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MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2

Class 9 Maths Chapter 2 MP Board Question 1.
Find the value of the polynomial 5x – 4x² + 3 at

1. x = 0
2. x = – 1
3. x = 2.

Solution:
Let f(x) = 5x – 4x² + 3

1. Value of f(x) at x = 0
= f(0) = 5(0) – 4(0)² + 3 = 3

2. Value of f(x) at x = – 1
= f(- 1) = 5(- 1) – 4(- 1)² + 3
= – 5 – 4 + 3 = – 6

3. Value of f(x) at x = 2
= f(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3 = – 3.

MP Board Class 9th Maths Chapter 2 Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:

1. p(y) = y² – y + 1
2. p(t) = 2 + t + 2t² – t³
3. p(x) = x³
4. p(x) = (x – 1)(x + 1)

Solution:
1. p(y) = y² – y + 1
P(0) = (0)2 – (0) + 1
= p(1) = (1)² – (1) + 1 = 1,
and p(2)² = (2)² – (2) + 1
= 4 – 2 + 1 = 3.

2. p(t) = 2 + t + 2t² – t³
p(0) = 2 + 0 + 2(0)² – (0)² = 2
p(1) = 2 + 1 + 2(1)² – (1)²
= 2 + 1 + 2 – 1 = 4
and p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4

3. p(x) = x³
p(0) = (0)² = 0,
p(1) = (1)² = 1,
and p(2) = (2)³ = 8.

4. p(x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1
P(1) = (1 – 1) (1 + 1) = (0) (2) = 0 and
p(2) = (2 – 1) (2 + 1) = (1) (3) = 3

MP Board Class 9 Maths Chapter 2 Question 3.
Verify whether the following are zero of the polynomial, written against them.

1. p(x) = 3x + 1, x = – \(\frac{1}{3}\)
2. p(x) = 5x – π, x = \(\frac{4}{5}\)
3. p(x) = x2 -1, x = 1, – 1
4. p(x) = (x + 1) (x – 2), x – 1, 2
5. p(x) = x², x = 0
6. p(x) = lx + m, x = – \(\frac{m}{l}\)
7. p(x) = 3x² – 1, x = – \(\frac{1}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)
8. p(x) = 2x + 1, x = \(\frac{1}{2}\)

Solution:
1. p(x) = 3x + 1
x = – \(\frac{1}{3}\)
p(- \(\frac{1}{3}\)) = 3(- \(\frac{1}{3}\)) + 1 = 0
= – 3 x \(\frac{1}{3}\) + 1 = – 1 + 1 = 0
As p(- \(\frac{1}{3}\)) = 0, – \(\frac{1}{3}\) is the zero of p(x).

2. p(x) = 5x – π
x = \(\frac{4}{5}\)
p(\(\frac{4}{5}\)) = 5 x \(\frac{4}{5}\) – π = 4 – π
As p(\(\frac{4}{5}\)) ≠ 0,
\(\frac{4}{5}\) is not the zero of p(x).

3 p(x) = x² – 1, x = 1, -1
P(1) = (1)² – 1 = 1 – 1 = 0
p(-1) = (-1)² – 1 = 1 – 1 = 0
As p(1) = 0 and p(-1) = 0
Both 1 and -1 are the zeros of p(x).

4. p(x) = (x + 1)(x – 2)
x = – 1, 2
p(- 1) = (-1 + 1) (- 1 – 2)
= (0) (-3) = 0
p( 2) = (2 + 1) (2 – 2)
= 3 x 0 = 0
As p(- 1) = 0 and p(2) = 0,
Both -1 and 2 are the zeros of p(x).

5. p(x) = x²
x = 0
P(0) = (0)² = 0
As p(0) = 0, 0 is a zero of p(x)

6. p(x) = lx + m, x = – \(\frac{m}{l}\)
p(- \(\frac{m}{l}\)) = l x – \(\frac{m}{l}\) + m
As p(- \(\frac{m}{l}\)) = 0, (-\(\frac{m}{l}\)) is a zero of p(x).

7.

8.

Class 9 Maths Chapter 2 Exercise 2.2 MP Board Question 4.
Find the zero of the polynomial in each of the following cases:

1. p(x) = x + 5
2. p(x) = x – 5
3. p(x) = 2x + 5
4. p(x) = 3x – 2
5. p(x) = 3x
6. p(x) = ax, a ≠ 0
7. p(x) = cx + d, c ≠ 0, c, d are real numbers.

Solution:
1. p(x) = x + 5
p(x) = 0
x + 5 = 0
x = – 5
∴ – 5 is a zero of the polynomial p(x).

2. P(x) = x – 5
p(x) = 0
x – 5 = 0 x = 5
∴ 5 is a zero of the polynomial p(x).

3. p(x) = 2x + 5
2x + 5 = 0
x = – \(\frac{5}{2}\)
∴ – \(\frac{5}{2}\) is the zero of p(x).

4. p(x) = 3x – 2
3x – 2 = 0
x = \(\frac{2}{3}\)
∴ \(\frac{2}{3}\) is the zero of p(x)

5. p(x) = 3x
3x = 0
x = \(\frac{0}{3}\)
∴ 0 is the zero of p(x).

6. p(x) = ax where a ≠ 0
ax = 0
x = \(\frac{0}{a}\)
∴ 0 is the zero of p(x).

7. p(x) = cx + d where c ≠ 0, c, d are real numbers
cx + d = 0
x = – \(\frac{d}{c}\)
∴ – \(\frac{d}{c}\) is the zero of p(x).

Remainder Theorem:
If a polynomial p{x) is divided by another polynomial (x – a), the remainder isp(a). Where p(x) is any polynomial of degree greater than or equal to one and also degree p(x) > degree p(a) ‘a’ be any real number.

Dividend = divisor x quotient + remainder

Proof:
Let p(x) be any polynomial with degree greater than or equal to 1, suppose that when p(x) is divided by x – a, the quotient is q(x) and remainder is r(x).
p(x) – (x – a) x q(x) + r(x)
p(x) = (x – a) x q(x) + r(x)…..(i)

Case I:
If r(x) = 0
Equation reduce to
p(x) = (x – a) q(x) ……(ii)
On putting x = a, in (ii),
We get p(a) = (a – a) q(a)
p(a) = 0 = Remainder

Case II:
If r(x) ≠ 0
Equation (i) reduced to
p(x) = (x – a) q(x) + r …..(iii)
On putting x = a in (iii), we get
p(a) =(a – a) x q(x) + r
p(a) = r = Remainder
∴ So the remainder is p(a) when p(x) is divided by (x – a).

MP Board Class 9th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 6 त्रिभुज और उसके गुण Ex 6.5

Class 7 Maths Exercise 6.5 Solutions In Hindi प्रश्न 1.
PQR एक त्रिभुज है जिसका P एक समकोण है। यदि PQ = 10 cm तथा PR = 24 cm तब QR ज्ञात कीजिए।
हल:
समकोण त्रिभुज PQR में पाइथागोरस प्रमेय का प्रयोग करने पर,
QR² = PQ² + PR²
OR² = (10)² + (24)²

या QR² = 100 + 576 = 676
या QR² = (26)²
∴ QR = 26 cm

Class 7 Maths Chapter 6 Exercise 6.5 In Hindi प्रश्न 2.
ABC एक त्रिभुज है जिसका ∠C एक समकोण है। यदि AB = 25 cm तथा AC = 7 cm तब BC ज्ञात कीजिए।
हल:
समकोण त्रिभुज ABC में पाइथागोरस प्रमेय का प्रयोग करने पर,

AC² + BC² = AB²
(7)² + x² = (25)²
या 49 + x² = 625
या x² = 625 – 49 = 576
या x² = 242 ⇒ x = 24
∴ BC = 24 cm

Class 7 Maths 6.5 In Hindi प्रश्न 3.
दीवार के सहारे उसके पैर कुछ दूरी पर टिका कर 15 m लम्बी एक सीढ़ी भूमि से 12 m ऊँचाई पर स्थित खिड़की तक पहुँच जाती है। दीवार से सीढ़ी के पैर की दूरी ज्ञात कीजिए।
हल:
माना कि सीढ़ी के पैर दीवार से am की दूरी पर हैं।

∴ पाइथागोरस प्रमेय से,
a² + 122 = 152
या a² + 144 = 225
या a² = 225 – 144 = 81
शस a² = (9)² ⇒ a = 9m
अत: सीढ़ी के पैर की दीवार से अभीष्ट दूरी = 9 m

Class 7 Maths Chapter 6.5 In Hindi प्रश्न 4.
निम्नलिखित में भुजाओं के कौन-से समूह एक समकोण त्रिभुज बना सकते हैं?
(i) 2.5 cm, 6-5 cm,6cm
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
हल:
(i) माना, त्रिभुज की भुजाएँ x = 2.5 cm, y = 6.5 cm, z = 6 cm हैं।
यहाँ, सबसे बड़ी भुजा की लम्बाई y = 6.5 cm
अब, x² + z² = (2.5)² + (6)² = 6.25 + 36
= 42.25
तथा, y² = (6.5)² = 42.25,
∵ x² + z² = y²
अतः दी गई भुजाएँ समकोण त्रिभुज बना सकती हैं और भुजा 6.5 cm के सामने का कोण समकोण होगा।

(ii) माना त्रिभुज की भुजाएँ x = 2 cm, y = 2 cm, z = 5 cm है।
यहाँ, सबसे बड़ी भुजा की लम्बाई z = 5 cm है।
अब, x² + y² = (2)² + (2)²
= 4 + 4 = 8
तथा z² = (5)² = 25
अत: अत: दी गई भुजाएँ समकोण त्रिभुज नहीं बना सकती हैं।

(iii) माना, त्रिभुज की भुजाएँ x = 1.5 cm, y = 2 cm और z = 2.5 cm हैं।
यहाँ, सबसे बड़ी भुजा की लम्बाई z = 2.5 cm है।
अब, x² + y² = (1.5)² + (2)² = 2.25 + 4.00
= 6.25
तथा z² = (2.5)² = 6.25,
∵ x² + y² = z²
अतः दी गई भुजाएँ समकोण त्रिभुज बना सकती हैं और भुजा 2.5cm के सामने का कोण समकोण होगा।

6.5 Class 7 In Hindi प्रश्न 5.
एक पेड़ भूमि से 5m की ऊँचाई पर टूट गया है और उसका ऊपरी सिरा भूमि को उसके आधार से 12 m की दूरी पर छूता है। पेड़ की पूरी ऊँचाई ज्ञात कीजिए।
हल:
माना कि पेड़ BD बिन्दु C से टूटा है,
इस प्रकार, CD = CA

अब पाइथागोरस प्रमेय से, ∆ABC में,
AB² + BC² = AC²
12² + 5² = AC²
या 144 + 25 = AC²
या AC² = 169 = 13²
या AC = 13 m
अब, पेड़ की ऊँचाई = BD = BC + CD
= BC + AC (∵AC = CD)
= 5 m + 13 m = 18 m
अतः पेड़ की अभीष्ट लम्बाई = 18 m

Class 7 Maths Chapter 6 Exercise 6.5 Solutions In Hindi प्रश्न 6.
त्रिभुज PQR में कोण Q = 25° तथा कोण R = 25° है। अग्रलिखित में कौन-सा कथन सत्य है ?

(i) PQ² + QR = RP
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²
हल:
∆PQR में,
∠P + ∠Q + ∠R = 180°
∠P + 25° + 65° = 180°
या ∠P + 90° = 180°
या ∠P = 180° – 90° = 90°
अत: ∆PQR समकोण त्रिभुज है, जिसका कोण P समकोण है।
अब, कर्ण = कोण P के सामने की भुजा = QR पाइथागोरस प्रमेय द्वारा
QR² = PQ² + PR²
अतः सम्बन्ध (ii) PQ² + RP² = QR² सत्य है।

Class 7 Maths Chapter 6 Exercise 6.5 MP Board प्रश्न 7.
एक आयत की लम्बाई 40 cm है तथा उसका एक विकर्ण 41 cm है। इसका परिमाप ज्ञात कीजिए।
हल:
आयत की लम्बाई = 40 cm, विकर्ण = 41 cm
माना कि आयत की चौड़ाई = x cm

समकोण त्रिभुज BAD से,
AB² + AD² = BD²
40² + AD² = 41²
या AD² = 41² – 40² = 1681 – 1600
= 81 = 9²
∴ चौड़ाई x = 9 cm
अब परिमाप = 2 (लम्बाई + चौड़ाई)
= 2 (40 + 9) = 2 × 49 = 98 cm
अत: आयत का परिमाप = 98 cm

Ex 6.5 Class 7 In Hindi प्रश्न 8.
एक समचतुर्भुज के विकर्ण 16 cm तथा 30 cm हैं। इसका परिमाप ज्ञात कीजिए।
हल:
माना कि ABCD एक समचतुर्भुज है, जिसमें AC = 30 cm और BD = 16 cm.

हम जानते हैं कि समचतुर्भुज के विकर्ण एक-दूसरे को समकोण पर काटते हैं। (यहाँ ये O पर काटते हैं।)
∠AOB = 90°

अब समकोण ∆AOB में,
AB² = 40 + BO²
= 15² + 8²
= 225 + 64 = 289
या AB² = 17² ⇒ AB = 17
∴ समचतुर्भुज का परिमाप = 4 × AB = 4 × 17 cm
= 68 cm

पाठ्य-पुस्तक पृष्ठ संख्या # 143

सोचिए, चर्चा कीजिए एवं लिखिए

Class 7 Maths Exercise 6.5 Solutions In Hindi Medium प्रश्न 1.
त्रिभुज PQR का कोण P एक समकोण है। इसकी सबसे लम्बी भुजा कौन-सी है?
हल:
∵ शीर्ष P पर कोण 90° बनता है।
∴ समकोण बनाने वाली भुजाएँ PQ और PR हैं।
∴ कर्ण = QR
अत: त्रिभुज की सबसे लम्बी भुजा QR है।

MP Board Class 7th Maths English Medium प्रश्न 2.
त्रिभुज ARC का कोण B एक समकोण है। इसकी सबसे लम्बी भुजा कौन-सी है ?
हल:
शीर्ष B पर कोण 90° बनता है।
∴ ∆ABC की समकोण बनाने वाली भुजाएँ AB और BC हैं।
∴ कर्ण = AC
अतः त्रिभुज की सबसे लम्बी भुजा AC है।

Ncert Class 7 Maths Exercise 6.5 Solutions प्रश्न 3.
किसी समकोण त्रिभुज में सबसे लम्बी भुजा कौन-सी होती है ?
हल:
समकोण त्रिभुज में सबसे लम्बी भुजा कर्ण होती है।

Maths Class 7 Chapter 6 Exercise 6.5 प्रश्न 4.
किसी आयत में विकर्ण पर बने वर्ग का क्षेत्रफल उसकी लम्बाई तथा चौड़ाई पर बने वर्गों के क्षेत्रफल के योग के बराबर होता है। यह बौधायन का प्रमेय है। इसकी पाइथागोरस गुण से तुलना कीजिए।
हल:
पाइथागोरस प्रमेय के अनुसार, “समकोण त्रिभुज में कर्ण का वर्ग शेष दो भुजाओं के वर्गों के योग के बराबर होता है।” बौधायन ने सुलभ सूत्र में कहा है कि “आयत के कर्ण द्वारा बनाया गया क्षेत्रफल उसकी दोनों भुजाओं के द्वारा बनाये गये क्षेत्रफल के बराबर होता है।”

अब, माना कि आयत की लम्बाई और चौड़ाई क्रमशः a और b है तथा इसका विकर्ण c है। इसलिए विकर्ण पर बने वर्ग का क्षेत्रफल = c × c = c² तथा आयत की भुजाओं पर बने वर्गों के क्षेत्रफल a² और b² हैं।
∴ c² = a² + b² (बौधायन सुलभ सूत्र)
अतः स्पष्ट है कि पूर्व में बौधायन कथन ही वर्तमान में पाइथागोरस प्रमेय है।

