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MP Board Class 8th Maths Solutions Chapter 3 चतुर्भुजों को समझना Ex 3.3

MP Board Class 8 Maths Chapter 3 Exercise 3.3 प्रश्न 1.
ABCD एक समान्तर चतुर्भुज है। प्रत्येक कथन को परिभाषा या प्रयोग किए गए गुण द्वारा पूरा कीजिए –

1. AD = …..
2. ∠DCB = …..
3. OC = …..
4. m∠DAB + m∠CDA = …..

हल:

1. AD = BC; (सम्मुख भुजाएँ बराबर होती हैं।)
2. ∠DCB = ∠DAB; (सम्मुख कोण बराबर होते हैं।)
3. OC = OA; (विकर्ण परस्पर समद्विभाजित करते हैं।)
4. m∠DAB + m∠CDA = 180°; तिर्यक रेखा के एक ही ओर के अन्तः कोण, क्योंकि \(\overline { AB } \) || \(\overline { CD } \).

MP Board Class 8th Maths Chapter 3 प्रश्न 2.
निम्न समान्तर चतुर्भुजों में अज्ञात x, y, z के मानों को ज्ञात कीजिए –

हल:
(i) समान्तर चतुर्भुज के किन्हीं दो आसन्न कोणों का योग 180° होता है, इसलिए,
x + 100° = 180°
या x = 180° – 100° = 80°
x + y = 180°
या y = 180° – x = 180 – 80°
= 100°
y + z = 180°
या 100° + z = 180°
या z = 180° – 100° = 80°
यहाँ, x = 80°, y= 100° और 2 = 80°

(ii) चूँकि ABCD एक समान्तर चतुर्भुज है, अतः AB || DC ओर AD = BC
अब, AB || DC और तिर्यक रेखा BC इन्हें प्रतिच्छेद करती है।
∴ z = y (∵ एकान्तर कोण बराबर होते हैं।)
और AD || BC और तिर्यक रेखा इन्हें काटती हैं।
∴ z =x (∵ संगत कोण बराबर होते हैं।)
∴ x = y =z
चूँकि समान्तर चतुर्भुज के किन्हीं दो संगत कोणों का योग = 180°
अतः y + 50° = 180°
y = 180° – 50° = 130°
इस प्रकार x = y = z = 130°

(iii) आकृति से स्पष्ट है कि,
x = 90°
∆DOC में, हम प्राप्त करते हैं,
∠DOC + ∠OCD + ∠CDO = 180°
या x + 30° + y = 180°
या 90° + 30° + y = 180°
या 120° + y = 180°
y = 180° – 120° = 60°
अब, समान्तर चतुर्भुज ABCD में, AB || DC और BD इन्हें प्रतिच्छेद करती है।
∴ z = y (एकान्तर कोण है)
z = 60° (∵ y = 60°)
अतः x = 90°, y = 60 और z = 60

(iv) ∵ ∠A + ∠B = 180° ⇒ x + 80° = 180°
या x = 180° – 80° = 100°
∠A + ∠D = 180° ⇒ x + y = 180°
y = 180° – x = 180° – 100° =80
∠D + ∠C = 180°
या y + ∠C = 180°
या 80° + ∠C = 180°
अब ∠C = 180 – 80 = 100°
∠C + z = 180°
100° + z = 180°
z = 180° – 100° = 80°
अतः x = 100°, y = 80° और z = 80°

(v) ∵ समान्तर चतुर्भुज में सम्मुख कोण बराबर माप के होते हैं।
इसलिए, y = 112°
∆ACD से,
x + y + 40° = 180°
x + 112° + 40° = 180°
x = 180° – 152° = 28°
समान्तर चतुर्भुज ABCD में, AB || DC और तिर्यक रेखा AC इन्हें प्रतिच्छेद करती है।
∴ z = x (एकान्तर कोण हैं।)
∴ z = 28°
अतः x = 28°
y= 112
z = 28°

Class 8 Maths Chapter 3 Exercise 3.3 Solutions In Hindi Medium प्रश्न 3.
क्या एक चतुर्भुज ABCD समान्तर चतुर्भुज हो सकता है, यदि –

1. ∠D + ∠B = 180° ?
2. AB = DC = 8cm, AD = 4cm और BC = 4.4cm?
3. ∠A = 70° और ∠C = 65°

हल:

1. ∠D + ∠B = 180° समान्तर चतुर्भुज हो सकता है, परन्तु यह आवश्यक नहीं है।
2. AB = DC = 8 cm, AD = 4 cm, BC = 4.4 cm. यह समान्तर चतुर्भुज नहीं हो सकता है।
   समान्तर चतुर्भुज में सम्मुख भुजाएँ बराबर होती हैं, परन्तु यहाँ AD ≠ BC.
3. ∠A = 70° और ∠C = 65°
   समान्तर चतुर्भुज नहीं हो सकता है। समान्तर चतुर्भुज के सम्मुख कोण बराबर होते हैं, परन्तु यहाँ ∠A ≠∠C.

MP Board Class 8 Maths प्रश्न 4.
एक चतुर्भुज की कच्ची (Rough) आकृति खींचिए जो समान्तर चतुर्भुज न हो परन्तु जिसके दो सम्मुख कोणों की माप बराबर हो।
हल:
चतुर्भुज की कच्ची आकृति जो समान्तर चतुर्भुज नहीं है।

यहाँ, ∠A = ∠C, उदाहरण-पतंग आकृति का चतुर्भुज।

Class 8 Maths 3.3 In Hindi प्रश्न 5.
किसी समान्तर चतुर्भुज के दो आसन्न कोणों का अनुपात 3 : 2 है। समान्तर चतुर्भुज के सभी कोणों की माप ज्ञात कीजिए।
हल:
माना कि दो आसन्न कोण ∠A और ∠B हैं, जो 3 : 2 के अनुपात में हैं।
∴ ∠A = 3x
∠B = 2x
∴ समान्तर चतुर्भुज के आसन्न कोण सम्पूरक होते हैं।
∠A + ∠B = 180°
या 3x + 2x = 180°
या 5x= 180°
या x = \(\frac{180°}{5}\) = 36°
∴ ∠A = 3 x 36° = 108°
और ∠B = 2 x 36° = 72°
∴ समान्तर चतुर्भुज के सम्मुख कोण बराबर होते हैं।
∴ ∠C = ∠A = 108° और ∠D = ∠B = 72°
अत: ∠A = 108°, ∠B = 72°, ∠C = 108°, ∠D = 72°

MP Board Class 8 Maths Book Pdf प्रश्न 6.
किसी समान्तर चतुर्भुज के आसन्न कोणों की माप बराबर है। समान्तर चतुर्भुज के सभी कोणों की माप ज्ञात कीजिए।
हल:
मानाकि समान्तर चतुर्भुज ABCD के दो आसन्न कोण A और B में प्रत्येक की माप x है।
∴ समान्तर चतुर्भुज के आसन्न कोण सम्पूरक होते हैं।
∠A + ∠B = 180°
या x + x = 180°
या 2x = 180°
∴ x = \(\frac{180°}{2}\) = 90°
अर्थात् ∠A= ∠B = 90°
चूँकि समान्तर चतुर्भुज के सम्मुख कोण बराबर होते हैं।
∠C = ∠A = 90°
∠D = ∠B = 90°
अतः ∠A = ∠B = ∠C = ∠D = 90°

MP Board Class 8th Maths Solutions प्रश्न 7.
संलग्न आकृति HOPE एक समान्तर चतुर्भुज है। x,और कोणों की माप ज्ञात कीजिए। ज्ञात करने में प्रयोग किए गए गुणों को बताइए।

हल:
∴ HOPE एक समान्तर चतुर्भुज है।
∴ HE || OP और HO || EP
अब HE || OP और HO तिर्यक रेखा इन्हें काटती है।
∠EHO = ∠POX
(∵ संगत कोण बराबर होते हैं)
∴ 40° + z = 70°
या z = 70° – 40° = 30°
पुनः HE ||OP और तिर्यक रेखा इन्हें काटती है।
∠OPH = ∠EHP (∴ एकान्तर कोण बराबर होते हैं)
∴ y = 40°
∴ समान्तर चतुर्भुज के सम्मुख कोण बराबर होते हैं।
∴ ∠HEP = ∠HOP
या x = 180° – ∠POX
या x = 180° – 70° = 110°
अतः x = 110°, y= 40°, z = 30°

MP Board Class 8 Maths Chapter 3.3 प्रश्न 8.
निम्न आकृतियाँ GUNS और RUNS समान्तर चतुर्भुज है। x तथा y ज्ञात कीजिए (लम्बाई cm में है)।

हल:
(i) ∴ चतुर्भुज GUNS समान्तर चतुर्भुज है। इसलिए इसकी सम्मुख भुजाएँ बराबर होंगी।
∴ 3x = 18
या x = \(\frac{18}{3}\) = 6 cm
ओर 3y – 1 = 26
या 3y = 26 + 1 = 27
या y = \(\frac{27}{3}\) = 9 cm
अतः x = 6 cm और y = 9 cm

(ii) ∴ समान्तर चतुर्भुज में विकर्ण एक-दूसरे को समद्विभाजित करते हैं।
∴ OR = ON
अर्थात् 16 = x + y ……(1)
और OU = OS
अर्थात् y + 7= 20 ……(2)

समीकरण (2) से,
y = 20 – 7 = 13
समीकरण (1) में, y = 13 रखने पर, हम प्राप्त करते हैं।
16 = x + 13
या x = 16 – 13 = 3
अतः x = 3 cm और y = 13 cm

MP Board Class 8 Math Chapter 3 प्रश्न 9.
दी हुई आकृति में RISK तथा CLUE दोनों समान्तर चतुर्भुज हैं, x का मान ज्ञात कीजिए।

हल:
समान्तर चतुर्भुज RISK से,
∠RIS = ∠RKS = 120°
(∴ सम्मुख कोण बराबर होते हैं)
और ∠RIS + ∠SIC = 180° (रेखीय युग्म)
या ∠SIC = 180° – 120° = 60° = ∠OIC
समान्तर चतुर्भुज CLUS से,
CE || LU और ICL इन्हें प्रतिच्छेद करती हैं।
∠OCI = ∠ULC (∴ संगत कोण है)
∠OCI = 70°
अब, ∆OLC से,
∠OIC + ∠OCI + ∠IOC = 180°
या 60° + 70° + ∠IOC = 180°
∠IOC = 180° – 130° = 50°
x = ∠IOC = 50°
(∴ शीर्षाभिमुख कोण हैं।)

MP Board Class 8th Maths Solutions English Medium प्रश्न 10.
बताइए कैसे यह आकृति एक समलम्ब है। इसकी कौन-सी दो भुजाएँ समान्तर हैं?

