] में संतत फलन है।
2. f'(x) = \(\frac{1}{x}\), में परिभाषित है इसलिए f (x), (1, e) में अवकलनीय है।
3. f (1) = log 1 = 0, f (e) = log e = 1

अतः मध्यमान प्रमेय की दोनों शर्त संतुष्ट होती हैं।
∴ \(\frac { f(e)-f(1) }{ e-1 } \) = f'(c)
⇒ \(\frac { 1-0 }{ e-1 } \) = \(\frac{1}{e}\)
⇒ c = e – 1 ∈ (1, e)
अत: मध्यमान प्रमेय सत्यापित होता है।

प्रश्न 25.
मध्यमान प्रमेय के प्रयोग से अंतराल [2, 3] में परिभाषित वक्र y = \(\sqrt { x-2 } \) पर एक बिन्दु ज्ञात कीजिए जबकि स्पर्श रेखा वक्र के बिन्दुओं को मिलाने वाली जीवा के समांतर है।
हल:
f (x) = \(\sqrt { x-2 } \), a = 2, b = 3

1. चूँकि f(x) = \(\sqrt { x-2 } \), x ∈ [2,3] के लिए परिभाषित है इसलिए f (x), [2, 3] में संतत फलन है।
2. f'(x) = \(\frac { 1 }{ 2\sqrt { x-2 } } \) विवृत अंतराल (2, 3) में परिभाषित है इसलिए f (x), (2, 3) में अवकलनीय है।
3. f (2) = 0, f (3) = 1

अतः मध्यमान प्रमेय के दोनों प्रतिबंध संतुष्ट होते हैं।
∴ \(\frac { f(3)-f(2) }{ 3-2 } \) = f’ (c)

अभीष्ट निर्देशांक ( \(\frac{9}{4}\), \(\frac{1}{2}\) ) हैं।

अवकलन दीर्घ उत्तरीय प्रश्न – II

प्रश्न 1.
यदि y = sin^-1 \(\frac { 2^{ x+1 } }{ 1+4^{ x } } \) हो, तो \(\frac{dy}{dx}\) ज्ञात कीजिए। (NCERT)
हल:
दिया है:
y = sin^-1 \(\frac { 2^{ x+1 } }{ 1+4^{ x } } \)

प्रश्न 2.
यदि y = sin^-1 x हो, तो सिद्ध कीजिए कि (1 – x²) \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) – x \(\frac{dy}{dx}\) = 0? (NCERT)
हल:
दिया है:
y = sin^-1 x
\(\frac{dy}{dx}\) = \(\frac{d}{dx}\) (sin^-1 x)
\(\frac{dy}{dx}\) = \(\frac { 1 }{ \sqrt { 1-x^{ 2 } } } \)
\(\frac{d}{dx}\) ( \(\frac{dy}{dx}\) ) = \(\frac{d}{dx}\) (1 – x²)^-1/2
1 – x² = t रखने पर,

प्रश्न 3.
यदि y = tan x + sec x हो, तो सिद्ध कीजिए कि \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = \(\frac { cosx }{ (1-sinx)^{ 2 } } \)
हल:
दिया है:
y = tan x + sec x
\(\frac{dy}{dx}\) = sec² x + sec x tan x
⇒ \(\frac{dy}{dx}\) = sec x (sec x + tan x)
⇒ \(\frac{dy}{dx}\) = \(\frac{1}{cosx}\) = [ \(\frac { 1 }{ cosx } +\frac { sinx }{ cosx } \) ]

प्रश्न 4.
यदि y = sin (sin x) हो, तो सिद्ध कीजिए कि
y₂ + y₁ tan x + y cos² x = 0? (CBSE 2018)
हल:
y = sin (sin x) ………………………. (1)
x के सापेक्ष अवकलन करने पर,
y₁ = cos (sin x). cos x
पुनः x के सापेक्ष अवकलन करने पर,
y₂ = cos(sin x)\(\frac { d }{ dx } \) (cos x) + cos x \(\frac { d }{ dx } \) {cos(sin x)}
= cos(sin x)(- sin x) + (cos x) [- sin(sin x)]cos x
⇒ y₂ = – sin x cos(sin x) – cos² x sin(sin x)
⇒ y₂ = – sin x cos(sin x) – y cos² x, [ समी. (1) से ]
⇒ y₂ = [ – \(\frac { sinx }{ cosx } .cosx\) ] cos(sin x) – y cos² x
⇒ y₂ = (- tan x) y₁ – y cos² x, [ समी. (2) से ]
⇒ y₂ + y₁ tan x + y cos² x = 0. यही सिद्ध करना था।