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 3 आँकड़ो का प्रबंधन Ex 3.1

MP Board Class 7th Maths Chapter 3 प्रश्न 1.
अपनी कक्षा के किन्हीं दस (10) विद्यार्थियों की ऊँचाई का परिसर ज्ञात कीजिए।
हल:
माना कि कक्षा के दस विद्यार्थियों की ऊँचाई निम्नलिखित है –
120 सेमी 130 सेमी 110 सेमी 140 सेमी 118 सेमी
118 सेमी 122 सेमी 120 सेमी 132 सेमी 128 सेमी
ऊँचाइयों को आरोही क्रम में लिखने पर,
110 सेमी 118 सेमी 118 सेमी 120 सेमी 120 सेमी
122 सेमी 128 सेमी 130 सेमी 132 सेमी 140 सेमी
अधिकतम ऊँचाई = 140 सेमी,
न्यूनतम ऊँचाई = 110 सेमी
अतः परिसर = अधिकतम ऊँचाई – न्यूनतम ऊँचाई
= 140 सेमी – 110 सेमी
= 30 सेमी

MP Board Class 7 Maths Solutions प्रश्न 2.
कक्षा के एक मूल्यांकन में प्राप्त किए गए निम्नलिखित अंकों को सारणीबद्ध रूप में संगठित कीजिए-
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
(i) सबसे बड़ा अंक कौन-सा है ?
(ii) सबसे छोटा अंक कौन-सा है ?
(iii) इन आँकड़ों का परिसर क्या है ?
(iv) अंकगणितीय माध्य ज्ञात कीजिए।
हल:

(i) सबसे बड़ा अंक = 9
(ii) सबसे छोटा अंक = 1
(iii) परिसर = सबसे बड़ा अंक – सबसे छोटा अंक = 9 -1 = 8

MP Board Class 7 Maths Solutions Chapter 3 प्रश्न 3.
प्रथम 5 पूर्ण संख्याओं का माध्य ज्ञात कीजिए।
हल:
प्रथम 5 पूर्ण संख्याएँ हैं- 0, 1, 2, 3 और 4.
संख्याओं का योग = 0 + 1 + 2 + 3 + 4 = 10

Class 7th Maths MP Board प्रश्न 4.
एक क्रिकेट खिलाड़ी ने 8 पारियों में निम्नलिखित रन बनाए-
58, 76, 40, 35, 46, 50, 0, 100
उसका माध्य स्कोर (Score) या रन ज्ञात कीजिए।
हल:

आंकड़ों का प्रबंधन कक्षा 7 Solution प्रश्न 5.
निम्नलिखित सारणी प्रत्येक खिलाड़ी द्वारा चार खेलों में अर्जित किए गए अंकों को दर्शाती है-

अब निम्नलिखित प्रश्नों के उत्तर दीजिए-
(i) प्रत्येक खेल में A द्वारा अर्जित औसत अंक ज्ञात करने के लिए, माध्य ज्ञात कीजिए।
(ii) प्रत्येक खेल में C द्वारा अर्जित माध्य अंक ज्ञात करने के लिए आप कुल अंकों को 3 से भाग देंगे या 4 से, क्यों ?
(iii) B ने सभी चार खेलों में भाग लिया है। आप इसके अंकों का माध्य किस प्रकार ज्ञात करेंगे?
(iv) किसका प्रदर्शन सबसे अच्छा है ?
हल:
(i) A द्वारा अर्जित अंकों का माध्य

∴A द्वारा अर्जित औसत अंक = 12.5.

(ii) ∵ C ने केवल 3 खेल खेले हैं, उसने तीसरा खेल नहीं खेला है।
∴ योग को 3 से भाग देंगे। क्योंकि C ने 4 खेल में से केवल 3 खेल में ही भाग लिया है।

अतः C के अंकों का माध्य = 10.67
यहाँ, A का माध्य सबसे अधिक (12-5) है। अत: A का प्रदर्शन सबसे अच्छा है।

Class 7 Maths MP Board प्रश्न 6.
विज्ञान की एक परीक्षा में विद्यार्थियों के एक समूह द्वारा (100 अंकों में से) प्राप्त किए गए अंक
85,76, 90,85, 39, 48, 56, 95, 81 और 75 हैं। ज्ञात कीजिए
(i) विद्यार्थियों द्वारा प्राप्त सबसे अधिक अंक और सबसे कम अंक।
(ii) प्राप्त अंकों का परिसर।
(iii) समूह द्वारा प्राप्त माध्य अंक।
हल:
विद्यार्थियों द्वारा प्राप्त अंकों का आरोही क्रम
39,48, 56, 75, 76, 81, 85, 85, 90, 95

(i) विद्यार्थियों द्वारा प्राप्त अधिकतम अंक = 95
विद्यार्थियों द्वारा प्राप्त न्यूनतम अंक = 39

(ii) परिसर = अधिकतम अंक – न्यूनतम अंक
= 95 – 39 = 56

अतः समूह द्वारा प्राप्त माध्य अंक = 73.

MP Board Class 7 Maths Solutions English Medium प्रश्न 7.
छह क्रमागत वर्षों में एक स्कूल में विद्यार्थियों की संख्या निम्नलिखित थी-
1555, 1670, 1750, 2013, 2540, 2820
इस समयकाल में विद्यार्थियों की माध्य संख्या ज्ञात कीजिए।
हल:
विद्यार्थियों की माध्य संख्या

अतः विद्यार्थियों की माध्य संख्या = 2058 प्रतिवर्ष

MP Board 7th Class Maths Book Solutions प्रश्न 8.
एक नगर में किसी विशेष सप्ताह के सात दिनों में हुई वर्षा (mm में) निम्नलिखित रूप में रिकॉर्ड की गई-

(i) उपर्युक्त आँकड़ों से वर्षा का परिसर ज्ञात कीजिए।
(ii) इस सप्ताह की माध्य वर्षा ज्ञात कीजिए।
(iii) कितने दिन वर्षा, माध्य वर्षा से कम रही?
हल:
(i) परिसर = अधिकतम वर्षा – न्यूनतम वर्षा = 20.5 – 0.0 = 20.5 mm

(iii) पाँच दिन (सोमवार, बुधवार, बृहस्पतिवार, शनिवार और रविवार) वर्षा माध्य वर्षा से कम रही।

Class 7 Math MP Board प्रश्न 9.
10 लड़कियों की ऊँचाई cm में मापी गई और निम्नलिखित परिणाम प्राप्त हुए-
135, 150, 139, 128, 151, 132, 146, 149, 143,141
(i) सबसे लम्बी लड़की की लम्बाई क्या है ?
(ii) सबसे छोटी लड़की की लम्बाई क्या है ?
(iii) इन आँकड़ों का परिसर क्या है ?
(iv) लड़कियों की माध्य ऊँचाई (लम्बाई) क्या है ?
(v) कितनी लड़कियों की लम्बाई माध्य लम्बाई से अधिक है ?
हल:
लड़कियों की लम्बाई का आरोही क्रम-
128, 132, 135, 139, 141, 143, 146, 149, 150, 151.
(i) सबसे लम्बी लड़की की लम्बाई 151 cm है।
(ii) सबसे छोटी लड़की की लम्बाई 128 cm है।
(iii) परिसर = अधिकतम लम्बाई – न्यूनतम लम्बाई
= 151 cm – 128 cm
= 23 cm

अतः लड़कियों की माध्य लम्बाई = 141.4 cm है।

(v) ∵ माध्य लम्बाई 141-4 cm से 143 cm, 146 cm, 149 cm, 150 cm, 151 cm अधिक है।
अतः 5 लड़कियों की लम्बाई माध्य लम्बाई से अधिक है।

पाठ्य-पुस्तक पृष्ठ संख्या # 70
प्रयास कीजिए

MP Board Solutions Class 7 Maths प्रश्न 1.
निम्नलिखित के बहुलक ज्ञात कीजिए-
(i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
(ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18,4
हल:
(i) आँकड़ों को आरोही क्रम में व्यवस्थित करने पर,
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6
यहाँ 2, 3 और 4 सबसे अधिक बार (3 बार) आये हैं। अतः बहलक 2, 3 और 4 हैं।
(ii) आँकड़ों को आरोही क्रम में व्यवस्थित करने पर,
2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18
यहाँ, 14 सबसे अधिक बार (6 बार) आया है।
अंत: बहुलक = 14

पाठ्य-पुस्तक पृष्ठ संख्या # 71

MP Board Class 7th Math Solution प्रश्न 1.
संख्याओं के एक समूह में दो बहुलक हो सकते हैं?
हल:
हाँ, संख्याओं के एक समूह में दो बहुलक हो सकते
इन्हें कीजिए

MP Board Maths Class 7 प्रश्न 1.
अपनी कक्षा के साथियों की वर्षों में आयु रिकॉर्ड कीजिए और फिर उनका बहुलक ज्ञात कीजिए।
हल:
साथियों की आयु इस प्रकार है-
12, 14, 13, 15, 14, 12, 13, 14, 14, 13, 14, 12, 15, 14, 12, 14, 14, 14, 12, 14.
आरोही क्रम में व्यवस्थित करने पर,
12, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15
आँकड़ों को सारणी के रूप में लिखने पर,

यहाँ, 14 सबसे अधिक बार (10 बार) आया है।
अतः बहुलक = 14.

MP Board Class 7th Math प्रश्न 2.
अपनी कक्षा के साथियों की cm में लम्बाइयाँ रिकार्ड कीजिए और उनका बहुलक ज्ञात कीजिए।
हल:
कक्षा के साथियों की लम्बाइयाँ (cm में) इस प्रकार हैं-
107, 111, 110, 105, 112, 112, 112, 113, 110, 108, 112, 105, 105, 110, 107, 108, 112, 106, 112, 107
आरोही क्रम में व्यवस्थित करने पर,
105, 105, 105, 106, 107, 107, 107, 108, 108, 110, 110, 110, 111, 112, 112, 112, 112, 112, 112, 113.
आँकड़ों को सारणी रूप में रखने पर,

यहाँ, 112 सबसे अधिक बार (6 बार) आया है।
अतः बहुलक = 112
उत्तर प्रयास कीजिए

MP Board Solution Class 7 Maths प्रश्न 1.
निम्नलिखित आँकड़ों का बहुलक ज्ञात कीजिए-
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14.
हल:
आँकड़ों को आरोही क्रम में व्यवस्थित करने पर_12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 18, 19
आँकड़ों को सारणी रूप में रखने पर,

यहाँ, सबसे अधिक बारम्बारता 10 है, जो कि संख्या 15 की है।
अतः अभीष्ट बहुलक = 15

MP Board 7th Class Maths Book Solutions English Medium प्रश्न 2.
25 बच्चों की ऊँचाइयों (cm में) नीचे दी गयी है।
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 160, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162.
उनकी लम्बाइयों का बहुलक क्या है ? यहाँ बहुलक से हम क्या समझते हैं ?
हल:
आँकड़ों को आरोही क्रम में व्यवस्थित करने पर
160, 160, 160, 160, 161, 162, 162, 162, 162, 163, 163, 163, 163, 163, 163, 163, 163, 163, 164, 164, 164, 165, 165, 165, 168.
आँकड़ों को सारणी रूप में रखने पर,

यहाँ 163 सबसे अधिक बार (9 बार) आया है।
अत: अभीष्ट बहुलक = 163
यहाँ बहुलक से हमारा तात्पर्य है कि अधिकांश विद्यार्थियों की ऊँचाई 163 cm है।

पाठ्य-पुस्तक पृष्ठ संख्या # 72
प्रयास कीजिए

Class 7 Maths Exercise 3.1 Solutions In Hindi प्रश्न 1.
अपने मित्रों से चर्चा कीजिए और
(a) दो स्थितियाँ दीजिए जहाँ प्रतिनिधि मान के रूप में माध्य का प्रयोग उपयुक्त होगा।
(b) दो स्थितियाँ दीजिए जहाँ प्रतिनिधि मान के रूप में बहुलक का प्रयोग उपयुक्त होगा।
हल:
(a) (i) कक्षा के विद्यार्थियों की ऊँचाई मापने में, और
(ii) समान आकार के ड्रमों में अनाज के भार के लिए, माध्य का प्रयोग उपयुक्त होगा।
(b) (i) जूता बेचने वाले एक दुकानदार को एक विशेष आयु वर्ग के लिए जूतों की आपूर्ति के लिए, और
(ii) कमीज बेचने वाले एक दुकानदार को अपने स्टॉक की आपूर्ति के लिए कमीजों के साइज की जानकारी के लिए बहुलक का प्रयोग करना चाहिए।

पाठ्य-पुस्तक पृष्ठ संख्या # 73 – 74
प्रयास कीजिए

Class 7 Maths Chapter 3 Exercise 3.1 Solutions In Hindi प्रश्न 1.
आपके एक मित्र ने दिए हुए आँकड़ों के माध्यक और बहुलक ज्ञात किए। उस मित्र द्वारा की गई त्रुटि यदि कोई हो तो बताइए और सही कीजिए-
35, 32, 35, 42, 38, 32, 34
माध्यक = 42,
बहुलक = 32.
हल:
आँकड़ों 35, 32, 35, 42, 38, 32, 34 को आरोही क्रम में रखने पर,
32, 32, 34, 35, 35, 38, 42

(i) यहाँ n = 7 (विषम)
अतः माध्यक = (\(\frac { n+ 1 }{ 2 }\)) वें पद का मान
= \(\frac { 7+1 }{ 2 }\) वें पद का नाम
= 35
अतः सही माध्यक 35 है।

(ii) ∵आँकड़ों में 32 और 35 अधिक बार (2 बार) आये हैं।
∴ सही बहुलक 32 और 35 हैं।

MP Board Class 7th Maths Solutions

MP Board Class 6th Sanskrit Solutions Surbhi Chapter 14 जन्तुशाला

MP Board Class 6th Sanskrit Chapter 14 अभ्यासः

MP Board Class 6 Sanskrit Chapter 14 प्रश्न 1.
एक वाक्येन उत्तरं लिखत (एक वाक्य में उत्तर लिखो)
(क) वानरः कुत्र भ्रमति? (बन्दर कहाँ घूमता है?)
उत्तर:
वानरः पञ्जरे भ्रमति। (बन्दर पिंजड़े में घूमता है।)

(ख) कः उच्चैः गर्जति? (ऊँचे स्वर में कौन गरजता है?)
उत्तर:
सिंहः उच्चैः गर्जति। (शेर ऊँचे स्वर में गर्जना करता है।)

(ग) मकरः कुत्र वसति? (मगरमच्छ कहाँ रहता है?)
उत्तर:
मकरः जले वसति। (मगरमच्छ जल में रहता है।)

(घ) जन्तुशालायां कः नृत्यति? (जन्तुशाला में कौन नाचता है?)
उत्तर:
जन्तुशालायां मयूरः नृत्यति। (जन्तुशाला में मोर नाचता है।)

Class 6 Sanskrit Chapter 14 MP Board प्रश्न 2.
रिक्तस्थानं पूरयत (रिक्त स्थान को पूरा करो)
(क) तव ………….. नमः।
(ख) मयूरः अस्माकं ………….. पक्षी अस्ति।
(ग) जन्तवः अस्माकं …………… सन्ति।
(घ) सिंहः ……….. गर्जति।
उत्तर:
(क) जनकाय जनन्यै च
(ख) राष्ट्रियः
(ग) मित्राणि
(घ) उच्चैः।

Class 6 Sanskrit Chapter 14 प्रश्न 3.
निम्नलिखिततालिकया वाक्यरचनां कुरुत (निम्नलिखित तालिका से वाक्य रचना कीजिए)

उत्तर:

1. वानरः वृक्षे वसति।
2. सिंहः उच्चैः गर्जति।
3. अहम् खगान् अपश्यम्।
4. सरोवरे कमलानि विकसन्ति।
5. वृक्षेषु फलानि सन्ति।

Chapter 14 Sanskrit Class 6 Hindi Translation प्रश्न 4.
रिक्तस्थानं पूरयत (रिक्त स्थान को पूरा करे)

उत्तर:

1. तरुः = (क) तरोः, (ख) तरुणाम्, (ग) तरौ।
2. भानुः = (क) भानोः, (ख) भानूनाम्, (ग) भानौ।
3. शिशुः = (क) शिशोः, (ख) शिशूनाम्, (ग) शिशौ।

Class 6 Sanskrit Chapter 14 Hindi Translation प्रश्न 5.
त्वं संदीपः असि। तव मित्रम् अस्ति सुनीलः। तव विद्यालयस्य वार्षिकोत्सवस्य वर्णनं कृत्वा मित्राय एकं पत्रं लिख। (तुम संदीप हो। तुम्हारा मित्र सुनील है। अपने विद्यालय के वार्षिक उत्सव का वर्णन करके मित्र को एक पत्र लिखो।)
उत्तर:

पत्रम् मित्रम् प्रति

उज्जयिनीतः
१४-०२-….