हल:
∴ ∠MLK + ∠NML = 180°
अर्थात् ये सम्पूरक कोणों का युग्म हैं।
अतः KL || NM
अतः KLMN एक समलम्ब है। .

MP Board 8th Class Maths Book Solutions प्रश्न 11.
संलग्न आकृति में m∠C ज्ञात कीजिए यदि AB || DC है।
हल:
∵ AB || DC और तिर्यक रेखा BC इन्हें प्रतिच्छेद करती है।

∠B + ∠C = 180°
या 120° + ∠C = 180°
∠C = 180° – 120° = 60°
अतः m∠C = 60°

Class 8th Maths Solutions MP Board प्रश्न 12.
संलग्न आकृति में ∠P तथा ∠S की माप ज्ञात कीजिए यदि \(\overline { SP } \) || \(\overline { RQ } \) (यदि आप m∠R ज्ञात करते हैं, तो क्या m∠P को ज्ञात करने की एक से अधिक विधि है?

हल:
∴ \(\overline { SP } \) || \(\overline { RQ } \) और PQ एक तिर्यक रेखा है जो P और Q पर काटती है।
∠P + ∠Q = 180°
(∵ अंत:कोणों का योग = 180°)
या ∠P + 130° = 180°
∠P = 180° – 130° = 50°
पुनः \(\overline { SP } \) || \(\overline { RQ } \) और SR एक तिर्यक रेखा है जो इन्हें S और R पर काटती है।
∴ ∠S + ∠R = 180°
या∠S + 90° = 180°
या ∠S = 180° – 90° = 90°
हाँ, हम m∠P को दूसरी विधि से भी ज्ञात कर सकते हैं।
m∠P + m∠Q + m∠R + m∠S = 360°
या m∠P + 130° + 90° + 90° = 360°
या m∠P + 310° = 360°
या m∠P = 360° – 310° = 50°

पाठ्य-पुस्तक पृष्ठ संख्या # 57

इन्हें कीजिए

Maths MP Board Class 8 प्रश्न 1.
समचतुर्भुज की एक प्रतिलिपि लीजिए। पेपर को मोड़कर जाँच कीजिए कि क्या प्रतिच्छेदी बिन्दु प्रत्येक विकर्ण का मध्य बिन्दु है। आप एक सेट स्क्वेयर के किनारे का उपयोग करके जाँच कर सकते हैं कि वे एक दूसरे को समकोण पर प्रतिच्छेद करते हैं।
हल:
हाँ, प्रतिच्छेदी बिन्दु प्रत्येक विकर्ण का मध्य बिन्दु है। सेट स्क्वे यर का उपयोग करने पर हम पाते हैं कि समचतुर्भुज के विकर्ण समकोण पर प्रतिच्छेद करते हैं।

पाठ्य-पुस्तक पृष्ठ संख्या # 58

एक आयत

Class 8 Maths Chapter 3.3 In Hindi प्रश्न 1.
आयत एक समान्तर चतुर्भुज है जिसके सभीकोण समान माप के होते हैं। इस परिभाषा का पूर्ण अर्थ क्या है? इसकी चर्चा अपने मित्रों के साथ कीजिए। यदि आयत समकोणिक हो तो प्रत्येक कोण की माप क्या होगी?
हल:
आयत एक समान्तर चतुर्भुज है जो समकोणिक होता है। आयत का प्रत्येक कोण समकोण होता है। इसकी सम्मुख भुजाएँ समान लम्बाई की होती हैं तथा इसके विकर्ण एक दूसरे को समद्विभाजित करते हैं।
माना कि आयत के प्रत्येक कोण की माप =x° है।
तब 4x° = 360°
इसलिए x° = \(\frac{360°}{4}\) = 90°
अतः आयत का प्रत्येक कोण 90° का होता है।

पाठ्य-पुस्तक पृष्ठ संख्या # 60

इन्हें कीजिए

MP Board Class 8 Maths Solutions English Medium प्रश्न 1.
एक वर्गाकारशीट PQRS (आकृति : 3.39) लीजिए। दोनों विकर्णों के अनुदिश तह (fold) लगाइए क्या उनके मध्य बिन्दु समान ही हैं?

सेट स्क्वेयर का उपयोग करके जाँच कीजिए, क्या o पर बना कोण 90° का है? यह ऊपर बताए गुणधर्म को सिद्ध करता है।
हल:
हाँ, दोनों विकर्णों के मध्य बिन्दु समान हैं। हाँ, ‘o’ पर बना कोण 90° का है। हाँ, यह वर्ग के गुणधर्म को सिद्ध करता है कि –

1. वर्ग की सभी भुजाएँ समान लम्बाई की होती है।
2. विकर्ण समान लम्बाई के होते हैं तथा एक-दूसरे को समकोण पर समद्विभाजित करते हैं।

MP Board Class 8th Maths Solutions

In this article, we will share MP Board Class 10th Science Solutions Chapter 10 Light Reflection and Refraction Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Science Solutions Chapter 10 Light Reflection and Refraction

MP Board Class 10th Science Chapter 10 Intext Questions

Class 10th Science Chapter 10 Intext Questions Page No. 168

Question 1.
Define the principal focus of a concave mirror.
Answer:
The number of rays parallel to the principal axis are falling on a concave mirror which meat at a point is called principal focus of the concave mirror.

(or)

Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of concave mirror.

MP Board Class 10th Science Chapter 10 Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 2f Here R = 20 cm
20 = 2f
∴ \(f=\frac { 20 }{ 2 } =10\)
∴ Focal length = 10 cm.

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object when object is placed between the pole and principal focus.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror when fitted at rear-view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him. Convex mirrors give a virtual, erect and diminished image of the objects in front of it. So, we prefer a convex mirror as a rear-view mirror in vehicles.

Class 10th Science Chapter 10 Intext Questions Page No. 171

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Radius of curvature, R = 32 cm
Radius of curvature = 2f
\(R=2f=\frac { R }{ 2 } =\frac { 32 }{ 2 } =16\)
∴ Convex mirror focal length is = 16cm

Class 10 Science Chapter 10 Question Answer Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:

Let the height of the object = h₀ = h
Then, height of the image, h₁ = -3h (Image formed is real)
= \(\frac { -3h }{ h } \) = \(\frac { -v }{ u } \)
Object-distance, u = – 10 cm
v = 3 × (- 10)
= – 30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror.

Class 10th Science Chapter 10 Intext Questions Page No. 176

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
Lightray bend towards normal. Because when a ray of light enters from rearer medium to denser medium, it changes its direction in the second medium.

Class 10th Physics Chapter 10 Question Answer Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 ms-1.
Answer:
Refractive index of a medium:
µm = Speed of light in vacuum/Speed of light in the medium
Speed of light in vacuum, c = 3 × 10⁸ ms^-1
Refractive index of glass, µ_g = 1.50
Speed of light in the glass,
v = Speed of light in vacuum / Refractive index of glass
= c/µ_g
= 3 × 10⁸/1.50
= 2 × 10⁸ ms^-1.

Class 10th Science Chapter 10 Question 3.
Find out, from Table the medium having highest optical density. Also find the medium w ith lowest optical density.
Table:

Answer:
Highest optical density = Diamond.
Lowest optical density = Air.

Optical density of a medium is proportional to the refractive index. Hence, medium with highest refractive index will have the highest optical density and vice-versa. It can be observed from table that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Science Class 10 Chapter 10 Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in table.

Answer:
Light travel faster in water when compared to kerosene and turpentine, since the refractive index of water is lower than kerosene and turpentine. The speed of light is inversely proportional to the refractive index.

Activity 10.9 Class 10 Science Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It means Ratio of velocity of light in air and velocity of air in diamond is 2.42.

Class 10th Science Chapter 10 Intext Questions Page No. 184

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of lens whose focal length is 1 metre 1 D = 1 m^-1

Class 10 Subject Science Chapter 10 Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Image of Needle is real and inverted means this is real image it is 2f
Image is at a distance of 50 cm
Hence needle is kept 50 cm in front of convex lens.
Distance of object, u = – 50 cm.
Distance of image v = 50 cm
Focal length f = ?
As per lens formula.

f = 25 cm = 0.25 m
Power of the lens

Power of the lens P = + 4D.

Chapter 10 Class 10 Science Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:
Focal length of concave lens, f = 2 m
Power of lens, P = \(\frac { 1 }{ f } \) = \(\frac { 1 }{ (-2) } \) = -0.5D

MP Board Class 10th Science Chapter 10 NCERT Textbook Exercises

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) A lens allows light to pass through it, but clay does not have that property.

Class 10 Science Ch 10 Question Answer Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens.

Science Class 10 MP Board Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be:
(а) both concave
(b) both convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex, but the lens is concave
Answer:
(а) both concave

Question 5.
No matter how far you stand from a mirror, your image appears erpct. The mirror is likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer:
(d) either plane or convex

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer:
(c)

Activity 10.7 Class 10 Science Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:
Range of the distance of the object = 0 cm to 15 cm.
Nature of the image = virtual, erect and larger than the object.

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirror: Concave mirrors can produce powerful parallel ’ beam of light when the light source is placed at their principal focus. Hence, we can visualize ways easily in little light.

(b) Convex mirror: A convex mirror when fitted at rear view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him.

(c) Concave mirror: They are converging mirrors. This is because it concentrates the parallel rays of sun at principal focus and increase intensity of light falling on it.