प्रश्न 5.
\(\frac{dy}{dx}\) का मान ज्ञात कीजिए यदि (x² + y²)² = xy? (CBSE 2018)
हल:
(x² + y²)² = xy
दोनों पक्षों के प्रत्येक पद का x के सापेक्ष अवकलन करने पर,

प्रश्न 6.
यदि y = 500e^7x + 600e^-7x हो, तो सिद्ध कीजिए कि \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 49 y? (NCERT)
हल:
दिया है:
y = 500e^7x + 600e^-7x

प्रश्न 7.
यदि y = (tan^-1 x)² हो, तो दर्शाइए कि (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2 है। (NCERT)
हल:
दिया है:
y = (tan^-1 x)²
tan^-1 x = t रखने पर,

प्रश्न 8.
sec^-1 ( \(\frac { 1 }{ 2x^{ 2 }-1 } \) ) के सापेक्ष अवकल गुणांक ज्ञात कीजिए।
हल:
माना y₁ = sec^-1 = ( \(\frac { 1 }{ 2x^{ 2 }-1 } \) )
⇒ y₁ = cos^-1 (2x² – 1)

प्रश्न 9.
tan^-1 ( \(\frac { 2x }{ 1-x^{ 2 } } \) ) का sin^-1 ( \(\frac { 2x }{ 1+x^{ 2 } } \) ) के सापेक्ष अवकल गुणांक ज्ञात कीजिए।
हल:
माना y₁ = tan^-1 ( \(\frac { 2x }{ 1-x^{ 2 } } \) ) तथा y₂ = sin^-1 ( \(\frac { 2x }{ 1+x^{ 2 } } \) )
माना x = tan θ, अत: θ = tan^-1 x

प्रश्न 10.
tan^-1 ( \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ) का tan^-1 x के सापेक्ष अवकल गुणांक ज्ञात कीजिए।
हल:
माना y₁ = tan^-1 ( \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) )
x = tan θ रखने पर,

प्रश्न 11.
यदि x \(\sqrt { 1+y } \) + y \(\sqrt { 1+x } \) = 0 हो, तो सिद्ध कीजिए कि
\(\frac{dy}{dx}\) = – (1 + x)^-2
हल:
दिया है:
x \(\sqrt { 1+y } \) + y \(\sqrt { 1+x } \) = 0
⇒ x \(\sqrt { 1+y } \) = – y \(\sqrt { 1+x } \)
दोनों पक्षों का वर्ग करने पर,
x² (1 + y) = y² (1 + x)
⇒ x² + x²y = xy² + y²
⇒ x² – y² + x²y – xy² = 0
⇒ (x – y)(x + y) + xy (x – y) = 0
⇒ (x – y)(x + y + xy) = 0
अतः था तो  x – y = 0
⇒ x = y
किंतु  x ≠ y
∴ x + y+ xy = 0
⇒ y(1 + x) = -x
∴ y = –\(\frac { x }{ x+1 } \)
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

प्रश्न 12.
यदि y \(\sqrt { 1-x^{ 2 } } \) + x \(\sqrt { 1-y^{ 2 } } \) = 1 हो, तो सिद्ध कीजिए कि
\(\frac{dy}{dx}\) + \(\sqrt { \frac { 1-y^{ 2 } }{ 1-x^{ 2 } } } \) = 0?
हल:
दिया है:
y \(\sqrt { 1-x^{ 2 } } \) + x \(\sqrt { 1-y^{ 2 } } \) = 1
अब x = sin θ तथा y = sin ∅ लेने पर,

प्रश्न 13.
(A) यदि y = x^sin-1x + x^x
हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
हल:
दिया है:
y = x^sin-1x + x^x

(B)
यदि y = x^tan-1x + x^x हो, तो \(\frac{dy}{dx}\) का मान ज्ञात कीजिए।
हल:
प्रश्न क्र. 13 (A) की भाँति हल करें।