प्रिय मित्र सुनीलः।
सस्नेह नमस्ते।
अत्र सर्वं कुशलं तत्रापि कुशलं भवतु । मम विद्यालये जनवरीमासस्य षड्विंशतिः तारिकायां वार्षिकोत्सवः सम्पन्नः अभवत्। दिसम्बर मासे अर्द्धवार्षिकी परीक्षा समाप्ता जाता। प्रत्येक कक्षा-वर्गस्य छात्रा: विविध कार्यक्रमेषु-क्रिकेट खेलम्, कन्दुक खेलम्, धावन प्रतियोगिताषु-प्रतिभागिनः आसन् सायंकाले विद्यालयस्य विशालकक्षे एकम् सांस्कृतिकम् कार्यक्रमः अपि संजातः। गीतानि, भाषणानि नाटकानि आदि कार्यक्रमानि प्रस्तुतानि छात्रैः। कार्यक्रमस्य अन्ते अस्य उत्सवस्य मुख्यः अतिथि: विद्यालयस्य प्राचार्यः आसीत्। तेन प्रतिभागिनेभ्यः छात्रेभ्यः विविधपुरस्कारैः पुरस्कृताः। सर्वाणि कार्यक्रमाणि अतिरुचिकराणि आसन्। अहं इच्छामि यत् त्वम् स्वपरीक्षानन्तरे – मम गृहं आगच्छः।

तव जनकाय जनन्यै च नमः। स्वस्ति अनुजाय। पत्रोत्तरम्। शीघ्रम् लिखतु।

तत्र मित्रम्
सन्दीपः

योग्यताविस्तारः

1. योग्यशब्दं चित्वा उचितस्थाने लिखत (उचित शब्द चुनकर उचित स्थान पर लिखो)-
काकः, पिकः, गजः, व्याघ्रः, धेनुः, शशकः, मयूरः, सिंहः, अजा, मार्जार, सर्पः, अश्वः, हरिणः, भल्लूकः, वानरः, श्वानः, महिषी, चटका।

2. पशुपक्षीणां चित्रावल्याः निर्माणं कुरुत। तेषां नामानि संस्कृते लिखत। (पशुपक्षियों की चित्रावली का निर्माण करो। उनके नाम संस्कृत में लिखिये)
उत्तर:
चित्रावली स्वयं निर्मित करें। पशुपक्षियों के संस्कृत में नाम-वन्य पशवः तथा ग्राम्य पशवः तालिका से लिखें।

जन्तुशाला हिन्दी अनुवाद

भोपालनगरात्
२२-१०-०५

प्रिय मित्र गोपाल!
सस्नेह नमस्ते।
अत्र सर्वं कुशलं तत्रापि कुशलं भवतु। मम मासिकमूल्याङ्कनं समाप्तम्। अहं जनकेन सह जन्तुशालाम् अगच्छम्। किं त्वं जानासि ? जन्तवः अस्माकं मित्राणि सन्ति। जन्तुशालायां जन्तवः अस्माकं मनांसि रञ्जयन्ति। तत्र अहं व्याघ्र जम्बूकं,ऋक्षं, गज, सिंह, वानरं, मकर,शशकं, हरिणं च अपश्यम्। अहं तत्र शुकं, बकं, सारसं, मयूर, हंसं तथा च अन्यान् खगान् अपि अपश्यम्।

अनुवाद :

भोपाल नगर से
२२-१०-०५

प्रिय मित्र गोपाल!
स्नेहपूर्वक नमस्ते।
यहाँ सब कुशल हैं, वहाँ भी कुशल होंगे। मेरी तिमाही परीक्षा समाप्त हो गयी हैं। मैं पिता के साथ जन्तुशाला (चिड़ियाघर) गया था। क्या तुम जानते हो ? जन्तु (प्राणी) हमारे मित्र हैं। जन्तुशाला में जन्तु हमारे मन को प्रसन्न करते हैं। वहाँ मैंने व्याघ्र, जम्बूक (सियार), रीछ, हाथी, सिंह (शेर), बन्दर, मगरमच्छ, खरगोश और हिरण देखे। मैंने वहाँ तोता, बगुला, सारस, मोर, हंस तथा अन्य पक्षियों को भी देखा।

सिंहः उच्चैः गर्जति। वानरः पञ्जरे भ्रमति, उत्पतति च। तस्य पुच्छं दीर्घं भवति। मकरः जले निवसति। मयूरः नृत्यति। सः अस्माकं राष्ट्रीयः पक्षी अस्ति। अलम् अतिविस्तरेण, त्वं अत्र स्थितस्य प्राणिसंग्रहालयस्य अवलोकनार्थं भोपालनगरे मम गृहं आगच्छ। तव जनकाय जनन्यै च नमः। स्वस्ति अनुजाय। पत्रोत्तरं शीघ्रं लिखतु।

तव मित्रम्
सुरेशः

अनुवाद :
सिंह जोर से (ऊँची) गरजना करता है। बन्दर पिंजड़े में घूमता है और उछलता है। उसकी पूँछ लम्बी होती है। मगरमच्छ जल में रहता है। मोर नाचता है। वह हमारा राष्ट्रीय पक्षी है। अधिक विस्तार की आवश्यकता नहीं है। तुम यहाँ स्थित प्राणि संग्रहालय को देखने के लिए भोपाल नगर में मेरे घर आ जाओ। तुम्हारे पिता और माता को नमस्कार। छोटे भाई को स्वस्ति (कल्याण हो)। पत्र का उत्तर शीघ्र लिखो।

तुम्हारा मित्र
सुरेश।

जन्तुशाला शब्दार्थाः

जन्तुशाला = चिड़ियाघर। अस्माकं = हमारा। शुकं = तोता को। रञ्जयन्ति = मनोरंजन करते हैं। बकं = बगुला को। व्याघ्नं = बाघ को। जम्बूकं = सियार को। उच्चैः = जोर से। ऋक्षं = रीछ को। गर्जति = गरजता है। खगान = पक्षियों को। उत्पतति = उछलता है। गज = हाथी को। निवसति = रहता है। वानरं = बन्दर को। अलं = बस। मकरं = मगर को। आगच्छ = आओ। शशकं = खरगोश को। नेष्यामि = ले जाऊँगा। हरिणं = हरिण को। अनुजाय = छोटे भाई को।

MP Board Class 6th Sanskrit Solutions

In this article, we will share MP Board Class 7th Sanskrit Solutions Chapter 18 सुभाषितानि Pdf, Subhashitani Class 7 Sanskrit, These solutions are solved subject experts from the latest edition books.

MP Board Class 7th Sanskrit Solutions Surbhi Chapter 18 सुभाषितानि

MP Board Class 7th Sanskrit Chapter 18 अभ्यासः

Class 7 Sanskrit Chapter 18 प्रश्न 1.
एक शब्द में उत्तर लिखो
(क) कति जनाः नित्यदुःखिता? [कितने लोग नित्य ही दुखी होते हैं?]
उत्तर:
षट्

(ख) कस्य विद्या विवादाय भवति? [किसकी विद्या विवाद के लिए होती है?]
उत्तर:
खलस्य

(ग) साधोः शक्तिः किमर्थं भवति? [साधु की शक्ति किसलिए होती है?]
उत्तर:
रक्षणाय।

Subhasitani Sloka In Sanskrit Class 7 प्रश्न 2.
एक वाक्य में उत्तर लिखो-
(क) देवताः कुत्र रमन्ते? [देवता कहाँ निवास करते हैं?]
उत्तर:
यत्र नार्यः तु पूज्यन्ते, तत्र देवताः रमन्ते। [जहाँ स्त्रियों की पूजा की जाती है, वहाँ देवता निवास करते हैं।]

(ख) कति पुराणानि सन्ति? [पुराण कितने होते हैं?]
उत्तर:
अष्टादश पुराणानि सन्ति। [पुराण अठारह होते हैं।]

(ग) परोपकारः किमर्थं भवति? [परोपकार किसके लिए होता है?]
उत्तर:
परोपकारः पुण्याय भवति। [परोपकार पुण्य के लिए होता है।]

Subhasitani Sloka Class 7 प्रश्न 3.
रिक्त स्थानों को पूरा करो
(क) …………. जनाः नित्युदुखिताः। (पञ्च/षड्)
(ख) साधौ धनं …………. भवति। (दानाय/मदाय)
(ग) …………. पुराणेषु व्यासस्य वचनद्वयम्। (नवदश/अष्टादश)
(घ) …………. परपीडनम्। (पुण्याय/पापाय)
(ङ) परोपकारः ………….. भवति। (पापाय/पुण्याय)
उत्तर:
(क) षड्
(ख) दानाय
(ग) अष्टादश
(घ) पापाय
(ङ) पुण्याय।

Subhasitani Sloka In Hindi Class 7 प्रश्न 4.
श्लोक को पूरा करो
(क) राष्ट्र मम ………….. सुखम्।
……………. स्वराष्ट्रकम्॥
(ख) अष्टादश…………।
…………. परपीडनम् ॥
उत्तर:
(क) राष्ट्रं मम पिता माता प्राणाः स्वामी धन सुखम्।
बन्धुराप्तः सखा भ्राता सर्वस्वं मे स्वराष्ट्रकम्।।
(ख) अष्टादश पुराणेषु व्यासस्य वचनद्वयम्।
परोपकारः पुण्याय पापाय परपीडनम्।।

Sanskrit Vilom Shabd Class 7 प्रश्न 5.
विलोम शब्दों को मिलाओ

उत्तर:
(क) → (4)
(ख) → (12)
(ग) → (2)
(घ) → (9)
(ङ) → (1)
(च) → (3)
(छ) → (10)
(ज) → (11)
(झ) → (6)
(ञ) → (14)
(ट) → (13)
(ठ) → (5)
(ड) → (7)
(ढ) → (15)
(ण) → (8)

Subhashitani Class 7 Solutions प्रश्न 6.
कोष्ठक से चुनकर वाक्य बनाओ-
[एतत्, एते, एषा, एषः, एतानि, एताः, एतौ]
उत्तर:
एतत् = एतत् फलम् मधुरम्।
एते = एते बालिके पठतः।
एषा = एषा बालिका लिखति।
एषः = एषः बालकः क्रीडति।
एतानि = एतानि फलानि मधुराणि।
एताः = एताः बालिकाः गच्छन्ति।
एतौ = एतौ बालकौ धावतः।

सुभाषितानि हिन्दी अनुवाद

यथा यथा हि पुरुषः कल्याणे कुरुते मनः।
तथा तथास्य सर्वार्थाः सिद्धयन्ते नात्र संशयः॥१॥

अनुवाद :
जैसे-जैसे पुरुष कल्याण में मन लगाता जाता है, वैसे ही वैसे उसके सभी कार्य सिद्ध होते जाते हैं। इसमें कोई संशय नहीं है।

ईर्ष्या घृणी न सन्तुष्टः क्रोधनो नित्यशङ्कितः।
परभाग्योपजीवी च षडेते नित्यदुःखिताः॥२॥

अनुवाद :
ईर्ष्या करने वाले, घृणा करने वाले, सन्तुष्ट न रहने वाले, क्रोध करने वाले तथा नित्य शङ्कालु और दूसरे के भाग्य पर जीवित रहने वाले-ये छः (प्रकार के लोग) प्रतिदिन ही दुःख पाते रहते हैं।

विद्या विवादाय धनं मदाय, शक्तिः परेषां परिपीडनाय।
खलस्य साधोः विपरीतमेतत्, ज्ञानाय दानाय च रक्षणाय॥३॥

अनुवाद :
दुष्ट की विद्या वाद-विवाद के लिए, दुष्ट का धन घमण्ड करने के लिए, दुष्ट की शक्ति दूसरों को पीड़ा पहुँचाने के लिए हुआ करती है, परन्तु सज्जन की ये सभी वस्तुएँ इसकी (दुष्ट की) वस्तुओं से विपरीत होती हैं। सज्जन की विद्या ज्ञान के लिए, उसका धन दान के लिए, उसकी शक्ति दूसरों की रक्षा करने के लिए हुआ करती हैं।

राष्ट्रं मम पिता माता प्राणाः स्वामी धनं सुखम्।
बन्धुराप्तः सखा भ्राताः सर्वस्वं मे स्वराष्ट्रकम्॥४॥

अनुवाद :
राष्ट्र ही मेरा पिता, मेरी माता, मेरे प्राण, मेरा स्वामी, मेरा धन व सुख है। राष्ट्र के निवासी ही मेरे बन्धु, आप्तजन, सखा, भाई हैं। इस तरह अपने राष्ट्र के मेरे सर्वस्व हैं।

यत्र नार्यस्तु पूज्यन्ते रमन्ते तत्र देवताः।
यत्रतास्तु न पूज्यन्ते सर्वास्तत्राफलाः क्रियाः॥५॥

अनुवाद :
जहाँ स्त्रियों की पूजा की जाती है (सम्मान किया जाता है) वहाँ देवता रमण करते हैं (निवास करते हैं)। जहाँ इनकी पूजा नहीं की जाती है, वहाँ सम्पूर्ण क्रियाएँ निष्फल जाती हैं।

सत्यं माता पिता ज्ञानं, धर्मो भ्राता दया सखा।
शान्तिः पत्नी क्षमा पुत्रः, षडेते मम बान्धवाः॥६॥

अनुवाद :
माता सत्य स्वरूप होती है, पिता ज्ञान स्वरूप होता है, धर्म भाई के समान तथा दयालुता सखा (मित्र) के समान होती है। पत्नी शान्ति सदृश होती है तथा क्षमा का गुण पुत्र तुल्य होता है। ये सभी छ: तो मेरे बान्धव हैं (बान्धव परिवारीजन होते हैं)।

वरमेको गुणी पुत्रो न च मूर्ख-शतान्यपि।
एकश्चन्द्रस्तमो हन्ति, न च तारागणा अपि॥७॥

अनुवाद-एक गुणवान् पुत्र श्रेष्ठ होता है परन्तु सौ मूर्ख पुत्र (अच्छे) नहीं होते। अकेला चन्द्रमा (रात्रि के) अन्धकार को नष्ट कर देता है, लेकिन तारों का समूह (अन्धकार को) नष्ट नहीं कर सकता।

अष्टादशपुराणेषु व्यासस्य वचनद्वयम्।
परोपकारः पुण्याय पापाय परपीडनम्॥८॥

अनुवाद :
अठारह पुराणों में व्यास के दो वचन ही श्रेष्ठ हैं। परोपकार से पुण्य लाभ होता है और दूसरों को पीड़ा (दुःख) पहुँचाने से पाप लगता है।

सुभाषितानि शब्दार्थाः

ईर्ष्या = ईर्ष्या करने वाला। घृणी = घृणा करने वाला। खलस्य = दुष्ट का। रमन्ते = निवास करते हैं। वरम् = श्रेष्ठ। तमः = अन्धकार। क्रोधनः = क्रोधित रहने वाला।

MP Board Class 7th Sanskrit Solutions

In this article, we will share MP Board Class 8th Social Science Solutions Chapter 3 Struggle Against the British Rule Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Social Science Solutions Chapter 3 Struggle Against the British Rule

MP Board Class 8th Social Science Chapter 3 Text Book Exercise

chose the correct option of the following

MP Board Class 8 Civics Chapter 3 Question 1.
Which uas nottriba1 uprising
(a) Vellore uprising
(h) Bull uprising
(c) Santhal upring
(d) Mundda uprising
Answer:
(a) Vellore uprising

MP Board Class 8 History Chapter 3 Question 2.
Who did not join 1857 struggle
(a) Rani Laxmi BaI
(b) Tatya Tope
(c) Bahadur Sháh Zafar
(d) Dileep Singh
Answer:
(d) Dileep Singh

Class 8 History Chapter 3 MP Board Question 3.
Which of the British policy Thred Bhil? to revolt?
(a) industrial policy
(b) Agriculture policy
(c) religious policy
(d) State’s inter ference
Answer:
(b) Agriculture policy

MP Board Class 8th Social Science Chapter 3 Very Short Answer Type Questions

MP Board Class 8th History Chapter 3 Question 1.
Which book mentions about San yasi struggle?
Answer:
Bankim Chandra Çhatarjee mentioned about Sanyasi uprising in his book. “Anand Math”

MP Board Class 8 Geography Chapter 3 Agriculture Question 2
Who Iedthe SanUil uprising”
Answer:
The Sauthal uprising was led by leaders Sido and Kanhu.