Science Chapter 10 Class 10th Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes, the convex lens will form complete image of the object, even if its one half is covered with black paper. Following two cases can better explain it:

Case I : When the upper half of the lens is covered.
In this case, a ray of light coming from the object is being refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Case II : If the lower half of the lens is covered.
In this case, a ray of light coming from the object is being refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Shivlal Guide 10th Class Science Question 10.
An object 5 cm in length.is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1 : 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows;

1. Draw a horizontal line to represent ‘die principal axis of the convex lens.
2. Centre line is shown by DE.
3. Mark two foci F and F’ on two sides of the lens, each at a distance of 2 cm from the lens.
4. Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens.
5. Draw a line AD parallel to principal axis and then allow it to pass straight through the focus (F’) on the right side of the lens.
   
6. Draw a line from A to C (centre of the lens), which goes straight without deviation.
7. Let the two lines starting from A meet at A’.
8. Draw A’B’, perpendicular to the principal axis from A’.
9. Now A’B represents the real, but inverted image of the object AB.
10. Then, measure CB’ and A’B’. It is found that CB’ = 3.3 cm and A’B’ = 0.7 cm.
11. Thus the final position, nature and size of the image A’B’ are
    • (a) Position of image A’B’ = 3.3 cm × 5 = 16.5 cm from the lens on opposite side.
    • (b) Nature of image A’B’: Real and inverted.
    • (c) Height of image A’B’: 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object.

10th Class Science 10 Chapter Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer:
Focal length of concave lens f = – 15 cm
Image distance, v = – 10 cm
According to the lens formula,
\(\frac { 1 }{ v } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)

On solving we get, u = – 30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the above ray diagram.

10th Science Chapter 10 Question Answer Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Focal length of convex mirror,
f = +15 cm
Object distance, u = -10 cm
As per lens formula

Magnification \(=\frac{v}{u}=\frac{-6}{-10}=0.6\)
Virtual image is formed at the distance of 6 cm and it is erect.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
The positive sign means image formed by a plane mirror is virtual and erect. Since, the magnification is I, it means that the size of the image is equal to the size of the object.

Activity 10.12 Class 10 Science Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object distance, u = – 20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × focal length
R = 2f ⇒ f = 15 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
The positive value indicates that the image is formed behind the mirror.

The positive value of image height indicates image is virtual, erect and smaller in size.

Question 15
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Object-distance, u = – 27 cm
Object-height, h = 7 cm
Focal length, f = – 18 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
Putting values, \(\frac { 1 }{ v } \) + \(\frac { 1 }{ (-27) } \) = \(\frac { 1 }{ (-18) } \)
So, v = – 54 cm
The screen should be placed at a distance of 54 cm in front of the given mirror
and \(\frac{h_{2}}{h_{1}}=\frac{-v}{u}\)
h₂ = -14 cm
The negative value of image indicates that the image is inverted.

Mp Board Class 10th Science Solutions Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Given, P = -2D
Power of lens. p = \(\frac { 1 }{ f } \)
and, f = – \(\frac { 1 }{ 2 } \)
\(\frac{h_{2}}{h_{1}}=\frac{-v}{u}\)
= – 0.5 m
A concave lens, because it has a negative value of focal length.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Given, P = 1.5.D
Power of lens, P = \(\frac { 1 }{ f } \)
and, focal length f = \(\frac { 1 }{ 1.5 } \)
= \(\frac { 10 }{ 15 } \) = 0.66 m
A convex lens, because it has a positive focal length. Lens is converging.

MP Board Class 10th Science Chapter 10 Additional Important Questions

MP Board Class 10th Science Chapter 10 Multiple Choice Questions

Question 1.
The image formed by a convex lens is virtual, erect and larger than the object. The position of the object must be:
(a) Between the lens and its focus
(b) At the focus
(c) At twice the focal length
(d) At infinity
Answer:
(a) Between the lens and its focus

Question 2.
A real image formed by a convex lens is always:
(a) On the same side of the lens as the object
(b) Erect
(c) Inverted
(d) Smaller than the object
Answer:
(c) Inverted

Class 10 MP Board Science Question 3.
If an object is moved towards a convex lens, the size of its image:
(a) Decreases
(b) Increases
(c) First decreases and then increases
(d) Remains the same
Ans.
(b) Increases

Question 4.
An object is placed at a distance of 30 cm from a concave mirror of focal length 15 cm. The image will be:
(a) Real and of same size
(b) Real and magnified
(c) Real and diminished
(d) virtual and magnified
Answer:
(a) Real and of same size

Class10 Science Ch 10 Question 5.
A concave mirror always forms real and inverted image except when the object is placed:
(a) At infinity
(b) Between F and C
(c) At F
(d) Between F and pole of the mirror
Answer:
(b) Between F and C

Question 6.
The mirror which has a wide field of view must be:
(a) Concave
(b) Convex
(c) Plane
(d) None of these
Ans.
(b) Convex

Question 7.
The image formed by a concave mirror:
(a) Is always real
(b) Is always virtual
(c) Can be both real and virtual
(d) None of these
Ans.
(c) Can be both real and virtual

MP Board Class 10 Science Solutions Question 8.
An object is placed 20 cm from a convex lens of focal length 10 cm. The image must be:
(a) Real and diminished
(b) Real and of same size
(c) Real and enlarged
(d) Virtual and enlarged
Answer:
(b) Real and of same size

Question 9.
The ratio of the focal length of spherical mirror to its radius of curvature is:
(a) 0.5
(b) 1
(c) 2
(d) 3
Answer:
(a) 0.5

10th Science MP Board Question 10.
A real and inverted image of the same size is formed by a concave mirror when the object is placed:
(a) Between the mirror and its focus.
(b) Between the focus and the centre of curvature.
(c) At the centre of curvature.
(d) Beyond the centre of curvature.
Ans.
(c) At the centre of curvature.

MP Board Class 10th Science Chapter 10 Very Short Answer Type Questions

Question 1.
What is a mirror? Mention the different types of mirrors commonly used.
Answer:
Mirror: A highly polished surface which is smooth enough to reflect a good fraction of light incident on it is called a mirror. The mirror may be a highly polished metal surface or an ordinary glass plate coated with a thin silver layer.

Question 2.
What is the number of images of an object held between two plane parallel mirrors?
Answer:
Infinity.

Question 3.
Does the refractive index for a given pair of media depend on the angle of incidence?
Answer:
No, it is independent of the angle of incidence.

Question 4.
The refractive index of water with respect to air is \(\frac { 4 }{ 3 } \). What is the refractive index of air with respect to water?
Answer:
Refractive index of air with respect to water = \(\frac { 3 }{ 4 } \)

Question 5.
Can absolute refractive index of a medium exceed unity?
Answer:
No, because speed of light is maximum in vacuum.

Question 6.
Why does a ray of light bend when it travels from one medium to another?
Answer:
The bending of light or refraction occurs due to the change in the speed of light as it passes from one medium to another due to change in the density of the medium.

Question 7.
What happens when a ray of light strikes the surface of separation between the two media at right angle?
Ans.
The ray of light passes undeflected from one medium to another.
Here, ∠i = ∠r = 0°

Question 8.
What do you mean by a magnification less than unity?
Answer:
It means that the size of the image is smaller than the size of the object.

Question 9.
Which spherical mirror has

1. a real focus and
2. a virtual focus?

Answer:

1. A concave mirror has a real focus.
2. A convex mirror has a virtual focus.

Question 10.
State the position of the object for which a concave mirror produces virtual magnified image.
Answer:
The object should be placed between F and P of the concave mirror.

MP Board Class 10th Science Chapter 10 Short Answer Type Questions

Question 1.
Name the type of mirror(s) that should be used to obtain:
(i) a magnified and virtual image.
(ii) a diminished and virtual image of an object.
Draw labelled diagrams to show the formation of the required image in each of the above two cases. Which of these mirrors could also form a magnified and real image of the object? State the position of object for which this could happen.
Answer:
(i) Concave mirror.

Fig. 10.3 : Concave mirror with the object between F and P.

(ii) Convex mirror.

Fig. 10.4: Convex mirror with the object between pole and infinity.

Question 2.
Explain the uses of concave and convex mirrors.
Answer:
Uses of concave mirrors:
1. Shaving mirror : A concave mirror is used as a shaving or make-up mirror because it forms erect and enlarged image of the face when it is held closer to the face.

2. As head mirror : E.N.T. specialists use a concave mirror on their forehead. The light from a lamp after reflection from the mirror is focussed into the throat, ear or nose of the patient making the affected part more visible.

3. In ophthalmoscope : It consists of a concave mirror with a small hole at its centre. The doctor looks through the hole from behind the mirror while a beam of light from a lamp reflected from it, is directed into the pupil of patient’s eye which makes the retina visible.

4. In headlights : Concave mirrors are used as reflectors in headlights of motor vehicles railway engines, torch lights etc. The source is placed at the focus of the concave mirror. The light rays after reflection travel over a. large distance as a parallel beam of high intensity.

5. In astronomical telescopes : A concave mirror of large diameter (5 m or more) is used as objective in an astronomical telescope. It collects light from the sky and makes visible even those faint stars which cannot be seen with naked eye.

6. In solar furnaces : Large concave mirrors are used to concentrate sunlight to produce heat in solar furnace.

Uses of convex mirrors : Drivers use convex mirror as a rear-view mirror in automobiles because of the following two reasons:

1. A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
2. A convex mirror has a wider field of view than a plane mirror of the same size as shown in Fig. 10.5.

Fig. 10.5: Field of view of (a) a plane mirror, (b) a convex mirror.

Thus, convex mirrors enable the driver to view much larger traffic behind him than would be possible with a plane mirror. The main disadvantage of a convex mirror is that it does not give the correct distance and the speed of the vehicle approaching from behind.

Question 3.
State the characteristics of the image formed by a convex mirror. What is the value of angle of incidence and angle of reflection when a ray of light retraces its path after reflection from a convex mirror? Illustrate with the help of a ray diagram.
Answer:
Properties of the image formed by a convex mirror:

(a) The image is always virtual and erect.
(b) The image is highly diminished or point sized.
(c) It is always formed between F and P.
(d) As the object is moved towards the pole of a convex mirror, image also moves towards its pole and gradually increases in size till its size becomes almost equal to that of the object.

When array of light retraces its path, ∠i = ∠r = 0°.