प्रश्न 14.
यदि sin y = x sin (a + y) हो, तो सिद्ध कीजिए कि,
\(\frac{dy}{dx}\) = \(\frac { sin^{ 2 }(a+y) }{ sina } \)
हल:
दिया है:
sin y = x sin (a + y)
x = \(\frac { siny }{ sin(a+y) } \)

प्रश्न 15.
यदि x^y = e^x-y हो, तो सिद्ध कीजिए कि \(\frac{dy}{dx}\) = \(\frac { logx }{ (1+logx)^{ 2 } } \)?
हल:
दिया है:
x^y = e^x-y
दोनों पक्षों का लघुगणक लेने पर,
y log x = (x – y) log_e e
⇒ y log x = (x – y).1 = x – y
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

प्रश्न 16.
यदि x^y = e^y-x हो, तो सिद्ध कीजिए कि
\(\frac{dy}{dx}\) = \(\frac { 2-log_{ e }x }{ (i-log_{ e }x)^{ 2 } } \)?
हल:
दिया है:
x^y = e^y-x
दोनों पक्षों में log लेने पर,
∴ log_e x^y = log_e(e^y-x)
⇒ y log_e x = (y – x) log_e e
⇒ y log_e x = y – x,
⇒ y(1 – log_e x) = x
⇒ y = \(\frac { x }{ 1-log_{ e }x } \)
अब दोनों पक्षों का x के सापेक्ष अवकलन करने पर,

MP Board Class 12 Maths Important Questions

MP Board Class 12th Physics Important Questions Chapter 12 Atoms

Atoms Important Questions

Atoms Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
According to Ruther ford’s atomic model the electrons inside the atom are:
(a) Stationary
(b) Centralized
(c) Revolving
(d) None of these.
Answer:
(c) Revolving

Question 2.
The size of atom is of order of:
(a) 10^-15 m
(b) 10^-15 cm
(c) 10^-10 m
(d) 10^-10 cm.
Answer:
(c) 10^-10 m

Question 3.
In Bohr’s model in an stable orbit the speed of electron vn is related with principle quantum number as:
(a) v_n = \(\frac {c}{n}\)
(b) v = \(\frac {n}{c}\)
(c) v = n.c
(d) \(\frac { c }{ { n }^{ 2 } }\)
Answer:
(a) v_n = \(\frac {c}{n}\)

Question 4.
In terms of Rydberg constant R the wave number of first spectral line of Balmar series is :
(a) R
(b) 3R
(c) \(\frac {5R}{18}\)
(d) \(\frac {8R}{9}\)
Answer:
(c) \(\frac {5R}{18}\)

Question 5.
Hydrogen atom is excited from ground state to n = 4. The number of spectral lines observed be:
(a) 3
(b) 6
(c) 5
(d) 2
Answer:
(b) 6

Question 6.
The value of Bohr’s radius is :
(a) 5.3 x 10^-11 nm
(b) 3.5 x 10^-10 m
(c) 53.3 x 10^-11 nm
(d) None of these.
Answer:
(a) 5.3 x 10^-11 nm

Question 7.
The value of Rydberg constant R is :
(a) 1.097 x 10^-11 nm
(b) 1.097 x 10^-8 m^-1
(c) 1.097 x 10⁸ m^-1
(d) 1.097 x 10⁷ m^-1.
Answer:
(d) 1.097 x 10⁷ m^-1.

Question 8.
According to Bohr the speed of electron of first orbit is :
(a) \(\frac {137}{c}\)
(b) 137 x c
(c) 137 + c
(d) \(\frac {c}{137}\)
Answer:
(d) \(\frac {c}{137}\)

Question 2.
Fill in the blanks :

1. In the Geiger Marsdan’s α – particle experiment the energy of a particle was ……………
2. According to Rutherford’s atomic model the total energy of energy was …………… (Positive/Negative)
3. The spectral lines of visible region of hydrogen spectrum was first seen by ……………
4. The value of Rydberg constant R is ……………
5. The electron revolve round the nucleus in such orbits in which the value of angular momentum is integral multiple of ……………
6. The maximum ionisation energy of hydrogen atom is ……………
7. In first excited state of hydrogen atom the total energy of electron is about …………… eV.

Answer:

1. 5.5 MeV
2. Negative
3. Balmar
4. 1.097 x 10⁷m^-1
5. \(\frac {h}{2π}\)
6. 13.6 MeV
7. -3.4 eV.