MP Board Class 8 Geography Chapter 3 Question 3.
Who was the leader of Wahabi upring?
Answer:
Sayyed Ahmed of Raibareily led the Wahabi streiggie.

MP Board Class 8th Social Science Chapter 3 Short Answer Type Questions

Class 8 Social Science Chapter 3 Question Answer Question 1.
When and where Khol uprising took place?
Answer:
The kohl led Budho Bhagat Bharat lanched revolit in 1831 in singhbham ranchi palamu hajari bagh manbhumi and other pleces the kohls protested against their landas given to muslim and sikh peasants.

MP Board Class Questions 2.
Name the places of India where 1857 struggle was ‘widespread?
Answer:
The places where the 1857 struggle was wide spread were Delhi. Awadh. Rohilkhand. Bundelkhand Allbhahad, Agra Meerut. Kanpur and jhansi.

MP Board Class 8th Social Science Chapter 3 Long Answer Type Questions

MP Board Class 8 Civics Chapter 1 Question 1.
Describe the Santh:il uprising?’
Answer:
The Snthal revolt was led by Santhal leaders and sido and Kanhu in 1856. It was one of the most power ful uprisings. ‘The opposed the maltreatment up-rings they opposed the maltrrement of lands revenu offioal in daman -e- koh area this area lies between Bhagalpur and Rajmahahal the santhals declared end of the company rule in their region and revolt and were later suppressed by Britisher’s.

Class 8 Atit Chapter 3 Question 2.
What were the causes of resentment among sepoys
Answer:
The causes of discontent among the Indian soldiers of the British army were as under:

• the salaries of Indian soldiers were very low and were practically for hem the higher posts in the army were reserved for European officers only .
• there was great disparity between the salaries of Indian and European soldiers.
• the Indian splicer felt the injustices which the old Royal families in Indian had been suffering the increased misery which the conman people had to suffer affected the soldiers directly as they were part and parce of the Indian society
• the cartridges of the newly introduced rifle were greased with cows and pigs fat their covers had to be cut with teeth before loading the rifle. the use of these greased cartage offended the religions sentiments of both the Hindu and the Muslim soldiers it became the immediate cause of the Revolt.

Class 8 Social Science Chapter 3 Question 3.
What were the major cuses of the failure of 1857 struggle
Answer:
There were many causes that led to the failure of the revolt of 1857:

• Not Planned –  The revolt of 1857 was not well planned. it sparked off abruptly
•  No Single or Common Object – The leaders of revolt in varies part of the country had different and limited Sanous parts of the objectives before them
• No Common Leadership – There was no common leader for the whole upping each region had its own leader.
• No Coordination – The uprising suffered from lack of coordination. There was no coordination among the rebel in various religions of the country
• Localized – It was Wotan all-India revolt It was localized in certain areas. it dd not start in South and Punjab
• Led by some experiences and Landlords – it was led by some experiences and landlord who were deprived of territories.
• Not People’ Revolt – it was not people revolt it was only at one two pleases where conman people participated
• New Emerging middle class did not participated
• some Indian rulers are sided with the British: some Indian rulers kept away while some rulers sided with the English
• lack of resource – the rebel did not have enough financial resources and weapons of war.

Class 8 Political Science Chapter 3 Question 4.
Write notes on:
1. sanyasis upring
2. Whabi upring
Answer:
1. sanyasis upring:
this upring was important among other civil up-rings. the apathy of British government after 1770 famine in Bengal economic loot and ban on religious they attacked fractious colonies and forts of the company governor General Warren Hastings could brought the revolt under control.

2. Whabi upspring :
sayyed Ahmed of Railbareilly led the Wahhabi And changes in Islam sayyed Ahwed launned revolt against sikh rule in Punjab and when company took control over Punjab in 1849 Wahhabi raised rebellion against British rule got most powerful challenge from Wahhabi struggle.

Science Chapter 3 Class 8 MP Board Question 5.
make an album with the help of leaders who participated in 1857 struggle.
Answer:

MP Board Class 8th Science Chapter 3 Question 6.
South major regions of 1857 struggle in the out line map of India.
Answer:

MP Board Class 6th Science Solutions Chapter 5 Separation of Substances

Separation of Substances Text Book Exercises

Question 1.
Why do we need to separate different components of a mixture? Give two examples.
Answer:
It is essential to purify foodstuffs to maintain good health. Researchers, chemists, and technologists also need pure substances. Thus, the separation of different components of a mixture is essential. The main purposes are:

1. To remove impurities for getting a pure sample. For example, salt from seawater.
2. To obtain the useful components for getting wheat or rice grains after separating.
3. To remove the harmful component.

MP Board Class 6 Science Chapter 5 Question 2.
What is winnowing? Where it is used?
Answer:
The process of separating heavier and lighter components of a mixture by wind or by blowing air is called winnowing? This method is commonly used by farmers to separate lighter husk particles from heavier seeds of grain. The farmer allows the mixture of grain and the husk to fall from a height.

The grains which are heavier fall vertically down on the ground and form a heap near the platform of winnowing. The husk which is higher, is carried away by the wind and forms a separate heap at a short distance from the heap of grains as shown in the figure.

The separated husk is used for many purposes such as fodder for catels.

Question 3.
How will you separate husk or dirt particles from a given sample of pulses before cooking?
Answer:
The husk or dirt particles from a given sample of pulses before cooking can be separated by hand picking. For example, when the components of a mixture are different in size, shape and color, they can be easily separated by hand picking. Housewife often clean pulses and spices by this method.

They remove small dust, dirt, pebbles and other unwanted materials. The quantity of such impurities is usually not very large. In such situations, we find that hand picking is a convenient method of separating substances.

MP Board Class 6 Science Chapter 3 Separation Of Substances Question 4.
What is sieving? Where is it used?
Answer:
Sieving is a method used to separate the components of a mixture which are of different size. Sieving allows the fine flour particles to pass through the holes of the sieve while the bigger impurities remain on the sieve In a flour mill, impurities like husk and stones are removed from wheat before grinding it.

Usually, a bagful of wheat is poured on a slanting sieve. The sieving removes pieces of stones, stalk and husk that may still remain with wheat after threshing and winnowing. You may have also noticed similar sieves being used at construction sites to separate pebbles and stones from sand.

Class 6 Science Chapter 5 Short Question Answer Question 5.
How will you separate sand and water from their mixture?
Answer:
A mixture of sand and water can be separated by sedimentation method or decantation method. Take the mixture of sand and water in a beaker. Allow the mixture to stand for sometime undisturbed.

You will observe that sand particles start settling down at the bottom of the beaker forming sediment (Fig. (a). The process of settling down of heavier, insoluble particles from a mixture is called sedimentation.

The clear liquid above the sediment is gently poured off into another beaker, without disturbing the sediment (Fig. b). This process of transferring the clear liquid without disturbing sediments is known as decantation.

Question 6.
Is it possible to separate sugar mixed with wheat flour? If yes, how will you do it?
Answer:
Yes, it is possible to separate sugar mixed with wheat flour. This can be separated by sieving. Sieving allows the fine wheat flour particles to pass through the holes of the sieve while the sugar remain on the sieve.

Class 6 Science Chapter 5 Question Answer Question 7.
How would you obtain clear water from a sample of muddy water?
Answer:
We can separate clear water from a sample of muddy water by loading process. Take a beaker half filled with muddy water and take a piece of alum. The loading of suspended particles is carried out by alum. The alum crystal is slowly moved in water.

The dissolved particles of alum in water loads on the fine dust particles. The particles become heavy and settles down to the bottom. The clear water is decanted as shown in Fig. The suspended particles remains at bottom as sediment.

Question 8.
Fill up the blanks:

1. The method of separating seeds of paddy from its stalks is called ………………………
2. When milk, cooled after boiling, is poured onto a piece of cloth the cream (malai) is left behind on it. This process of separating cream from milk is an example ……………………………..
3. Salt is obtained from seawater by the process of ……………………….
4. Impurities settled at the bottom when muddy water was kept overnight in a bucket. The clear water was then poured off from the top. The process of separation used in this example is called ………………………..

Answer:

1. Threshing
2. Filtration
3. Evaporation
4. Decantation.

Class 6 Science Chapter 5 Question Answer In English Question 9.
True or false:

1. A mixture of milk and water can be separated by filtration.
2. A mixture of powdered salt and sugar can be separated by the process of winnowing.
3. Separation of sugar from tea can be done with filtration.
4. Grain and husk can be separated with the process of decantation.

Answer:

1. False
2. False
3. False
4. False.

Question 10.
Lemonade is prepared by mixing lemon juice and sugar in water. You wish to add ice to cool it. Should you add ice to the lemonade before or after dissolving sugar? In which case would it be possible to dissolve more sugar?
Answer:
We should add ice in lemonade after dissolving sugar in it, because sugar dissolve more quickly before adding ice. It would be possible to dissolve more sugar before adding ice in the lemonade, because it dissolve more into hot than in cold.

Projects and Activities

Activity 1.
You have tried a number of methods to separate impurities like mud from water. Sometimes, the water obtained after employing all these processes could still be a little muddy. Let us see if we can remove even this impurity completely. Take this filtered water in a glass.

Tie a thread to a small piece of alum. Suspend the piece of alum in the water and swirl. Did the water become clear? What happened to the mud? This process is called loading. Talk to some elders in your family to find out whether they have seen or used this process.
Answer:
We can separate clear water from a sample of muddy water by loading process. Take a beaker half filled with muddy water and take a piece of alum. The loading of suspended particles is carried out by alum. The alum crystal is slowly moved in water.

The dissolved particles of alum in water loads on the fine dust particles. The particles become heavy and settles down to the bottom. The clear water is decanted as shown in Fig. The suspended particles remains at bottom as sediment.

Activity 2.
Match the each process with its purpose and the way separated components are used:
Table: Why do we separate substances?

Activity 3.
Complete the following table:

Answer:

Activity 4.
To separate common salt from common salt solution (brine)?
Method:
Take brine in a china dish. Heat it by placing it over a wire gauze on a tripod stand as shown in Fig. After sometime the water starts evaporating. Go on heating till all the water present in the solution evaporates. You will see that some white solid substance is left behind as a residue. This is a sample of pure common salt.

Activity 5.
To separate a mixture of kerosene and water?
Method:
Prepare a mixture of kerosene and water in 200ml. Beaker. Carefully transfer to a separating funnel through a funnel. Stopper and shaken well. Now the mixture is allowed to settle. Note that kerosene is the top layer and water is the bottom layer.

This is due to the density differences between the two liquids Place the separating funnel in a ring stand and place a beaker below it to collect the components. Carefully looking at the layers of separation, drain kerosene and water in different containers.

Separation of Substances Additional Important Questions

Separation of Substances Objective Type Questions

Question 1.
Choose the correct answer:

Class 6 Science Chapter 5 Exercise Question (a)
A mixture of iodine and sand can be separated by –
(a) Decantation
(b) Centrifugation
(c) Filtration
(d) Sublimation.
Answer:
(d) Sublimation.

Question (b)
A mixture of leaves of tea and iron fillings are separated by –
(a) Filtration
(b) Handpicking
(c) Magnetic separation
(d) Sieving.
Answer:
(c) Magnetic separation

Sieving Definition Class 6 Question (c)
A mixture of mustard oil and kerosene oil can be separated by –
(a) Sublimation
(b) Evaporation
(c) Separating funnel
(d) Filtration.
Answer:
(c) Separating funnel

Question (d)
When two elements are brought together, the always form
(a) A mixture
(b) An element
(c) A compound
(d) None of these.
Answer:
(a) A mixture

Class 6th Science Chapter 5 Question (e)
In nature most elements occur as –
(a) Pure elements
(b) Mixture of elements
(c) Compound form
(d) None of these.
Answer:
(c) Compound form

Question (f)
A solution of salt in water is a –
(а) Compound
(b) Homogeneous mixture
(c) Heterogeneous mixture
(d) None of these.
Answer:
(b) Homogeneous mixture

Std 6 Science Chapter 5 Question Answer Question 2.
Fill in the blanks:

1. A mixture of mustard oil and water can be separated by using ………………………
2. Ammonium chloride and salt are separated by …………………………
3. Common salt is obtained from seawater by …………………….
4. A mixture of chalk powder and water is separated by ………………………
5. Husk is separated from rice by …………………………
6. Cream is separated from milk by ……………………….
7. Naphthalene is separated from common salt by ……………………..
8. Insects are separated from wheat by ……………………..

Answer:

1. Separating funnel
2. Sublimation
3. Evaporation
4. Filtration
5. Winnowing
6. Centrifugation
7. Sublimation
8. Hand-picking.

Class 6 Science Chapter 3 Separation Of Substances Question Answer Question 3.
Which of the following statements are true (T) or false (F):

1. Sherbet is a pure substance.
2. Rock salt is impure substance.
3. The process of winnowing is used to remove small pieces of stone from grams.
4. A pure sample of a substance consists of only one kind of particles.
5. Sugar is separated from its solution in water by decantation.
6. Butter is separated from milk by the. method of crystallization.

Answer:

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

Ch 5 Science Class 6 Question 4.
Match the items in Column A with the Column B:

Answer:

(i) – (c)
(ii) – (a)
(iii) – (b)
(iv) – (d).

Separation of Substances Very Short Answer Type Questions

Question 1.
Name the process used in flour mill.
Answer:
Sieving.

Class 6th Science Chapter 5 Question Answer Question 2.
Name the method of separating iron from waste material.
Answer:
Magnetic separation.

Question 3.
Name the process to separate two unmixible liquids.
Answer:
Decantation.

Class 6 General Science Chapter 5 Question Answer Question 4.
Name the apparatus used to separate two unmixible liquids.
Answer:
Separating funnel.

Question 5.
Some Suji has been stored in a rusty tin. When taken out it contains some flakes of rust. How will you remove them?
Answer:
We can remove flakes of rust by sieving the mixture.

MP Board Class 6th Question 6.
What is the advantage of distillation over evaporation?
Answer:
In distillation liquid can be recovered while in evaporation it is lost.

Question 7.
Which substance is used for loading?
Answer:
Alum.

Separation Of Substances Class 6 Table 5.2 Question 8.
What is the use of alum in loading?
Answer:
Alum is used to make the decantation faster.

Question 9.
Name the process to obtain salt from seawater.
Answer:
Evaporation.

Science Ch 5 Class 6 Question 10.
Name two sublime substances.
Answer:

1. Iodine
2. Ammonium chloride.

Question 11.
What happens during sublimation?
Answer:
Solids are directly converted into their vapours without undergoing liquid state.

Question 12.
How does doctor get distilled water?
Answer:
By distillation of water.

Question 13.
Can you say that a mosquito net is a filter?
Answer:
Yes, mosquito net is a filter. It filters out mosquito from the air.

Class 6th Science Chapter 3 Separation Of Substances Question 14.
How does jeweller separate pearls of different sizes?
Answer:
Jeweller separates pearls of different size by sieving.

Question 15.
Are all crystals of substances of same shape and colour?
Answer:
No, crystals of different substances are of different shape and colour.

Class 6 Science Ch 3 Separation Of Substances Question Answer Question 16.
Tea filter is missing, then what method would you use for separation of tea powder?
Answer:
Sedimentation.

Question 17.
Which organ of our body separates harmful components from blood?
Answer:
Kidney.

Question 18.
What happens when camphor is left open?
Answer:
When camphor is left open then a strong smell is produced.

Case Study Separation Of Substances Class 6 Question 19.
Name two pure substances that you know.
Answer:
All elements and compounds are pure substances like salt, water, carbon (graphite), diamond, silver, copper, etc.