Fig. 10.6: A ray directed towards C is reflected back along same path after reflection from a convex mirror.

Question 4.
State the new Cartesian sign convention followed for reflection of light by spherical mirrors.
Answer:
According to this convention:

1. The object is on the left of the mirror. So all the ray diagrams are drawn with the incident light travelling from left to right.
2. All the distances parallel to the principal axis are measured from the pole of the mirror.
3. All distances measured in the direction of incident light are taken as positive.
4. All distances measured in the opposite direction of incident light are taken as negative.
5. Heights measured upwards and perpendicular to the principal axis are taken positive.
6. Heights measured downwards and perpendicular to the principal axis are taken negative.

Fig. 10.7: New Cartesian sign convention for reflection of light by spherical mirrors.

Question 5.
State the type of mirror preferred as
(i) rear view mirror in vehicles
(ii) shaving mirror. Justify your answer giving two reasons in each case.
Answer:
(i) A convex mirror is preferred as a rear-view mirror because:

(a) It always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
(b) It has wider field of view.

(ii) A concave mirror is preferred as a shaving mirror because when it is held closer to the face, it forms:

(a) an enlarged image of the face.
(b) an erect image of the face.

Question 6.
State the laws of refraction of light.
Answer:
Laws of refraction of light: The refraction of light obeys the following two laws:
1st law: The incident ray, the refracted ray and normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

2nd law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.
Mathematically,

The ratio µ21 is called refractive index of the second medium with respect to the first medium. The second law of refraction is also called Snell’s law of refraction.

Question 7.
What is the physical significance of refractive index?
Answer:
The refractive index of any medium gives the ratio of the speed of light in vacuum to the speed of light in that medium. For example, the refractive index of water, µ_w = 1.33. This means that the ratio of the speed of light in vacuum or air to the speed of light in water is 1.33.

Question 8.
What do you mean by optically denser and optically rarer media? How is the speed of light related to optical density?
Answer:
The optical density of a medium represents its ability to refract light. A medium having larger refractive index is called optically denser medium than the other. The other medium having lower refractive index is called optically rarer medium.

The speed of light is higher in a rarer medium than a denser medium. Thus, a ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal.

When it travels from a denser medium to a rarer medium, it speeds up and bends away from the normal.

Table:
Refractive indices of some material media (with respect to vacuum)

MP Board Class 10th Science Chapter 10 Long Answer Type Questions

Question 1.
With the help of a ray diagram, state and explain the laws of reflection of light at a plane mirror. Mark the angles of incidence and reflection clearly on the diagram.
Answer:
As shown in Fig. 10.8, when a ray of light is incident on a mirror, it gets reflected in accordance with the following laws of reflection.
1st law: The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
2nd law: The angle of incidence (i) is equal to the angle of reflection (r) i.e.∠i = ∠r

Fig.10.8: Reflection in a plane mirror.

Question 2.
What is lateral inversion of an image? What is the cause of lateral inversion?
Answer:
Lateral inversion: If we stand before a plane mirror and move our right hand, our image appears to move its left hand. In fact, our entire image is reversed sideways. This sideways reversal of the image is known as lateral inversion.

Cause of lateral inversion: Lateral inversion is due to the fact that in a plane mirror the image is as far behind the mirror as the object is in front of it, and that the front of the image and the front of the object face each other. The laterally inverted image of the word PAPYRUS is as shown in Fig. 10.9. The images of symmetrical letters like A, H, I, M, O, T, U, V, W, X, Y, 8 are not affected by lateral inversion.

Fig.10.9: Lateral inversion before a mirror.

Question 3.
Define the following terms in connection with spherical mirrors:
(i) Angular aperture
(ii) Centre of curvature
(iii) Radius of curvature
(iv) Principal axis
(v) Linear aperture
(vi) Pole
(vii) Principal force
(viii) Focal length
(ix) Principle focus Focal plane.
Answer:
Definition in connection with spherical mirrors: In Fig 10.10, let APB be a principal section of a spherical mirror, i.e., the section cut by a plane passing through pole and centre of curvature of the mirror
(i) Angular aperture : It is the angle ACB subtended by the boundary of the spherical mirror at its centre of curvature.
(ii) Centre of curvature : It is the centre C of the sphere of which the mirror forms a part.
(iii) Radius of curvature : It is the radius R (= AC or BC) of the sphere of which the mirror forms a part.
(iv) Principal axis : The line passing through the pole and the centre of curvature of mirror is called its principal axis.
(v) Linear aperture : It is the diameter AB of the circular boundary of the spherical mirror.
(vi) Pole: It is the middle point P of the spherical mirror.

Fig. 10.10: Characteristics of a concave mirror.

Question 4.
Deduce a relation between focal length (f) and radius of curvature (R) for a concave mirror.
Answer:
Relation between f and R for a concave mirror: As shown in Fig. 10.11, consider a ray AB parallel to the principal axis and incident at the point B of a concave mirror. After reflection from the mirror, this ray passes through its focus F, obeying the laws of reflection. If C is the centre of curvature, then CP = R, is the radius of curvature and CB is normal to the mirror at point B.

Fig. 10.11: Relation betweenTand R for a concave mirror.

According to the law of reflection, ∠i = ∠r
As AB is parallel to CP, so ∠a = ∠i (Alternate angles)
∠a = ∠r
Thus, ∆ BCF is isosceles,
Hence, CF = FB.
If the aperture (or size) of the mirror is small, then B lies close to P, so that,
FB = FP
FP = CF = \(\frac { 1 }{ 2 } \) CP
or f = \(\frac { R }{ 2 } \)
or Focal lenght = \(\frac { 1 }{ 2 } \) × Radius of curvature
Thus, the principal focus of a spherical mirror lies midway between the pole and the centre of curvature.

Question 5.
What happens to the size of the image formed by a convex mirror, when an object is gradually moved towards the mirror?
Answer:
When the object is at position A1B1, its virtual image is at a1b1 When the object is at position A2B2, its virtual image is at a2b2. So, when an object is gradually moved towards the pole of a convex mirror, its image also moves towards its pole and gradually increases in size till it has a size almost equal to that of the object. However, the image is always formed between F and P.

MP Board Class 10th Science Chapter 10 NCERT Text Book Activities

Class 10 Science Activity 10.1 Page No. 161

• Take a large shining spoon. Try to view your face in its curved surface.
• Do you get the image? Is it smaller or larger?
• Move the spoon slowly away from your face. Observe the image. How docs it change?
• Reverse the spoon and repeat the Activity. How does the image look like now?
• Compare the characteristics of the image on the tw o surfaces.

Observations:

• The image is formed that is smaller in size.
• On moving the spoon away, the size of the image gradually decreases with increasing field of view.
• On reversing the spoon, spoon w hen dose forms elect and magnified image on the inner curved ..urface. As we move away, image gets inverted and gradually decrease in Its size.
• The image on outer surface of spoon is erect and gradually decreases in size on moving away. The image of spoon on inner surface is erect and gets inverted on moving away. The size also decreases when moved away.

Class 10 Science Activity 10.2 Page No. 162

Caution: Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?

Observations:

• The paper starts burning when the mirror and paper are held in the same position for a few minutes as light rays from the sun, sharply focuses on this point and due to which heat concentrates at a point resulting in burning because of intense heating.

Class 10 Science Activity 10.3 Pages No. 163,164

You have already learnt a way of determining the focal length of a concave mirror. In activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun. It was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror:

• Take a concave mirror. Find out its approximate focal length in the w ay described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines it equal to the focal length of the minor. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies midway between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle-(a) just beyond C, (b) at C, (c) between F and C. (d) at F, and (c) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case, Then, look for its virtual image in the mirror itself.
• Note down and tabulate your observations.

Observations:

Class 10 Science Activity 10.4 Page No. 166

• Draw neat ray diagrams for each position of the object shown in previous activity 10.3 observations.
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.1.
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observations:

Fig. 10.1: Ray diagrams for the image formation by a concave mirror.

Position of the object	Position of the. image	Size of the image	Nature of the image
At infinity	At the focus F, behind the mirror	Highly diminished, point-sized	Virtual and erect
Between infinity and the pole P of the mirror	Between P and F, behind the mirror	Diminished	Virtual and erect

Class 10 Science Activity 10.5 Page No. 167

• Take a convex mirror. Hold it in one hand.
• I Iold a pencil in the upright position in the other hand.
• Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
• Move the pencil away from the mirror slowly. Does the image become smaller or larger?
• Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the min or?

Observations:

• The image is erect and diminished. The image becomes smaller or moving pencil away.
• The image moves closer to the focus as the object is moved away from the mirror.

Class 10 Science Activity 10.6 Page No 167

• Observe the image of a distant object, say a distant tree, in a plane mirror.
• Could you see a full-length image?
• Try with plane minors of different sizes. Did you see the entire object in the image?
• Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
• Now try using a convex mirror. Did you succeed? Explain your observations with reason.

Observations:

• No, full – length image of a distant object is not seen in a plane mirror.
• The entire images of the objects were not seen.
• No, the mirror do not show full length image of the object.
• Yes, with the convex mirror we can see full length image of distant object with under field of view’ this is because the convex mirror are used as rear or side view mirrors in vehicles. The image formed is diminished, erect and virtual.

Class 10 Science Activity 10.7 Page No. 172

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Repeat the Activity. Why did you not succeed in doing it in one go?
• Ask your friends to do this. Compare your experience with theirs.

Observations:

• No, we cannot succeed in picking up the coin.
• This happens due to refraction, the coin appears to he at some other place from where it is actually present. The light rays coming out from water tends to bend creating this problem.

Class 10 Science Activity 10.8 Page No. 172

• Place a large shallow bowl on a Table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?

Observations:

• Yes, on pouring water, it again becomes visible and little raised due to refraction.

Class 10 Science Activity 10.9 Page No. 172

• Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
• Place a glass slab over the line in such a way that one of its edges make an angle with the line.
• Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges ?
• Next, place the glass slab such that it is normal to the line. What do you observe now’ Does the part of the line under the glass slab appear bent?
• Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?