Question 3.
Match the column:

Answer:

1. (e)
2. (a)
3. (d)
4. (b)
5. (c)

Question 4.
Write the answer in one word/sentence:

1. What is α – pariide?
2. Write an important conclusion of Thomson’s atomic model.
3. Which important part of the atom was discovered by Rutherford?
4. In which condition the Balmar spectral line in the hydrogen spectrum is obtained?
5. What is the order of radius of nucleus of an atom?
6. What is order of size of atom?
7. Which of the following particle is unstable? α – particle, β – particles, protons, neutrons.

Answer:

1. The helium nucleus ₂He⁴ is called as α – particle
2. The mass of atom and positive charge is uniformly distributed in the sphere
3. Nucleus
4. When the electrons jump from higher energy level to second energy level
5. 10^-14 m
6. 10^-10 m
7. Neutrons.

Atoms Very Short Answer Type Questions

Question 1.
What will be the value of scattering angle of α – par tide if impact parameter is zero?
Answer:
In the relation b ∝ cot\(\frac {θ}{2}\) then
cot\(\frac {θ}{2}\) = 0
or \(\frac {θ}{2}\) = 90°
or θ = 180°

Question 2.
In Bohr’s model the radius of first orbit is π₀. What will be the radius of second orbit?
Answer:
From the formula π_n = π₀ n²
If n = 2
Then π₂ = π₀ x 2² = 4π₀

Question 3.
Define stable orbit in Bohr’s atoms model.
Answer:
The stable orbits are those orbits in which the angular momentum of the electron (revolving in those orbits) is integral multiple of \(\frac {h}{2π}\)

Question 4.
State Bohr’s quantization condition.
Answer:
Angular momentum mvr = \(\frac {nh}{2π}\), where n = 1,2,3 …..

Question 5.
Write quantization condition for the angular momentum of electron revolving in second orbit.
Answer:
In the formula mvr = \(\frac {nh}{2π}\)
If n = 2
Then mvr = \(\frac {2h}{2π}\) = \(\frac {h}{π}\)

Question 6.
What is importance of negative energy of electrons in an orbit?
Answer:
The nucleus and electrons makes a bounded system together.

Question 7.
What would happen if the electrons were at rest inside the atom?
Answer:
If the electrons were stationary inside the atom, then they tends to attract to – wards the nucleus.

Question 8.
Why the electrons always revolve round the nucleus?
Answer:
In order to provide a stable configuration to the atoms.

Question 9.
What is Bohr’s radius?
Answer:
The radius of first orbit of a hydrogen atom is called Bohr’s radius. Its value is 0.53 Å.

Atoms Short Answer Type Questions

Question 1.
Why only gold leaf is taken for the scattering of α – particles?
Answer:
For the scattering of α – particles, the leaf should be very thin, so that α – particles gets scattered after collision. Moreover the nucleus should be heavy, so that the α – particles are scattered by large angles. Gold has both the properties moreover this foil of gold can easily be made.

Question 2.
What is angular momentum of electron revolving in first Bohr’s orbit?
Answer:
The angular momentum in nth orbit is
mvr = \(\frac {nh}{2π}\)
Put n = 1
The angular momentum in first orbit = \(\frac {h}{2π}\).

Question 3.
The spectrum of hydrogen atom has many lines though a hydrogen atom contains only one electron why?
Answer:
Large number of spectral lines are present in hydrogen atom spectrum because in the light source of hydrogen a large number of atoms are there. There is different transitions in different atoms. That is why hydrogen spectrum contains number of lines.

Question 4.
What is meant by transition? How much time it takes?
Answer:
The jump of an atom from one energy state (level) to other is called transition. It takes about 10^-8 sec.

Question 5.
What may be the maximum and minimum number of emission transitions between energy levels n = 5 to n = 1.
Answer:
Maximum numbers = 10
(5 → 4, 5 → 3, 5 → 2, 5 → 1, 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, 2 →1)
Minimum numbers = 1 (5 → 1).

Question 6.
What will happen if electron remain stationary in an orbit?
Answer:
If the electron remains stationary in an orbit it would fall into the nucleus due to electrostatic attraction of nucleus. So the atom would be unstable.