Question 20.
Solid substance is dissolved in water. Which of the following methods is used to separate it?

1. Filtration
2. Evaporation
3. Sublimation
4. Decantation.

Answer:
A solid substance is dissolved in water. It is separated by the method of evaporation.

Hand Picking Drawing Science Question 21.
Which method would you use for the separation of a solid substance dissolved in water? Both the components are to be collected.
Answer:
Distillation is used to separate solid substances and water from the mixture like salt solution.

Question 22.
Why does visibility increase after rains?
Answer:
This is because the dust particles that were present in air settle down due to loading by fain drops and we can see distant objects if the air dust is free and clean.

Question 23.
Name the two materials used in water filters.
Answer:

1. Filters made from ceramics are used in households.
2. In water-works for large supply of drinking water a material called resin, a kind of plastic is used.

What Is Sieving Where Is It Used For Class 6 Question 24.
Why is chlorine mixed with drinking water?
Answer:
In water – works, chlorine is used to kill the harmful bacteria present in water before supplied to consumers.

Question 25.
Name the method by which we can remove butter from milk.
Answer:
The process called centrifugation is used to remove butter from milk. Milk is churned in special machines like centrifuges. Butter being light in weight floats on the surface of milk and is removed easily.

Question 26.
Define evaporation.
Answer:
Evaporation method is used to separate solids dissolved in a liquid. This process is largely used to obtain common salt from sea or lake water. The process of converting a liquid into its vapor by heating is called evaporation.

Decantation Drawing Easy Question 27.
Define crystallization.
Answer:
Crystallization method is used to purify solid substances. The process of separating a pure substance in the form of crystals from its hot saturated solution by cooling is called crystallization.

Separation of Substances Short Answer Type Questions

Question 1.
Name three mixtures commonly found in nature.
Answer:
The three mixtures commonly found in nature are soil, air and seawater. Soil is a mixture of various inorganic and organic matter. Air is a mixture of various gases and dust particles. Sea water is a mixture of water and salts.

Question 2.
How would you separate water and mustard oil?
Answer:
The separation of water and mustard oil is based on the principle that water is heavier than oil and also the two are immiscible liquids. They form distinct layers. Take the mixture in separating funnels. Separate the water by opening the stop-clock at the base of the funnel in a beaker.

Question 3.
What is the process of separating salt and camphor?
Answer:
Camphor is a volatile substance. It sublimes on heating. When a mixture of salt and camphor is heated, the camphor being volatile sublimes and is collected as shown in figure. The salt being non-volatile is left behind.

Question 4.
Explain how would you make crystals of sugar.
Answer:
Prepare sugar solution in hot water. The beaker is allowed to cool. Filter the solution. Hang a crystal of sugar in this solution with the help of a thread and a glass rod or a pencil. Leave the solution uncovered and undisturbed for a few days. We observe that the crystal slowly begin to grow.

Question 5.
Write the name of methods which we use in our daily life for separating components of mixtures.
Answer:
The various processes involved in separation of a mixture components of a mixture are:

1. Winnowing
2. Hand-picking
3. Sieving
4. Decantation
5. Filtration
6. Magnetic separation
7. Loading
8. Distillation
9. Centrifugation
10. Crystallization
11. Sublimation.

Question 6.
How you separate a mixture of milk and cooking oil?
Answer:
Milk and cooking oil are two immiscible liquids. The cooking oil is lighter than milk and it will form upper layer in this mixture. We take the mixture in a separating funnel and allow the mixture to settle. When the separation of milk and cooking oil is complete, we open the stop – cock of the separating funnel. The lower layer of milk is decanted leaving behind oil in the separating funnel. The cooking oil is collection in a separate breaker.

Question 7.
Define distillation.
Answer:
Distillation is the process of evaporation followed by condensation. This method is used to separate a mixture of two miscible liquids with a difference in their boiling points. This process is also used to obtain pure substance from a solution. In evaporation the liquid is lost but in distillation both solute and solvent are obtained. Distilled water used by the doctor for injections is obtained by distillation. In Kuwait, sea water is distilled to get drinking water.

Separation of Substances Long Answer Type Questions

Question 1.
What is a mixture? Give examples of mixtures.
Answer:
A mixture is a heterogeneous material which has two or more kind of particles. A mixture may have solid, liquid or gas components. Examples:

1. Rock salt is a mixture of some salts, soil and sand.
2. Soil is a mixture of clay, sand and particles of grass and other dead plants.
3. Sherbet we drink in summer is a mixture of water, sugar and colouring and flavouring substances.
4. Sea water is a mixture of water and many salts.
5. Air is a mixture of oxygen, nitrogen, carbon dioxide, water vapour, dust particles and gases like argon.

Question 2.
You are given a mixture of sand, water and mustard oil. How will you separate the components of the mixture?
Answer:
Principle – The separation is based on the principle that sand and mustard oil are insoluble in water. But sand being heavy, so settles at bottom and mustard oil being lighter floats on water surface. The separation is carried out in two steps.

Step – I:
Take the mixture of water, sand and mustard oil, filter the mixture. The sand will be separated. The water and oil will pass through filter – paper and will be collected as filtrate.

Step – II:
Take the filtrate in a separating funnel. Allow the mixture to set. Then open the stopcock and decant of water. Mustard oil will be left in the separating funnel.

Question 3.
How will you obtain pure salt from its impure sample?
Answer:
Rock salt is a mixture of salt, clay and sand. Of these pure salt is soluble in water. Therefore, rock salt is first ground to a fine powder. The powdered rock salt is dissolved in water. On dissolving, salt being soluble in water so dissolves and the clay and sand is left behind:

The mixture is now filtered through a filter – paper. The clay and sand filtrate. The filtrate is evaporated in China dish. The water evaporates and pure salt is left behind.

Question 4.
How will you separate a mixture of iron filings, ammonium chloride and sand from their mixture?
Answer:
Principle – The separation is based on the principle that ammonium chloride is volatile, iron filings are magnetic and sand is neither volatile nor magnetic. The separation is carried out in two steps:

Step – I:
Take the mixture in a dish. Bring a strong bar magnet over the mixture. The iron filings are attracted to the magnet and are separated leaving the mixture of ammonium chloride and sand.

Step – II:
Take the mixture in a China dish. Cover the China dish with an. inverted funnel as shown in the fig. Heat the content of China dish with spirit lamp. The ammonium chloride being volatile sublimes and its vapours collects on the surface of funnel. Allow the funnel to cool. Scratch the ammonium chloride. The sand will remain in China Dish.

Question 5.
Define threshing with diagram?
Answer:
The process that is used to separate grain from stalks is threshing. In this process, the stalks are beaten to free the grain seeds (Fig.). Sometimes, threshing is done with the help of bullocks. Machines are also used to thresh large quantities of grain.

Question 6.
You know various methods of separating mixtures. Here four pictures are shown. Write the correct names of methods under each picture.

Answer:

1. Decantation.
2. Winnpwing
3. Sieving
4. Loading.

Question 7.
How will you separate pure water from a solution of salt in water?
Answer:
Take the mixture of salt in water in a flask. Set the arrangement for distillation. On heating water vapour rises and passes through the tube. They are condensed in the test – tube surrounded by the cold water. Thus we get pure water leaving behind salt in the flask.

MP Board Class 6th Science Solutions

MP Board Class 9th Science Solutions Chapter 5 The Fundamental Unit of Life

The Fundamental Unit of Life Intext Questions

The Fundamental Unit of Life Intext Questions Page No. 59

Question 1.
Who discovered cells and how?
Answer:
Robert Hooke (1665), by chance observed a slice of cork through a self designed microscope. He observed that it contained many little compartments, like a honey comb, which he named as cells. However, Leeuwenhoek (1674) discovered the free living cells in pond water for the first time, by his improved microscope.

MP Board Class 9th Science Chapter 5 Question 2.
Why is the cell called the structural and functional unit of life?
Answer:
The body of all organisms consists of one or many cells. Therefore, cell is called the structural unit of life. All processes associated with life such as respiration, digestion, excretion etc. are performed by cell. So, cell is also called as functional unit of life.

The Fundamental Unit of Life Intext Questions Page No. 61

Question 1.
How do substances like CO₂ and water move in and out of the cell? Discuss.
Answer:
When concentration of CO₂ is more inside the cell than outside, CO₂ diffuses from the cell to outside of cell. If CO₂ concentration inside the cell is less, CO₂ moves inside the cell from outside. The water move in and out of the cell by the process of osmosis. Osmosis is the passage of water from a region of high water concentration through semi – permeable membrane (cell membrane) to a region of low concentration of water.

MP Board Class 9 Science Chapter 5 Question 2.
Why is the plasma membrane called a selectively permeable membrane?
Answer:
Plasma membrane permits the entry and exit of some materials in and out of the cell. It also prevents movement of some other materials. So, it is called selectively permeable membrane.

The Fundamental Unit of Life Intext Questions Page No. 63

Question 1.
Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.

Prokaryotic Cell	Eukaryotic Ceil
1. Size: Generally, small (1 – 10 μm). (1 μm = 10-6m).	1. Size: Generally, large (5 – 100 μm).
2. Nuclear region: ……………… …………………………………………….. …………………… and known as …………. .	2. Nuclear region: Well – defined and surrounded by a nuclear membrane.
3. Chromosome: Single.	3. More than one chromosome.
4. Membrane – bound cell organelles absent.	4. ………………………………………….. .

Answer:
Prokaryotic cell:
2. Nuclear region is poorly defined due to absence of a nuclear membrane and is known as nucleoid.

Eukaryotic cell:
4. Membrane bound cell organelles are present.

The Fundamental Unit of Life Intext Questions Page No. 65

Question 1.
Can you name the two organelles we have studied that contain their own genetic material?
Answer:

1. Mitochondria and
2. Plastid (chloroplast).

Class 9 Science Chapter 5 MP Board Question 2.
If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Answer:
Each cell has got certain specific cell organelles. Each cell organelle performs a special function.
e.g.,

• making of new material, removal of waste from the cell, release of energy etc. If the organization of a cell is destroyed, the functioning of the cell organelles will be disturbed, control of the nucleus will be lost. Ultimately, cell will die.

Question 3.
Why are lysosomes known as suicide bags?
Answer:
Lysosome is a membrane bound bag – like cell organelle which contains powerful enzymes. If lysosome bursts, its enzymes eat up (digest) other organelles of its own cell. Therefore, they are known as ‘suicide bags’.

Golgi Apparatus Diagram Class 9 Question 4.
Where are proteins synthesised inside the cell?
Answer:
Ribosomes.

The Fundamental Unit of Life NCERT Textbook Exercises

Question 1.
Make a comparison and write down ways in which plant cells are different from animal cells?
Answer:
Difference between plant cell and animal cell is given below:

Plant Cell	Animal Cell
1. The outermost covering of the plant cell is the cell wall which is formed of cellulose.	1. The outermost covering is the plasma membrane.
2. Plastids (e.g., chloroplast) present.	2. Plastids absent.
3. Large vacuole present.	3. No or small vacuoles are present.
4.  Centrioles are absent but plastids caps are present.	4. Centrioles are present within centrosome.

Diagram Of Golgi Apparatus For Class 9 Question 2.
How is a prokaryotic cell different from a eukaryotic cell?
Answer:

Prokaryotic Cell	Eukaryotic Cell
1. Cell size is generally small. (1 – 10 mm).	1. Cell is generally large. (5 – 100 mm).
2. Nuclear region is called nucleoid is not surrounded by a nuclear membrane.	2. Nuclear material is surrounded by a nuclear membrane.
3. Only a single chromosome is present.	3. More than one chromosome are present.
4. Nucleolus is absent.	4. Nucleolus is present.
5. Membrane bound cell organelles are absent.	5. Cell organelles bounded by membrane are present.
6. Cell division by fission or budding (no mitosis).	6. Cell division mitotic or meiotic.

Question 3.
What would happen if the plasma membrane ruptures or breaks down?
Answer:
Plasma membrane regulates the movement of substances in and out of the cell. Due to rupture of plasma membrane, there will be no regulation of the movement of molecules. Secondly, contents of the cell may leak out. Because of these two reasons cell will die.

Golgi Body Diagram Class 9 Question 4.
What would happen to the life of a cell if there was no Golgi apparatus?
Answer:
Golgi apparatus is connected to ER. It collects simpler molecules from ER and convert them into more complex molecules. These are then packaged in small vesicles and are stored or transported inside or outside the cell. If Golgi apparatus is not present in the cell, all the above processes of modification, storage and tRansportation will not be possible.

Golgi apparatus is also involved in the formation of lysosomes. If there was no Golgi apparatus in the cell, lysosomes would not be formed and hence, foreign materials like bacteria could easily invade and destroy the cell.

Question 5.
Which organelle is known as the powerhouse of the cell? Why?
Answer:
Mitochondria are known as ‘Power House’ of the cell. They are said so because, the energy required for various life activities is released by mitochondria in the form of ATP molecules. The body uses energy stored in ATP for synthesis of new compounds and for mechanical work. As ATP instantly provide energy, they are called energy currency of the cell.

Golgi Apparatus Diagram Class 9 Easy Question 6.
Where do the lipids and proteins constituting the cell membrane get synthesised?
Answer:

1. Rough Endoplasmic Reticulum (RER) synthesizes proteins constituting cell membrane.
2. Smooth Endoplasmic Reticulum (SER) synthesizes and secretes lipids constituting cell membrane.

Question 7.
How does an Amoeba obtain its food?
Answer:
Amoeba has flexible cell membrane. It enables Amoeba to engulf in food by the process called endocytosis.

Question 8.
What is osmosis?
Answer:
The passage of water from a region of its high concentration through a semipermeable membrane to a region of low water concentration is known as osmosis.

Class 9 Science Chapter 5 Question Answer Question 9.
Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,
(a) Keep cup A empty.
(b) Put one teaspoon sugar in cup B.
(c) Put one teaspoon salt in cup C.
(d) Put one teaspoon sugar in the boiled potato cup D.

Keep these for two hours. Then observe the four potato cups and answer the following:

1. Explain why water gathers in the hollowed portion of B and C?
2. Why is potato A necessary for this experiment ?
3. Explain why water does not gather in the hollowed out portions of A and D.

Answer:

1. Water gathers in the hollowed portion of B and C because:
   • Living plasma membrane of potato cells act as semipermeable membrane.
   • There is higher concentration of water in trough than the hollowed portion of B and C.
   • So, water by the process of osmosis moves into hollowed portion of potato cups B and C.
2. Potato cup A is kept empty to act as control set – up.
3. • As the potato cup A, is empty, water does not gather in hollowed out portions of A.
   • In the potato cup D, the potato cell membrane lost quality of semi – permeability due to boiling. So, no water movement occurs from the trough into the potato cup D.

The Fundamental Unit of Life Additional Questions

The Fundamental Unit of Life Multiple Choice Questions

Question 1.
Who coined the term protoplasm for the fluid substance of the cell?
(a) J.E. Purkinje
(b) Robert Brown
(c) W. Flemming
(d) Robert Hooke.
Answer:
(a) J.E. Purkinje

Class 9th Science Chapter 5 Question 2.
Ribosomes are the centre for __________.
(a) Fat synthesis
(b) Protein synthesis
(c) Starch synthesis
(d) Sugar synthesis.
Answer:
(b) Protein synthesis

Question 3.
The complete break down of glucose in presence of oxygen in a cell takes place in __________.
(a) Mitochondria
(b) Ribosome
(c) Chloroplast
(d) None of these.
Answer:
(a) Mitochondria

Drawing Golgi Apparatus Diagram Class 9 Question 4.
Lysosomes contains __________.
(a) Fats
(b) Secretory glycoproteins
(c) Hydrolytic enzymes
(d) RNA.
Answer:
(c) Hydrolytic enzymes

Question 5.
Plant cells have large vacuoles each surrounded by a membrane known as __________.
(a) Plasma membrane
(b) Cell wall
(c) Leucoplast
(d) Tonoplast.
Answer:
(d) Tonoplast

Science Ch 5 Class 9 Question 6.
The process of selective movement of substances through semi – permeable membrane is called?
(a) Osmosis
(b) Diffusion
(c) Plasmolysis
(d) Imbibition.
Answer:
(a) Osmosis

Question 7.
What makes to withstand greater changes in the surrounding medium in animal cell?
(a) Plasma membrane
(b) Cell wall
(c) Vacuoles
(d) Plastids.
Answer:
(b) Cell wall

Class 9 Science Lesson 5 Question 8.
When an animal cell is placed in hypotonic solution, it __________.
(a) Swells up
(b) Shows plasmolysis
(c) Bursts due to over swelling
(d) Shows crenation.
Answer:
(c) Bursts due to over swelling

Question 9.
Which of the following maintains the basic structure (shape) of the plant cell after shrinkage of the cell content during plasmolysis?
(a) Plasma membrane
(b) Vacuole
(c) Plastids
(d) Mitochondria.
Answer:
(d) Mitochondria.