Observations:

• The line under the glass slab appear bent at the edges due to refraction.
• No, now it does not appear bend as a ray of light perpendicular to the plane of a refracting medium does not change angle during refraction.
• Yes, this is also due to refraction that apparent position of image of object seems nearer than its actual position.

Class 10 Science Activity 10.10 Page No. 173

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
• Join O and O’. Also produce EF up to P, as shown by a dotted line in Fig. 10.2.

Observations:

Fig. 10.2: Refraction of light through a rectangular glass slab.

Class 10 Science Activity 10.11 Page No. 177

Caution: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.
• Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 10.2.

Observations:

• This is due to sharp focusing of all rays at a single point after passing through the lens. The concentration of light rays increases at a point resulting in burning of paper.

Class 10 Science Activity 10.12 Page No. 178

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on cither side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂ and 2F₂, respectively.
• Place a burning candle, far beyond 2F, to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing object just behind 2F₁ between F₁ and 2F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observations:

Class 10 Science Activity 10.13 Page No. 179

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the image.
• Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Observations:

MP Board Class 9th Sanskrit Solutions Durva Chapter 18 पुरुषोत्तमः (पद्यम्)

MP Board Class 9th Sanskrit Chapter 18 पाठ्य पुस्तक के प्रश्न

Class 9 Sanskrit Chapter 3 Purushottam Ram प्रश्न 1.
एकपदेन उत्तरं लिखत (एक शब्द में उत्तर लिखो)
(क) नियतात्माः महावीर्यः कः? (मन को वश में रखने वाला कौन है?)
उत्तर:
प्रभवो रामः। (प्रभु राम)

(ख) रामः किमर्थं वनं गतः? (राम किसलिए वन को गए?)
उत्तर:
पितुर्वचननिर्देशात्। (पिता के वचनों का पालन करने के लिए।)

(ग) सुग्रीववचनात् रामः कं हतवान्? (सुग्रीव के वचन से राम ने किसका वध किया?)
उत्तर:
बालिन। (बालि का)

(घ) महाकपिः कां दग्धवान्? (हनुमान ने किसे जलाया?)
उत्तर:
लङ्कां। (लंका को)

(ङ) सीता दृष्टा इति कः रामं न्यवेदयत? (सीता को देखकर किसने राम को सम्पूर्ण कथा सुनायी?)
उत्तर:
महाकपि हनुमानः (महावीर हनुमान ने)

Purushottam Rama Class 9 प्रश्न 2.
एकवाक्येन उत्तरं लिखत (एक वाक्य में उत्तर लिखो)
(क) सीता लक्ष्मणश्च केन सह वनम् अगच्छताम्? (सीता और लक्ष्मण के साथ कौन वन को गए?)
उत्तर:
सीता लक्ष्मणश्च रामेण सह वनम् अगच्छताम्। (सीता और लक्ष्मण के साथ राम वन को गए।)

(ख) रामः कस्मिन् वंशे उत्पन्नः अभूत? (राम किस वंश में उत्पन्न हुए?)
उत्तर:
राम इक्ष्वाकुवंशे उत्पन्नः अभूत्। (राम इक्ष्वाकुवंश में उत्पन्न हुए।)

(ग) केन दशरथस्य मरणमभवत्? (किसके कारण दशरथ की मृत्यु हुई?)
उत्तर:
पुत्र शोकात् दशरथस्य मरणमभवत्।। (पुत्र शोक के कारण दशरथ की मृत्यु हुई।)

(घ) मायावी रावणः सीतां कथं जहार? (मायावी रावण ने सीता का कैसे हरण किया?)
उत्तर:
मायावी रावणः सीतां नृपात्मजौ दूरम् अपवाह्य रामस्य भार्यां जहार। (मायावी रावण ने सीता को राजकुमारों (राम-लक्ष्मण) को दूर भेजकर)

(ङ) श्रीराम बालिनः राज्ये कं प्रत्यपादयत्? (श्रीराम ने बालि का राज्य किसको प्रदान किया?)
उत्तर:
श्रीराम बालिनः राज्ये सुग्रीवमेव प्रत्यपादयत्। (श्रीराम ने बालि का राज्य सुग्रीव को प्रदान किया।)

Purushottam Ram Class 9 Questions And Answers Pdf प्रश्न 3.
अधोलिखितप्रश्नानाम् उत्तराणि लिखत
(क) रामस्य विशिष्टगुणाः के? (राम के विशिष्ट गुण कौन-कौन से हैं?)
उत्तर:
रामस्य विशिष्टगुणाः महावीर्यं द्युतिमान् धृतिमान् श्च आसीत्। (राम के विशिष्ट गुण महान बलशाली, वीर और कांति युक्त थे।)

(ख) रामस्य वनगमनसमये के दूरम् अनुगतः? (राम के वनगमन के समय कौन दूर तक अनुगमन किया?)
उत्तर:
रामस्य वनगपन समये पित्रा दशरथेन पौरेः च दूरम् अनुगतः। (राम के वनगमन के समय पिता दशरथ और ग्रामवासी दूर तक अनुगमन किए।)

(ग) वने रामः शूर्पणखावाक्यादुधुक्तान् कान्-निजधान? (वन में राम शूर्पणखा द्वारा उत्तेजित किए गए किन-किन राक्षसों को मारा?)
उत्तर:
वने रामः शूर्पणखावाक्यादुधुक्तान् सर्वराक्षसान् खरं, त्रिशिरसं, दूषणं च निजधान। (वन में राम ने शूर्पणखा द्वारा उत्तेजित किए गए राक्षसों खर, त्रिशला, और दूषण आदि को मारा।)

(घ) हनुमान कं प्रियम् आख्यातुम आयात्! (हनुमान किसको प्रिय कथा सुनाने आये?)
उत्तर:
हनुमान रामं प्रियम् आख्यातुम् आधात्। (हनुमान राम को प्रिय कथा सुनाने आये।)

(ङ) रावणस्य वधानन्तरं रामः किं कृतवान्? (रावण का वध करके राम ने क्या किया?)
उत्तर:
रावणस्य वधानन्तरं रामं नदीग्रामे जटां हित्वा भ्रातृभिः सहितः सीताम् अनुप्राप्य राजं पुनः आप्तवान्। (रावण के वध के अनन्तर राम नंदी ग्राम में जटाओं को कटवाकर भाइयों सहित सीता के साथ राज्यभार पुनः प्राप्त किया।)

Class 9 Purushottam Ram प्रश्न 4.
युग्ममेलनं कुरुत-

Class 9 Hindi Purushottam Ram प्रश्न 5.
शुद्धवाक्यानां समक्षम् “आम्” अशुद्धवाक्यानां समक्षं ‘न’ इति लिखत-
यथा-
श्रीरामः धैर्येण हिमवान इवं अस्ति।। – आम्
आम् सुग्रीवेण सह रामः मित्रतां न कृतवान् – न
(क) कैकेय्याः प्रियकारणात् रामः वनं गतः।
(ख) सुमित्रानन्दवर्धनः लक्ष्मणः वनं न गतः।
(ग) रामः चतुर्दशसह राक्षसान् हतवान्।
(घ) सुग्रीवः बालिनं हतवान्।
(ङ) रामः रावणं न हतवान।
उत्तर:
(क) आम्
(ख) न
(ग) आम्
(घ) न
(ङ) न

Class 9 Hindi Anivarya Sanskrit Chapter 3 प्रश्न 6.
उचितैः शब्दैः रिक्तस्थानानि पूरयत
(क) रामः समुद्रइव गाम्भीर्ये अस्ति।
(ख) वनं गत्वा रामः राक्षसान् निजधान।
(ग) सुग्रीवेण पम्पातीरे सह रामस्य मित्रता सम्पन्नता।
(छ) लङ्का गत्वा रामः रावणं हतवान्।
(ङ) रामः नन्दिग्रामे जटां त्यक्तवान्।

Purushottam Ram Class 9 Hindi Anuvad प्रश्न 7.
अधोलिखितवाक्यानि भूतकाले परिवर्तयत
(क) राक्षसान् मारणाय रामः वनं गच्छति।
उत्तर:
राक्षसान मारणाय रामः वनं आगच्छत्।।

(ख) समेण सह सीता लक्ष्मणश्च वनं गच्छतः।
उत्तर:
रामेण सह सीता लक्ष्मणश्च वनं गतवन्तौ।

(ग) रामः सुग्रीवेण सह मैत्री करोति।”
उत्तर:
रामः सुग्रीवेण सह मैत्री अकरोत्।

(ध) वाल्मीकिः रामचरित लिखति।
उत्तर:
वाल्मीकिः रामचरितं अलिखत्।

(ङ) श्रीरामः विभीषणाय लङ्काराज्यं ददाति।
उत्तर:
श्रीरामः विभीषणाय लङ्काराज्यं अद्दात्/दत्ता।

Class 9 Anivarya Sanskrit प्रश्न 8.
उदाहरणानुसारं शब्दानां मूलशब्दं विभक्तिं वचनं च लिखत-

Class 9 Anivarya Sanskrit Chapter 3 प्रश्न 9.
अधोलिखित शब्दानां समानार्थशब्दान् लिखत
उदाहरणं यथा- सुतः – पुत्रः
रामः – सीतापतिः।
हनमुान् – अंजनि पुत्रः।
शशी – मयङ्क
राजा – नृपः
समुद्रः – रत्नाकरः
वनम् – अरण्यम्।

Rajiv Prakashan Sanskrit Book Class 9 Solutions प्रश्न 10.
निम्नलिखित अव्ययानि प्रयुज्य वाक्यनिमाणं कुरुत-
इव – रामस्य गम्भीरता समुद्रः इव अस्ति।
च – कौशल्यानन्द स्य गंभीरता समुद्रः इव च धैर्येण हिमवान् इव आसीत्।
तत: – ततः रामः परिवार सहितेन वनं जगाम।
अपि – अहं अपि नगरं गमिष्यामि।
एव – गीता एव मधुरं गीतं गायति।

अनिवार्य संस्कृत कक्षा 9 पाठ 3 Question Answer प्रश्न 11.
निम्नलिखितेषु प्रत्ययं च पृथक् कुरुत-
दृष्ट्वा – दृशृ + क्त्वा
कृत्वा – कृ + क्त्वा
श्रुत्वा – श्रु + क्त्वा
हत्वा – हनू + क्त्वा
दग्ध्वा – दग्ध् + क्त्वा।

पुरुषोत्तमः पाठ-सन्दर्भ/प्रतिपाद्य

वाल्मीकि (चित रामायण विश्व प्रसिद्ध महाकाव्य है। उसमें से उद्धृत कुछ गीत काव्य के महानायक राम के चरित्र को व्यक्त करते हैं।

पुरुषोत्तमः पाठ का हिन्दी अर्थ

1. इक्वाकुवंशप्रभवो रामो नाम जनैः श्रुतः।
नियतात्मा महावीर्यो द्युतिमान् धृतिमान् वशी॥

शब्दार्थः :
इक्ष्वाकुवंश प्रभवो-इक्ष्वाकु-family; श्रुतः-सुना गया-listen; नियतात्मा-मन को वश में रखने वाले -controller of mind /who under the mind; द्युतिमान-कान्तिमान्-fame full.