Question 7.
With the help of fundamental constants e, me and h establish a quantity in the dimension of length.
Answer:
This quantity is Bohr’s radius

Question 8.
If the elements with principal quantum number n > 4 are not included in nature then what will be the number of such possible elements?
Answer:
For principal quantum n = 4 the atom will have only K, L, M and N shell. The maximum number of electrons in these shell according to 2n² will be, 2, 8 18 and 32 i.e., the maximum number of electrons in these shell will be 2 + 8 – 18 – 32 = 60. Thus the number of such elements be 60.

Question 9.
What is the significance of the negative energy of electron in the orbit? If it be positive then what does it mean?
Answer:
Total energy of electron E = K + U = \(\frac { { -e }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }r }\)
Here negative sign shows that the electron is bounded with nucleus if the negative energy be positive then it means that the electron will not revolve round the nucleus in a bounded orbit.

Question 10.
Bohr’s orbit is called stable orbit. Why?
Answer:
Bohr’s orbit is called as stable orbit because the electron revolving in this orbit neither emit the energy nor absorb.

Question 11.
The nuclear density is more than that of atom. Why?
Answer:
we know that density = \(\frac {Mass}{Volume}\).
The mass of nucleus and atoms are almost same but the radius of nucleus is about of order of 10^-15 m and that of atom is about of order of 10^-10 m. Hence the volume of nucleus is of order of 10^-45 m³ and that of atom is of order of 10^-30 m³. Obviously the volume of nucleus is about 10^-15 times that of atom, so the nuclear density is more than that of atom.

Question 12.
What are Bohr’s quantum condition and frequency conditions?
Answer:
1. The electrons can revolve in those orbits in which the angular momentum of electron is integral multiple of \(\frac {h}{2π}\). Where h is Planck’s constant i.e.
Angular momentum L = mvr = \(\frac {nh}{2π}\), where n = 1,2,3, …..
This is called as Bohr’s quantum condition.

2. When an electron jumps from energy level of higher energy E₂ to the energy level having energy E₁ then a photon is emitted whose energy is equal to the difference of energies of the two levels i.e., E₂ – E₁ = hu.
Where o is frequency of emitted photon. This is called Bohr’s frequency condition.

Question 13.
Calculate the radius of first orbit of hydrogen atom. Prove that the velocity of electron in first orbit is \(\frac {1}{137}\) times the velocity of light. (NCERT)
Solution:

Question 14.
According to the equation \(\frac {1}{λ}\) = R(\(\frac { 1 }{ { n }_{ 2 }^{ 2 } } -\frac { 1 }{ { n }_{ 1 }^{ 2 } }\)),
what will be the shortest wavelength which will either be absorbed or be emitted?
Solution:
For the shortest wavelength the R.H.S. of the equation should be greatest i.e.,
when n₁ = ∞ and n₂ = 1, then \(\frac {1}{λ}\) = R

Question 15.
Explain the Bohr’s quantization condition of angular momentum.
Answer:
Bohr’s quantization condition: The electron can revolve only in those orbits in which the angular momentum of electron is integral multiple of \(\frac {h}{2π}\) and h is Planck’s constant.

Atoms Long Answer Type Questions

Question 1.
Describe Rutherford’s experiment related to α – particles scattering. Also write conditions drawn from it
Answer:
The experimental arrangement of Rutherford’s α – particles scattering experiment is shown in fig. A radioactive substance radon as ‘Po’ is kept in a lead box from the hole ‘O’ the α – particles are emitted out with high speed. After passing through D₁ and D₂ these α – particles are incident on gold leaf in the form of a beam.

The thickness of gold , foil is about 10 cm. The scattered α – particles falls on screen S in which ZnS is coated α – particles produce flash of light on this screen which can be observed by a microscope M.

Conclusion:

1. Most of the α – particles pass through the foil straight undeflected. Thus it can be concluded that most of the part of atom is hollow.
2. α – particles are repelled so, there should be positive charge on nucleus.
3. The entire positive charge should be on the centre of atom (or nucleus).

Importance:
By the Rutherford’s experiment it is come to know that the positive charge of the atom is concentrated at the central core of atom, which is called as nucleus.

Question 2.
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV :

1. What is the kinetic energy of the electron in this state?
2. What is the potential energy of the electron in this state?
3. Which of the answers above would change, if the choice of the zero of potential energy is changed?