Question 10.
Plasmolysis occurs due to __________.
(a) Endosmosis
(b) Exosmosis
(c) Absorption
(d) None of these.
Answer:
(b) Exosmosis

Class 9th Subject Science Chapter 5 Question Answer Question 11.
Who among the following scientists coined the term ‘cell’?
(a) Leeuwenhoek
(b) J.E. Purkinje
(c) Robert Hooke
(d) Robert Brown.
Answer:
(c) Robert Hooke

Question 12.
Which is the largest cell in human body?
(a) Muscle cell
(b) Nerve cell
(c) Kidney cell
(d) Liver cell.
Answer:
(b) Nerve cell

MP Board Class 9th Science Solution Question 13.
Iodine solution is used to __________.
(a) Stain onion peel cells
(b) Stain human cheek cells
(c) Mount onion peel cells
(d) Mount human cheek cells.
Answer:
(a) Stain onion peel cells

Question 14.
The barrier between the cytoplasm and the outer environment in an animal cell is __________.
(a) Tonoplast
(b) Nuclear membrane
(c) Cell wall
(d) Plasma membrane.
Answer:
(d) Plasma membrane.

Question 15.
The power house of a cell is known as __________.
(a) Golgi apparatus
(b) Chloroplast
(c) Mitochondrion
(d) Vacuole.
Answer:
(c) Mitochondrion

9th Standard Science Lesson Number 5 Question 16.
Which of the following organelles possesses its own DNA and Ribosomes?
(a) Mitochondria
(b) Lysosomes
(c) Golgi apparatus
(d) Endoplasmic reticulum.
Answer:
(a) Mitochondria

Question 17.
The functions of which of die organelle include the storage, modification and packaging of products in vesicles:
(a) Lysosome
(b) Vacuoles
(c) Golgi apparatus
(d) SER.
Answer:
(c) Golgi apparatus

Science 5 Chapter Class 9 Question 18.
Which organelles, like mitochondria, have their own DNA and ribosomes?
(a) Golgi apparatus
(b) Vacuoles
(c) Endoplasmic reticulum
(d) Plastids.
Answer:
(d) Plastids.

Question 19.
Mitochondria and plastids are able to synthesis some of their own proteins because they have?
(a) DNA and nucleolus
(b) RNA and lysosomes
(c) DNA and ribosomes
(d) RNA and ribosomes.
Answer:
(c) DNA and ribosomes

Class 9 Science Chapter 5 Question Answer In English Question 20.
Stroma is present in __________.
(a) Mitochondria
(b) Leucoplast
(c) Endoplasmic reticulum
(d) Lysosomes.
Answer:
(b) Leucoplast

Question 21.
Which of the following cell organelle is called suicide bag of a cell?
(a) Mitochondria
(b) Lysosome
(c) Plastids
(d) Golgi apparatus.
Answer:
(b) Lysosome

Cell Notes Class 9 Question 22.
Which of the following molecules is known as the energy currency of the cell?
(a) RNA
(b) DNA
(c) ATP
(d) Aminoacid.
Answer:
(c) ATP

Question 23.
ATP stands for __________.
(a) Adenosine triphosphate
(b) Amino triphosphate
(c) Amino triglycerophosphate
(d) Adeninetri – phosphoglyceride.
Answer:
(a) Adenosine triphosphate

Mitochondria Diagram Class 9 Question 24.
Which cell organelle is involved in the formation of lysosomes?
(a) Mitochondria
(b) Golgi apparatus
(c) Plastids
(d) Endoplasmic reticulum.
Answer:
(d) Endoplasmic reticulum.

Question 25.
Which cell organelle plays a crucial role in detoxifying many poisons and drugs?
(a) RER
(b) SER
(c) RNA
(d) DNA.
Answer:
(b) SER

The Fundamental Unit of Life Very Short Answer Type Questions

Question 1.
Which is the longest flowering plant cell?
Answer:
The fibre.

Plant Cell Class 9 Question 2.
Which is the longest animal cell?
Answer:
The nerve cell.

Question 3.
Name any two single celled animals.
Answer:
Amoeba, paramoecium and euglena etc.

Question 4.
Which is the largest animal cell?
Answer:
An ostrich egg.

Question 5.
Which is the smallest animal cell?
Answer:
Pleuro Pneumonia like organisms (PPLO).

Diagram Of Golgi Apparatus Class 9 Question 6.
What are cell organelles?
Answer:
The functional units of cell.

Question 7.
Name the process involved in the movement of O₂ in and out of the cell.
Answer:
Diffusion.

Question 8.
Which is the outer most covering of the plant cell?
Answer:
Cell wall.

Question 9.
Name is the outer most covering of an animal cell.
Answer:
Plasma membrane.

Plant Cell Diagram Class 9 Ncert Question 10.
Which is the largest plant cell?
Answer:
Acetabalaria.

Question 11.
Is plasma membrane living or dead?
Answer:
Living.

Question 12.
Name the process involved in the movement of water from outside into the cell.
Answer:
Osmosis.

Science Diagram Class 9 Question 13.
By which process CO₂ and water (H₂O) move in and out of the cell?
Answer:
CO₂ by diffusion and water (H₂O) by osmosis.

Question 14.
Is cell wall living or dead?
Answer:
Dead.

Question 15.
Name the control centre of the cell.
Answer:
Nucleus is the control centre of the cell.

Question 16.
What is the function of ribosomes?
Answer:
Ribosomes are involved in protein synthesis.

Mitochondria Diagram For Class 9 Question 17.
Name the site of photosynthesis.
Answer:
Chloroplast.

Question 18.
Which organelles are called

1. powerhouse of cell
2. suicide bags
3. Kitchen of the cell?

Answer:

1. Mitochondria
2. Lysosome and
3. Chloroplasts.

Question 19.
Which organelle makes endoplasmic reticulum rough?
Answer:
Ribosomes.

Question 20.
Name the scientists who formed primitive microscope in the 17th century.
Answer:
Robert Hooke (1665) formed primitive microscope.

Question 21.
Name the plastid which stores starch, oils and protein granules.
Answer:
Leucoplast.

Question 22.
Name the main substance which nucleolus consists of.
Answer:
Nucleolus, found in nucleus, mainly consists of RNA (Ribonucleic acids).

Question 23.
What are genes?
Answer:
Genes are segments of DNA present linearly on chromosomes. They are the functional units of chromosomes.

Question 24.
What is cytoplasm?
Answer:
The cytoplasm is the fluid content inside the plasma membrane. It contains many specialised cell organelles.

Question 25.
Where are centrioles found?
Answer:
Centrioles are found in centrosome of an animal cell only.

The Fundamental Unit of Life Short Answer Type Questions

Question 1.
Define diffusion.
Answer:
The spontaneous movement of a substance from a region of its high concentration to Che region where its concentration is low, is called diffusion.

Question 2.
Observe the following diagram and answer the questions given below:

1. What is the name of this instrument?
2. What are the names of the parts marked as 1, 2 and 3 ?
3. There are two objectives, one is marked as 45X and second one as 10X. Under which objective you would first view the onion peel cells ?

Answer:

1. Compound microscope.
2. • Eye piece
   • Fine adjustment screw
   • Coarse adjustment screw.
3. Under 10X objective (low power objective).

Question 3.
What is the function of cell wall and plasma membrane?
Answer:

1. Cell wall: Gives rigidity, shape and protection to plant cell.
2. Cell membrane: Allows only selected materials to move in and out of the cell.

Question 4.
Define osmosis.
Answer:
The passage of water from a region of higher concentration to a region of lower concentration through a semipermeable membrane is called osmosis.

Question 5.
What is chromatin material?
Answer:
Inside the nucleus a tangled mass of thread – like structure is present which is called chromatin material. It mainly consists of DNA (deoxyribonucleic acid) and protein.

Question 6.
What is the function of centrosome? Name the structure found in plant cell to do the same function.
Answer:
Centrosome helps in cell division in animal cells. The polar caps perform the same function in a plant cell.

Question 7.
What is the function of Mitochondria?
Answer:
In mitochondria, digested food is broken down to release energy. This released energy is stored in the form of ATP.

Question 8.
What is plasmolysis?
Answer:
When a living plant cell loses water through osmosis, there is shrinkage or contraction of the protoplasm away from the cell wall. This phenomenon is called plasmolysis.

Question 9.
What is endocytosis? Give one example.
Answer:
The process in which a cell due to flexible cell membrane engulfs in food and other material from its external environment is known as endocytosis. Amoeba acquires its food through this process.

Question 10.
Who discovered cells in living organisms? Give an example of unicellular organism.
Answer:
Leeuwenhoek (1674) observed free living cells in pond water for the first time. Example of unicellular organisms: Amoeba, chlamydomonas, paramoecium, bacteria etc.

Question 11.
What is the location and main component of cell wall?
Answer:
Cell wall is a rigid structure found outside the plasma membrane of the plant cells. Cellulose (a fibrous polysaccharide) is the main constituent of the cell wall.

Question 12.
What are chromosomes?
Answer:
Chromosomes are rod – shaped structures found in the nucleus of the cell. Chromosome contain information for inheritance of features from parents to next generation in the form of DNA molecule.

Question 13.
What is a eukaryotic cell?
Answer:
A cell that has a definite nucleus with nuclear membrane and also contains membrane bound organelles in its cytoplasm is called eukaryotic cell.

Question 14.
What is nucleoid?
Answer:
In prokaryotes, due to absence of nuclear membrane, nuclear region of the cell is poorly defined. This kind of undefined nuclear region containing only nucleic acid is called a nucleoid.

Question 15.
Give two examples of prokaryotic organisms.
Answer:
Bacteria and cyanobacteria (blue green algae)
e.g.,

• Nostoc are prokaryotic unicellular organisms.

Question 16.
What is a prokaryotic cell?
Answer:
A cell which lacks nuclear membrane, contains a single chromosome and also lacks membrane bound organelles is called prokaryotic cell.

Question 17.
What is cytoplasm?
Answer:
The part of protoplasm which remains after excluding nucleus, is known as cytoplasm. Cytoplasm contains an aggregate molecules of various chemicals and cell organelles. Most of the biochemical reactions such as protein synthesis, release of energy etc. take place in cytoplasm or organelles present in the cytoplasm.

Question 18.
Write three main points of cell theory as expressed by Schleiden, Schwann and Virchow.
Answer:
Cell Theory: Schleiden and Schwann proposed that:

1. all the plants and animals are composed of cells and that
2. the cell is the basic unit of life. Later on, Virchow added that
3. all cells arise from pre – existing cells.

Question 19.
List four major functions of a cell.
Answer:
Functions of Cell:

1. Synthesis of substances.
2. Digestion (lysis) of substances.
3. Generation of energy for vital functions.
4. Secretion.

Question 20.
What is the function of an endoplasmic reticulum?
Answer:
Endoplasmic reticulum helps in the transport of substances between various regions of the cytoplasm or between the cytoplasm and the nucleus. Endoplasmic reticulum also functions as a cytoplasmic framework providing a surface for some of the biochemical activities of the cell. In vertebrates, SER present in the liver cells also plays an important role in detoxifying poisons and drugs.

Question 21.
What is lacking in a virus which makes it dependent on a living cell to multiply?
Answer:
Viruses lack selectively .permeable plasma membrane and cell organelles. Thus, they lack a basic structural organisation to perform various life processes effectively and in their own way. After entering a living cell, a virus utilises its own genetic material and machinery of host cell to multiply.

Question 22.
Write one function of each:

1. rough endoplasmic reticulum and
2. smooth endoplasmic reticulum.

Answer:

1. Rough endoplasmic reticulum (RER) has ribosomes on its surface for synthesizing proteins.
2. Smooth endoplasmic reticulum (SER (without ribosomes) secrete lipids or fats. Some of the proteins and lipids help in building cell membranes.

Question 23.
What is membrane biogenesis?
Answer:
The SER (Smooth Endoplasmic Reticulum) synthesises certain proteins and lipids which help in building the plasma membrane. This building of plasma membrane by SER products is called membrane biogenesis.

Question 24.
What are the leucoplasts and their functions?
Answer:

1. Leucoplasts: These are colourless plastids which do not contain any pigment.
2. Function: Leucoplasts are involved in formation and storage of starch, oil drops and proteins granules.
   • For example: Amyloplasts found in potato tuber store starch.

Question 25.
Why mitochondria are referred to as strange organelles?
Answer:
Mitochondria have their own DNA and ribosomes. Therefore, they can make some of their own proteins. So, sometimes they are referred as ‘strange organelles’. On the other hand, they help in cellular respiration and formation of ATP.

Question 26.
Expand the term ATP. What is use of ATP?
Answer:
ATP stands for Adenosine triphosphate.
Use:
ATP molecules are rich in chemical energy. The body cells use the energy stored in ATP for synthesis of new chemical compounds, their transport and for mechanical work done by cells / tissues.

Question 27.
What are plastids?
Answer:
Plastids are double membrane bounded structures which are found in plant cells only. Inside the plastid, a fluid material called stroma is present. In stroma, numerous membrane layers remain scattered. Plastids like mitochondria do not have cristae. They also have extra – nuclear DNA. There are two types of plastids: Leucoplasts and Chromoplasts.

Question 28.
What type of enzymes are present in the lysosomes? What is their functions? Which organelles membrane manufacture these enzymes?
Answer:

1. Lysosomes contain powerful digestive enzymes capable of breaking down all organic material.
2. Lysosomes help to keep the cell clean by digesting worn out cell organelles and foreign material such as bacteria or food.
3. RER (Rough Endoplasmic Reticulum) makes the digestive enzymes present in the lysosomes.

Question 29.
What are the three functional regions of cell?
Answer:
The functional regions of all cells are:

1. The plasma membrane
2. The nucleus and
3. The cytoplasm.

Question 30.
What are vacuoles?
Answer:
Vacuoles are fluid – filled structures surrounded by a membrane known as tonoplast. The fluid in the vacuoles is called cell sap. In animal cells either they are absent or are very small in size but can be many. In most plant cells, vacuole is big and centrally placed which provides turgidity and rigidity to cells.

The Fundamental Unit of Life Long Answer Type Questions

Question 1.
What will happen if an animal or plant cell is put into a solution of sugar and water?
Answer:
The following three things could happen:

1. If the solution surrounding the cell is very dilute than cytoplasm, the water will move into the cell, i.e., the cell will gain water.
2. If the solution has exactly similar water concentration as that of cytoplasm of cell, there will be no net movement of water across the cell membrane, i.e., no gain or loss of water from the cell.
3. If the medium (solution) has a lower concentration of water than the cell i.e., the solution is concentrated, the cell will lose water by osmosis.

Question 2.
What will happen when

1. an egg without shell is placed in concentrated salt solution for 5 minutes?
2. an egg without shell is placed in pure I distilled water for 5 minutes? Give reason in brief.

Answer:

1. The egg loses water and shrinks.
   • Reason: The egg membrane acts as semipermeable membrane. The concentrated salt solution present outside the egg is hypertonic so the egg loses water by osmosis.
2. The egg gains water and swells.
   • Reason: The pure water is hypotonic (dilute) than egg cells, so water from the outside moves into the egg cells by osmosis. The gain of water results in swelling of egg.

Question 3.
What is the significance of cell wall in plant cell?
Answer:
It performs the following functions in the plant cell:

1. It gives a definite shape to the cell.
2. It provides rigidity and strength to the cell.
3. It protects the plasma membrane and inner cell organelles by bounding the cell from outside.
4. It allows cells of plants, fungi and bacteria to withstand very dilute external media without bursting.
5. It helps in transport of the material in and out of the cell.
6. Cell wall maintains the shape of the cell when cytoplasm shrinks away from cell wall during plasmolysis.
7. It prevents desiccation of cells.