हिन्दी अर्थः :
इक्ष्वाकु वंश में उत्पन्न, अपने मन को वश में रखने वाले महावीर, कान्तिमान, धैर्यवान, इन्द्रियजयी राम का नाम सभी लोगों न सुना है।

2. स च सर्वगुणोपेतः कौसल्यानन्दवर्धनः
समुद्र इव गाम्भीर्ये धैर्येण हिमवानिव ॥

शब्दार्थः :
कौसल्यानन्दवर्धनः-कौसल्या के आनन्द को बढ़ाने वाला-who incresing of Kaushalya’s enjoyment; इव-के समान-like you.

हिन्दी अर्थः :
वे कौसल्या के आनन्द को बढ़ाने वाले सर्वगुण संपन्न, समुद्र के समान गम्भीर, हिमालय के समान धैर्यवान हैं।

3. स जैगाम वनं वीरः प्रतिज्ञामनुपालयन्।
पितुर्वचननिर्देशात् कैकेय्याः प्रियकारणात् ॥

शब्दार्थः :
पितुर्वचननिर्देशात्-पिता के वचनों के निर्देश से-orders of father’s sentence. कैकेयाप्रियकारणात्-कैकेयी को प्रिय लगने के कारण-reason of loving Kaikeyi.

हिन्दी अर्थ :
वे कैकेयी के अभीष्ट सिद्धि एवं पिता की प्रतिज्ञा के पालनार्थ उनकी आज्ञा से वन गए।

4. तं व्रजन्तं प्रियो भ्राता लक्ष्मणेऽनुजगाम ह।
स्नेहाद् विनयसम्पन्नः सुमित्रानन्दवर्धनः॥

शब्दार्थ :
भ्राता-भाई-Brother; विनयसम्पन्नः-विनय से युक्त-Full of respect; सुमित्रानन्दवर्धनः-सुमित्रा के आनन्द को बढ़ाने वाला-for Sumitra and Anand; अनुजगाम-अनुसरण किया-followed.

हिन्दी अर्थ :
उनको जाते देख उनके प्रिय भाई स्नेह व विनय से संपन्न, सुमित्रा के आनन्द को बढ़ाने वाले लक्ष्मण ने भी उनका अनुसरण किया।

5. सीताप्यनुगता रामं शशिनं रोहिणी यथा।
पौरैरनुगतो दूरं पित्रा दशरथेन च॥

शब्दार्थः :
पौरैः-ग्रामवासी-villagers; अनुगतः-अनुकरण-follower; शशिं-चन्द्रमा-Moon.

हिन्दी अर्थ :
वन जाते समय पिता दशरथ एवं नगरवासियों द्वारा उनका बहुत दूर तक अनुगमन किया। सीता भी चन्द्रमा-रोहिणी के समान उनका अनुगमन किया अर्थात् उनके साथ वन को गईं।

6. चित्रकूटं गते रामे पुत्रशोकातुरस्तदा।
राजा दशरथः स्वर्गे जगाम विलपन् सुतम्॥

शब्दार्थः :
गते-गए हुए-went; जगाम्-गया-went. विलाप-रोना-weep.

हिन्दी अर्थ :
राम के चित्रकूट चले जाने पर पुत्र शोक से विह्वल महाराज दशरथ विलाप करते हुए स्वर्गारोहण किया।

7. ततः शूर्पणखावाक्यादुधुक्तान् सर्वराक्षसान्।
खरं त्रिशिरसं चैव दूषणं चैव राक्षसम्
निजघान रणे रामस्तेषां चैव पदानुगान्॥

शब्दार्थः :
ततः-तब-then; चैव-और भी-and also; विजधान्-विजय को-to receive Victory.

हिन्दी अर्थः :
इसके बाद (रावण की बहन) शूर्पणखा द्वारा उत्तेजित किए जाने वाले सभी राक्षस गण-खर, दूषण, त्रिसिरा और उनके सहयोगी सभी राम द्वारा युद्ध में मारे गए।

8. वने तस्मिन् निवसता जनस्थाननिवासिनाम्।
रक्षसां निहतान्यासन् सहस्त्राणि चतुर्दश।
ततो ज्ञातिवधं श्रुत्वा रावणः क्रोधमूर्छितः॥

शब्दार्थः :
तस्मिन्-उस-His; जनस्थानं-लोक स्थान-public place; सहस्राणि-हजारों-thousands; ततो-तब-then; श्रुत्वा-सुनकर-to listen.

हिन्दी अर्थः :
वन में रहते हुए जनस्थान के निवासियों की रक्षार्थ चौदह हजार राक्षस राम के द्वारा मारे गए। यह जानकर कि (जनस्थान में खर-दूषण आदि सारे राक्षस मारे गए) रावण क्रोधान्ध हो गया।

9. तेन मायाविना दूरमपवाह्य नृपात्मजौ।
जहार भार्यां रामस्य गृधं हत्वा जटायुषम्॥

शब्दार्थः :
तेन-उससे-his,him;जहार-अपहरण-kiddnap; गृधं-गिद्ध (जटायु)-A big bird; हत्वा-मारकर-after kill.

हिन्दी अर्थः :
तब रावण ने (अपने मामा) मायावी मारीच राजपुत्रों (राम-लक्ष्मण) को दूर ले जाकर राम की भार्या सीता का अपहरण कर लिया और जटायु नामक गिद्ध को मार दिया।

10. पम्पातीरे हनुमता सङ्गतो वानरेण ह।
हनुमद्वैचनाच्चैव सुग्रीवेण समागतः॥

शब्दार्थः :
पम्पा तीरे-पम्पा नामक सरोवर के किनारे-Near by the Pumpa; सङ्गतो-मित्रता–friendship; वानरेण-वानर के द्वारा-By monkey; सुग्रीवेण-सुग्रीव के ART-By Sugreev.

हिन्दी अर्थ :
उसके बाद पम्पासर के तट पर हनुमान नामक वानर के साथ मित्रता हुई। हनुमान के कहने पर सुग्रीव से उनकी मित्रता हुई।

11. ततः सुग्रीववचनाद्धत्वा वालिनमाहवे।
सुग्रीवमेव तद्राज्ये राघवः प्रत्यपादयत्॥

शब्दार्थः :
ततः-तब-then; सुग्रीववचनात्-सुग्रीव के वचन से;प्रत्ययादयत्-बैठा दिया-to sat.

हिन्दी अर्थ :
फिर सुग्रीव के कहने पर (राम ने) बाली का वध कर सुग्रीव को उस राज्य (किष्किंधा) का राजा नियुक्त किया।

12. ततो दग्ध्वा पुरीं लङ्कामृते सीतां च मैथिलीम्।
रामाय प्रियमाख्यातुं पुनरायात्महाकपिः॥

शब्दार्थः :
दग्ध्वा-जलाकर-to burn; मैथिलीम-सीता को-to Sita; प्रियमाख्यातुं-शुभ समाचार-Good news.

हिन्दी अर्थ :
उसके बाद हनुमान ने समुद्र को लांघकर, लंकापुरी को भस्मसात कर मैथिली सीता का समाचार राम को सुनाया।

13. सोभिगम्य महात्मानं कृत्वा रामं प्रदक्षिणम्।
न्यवेदयदमेयात्मा दृष्टा सीतेति तत्त्वतः॥

शब्दार्थ :
सोभिगम्य-बहुत अधिक-Highly; महात्मानं-महात्मा को-to saint; कृत्वा-करके-doing.

हिन्दी अर्थ :
अपरिमित, अतुलित बलशाली हनुमान ने आकर भगवान राम की प्रदक्षिणा करके माँ सीता का समाचार उन्हें सुनाया और कहा–हमने माँ सीता को देखा है।

14. तेन गत्वा पुरीं लङ्कां हत्वा रावणमाहवे।
रातः सीतामनुप्राप्य परां व्रीडामुपागमत्।।

शब्दार्थः :
तेन-उससे-his, her; गत्वा-जाकर-going; परां-बहुत अधिक-very much.

हिन्दी अर्थ :
तब राम ने समुद्र में सेतु का निर्माण कर लंका पुरी में जाकर रावण को युद्ध में मार कर सीता को प्राप्त कर लज्जित हुए।

15. नन्दिग्रामे जटां हित्वा भ्रातृभिः सहितोऽनघः।
रामः सीतामनुप्राप्य राज्यं पुनरवाप्तवान्॥

शब्दार्थः :
नन्दिग्रामे-नंदिग्राम में-In Nandigram; भ्रातृभिः-भाईयों के साथ-with brothers; पुनःप्राप्तवान्-पुनः प्राप्त किया-re-received.