Answer:

Also U = – 2 K

1. Here £ = – 3.4 eV
Thus K = – E = -(-3.4) eV
= 3.4 eV.

2. Here U = – 2K
= – 2 (3.4) eV
= – 6.8 eV.

3. E = – 3.4 eV only when the zero of potential energy is at infinity. If the zero of the potential energy is changed, only potential energy (hence total energy) would change. However kinetic energy remain unaffected.

Question 3.
Prove that the radius of n^th Bohr’s orbit of an atom is proportional to n^th, where n is principal quantum number.
Answer:
The angular momentum of electron in n stable orbit is integral multiple of \(\frac {h}{2π}\)
mvr = \(\frac {nh}{2π}\)
or v = \(\frac {nh}{2πmr}\) … (1)
The equation essential for circular motion of electron is

In the above expression except n all the quantities are constant.
∴ r ∝ n².

Question 4.
In the Balmer series of hydrogen spectrum the two lines are at 4102 Å and 4861Å, if the difference of wave number of these lines is same as the wave number of either spectral lines, then which series will be represented by this line? What will be the wavelength of this line?
Answer:
Let n₁ = 2 and n₂ = x (for 4102 Å) and for line 4861 Å n₁ = 2 and n₂ = y.

Solving equation (1) and (2) we get,
x = 6 and y = 4
when the electron jumps from n₂ = 6 to n₁ = 4, then we get a third spectral line which belongs to Brackett’s series.
The wavelength of this spectral line be

Question 5.
State main postulates of Bohr’s atomic model.
Answer:
The main postulates of Bohr’s atomic model are as follows :
1. Entire mass and entire positive charge of atom is concentrated at the nucleus.

2. The electrostatic attraction force between nucleus and electron provides the require centripetal force to electrons to revolve round the nucleus.

3. The electron does not revolve round the nucleus in all possible orbits, but they revolve in some specific orbits. These orbits are called as stable orbits.

4. The stable orbits are those orbits in which the angular momentum of electron is integral multiple of \(\frac {h}{2π}\), where h is Plank constant (h = 6.63 x 10^-34J-sec.).
If m is mass of electron r is radius of orbit and v is its speed then angular momentum
mvr = \(\frac {nh}{2π}\), where n = 1, 2, 3,
n is called as principal quantum number.

5. The electrons revolving in stable orbits neither absorb nor emit the energy. The emission or absorption of energy takes place only when they jump from one orbit to other.

Atoms Numerical Questions

Question 1.
The ionization energy of hydrogen atom is 13.6 eV. An electron which jumps from ground state to n = 4 state. Calculate the wavelength of photon.
Solution:
Energy in ground state of hydrogen atom E₁ = -136 eV
Energy in n = 4 state E₄ = \(\frac { -13.6 }{ { 4 }^{ 2 } }\) = – 0.85 eV
∴ Energy of photon = E₄ – E₁ = – 0.85 + 13.6 = 12.75 eV
∵ hν = E and c = νλ

⇒ λ = 973 x 10^-10 m = 973 Å.

Question 2.
If the value of energy for an impact parameter is increased then, will the scattering angle increase or decrease? (NCERT)
Solution:

If the energy is increased then in the same ratio, the value of cot(θ/2)should be increased i.e., the angle of scattering will decrease.

Question 3.
What should be the value of angle of scattering for impact parameter b = 0? (NCERT)
Solution:
We know that

If b = 0, then cot(θ/2) = 0 = cot(90°)
⇒ \(\frac {θ}{2}\) = 90°
or θ = 180°

Question 4.
In a Rutherford’s experiment the number of α – particle scattered at 90° is 28 per minute. Find out the number of particles scattered at 120°.
Solution:

Given θ₁ = 90°, θ₂ = 120° and N₁ = 28 per minute
Putting values in the formula

Question 5.
Up to what least distant from the nucleus of silver (Z = 47) or α – particles of energy 8 MeV can reach?
Solution:
Formula : r₀ = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2Z{ e }^{ 2 } }{ E }\)
Given : \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } }\) = 9 x 10⁹ Nm²c^-2, Z = 47, e = 1.6 x 10^-19C
E = 8MeV = 8 x 10⁶ x 1.6 x 10^-19J
Putting values in the formula

MP Board Class 12th Physics Important Questions