Question 4.
Explain the structure of nucleus.
Answer:
Nucleus is the control centre of the cell. It is covered by a double layered envelope called nuclear membrane. The nuclear membrane has pores which allow the transfer of material from inside the nucleus to cytoplasm. Inside the nuclear membrane an entangled mass of thread – like structures is present. This is called chromatin material (see Fig. given).

The chromatin material mainly formed of DNA (deoxyribonucleic acid) and proteins. When a cell starts to divide, chromatin material condenses into rod – shaped structures called chromosomes. The chromosomes contain stretches of DNA which are called genes.

Question 5.
Name the cell organelle which are known as:

1. Control centre of the cell.
2. Demolition squads / suicidal bags of the cell.
3. Export firms.
4. Power house of the cell.
5. Kitchen of the cell.
6. Internal transport system.

Answer:

1. Nucleus.
2. Lysosomes.
3. Golgi bodies.
4. Mitochondria.
5. Chloroplast.
6. Endoplasmic reticulum.

Question 6.
What is endoplasmic reticulum? Write its main functions.
Answer:
Endoplasmic reticulum. It is a membranous network. enclosing a fluid – filled lumen. Its main functions are:

1. Synthesis of proteins (Rough ER).
2. Synthesis of lipids and other metabolic products and their secretion (SER).
3. Helps in formation of cell plate and nuclear membrane during cell division.
4. ER also produces substance for new cellular parts (especially cell membrane).
5. ER provides internal support (Mechanical support) to the colloidal cytoplasmic matrix of the cell.
   

Question 7.
(a) What would happen to the life of a cell if there was no Golgi apparatus?
(b) Which cell organelle detoxifies poisons and drugs in the liver of vertebrates?
Answer:
(a) Golgi apparatus is connected to ER. It collects simpler molecules from ER and convert them into more complex molecules. These re then packaged in small vesicles and are stored or transported inside or outside the cell. If Golgi apparatus is not present in the cell, all the above processes of modification, storage and transportation will not be possible. Golgi apparatus is also involved in the formation of lysosomes. If there was no Golgi apparatus in the cell, lysosomes would not hr formed and hence foreign materials like bacteria could easily invade and destroy the cell.

(b) Try yourself.

Question 8.
Briefly explain the structure of Golgi apparatus.
Answer:
Golgi apparatus is also known as Golgi body or Golgi complex. It consists of a set of smooth, flattened membranous sac like structures called cisternae. These are placed one above the other (staked) in parallel rows.

On the outer edge of cisternae (stacks of membrane), tubules and vesicles (small sacs) are present. In plant cells, the Golgi apparatus consists of many unconnected units called dictyosomes.

Question 9.
What are the functions of Golgi apparatus?
Answer:
Functions of Golgi apparatus:

1. Golgi apparatus is the secretory organelle of the cell. It packages and dispatches the material synthesized in the cell to intracellular (plasma membrane and lysosomes) and extracellular targets.
2. Golgi complex is also involved in the formation of lysosomes.
3. Golgi apparatus is also involved in the synthesis of many substances such as polysaccharides, glycoproteins etc.

Question 10.
What are lysosomes?
Answer:
Lysosomes: These are membrane bound vesicles which bud off from Golgi apparatus. They contain powerful enzymes capable of digesting or breaking down all organic material. The enzymes present in lysosome are synthesised on RER.

The membrane that bound lysosomes does not allow the enclosed enzymes to pass freely into the cell cytoplasm. Thus, protects the cell from autolysis (self – dissolution or breakdown). They are absent in RBCs and a few plant cells
for example:

• yeast
• fungi and
• green algae.

Question 11.
Describe the role played by the Lysosomes. Why are these termed as suicidal bags? How do they perform their function?
Answer:
Functions of Lysosomes:

1. Extracellular digestion: Sometimes lysosome enzymes are released outside the cell to break down extracellular material.
2. Digestion of foreign material: Lysosomes also destroy any foreign material which enters inside the cell such as bacteria.
3. Cellular digestion: In damaged cells, ageing cells or dead cells lysosomes get ruptured and enzymes are released. These enzymes digest their own cell.

Lysosomes contain about 40 hydrolytic enzymes. When the cell gets damaged, lysosomes burst and their enzymes digest their own cell. So, lysosomes are called ‘suicide bags’. Foreign materials entering the cell, such as bacteria or food, as well as old organelles end up in the lysosomes, which break them up into small pieces.

Question 12.
Describe the structure of mitochondria.
Answer:
Mitochondria are rod – shaped cell organelle found in the cytoplasm. Each mitochondrion (singular of mitochondria) is a double membrane bound structure. The outer membrane of mitochondrion is smooth. But the inner membrane of the mitochondrion is folded inwardly, into the matrix of mitochondrion forming finger – like projections.

The inward finger – like projections of inner membrane are called cristae. Cristae greatly increase the surface area of the inner membrane. Mitochondria contain extra – nuclear DNA.

Question 13.
What are chromoplasts and leucoplasts? Give an example of chromoplasts which has green pigment.
Answer:
Chromoplasts are coloured plastids i.e., they may contain pigment of different colours. For example, Blue green chromoplasts found in blue green algae.

The most important chromoplasts are chloroplasts which contain green pigment, known as chlorophyll. Chloroplasts are very important for photosynthesis. Other non – green chromoplasts are responsible for characteristic colour of flowers and fruits.

1. Leucoplast: These are colourless plastids which do not contain any pigment.
2. Function: Leucoplasts are involved in formation and storage of starch, oil drops and proteins granules.
   • For example: Amyloplasts found in potato tuber store starch.

Question 14.
(a) Name the organelle which provides turgidity and rigidity to the plant cell. Name any two substances which are present in it.
(b) How are they useful in unicellular organisms?
Answer:
(a) Plant cells have big vacuoles full of cell sap that provide them turgidity and rigidity. Plant vacuoles store amino acids, sugars, various organic acids and some proteins.
(b) In unicellular organisms they may serve the following purposes:

1. Forming food vacuoles: In single celled organisms like Amoeba, the food vacuole contains the food items that the Amoeba has ingested. The food items are digested by the enzymes later on.
2. Removal of excess water and wastes: In some unicellular organisms, specialised vacuoles play important roles in expelling excess water and some wastes from the cell.

The Fundamental Unit of Life Higher Order Thinking Skills (HOTS)

Question 1.
When we take bath for long time, the skin of our fingers shrink. Give reason.
Answer:
The soaps used are hypertonic as compared to the osmotic concentration of our skin. Exosmosis in skin cells occurs resulting in shrinking of skin of fingers.

Question 2.
‘The Golgi apparatus is also called the secretary organelle of the cell’. Why?
Answer:
Golgi apparatus is called so as its main function is secretion. The secretary proteins and lipids are packed and released on the surface by Golgi via exocytosis.

The Fundamental Unit of Life Value Based Questions

Question 1.
Manjeet is a five year old boy who joined the swimming classes. After the first class, he was worried when he saw his wrinkled fingers. He asked his elder brother about the wrinkling and shrinking of his fingers. His brother explained Manjeet, why it was so.

1. Why did the fingers wrinkle after swimming?
2. What caused the shrinking / wrinkling of fingers?
3. What value of Manjeet is seen in the above case?

Answer:

1. Fingers wrinkled because the cells of the skin lost some water.
2. This happened because of the difference in the concentration of water in the skin cells and swimming pool’s water.
3. Manjeet showed the value of aware person and a good learner who clarifies the doubts.

Question 2.
Kalyan’s mother wanted to use some eggs for incubation. Kalyan helped his mother in separating rotten and spoilt eggs from the good ones. He took a bucket of water to separate them.

1. How can one separate the rotten eggs from the good ones using water?
2. What is the shell of an egg made up of?
3. What value of Kalyan is seen in this act?

Answer:

1. We can separate the rotten eggs by dipping them in water. The eggs that will float in water are rotten eggs and those that sinks are good one.
2. The shell of an egg is made up of calcium carbonate.
3. Kalyan showed the value of being helpful and responsible.

MP Board Class 9th Science Solutions

MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.2

Class 8 Maths 8.2 In Hindi प्रश्न 1.
एक व्यक्ति के वेतन में 10% वृद्धि होती है। यदि उसका नया वेतन ₹ 1,54,000 है तो उसका मूल वेतन ज्ञात कीजिए।
हल:
माना कि व्यक्ति का मूल वेतन र x है।
वेतन में वृद्धि = वेतन का 10%
= ₹ x × \(\frac{10}{100}\) = ₹ \(\frac{x}{10}\)
अब नया वेतन = ₹ x + ₹ \(\frac{x}{10}\) = ₹ \(\frac{11}{10}\) x
परन्तु नया वेतन = ₹ 1,54,000
अब, प्रश्नानुसार,
₹ \(\frac{11}{10}\) x = ₹ 1,54,000
x = ₹ \(\frac{154000×10}{11}\)
= ₹ 1,40,000
अतः उसका मूल वेतन ₹ 1,40,000 था।

MP Board Class 8 Maths Chapter 8 Exercise 8.2 प्रश्न 2.
रविवार को 845 व्यक्ति चिड़ियाघर गए। सोमवार को केवल 169 व्यक्ति गए। चिड़ियाघर की सैर करने वाले व्यक्तियों की संख्या में सोमवार को कितने प्रतिशत कमी हुई?
हल:
रविवार को चिड़ियाघर गए व्यक्ति = 845
सोमवार को चिड़ियाघर गए व्यक्ति = 169
∴ चिड़ियाघर की सैर करने गए व्यक्तियों की संख्या में कमी
= 845 – 169
= 676
∴ व्यक्तियों की संख्या में प्रतिशत कमी

=80%

8.2 Class 8 In Hindi प्रश्न 3.
एक दुकानदार ₹ 24,00 में 80 वस्तुएँ खरीदता है और उन्हें 16% लाभ पर बेचता है। एक वस्तु का विक्रय मूल्य ज्ञात कीजिए।
हल:
80 वस्तुओं का क्रय मूल्य = ₹ 24,00
लाभ = 16%
∴ 80 वस्तुओं का विक्रय मूल्य

= ₹ 2,784
अतः एक वस्तु का विक्रय मूल्य = \(\frac{2,784}{80}\)
= ₹ 34.80

Class 8 Maths Ex 8.2 Solutions In Hindi प्रश्न 4.
एक वस्तु का मूल्य ₹ 15,500 था। ₹ 450 इसकी मरम्मत पर खर्च किए गए थे। यदि उसे 15% लाभ पर बेचा जाता है तो उसका विक्रय मूल्य ज्ञात कीजिए।
हल:
एक वस्तु का क्रय मूल्य = ₹ 15,500
ऊपरी खर्चा (मरम्मत) = ₹ 450
वस्तु का कुल क्रय मूल्य = ₹ 15,500 + ₹ 450
= ₹ 15,950
लाभ% = 15%
कुल लाभ = ₹ 15,950 का 15%
= ₹ \(\frac{15}{100}\) x 15,950
= ₹ 2,392.50
∴ विक्रय मूल्य = ₹ 15950 + ₹ 2392.50
= ₹ 18,342.50

Class 8 Math Ex 8.2 In Hindi प्रश्न 5.
एक VCR और TV में से प्रत्येक को ₹ 8,000 में खरीदा गया। दुकानदार को VCR पर 4% हानि और TV पर 8% लाभ हुआ। इस पूरे लेन-देन में लाभ अथवा हानि प्रतिशत ज्ञात कीजिए।
हल:
VCR का क्रय मूल्य = ₹ 8,000 तथा हानि = 4%

= ₹ 96 x 80 = ₹ 7,680
TV का क्रय मूल्य = ₹ 8,000 तथा लाभ = 8%

= ₹ 108 x 80 = ₹ 8,640
कुल विक्रय मूल्य = ₹ 7,680 + ₹ 8,640
= ₹ 16,320
कुल क्रय मूल्य = ₹ 8,000 + ₹ 8,000 = ₹ 16,000
∴ विक्रय मूल्य > क्रय मूल्य
लाभ = विक्रय मूल्य – क्रय मूल्य
= ₹ 16320 – ₹ 16000 = ₹ 320

=2%

Prashnavali 8.2 Class 8 प्रश्न 6.
सेल के दौरान एक दुकान सभी वस्तुओं के अंकित मूल्य पर 10% बट्टा देती है। ₹ 1,450 अंकित मूल्य वाला एक जीन्स और दो कमीजें, जिनमें से प्रत्येक का अंकित मूल्य ₹ 850 है, को खरीदने के लिए किसी ग्राहक को कितना भुगतान करना पड़ेगा?
हल:
जीन्स पर अंकित मूल्य = ₹ 1,450
जीन्स पर बट्टा = ₹ 1450 का 10%
₹ \(\frac{1450×10}{100}\) = ₹ 145
∴ जीन्स के लिए किया गया भुगतान
= ₹ 1,450 – ₹ 145 = ₹ 1,305
दो कमीजों पर अंकित मूल्य = ₹ 850 x 2 = ₹ 1,700
कमीजों पर बट्टा = ₹ 1,700 का 10%
₹ \(\frac{1700×10}{100}\) = ₹ 170
कमीजों के लिए किया गया भुगतान = ₹ (1700 – 170)
= ₹ 1,530
ग्राहक द्वारा किया गया कुल भुगतान = ₹ 1305 + ₹ 1530
= ₹ 2,835
अतः ग्राहक को ₹ 2,835 का भुगतान करना पड़ेगा।

Class 8 Math Chapter 8.2 In Hindi प्रश्न 7.
एक दूध वाले ने अपनी दो भैंसों को ₹ 20,000 प्रति भैंस की दर से बेचा। एक भैंस पर उसे 5% लाभ हुआ और दूसरी भैंस पर उसे 10% हानि हुई। इस सौदे में उसका कुल लाभ या हानि ज्ञात कीजिए। (संकेत : पहले प्रत्येक का क्रय मूल्य ज्ञात कीजिए।)
हल:
भैंस का विक्रय मूल्य= ₹ 20,000 तथा लाभ = 5%

दूसरी भैंस का विक्रय मूल्य = ₹ 20,000 तथा हानि = 10%

दोनों भैंसों का कुल क्रय मूल्य

= ₹ 41,269.84
कुल विक्रय मूल्य = ₹ (20000 + 20000)
= ₹ 40,000
∵ विक्रय मूल्य < क्रय मूल्य हानि होगी
हानि = क्रय मूल्य – विक्रय मूल्य
= ₹ 41269.84 – 40000
= ₹ 1,269.84
अतः इस सौदे में दूध वाले को ₹ 1,269.84 हानि हुई।

8th Class Math 8.2 In Hindi प्रश्न 8.
एक टेलीविजन का मूल्य ₹ 13,000 है। इस पर 12% की दर से बिक्री कर वसूला जाता है। यदि विनोद इस टेलीविजन को खरीदता है तो उसके द्वारा भुगतान की जाने वाली राशि ज्ञात कीजिए।
हल:
टेलीविजन का मूल्य = ₹ 13000, बिक्री कर की दर
= 12%
∴ बिक्री कर = ₹ 13000 का 12%
= ₹ \(\frac{12}{100}\) x 13000
= ₹ 1560
∴ विनोद द्वारा भुगतान की जाने वाली राशि
= ₹ 13,000 + ₹ 1,560
= ₹ 1,4560
अतः विनोद द्वारा भुगतान की जाने वाली राशि = ₹ 14,560

MP Board Class 8 Maths Solutions English Medium प्रश्न 9.
अरुण एक जोड़ी स्केट्स (पहियेदार जूते) किसी सेल से खरीदकर लाया जिस पर बट्टे की दर 20% थी। यदि उसके द्वारा भुगतान की गई राशि ₹ 1,600 है तो अंकित मूल्य ज्ञात कीजिए।
हल:
माना कि स्केट्स का अंकित मूल्य = ₹ x है। बट्टे की दर = 20%
बट्टा = ₹ x का 20%
= ₹ \(\frac{x×20}{100}\) = ₹ \(\frac{x}{5}\)
विक्रय मूल्य = अंकित मूल्य – बट्टा
∴ ₹ 1,600 = ₹ x – ₹ \(\frac{x}{5}\)
या 1,600 = \(\frac{5x-x}{5}\)
या \(\frac{4x}{5}\) = 1600
या x = ₹ \(\frac{1600×5}{4}\)
= ₹ 2,000
अतः स्केट्स का अंकित मूल्य ₹ 2,000 है।