हिन्दी अर्थः :
पुनः पाप रहित राम ने नन्दिग्राम में आकर अपनी बढ़ी हुई जटाओं को कटवा कर अपने साथ सीता को प्राप्त कर पुनः राज्याभिषिक्त हुए।

MP Board Class 9th Sanskrit Solutions

MP Board Class 8th Sanskrit Solutions Surbhi Chapter 14 आचार्योपदेशाः

MP Board Class 8th Sanskrit Chapter 14 अभ्यासः

Class 8 Sanskrit Chapter 14 MP Board प्रश्न 1.
एकपदेन उत्तरं लिखत(एक शब्द में उत्तर लिखो-)
(क) पुष्पसजं कण्ठे कः समर्पयति? (पुष्पाहार गले में कौन समर्पित करता है?)
उत्तर:
शिवराजः। (शिवाजी)

(ख) वृत्तं केन रक्ष्यते? (चरित्र की रक्षा कैसे की जाती है?)
उत्तर:
धर्मभयेन। (धर्म के भय से)

(ग) अद्य मे किं निवृत्तम्? (आज मेरा क्या समाप्त हो गया है?)
उत्तर:
मोहावरणम्। (मोह का आवरण)

(घ) नृपः धर्मान् केन पालयेत्? (राजा धर्म का पालन कैसे कराये?)
उत्तर:
नियमेन। (नियम से)

(ङ) शिवराजम् भारतकवीर! इति शब्देन कः सम्बोधयति? (शिवाजी को ‘भारत का एक वीर!’ इस शब्द से कौन सम्बोधित करता है?)
उत्तर:
श्रीरामदासः। (श्रीरामदास)

MP Board Class 8 Sanskrit Chapter 14 प्रश्न 2.
एकवाक्येन उत्तरं लिखत(एक वाक्य में उत्तर लिखो-)
(क) शिवराजस्य गुरुः कः आसीत्? (शिवाजी के गुरु कौन थे?)
उत्तर:
शिवराजस्य गुरुः श्रीरामदासः आसीत्। (शिवाजी के गुरु श्रीरामदास थे।)

(ख) क्षत्रियस्य परो धर्मः किं अस्ति? (क्षत्रिय का परम धर्म क्या है?)
उत्तर:
क्षत्रियस्य परोधर्मः दुष्कृतां हिंसनं साधूनां च परित्राणम् अस्ति। (क्षत्रिय का परम धर्म दुष्कर्मियों को मारना और सज्जनों की सुरक्षा है।)

(ग) शिवराजस्य साहाय्यार्थं श्रीरामदास किम् करोति स्म? (शिवाजी की सहायता के लिए श्रीरामदास क्या कर रहे थे?)
उत्तर:
शिवराजस्य साहाय्यार्थं श्रीरामदासः प्रतिमठे राष्ट्रभावभावितान् शतशः युवगणान् निर्माति स्म। (शिवाजी की सहायता के लिए श्रीरामदास प्रत्येक मठ में सैकड़ों युवागणों का निर्माण कर रहे थे।)

कीदृशाः युवगणाः भाविरणे सहायाः भविष्यन्ति? (कैसे युवकों के समूह भविष्य में होने वाले युद्ध में सहायक होंगे?)
उत्तर:
राष्ट्रैकभक्ताः युवगणाः भाविरणे सहायाः भविष्यन्ति। (राष्ट्रभक्त युवकों के समूह भविष्य में होने वाले युद्ध में सहायक होंगे।)

(ङ) शिवराजस्य अभीष्टं का सम्पादयतु? (शिवराज की इच्छा को कौन पूरा करे?)
उत्तर:
शिवराजस्य अभीष्ट भगवती परदेवता सम्पादयतु। (शिवराज की इच्छा को भगवान् परमात्मा पूरा करें।)

Class 8 Sanskrit Chapter 13 MP Board प्रश्न 3.
रिक्तस्थानं पूरयत(रिक्त स्थान भरो-)
(क) वृत्तं यथा ……………. रक्ष्यते।
(ख) प्रजाहितज्ञो नियमेन ……………..।
(ग) मया ……….. राष्ट्रभावभाविताः।
(घ) अपितु त्वमसि मे …………।
(ङ) ……………. सम्पादयतु तवाभीष्टम्।
उत्तर:
(क) धर्मभयेन
(ख) पालयेत
(ग) निर्मीयन्ते
(घ) द्वितीयं हृदयम्
(ङ) भारतैकवीर।

Sanskrit Chapter 14 Class 8 MP Board प्रश्न 4.
सन्धि-विच्छेदं कुरुत
(सन्धि विच्छेद करो-)
(क) गमितोऽस्मि
(ख) त्वमसि
(ग) नृभिस्तथा
(घ) भगवतैवारब्धे
(ज) भारतैकवीरः
(च) सम्प्रत्यपि
(छ) प्रतिष्ठेऽहम्
(ज) तवाभीष्टम्
(झ) योगोपचिताः
(ण) राष्ट्रकभक्तेः।
उत्तर:
(क) गमितः + अस्मि
(ख) त्वम् + असि
(ग) नृभिः + तथा
(घ) भगवत् + एव+ आरब्धे
(ङ) भारत + एक + वीरः
(च) सम्प्रति + अपि
(छ) प्रतिष्ठे + अहम्
(ज) तव + अभीष्टम्
(झ) योग + उपचित
(ण) राष्ट्र + एक + भक्तेः।

MP Board Class 8th Sanskrit Chapter 14 प्रश्न 5.
सन्धिं कुरुत(सन्धि करो-)
उत्तर:
(क) शङ्कर + अंशेन + अवतीर्णस्य = शङ्करांशेनावतीर्णस्य।
(ख) वर्णाश्रमे + अस्मिन् = वर्णाश्रमेऽस्मिन्।
(ग) उत् + मूल्य = उन्मूल्य।
(घ) राष्ट्र + उद्धरण + उद्यमे = राष्ट्रोद्धरणोद्यमे
(ङ) उत् + ईक्ष्यते = उदीक्ष्यते।

MP Board Class 8 Sanskrit Chapter 13 प्रश्न 6.
श्लोकं पूरयत(श्लोक पूरा करो-)
उत्तर:
वृत्तं यथा धर्मभयेन रक्ष्यते नृभिस्तथा नैव नरेन्द्रशासनात्।
धर्मान् सदाचारपरानतो नृपः प्रजाहितज्ञो नियमेन पालयेत्॥

Sanskrit Class 8 Chapter 14 MP Board प्रश्न 7.
संस्कृतेन भावार्थं लिखत (संस्कृत में भावार्थ लिखो-)
व्यायामयोगोपचिताङ्गसत्त्वा विद्याकलादण्डनयप्रतिष्ठताः।
राष्ट्रकभक्ता उपधाविशोधिता भवन्तु ते भाविरणे सहायाः॥
उत्तर:
राष्ट्र प्रति एकभक्ताः, व्यायामेन योगेन च अङ्गनां शक्तिसम्पन्नाः, विद्यासु कलासु दण्डनीतेः कुशलाः, धर्मे अर्थे च संस्कारिताः, भविष्ये युद्धे शतशः युवगणाः तव सहायकाः भवन्तु।

Class 8 MP Board Sanskrit Chapter 14 प्रश्न 8.
निम्नाङ्कितशब्दान् आधृत्य वाक्यरचनां कुरुत(निम्न शब्दों के आधार पर वाक्य रचना करो-)
(क) दिष्ट्या
(ख) सदाचारः
(ग) परित्राणम्
(घ) धर्मशासनम्
(ङ) राष्ट्रियभावना।
उत्तर:

Class 8 Sanskrit Chapter 14 प्रश्न 9.
अर्थानुसारं युग्मनिमाणं कुरुत (अर्थ के अनुसार जोड़े बनाओ-)

उत्तर:
(क) → (iii)
(ख) → (v)
(ग) → (iv)
(घ) → (i)
(ङ) → (ii)

Sanskrit Class 8 Chapter 14 Pdf प्रश्न 10.
निम्नाङ्कितपदानां विलोमपदानि लिखत(नीचे लिखे शब्दों के विलोम शब्द लिखो-)
उत्तर:
पदानि – विलोमपदम्
(क) मया – त्वया
(ख) निवृत्तम् – संवृत्तम्
(ग) तव – मम
(घ) उत्थानम् – पतनम्
(ङ) अस्मिन् – तस्मिन्।

(संस्कृत में नाटकों की परम्परा अति प्राचीन है। यह परम्परा इस समय भी निर्बाध रूप से चल रही है। बीसवीं शताब्दी में गुजरात प्रदेश के श्री मूलशंकर मणिक लाल याज्ञिक ने भी अनेक पुस्तकें रचीं। उनमें संस्कृत भाषा में संयोगितास्वयम्वरम्, प्रतापविजयम् और छत्रपतिसाम्राज्यम् का वर्णन करते हैं।

‘छत्रपतिसाम्राज्यम्’ तो ऐतिहासिक नाटक है। इस नाटक में छत्रपति शिवाजी के शौर्यपूर्ण कार्यों का एवं तात्कालिक यवन सम्राट की दुर्नीति के विरुद्ध संघर्ष का और अन्त में स्वराज्य की स्थापना का चित्रण है।

यह प्रस्तुत नाट्य अंश ‘छत्रपतिसाम्राज्यम्’ इस नाटक से ही उद्धृत है। इसमें शिवाजी के गुरु श्रीरामदास के उपदेश हैं। राष्ट्रीय भक्ति की भावना से भरा यह अंश देखने योग्य है।)

आचार्योपदेशाः हिन्दी अनुवाद

(ततः प्रविशति रामदासेन सह शिवराजः)
शिवराज: :
(सप्रश्रयम्) दिष्ट्याद्य कृतार्थतां गमितोऽस्मि चिरप्रार्थितेन भगवत्-प्रसाद-अधिगमेन। (इति पुष्पस्रजंकण्ठे समर्प्य पादयोः पतति।)

श्रीरामदासः :
भारतैकवीर! उत्तिष्ठ। धर्मराज्यसंस्थापनार्थं शङ्कर-अंशेन-अवतीर्णस्य तव भवतु सर्वत्र अप्रतिहतो विजयः।

अनुवाद :
(उसके बाद रामदास के साथ शिवाजी प्रवेश करते हैं।)

शिवराज :
(विनम्रतापूर्वक) सौभाग्य से आज मैं बहुत समय से प्रार्थित (प्रार्थना करने पर) भगवान की कृपापूर्वक आने से सफलता को प्राप्त हुआ हूँ। (इस प्रकार पुष्पाहार गले में समर्पित करके पैरों में गिरते हैं।)

श्रीरामदास :
भारत के एक वीर! उठो। धर्म के राज्य की अच्छी प्रकार से स्थापना के लिए शंकर के अंश (भाग) के द्वारा अवतरित तुम्हारी सब जगह निर्विघ्न विजय हो।