Class 8 Math 8.2 Solution In Hindi प्रश्न 10.
मैंने एक हेयर ड्रायर 8% वैट सहित ₹ 5,400 में खरीदा। वैट को जोड़ने से पहले उसका मूल्य ज्ञात कीजिए।
हल:
माना कि वैट को जोड़ने से पहले हेयर ड्रायर का मूल्य = ₹ x है।
8% से ₹ x पर वैट = ₹ \(\frac{8x}{100}\) × x = ₹ \(\frac{8x}{100}\)
वैट जोड़ने के बाद हेयर ड्रायर का मूल्य

= ₹ 5,000
अतः वैट को जोड़ने से पहले हेयर ड्रायर का मूल्य ₹ 5,000 था।

पाठ्य-पुस्तक पृष्ठ संख्या # 135

प्रयास कीजिए (क्रमांक 8.6)

Class 8 Maths Chapter 8.2 Hindi Medium प्रश्न 1.
5% वार्षिक दर से ₹15,000 का 2 वर्ष के अन्त में ब्याज और भुगतान की जाने वाली कुल राशि ज्ञात कीजिए।
हल:
मूलधन = ₹ 15,000
समय = 2 वर्ष
दर = 5% वार्षिक

मिश्रधन = मूलधन + ब्याज
= ₹ 15,000 + ₹ 1,500
= ₹ 16,500
अत: 2 वर्ष के अन्त में ब्याज = ₹ 1,500 तथा भुगतान की गई कुल राशि = ₹ 16,500

पाठ्य-पुस्तक पृष्ठ संख्या # 138

प्रयास कीजिए (क्रमांक 8.7)

Class 8 Maths 8.2 Hindi Medium प्रश्न 1.
₹ 8,000 का 2 वर्ष के लिए 5% वार्षिक दर से चक्रवृद्धि ब्याज ज्ञात कीजिए यदि ब्याज वार्षिक संयोजित होता है।
हल:
मूलधन (P) = ₹ 8,000
समय (x) = 2 वर्ष
दर (R) = 5%

= ₹ 8,820
चक्रवृद्धि ब्याज = मिश्रधन (A) – मूलधन (P)
= ₹ 8,820 – ₹ 8,000 = ₹ 820
चक्रवृद्धि ब्याज = ₹ 820

पाठ्य-पुस्तक पृष्ठ संख्या # 139

प्रयास कीजिए (क्रमांक 8.8)

Class 8 Maths Chapter 8 Exercise 8.2 In Hindi प्रश्न 1.
निम्नलिखित में ब्याज संयोजन के लिए समय अवधि और दर ज्ञात कीजिए –

1. 1\(\frac{1}{2}\) वर्ष के लिए 8% वार्षिक दर पर उधार ली गई एक राशि पर ब्याज अर्धवार्षिक संयोजित किया जाता है।
2. 2 वर्ष के लिए 4% वार्षिक दर पर उधार ली गई एक राशि पर ब्याज अर्धवार्षिक संयोजित किया जाता है।

हल:
1. ब्याज दर = 8%
वार्षिक = \(\frac{8}{2}\) %=4%
प्रति अर्धवार्षिक (∴ दर आधी कर दी जाती है, तथा समय दुगुना हो जाता है।)
समय = 1\(\frac{1}{2}\)
वर्ष = \(\frac{3}{2}\) x 2 = 3 अर्धवर्ष
अतः समय = 3 अर्धवर्ष तथा ब्याज दर = 4% अर्धवार्षिक

2. समय = 2 वर्ष = 2 x 2 = 4 अर्धवर्ष
ब्याजदर = 4%
वार्षिक = \(\frac{4}{2}\)% = 2% प्रति अर्धवार्षिक

सोचिए चर्चा कीजिए और लिखिए (क्रमांक 8.2)

Exercise 8.2 Class 8 In Hindi प्रश्न 1.
एक राशि 16% वार्षिक दर पर 1 वर्ष के लिए उधार ली जाती है। यदि ब्याज प्रत्येक तीन महीने बाद संयोजित किया जाता है, तो एक वर्ष में ब्याज कितनी बार देय होगा?
हल:
दर = 16% वार्षिक, समय = 1 वर्ष
यदि ब्याज तिमाही संयोजित किया जाता है, तो
दर = वार्षिक दर का एक चौथाई तथा समयावधि 1 वर्ष में 4 तिमाही
ब्याज दर = 16% वार्षिक = \(\frac{14}{4}\)%
= 4% प्रति तिमाही
तथा समय = 1 वर्ष = 1 x 4 = 4 तिमाही
अतः 1 वर्ष में ब्याज 4 बार देय होगा।

पाठ्य-पुस्तक पृष्ठ संख्या # 140

प्रयास कीजिए (क्रमांक 8.9)

Ex 8.2 Class 8 Hindi Medium प्रश्न 1.
निम्नलिखित के लिए भुगतान की जाने वाली राशि ज्ञात कीजिए –

1. ₹ 2,400 पर 5% वार्षिक दर से ब्याज वार्षिक संयोजन करते हुए 2 वर्ष के अन्त में।
2. ₹ 1,800 पर. 8% वार्षिक दर से ब्याज तिमाही संयोजन करते हुए 1 वर्ष के अन्त में।

हल:
1. यहाँ, मूलधन = ₹ 2,400, दर = 5%
वार्षिक, समय = 2 वर्ष

= ₹ 2,646
अत: 2 वर्ष के अन्त में भुगतान की जाने वाली राशि
= ₹ 2,646

2. यहाँ, मूलधन = ₹ 1,800
दर =8%
वार्षिक, समय = 1 वर्ष
जब ब्याज तिमाही संयोजित होती है, तब
ब्याज की दर = वार्षिक दर की एक चौथाई = \(\frac{8}{4}\)%
= 2% प्रति तिमाही
समय = 1 वर्ष =4 रूपान्तरण अवधि =4 तिमाही

अतः 1 वर्ष के अन्त में भुगतान की जाने वाली राशि
= ₹ 1,948.38

पाठ्य-पुस्तक पृष्ठ संख्या # 142

प्रयास कीजिए (क्रमांक 8.10)

Class 8 Maths MP Board प्रश्न 1.
₹ 10,500 मूल्य की एक मशीन का 5% की दर से अवमूल्यन होता है। एक वर्ष पश्चात् इसका मूल्य ज्ञात कीजिए।
हल:
यहाँ, मशीन का मूल्य, P = ₹ 10,500
अवमूल्यन दर R = 5%
समय n = 1
वर्ष अवमूल्यन होने पर 1 वर्ष पश्चात् मशीन का मूल्य
= ₹ 10,500 x ( 1 – \(\frac{5}{100}\))¹
= ₹ 10,500 x \(\frac{19}{20}\)
=₹ 9,975
अतः 1 वर्ष पश्चात् मशीन का मूल्य = ₹ 9,975

MP Board Class 8 Maths प्रश्न 2.
एक शहर की वर्तमान जनसंख्या 12 लाख है। यदि वृद्धि की दर 4% है, तो 2 वर्ष पश्चात् शहर की जनसंख्या ज्ञात कीजिए।
हल:
वर्तमान जनसंख्या = 12 लाख = 1200000
वृद्धि दर = 4%
समय = 2 वर्ष

= 1920 x 676
= 1297920
अतः 2 वर्ष बाद जनसंख्या = 1297920

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 7 घन और घनमूल Ex 7.1

 निम्नलिखित में से कौन-सी संख्याएँ पूर्ण घन – नहीं हैं?

1. 216
2. 128
3. 1000
4. 100
5. 46656

हल:
1. 216 = 2 x 2 x 2 x 3 x 3 x 3
स्पष्ट है कि अभाज्य गुणनखण्ड समान गुणनखण्डों के त्रिक हैं और कोई गुणनखण्ड शेष नहीं है।
अत: 216 एक पूर्ण घन है।

2. 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
यहाँ अभाज्य गुणनखण्डों के त्रिक बनाने पर 2 शेष रहता है।
अतः 128 एक पूर्ण घन नहीं है।

3. 1000 = 2 x 2 x 2 x 5 x 5 x 5
स्पष्ट है कि अभाज्य गुणनखण्ड समान गुणनखण्डों के त्रिक हैं और कोई गुणनखण्ड शेष नहीं है।
अतः 1000 एक पूर्ण घन है।

4. 100 = 2 x 2 x 5 x 5
स्पष्ट है कि अभाज्य गुणनखण्ड समान गुणनखण्डों के त्रिक नहीं हैं तथा 2 x 2 x 5 x 5 शेष रहता है।
अतः 100 पूर्ण घन नहीं है।

5. 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
स्पष्ट है कि अभाज्य गुणनखण्ड समान गुणनखण्डों के त्रिक हैं तथा कोई गुणनखण्ड शेष नहीं है।
अतः 46656 एक पूर्ण घन संख्या है।

MP Board Class 8 Maths Chapter 7 प्रश्न 2.
वह सबसे छोटी संख्या ज्ञात कीजिए जिसमें निम्नलिखित संख्याओं को गुणा करने पर पूर्ण घन बन जाए:

1. 243
2. 256
3. 72
4. 675
5. 100.

हल:
1.

243 के अभाज्य गुणनखण्ड करने पर, 243 = 3 x 3 x 3 x 3 x 3
यहाँ अभाज्य गुणनखण्डों के तीन-तीन का समूह बनाने पर 3 के समूह का एक गुणनखण्ड कम है।
अतः 3 से गुणा करने पर संख्या 243 पूर्ण घन बन जाएगी।

2.

256 के अभाज्य गुणनखण्ड करने पर, 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
यहाँ अभाज्य गुणनखण्डों के तीन-तीन का समूह बनाने पर 2 के समूह का एक गुणनखण्ड कम है।
अतः 2 से गुणा करने पर संख्या 256 पूर्ण घन बन जाएगी।

3.

72 के अभाज्य गुणनखण्ड करने पर, 72 = 2 x 2 x 2 x 3 x 3
यहाँ अभाज्य गुणनखण्डों के तीन-तीन का समूह बनाने पर 3 के समूह का एक गुणनखण्ड कम है।
अतः 3 से गुणा करने पर संख्या 72 पूर्ण घन बन जाएगी।

4.

675 के अभाज्य गुणनखण्ड करने पर, 675 = 3 x 3 x 3 x 5 x 5
यहाँ अभाज्य गुणनखण्डों के तीन-तीन के समूह बनाने पर 5 के समूह का एक गुणनखण्ड कम है।
अतः 5 से गुणा करने पर संख्या 675 पूर्ण घन बन जाएगी।

5.

100 के अभाज्य गुणनखण्ड करने पर, 100 = 2 x 2 x 5 x 5
यहाँ अभाज्य गुणनखण्डों के तीन-तीन के समूह बनाने पर 2 के समूह का एक व 5 के समूह का एक गुणनखण्ड कम है।
अत: 2 x 5 = 10 से गुणा करने पर संख्या 100 पूर्ण घन बन जाएगी।

Class 8 Maths 7.1 In Hindi प्रश्न 3.
वह सबसे छोटी संख्या ज्ञात कीजिए जिससे निम्नलिखित संख्याओं को भाग देने पर भागफल एक पूर्ण घन प्राप्त हो जाए

1. 81
2. 128
3. 135
4. 192
5. 704.

हल:
1.

81 = 3 x 3 x 3 x 3
यहाँ अभाज्य गुणनखण्डों की तीन-तीन के समूह (त्रिक) बनाने पर गुणनखण्ड 3 अधिक है।
अत: 3 से भाग देने पर संख्या 81 पूर्ण घन बन जाएगी।

2.

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
यहाँ अभाज्य गुणनखण्डों के तीन-तीन के समूह (त्रिक) बनाने पर गुणनखण्ड 2 अधिक है।
अतः 2 से भाग देने पर संख्या 128 पूर्ण घन बन जाएगी।

3.

135 = 3 x 3 x 3 x 5
यहाँ अभाज्य गुणनखण्डों के तीन-तीन के समूह (त्रिक) बनाने पर गुणनखण्ड 5 अधिक है।
अत: 5 से भाग करने पर संख्या 135 पूर्ण घन बन जाएगी।

4.

192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
यहाँ अभाज्य गुणनखण्डों के तीन-तीन के समूह (त्रिक) बनाने पर गुणनखण्ड 3 अधिक है।
अतः 3 से भाग करने पर संख्या 192 पूर्ण घन बन जाएगी।

5.

704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
यहाँ अभाज्य गुणनखण्डों के तीन-तीन के समूह त्रिक बनाने पर गुणनखण्ड 11 अधिक है।
अतः 11 से भाग देने पर संख्या 704 पूर्ण घन बन जाएगी।

Class 8 Maths Chapter 7 Exercise 7.1 In Hindi प्रश्न 4.
परीक्षित प्लास्टिसिन का एक घनाभ बनाता है जिसकी भुजाएँ 5 cm, 2 cm और 5 cm हैं। एक घन बनाने के लिए ऐसे कितने घनाभों की आवश्यकता होगी?
हल:
घनाभ का आयतन = लम्बाई x चौड़ाई x ऊँचाई
5 x 2 x 5 सेमी³ = 2 x 5 x 5 सेमी³ घन बनाने के लिए आवश्यक घनाभों की संख्या
= 2 x 2 x 5 = 20 घनाभ

पाठ्य-पुस्तक पृष्ठ संख्या # 122

घनमूल

MP Board Class 8 Maths प्रश्न 1.
यदि किसी घन का आयतन 125 cm है, तो उसकी भुजा की लम्बाई क्या होगी?
हल:
घन का आयतन = 125 घन सेमी
घन की भुजा = \(\sqrt[3]{125}\)
\(\sqrt[3]{5 \times 5 \times 5}\)
= 5 सेमी

पाठ्य-पुस्तक पृष्ठ संख्या # 123

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 7.2)

Class 8 Maths Exercise 7.1 Solutions In Hindi प्रश्न 1.
बताइए कि सत्य है या असत्य: किसी पूर्णांक m के लिए, m² < m³ होता है। क्यों?
हल:
1. माना कि यदि m = 2, तब
m² = 2 x 2 = 4 तथा m³ = 2 x 2 x 2 = 8
स्पष्ट है कि 4 < 8 अर्थात् m² < m³

2. यदि m = 3, तब
m² = 3 x 3 = 9 तथा m³ = 3 x 3 x 3 = 27
स्पष्ट है कि 9 < 27 अर्थात् m² < m³

3. यदि m = 4, तब
m² = 4 x 4 = 16 तथा m³ = 4 x 4 x 4 = 64
स्पष्ट है कि 16 < 64 अर्थात् m² < m³

4. यदि m = 5, तब
m² = 5 x 5 = 25 तथा m² = 5 x 5 x 5= 125
स्पष्ट है कि 25 < 125 अर्थात् m² < m³

5. परन्तु यदि m = 1, तब
m² = 1 x 1 = 1 तथा m³ = 1 x 1 x 1 = 1
स्पष्ट है कि m² = m³

6. यदि m = – 1, तब
m² = (-1) x (-1) = 1
तथा m³ = (-1) x (-1) x (-1) = – 1
स्पष्ट है कि, 1 > – 1 अर्थात् m² > m³

7. यदि m = -2, तब
m² = (-2) x (-2) = 4
तथा m³ = (-2) (-2) (-2) = – 8
स्पष्ट है कि 4 > – 8 अर्थात् m² > m³

8. यदि m = – 3, तब
m² = (-3) x (-3)= 9
तथा m³ = (-3) (-3) (-3) = – 27
स्पष्ट है कि 9 > (-27) अर्थात् m² > m³

9. परन्तु यदि m = 0, तब
m² = 0 तथा m³ = 0
∴ m² = m³
अतः हम कह सकते हैं कि ऋणात्मक पूर्णांक m के लिए m² < m³ असत्य है।

MP Board Class 8th Maths Solutions