शिवराजः :
(उत्थाय) प्रतिगृहीताशीः।

श्रीरामदास: :
व्यवस्थितवर्णाश्रमे अस्मिन् भारते वर्षे दुष्कृतां हिंसनं साधूनां च परित्राणम् एव क्षत्रियस्य परो धर्मः। तत् नयमार्गम् अवलम्ब्य उत्पथगामिनो नृपाधमान् च उन्मूल्य प्रवर्तय स्व धर्मशासनम् यतः वृत्तं यथा धर्मभयेन रक्ष्यते नृभिस्तथा नैव नरेन्द्रशासनात्। धर्मान् सदाचारपरानतो नृपः प्रजाहितज्ञो नियमेन पालयेत्॥

अनुवाद :
शिवराज :
(उठकर) आशीर्वाद प्राप्त हो गया।

श्रीरामदास :
व्यवस्थित वर्णाश्रम में इस भारतवर्ष में दुष्कर्मियों को मारना और सज्जनों की सुरक्षा ही क्षत्रिय का परम धर्म है। इसलिए नीति के मार्ग का सहारा लेकर कुमार्ग गामी और अधम राजाओं को जड़ से उखाड़कर अपना धर्मराज्य स्थापित करो। क्योंकि-

‘जैसी मनुष्यों द्वारा धर्म के भय से चरित्र की रक्षा की जाती है वैसी राजा की आज्ञा से नहीं। सदाचारी प्रजा के हित को जानने वाला राजा नियम से धर्म का पालन कराये।’

शिवराजः :
भगवन्। तव अनुग्रहेण अद्य निवृत्तम् मे मोहावरणम्। नवीकृतश्च साम्राज्य-संस्थापनोत्साहः।

श्रीरामदासः :
वत्स! तव साहाय्यार्थं प्रतिमठं मया निर्मीयन्ते राष्ट्रभावभाविताः शतशो युवगणाः। तदिमेव्यायामयोगोपचिताङ्गसत्त्वा विद्याकलादण्डनयप्रतिष्ठताः। राष्ट्रकभक्ता उपधाविशोधिता भवन्तु ते भाविरणे सहायाः॥

अनुवाद :
शिवराज :
भगवन्! आपकी कृपा से आज मेरा मोह का आवरण (पर्दा) समाप्त हो गया है और साम्राज्य की स्थापना का उत्साह नया सा कर दिया गया है।

श्रीरामदास :
वत्स! तुम्हारी सहायता के लिए मेरे द्वारा प्रत्येक मठ (आश्रम) में राष्ट्रीय भावना वाले सैकड़ों युवाओं के समूह तैयार किये जा रहे हैं। इसलिये ये-

‘राष्ट्र के एक भक्त व्यायाम और योग से प्राप्त अंगों की शक्ति वाले, विद्याओं, कलाओं, दण्डनीति में कुशल, धर्म, अर्थ में संस्कारित भविष्य में होने वाले युद्ध में तुम्हारी सहायता करने वाले होवें।’

शिवराजः :
अहो, परमार्थतो भगवतैवारब्धे राष्ट्र-उद्धरण-उद्यमे अहं तु निमित्तमात्रमेव।

श्रीरामदासः :
वत्स! न केवलं शिष्य इति, त्वमसि मम प्रेमास्पदम्। अपितु त्वमसि मे द्वितीयं हृदयं त्वदधीनैवास्ति मे साध्यसिद्धि। तन्मया सततं सावधानेन उदीक्ष्यते त्वद् विजयध्वजप्रसरः। सम्प्रत्यपि त्वां निर्विण्णम् उपश्रुत्य संप्राप्तोऽस्मि अहं तव प्रोत्साहनार्थम् एतद् दुर्गराजम्। अथ त्वां स्वकर्मणि अभिप्रवृत्तं वीक्ष्य प्रतिष्ठेऽहं धर्मप्रवचनाय दुर्गान्तरम्।।

शिवराजः :
भगवतानुग्राह्यः अयं जनो भूयो दर्शनेन।

श्रीरामदासः :
भारतैकवीर! सम्पादयतु तवाभीष्टं भगवती परदेवता। (इति निष्क्रान्तः)

अनुवाद :
शिवराज :
अहो, वस्तुतः भगवान द्वारा ही आरम्भ किये गये राष्ट्र के उद्धार के कार्य में मैं तो निमित्त (कारण) मात्र ही हूँ।

श्रीरामदास :
वत्स! तुम न केवल मेरे शिष्य बल्कि प्रिय हो। अपितु तुम मेरे द्वितीय हृदय हो, तुम्हारे हाथ में ही मेरे लक्ष्य की प्राप्ति है। इसलिए मैं निरन्तर सावधानी से तुम्हारी विजय पताका का लहाराना सादर देखता हूँ। इस समय भी तुमको दुःखी सुनकर मैं तुम्हारे प्रोत्साहन के लिए इस विशाल किले में आया हूँ। अब तुमको अपने कार्य में लगा हुआ देखकर मैं धर्म के उपदेश देने के लिए दूसरे किले की ओर प्रस्थान करता हूँ।

शिवराज :
यह जन (शिवाजी) फिर (आपके द्वारा) दर्शन से कृपा करने योग्य है।

श्रीरामदास :
भारत के एक वीर! तुम्हारे इच्छित को भगवान् परमात्मा पूरा करें। (निकल जाते हैं)

आचार्योपदेशाः शब्दार्थाः

सप्रश्रयम् = विनम्रतापूर्वक। प्रतिगृहीताशीः = आशीर्वाद प्राप्त। दिष्ट्या = सौभाग्य से। दुष्कृताम् = निन्दित कर्म करने वालों का या दुष्कर्मियों का। कृतार्थताम् = सफलता को। अस्मिन् = इसमें। प्रसादाधिगमेन = कृपापूर्वक आने से। हिंसनम् = मारना। पुष्पस्रजम् = पुष्पहार। परित्राणम् = सुरक्षा।। समर्प्य = समर्पित करके। परोधर्मः = श्रेष्ठ धर्म। पादयोः = पैरों पर। नयमार्गम् = नीतिपथ। उत्तिष्ठ = उठो। अवलम्ब्य = सहारा लेकर। संस्थापनार्थम् = अच्छे प्रकार से स्थापना के लिए। उत्पथगामिनः = कुमार्ग गामी। नृपाधमान् = अधम । राजाओं को। अंशेन = अंश (या भाग) के द्वारा। धर्मशासनम् = धर्मराज्य। वृत्तम् = चरित्र को। अवतीर्णस्य = अवतरित का। नृभिः = मनुष्यों के द्वारा। अप्रतिहतः = निर्बाध, निर्विघ्न। सदाचारपरान् = सदाचार परायण या सदाचारी। उत्थाय = उठकर। प्रजाहितज्ञः = प्रजाहित का ज्ञाता या प्रजा के हित को जाने वाला। परमार्थतः = वस्तुतः। आरब्धे = आरम्भ किये गये। प्रेमास्पदम् = प्रिय। अनुग्रहेण = कृपा से।

साध्यसिद्धिः = लक्ष्य की प्राप्ति। निवृत्तम् = समाप्त। सततम् = निरन्तर। मोहावरणम् = मोह का आवरण। उदीक्ष्यते = सादर दिखाई देता है। नवीकृतः = नया कर दिया। विजयध्वजप्रसरः = विजय पताका का लहराना। प्रतिमठम् = प्रत्येक मठ (या आश्रम) में। सम्प्रत्यपि = इस समय भी। निर्मीयन्ते = तैयार किये जाते हैं। निर्विष्णम् = विरल हृदय को। राष्ट्रभावभाविताः = राष्ट्रिय भावना वाले। प्रोत्साहनार्थम् = उत्साह बढ़ाने के लिए। शतशः = सैकड़ों। दुर्गराजम् = विशाल दुर्ग या बड़ा किला। युवगणाः = युवक समूह। व्यायामयोगोपचित = व्यायाम और योग से प्राप्त। स्वकर्मणि = अपने कार्य में। अभिप्रवृत्तम् = लगा हुआ। अङ्गसत्वाः = अंगों की शक्ति वाले। वीक्ष्य = देखकर। विद्याकलादण्डनयप्रतिष्ठिताः = विद्याओं, कलाओं, दण्डनीति में कुशल। प्रतिष्ठेडहम् = मैं प्रस्थान करता हूँ। धर्मप्रवचनाय = धर्म के उपदेश देने के लिए। उपधाविशोधिता = धर्म, अर्थ में परीक्षित या संस्कारित। दुर्गान्तरम् = दूसरे दुर्ग (किला) को। अनुग्राह्यः = कृपा करने योग्य। भाविरणे= भविष्य में होने वाले समर में। भूयः = फिर। सम्पादयतु = पूरा करें। सहायाः = सहायता करने वाले। अभीष्टम् = इच्छित। भवन्तु = होवें। परमार्थतः = वस्तुतः।

MP Board Class 8th Sanskrit Solutions

In this article, we will share MP Board Class 8th Science Solutions Chapter 9 Reproduction in Animals PDF, Class 8th Science Chapter 6 Reproduction In Animals, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Science Solutions Chapter 9 Reproduction in Animals

MP Board Class 8 Science Chapter 9 Question 1.
Explain the importance of reproduction in organisms.
Answer:
The production of new individuals from their parents cells is known as reproduction. Reproduction is very important as it ensures the continuation of similar kinds of individuals, generation after generation. If reproduction process does not exist, the generation of living beings will be vanished from the earth.

MP Board Class 8 Science Chapter 6 Reproduction In Animals Question 2.
Describe the process of fertilization in human beings.
Answer:
In human beings, sexual reproduction occurs which is reproduction results from the fusion of male and female gametes. The male reproductive organs are a pair of testis, sperm ducts and penis. The testis produces the male gametes called sperms. Millions of sperms are produced by the testis. Though the sperms are very small in size, each has a head, a middle piece and a tail. Each sperm is a single cell with all the usual cell components.

The female reproductive organs are a pair of ovaries, oviducts and the uterus. Ovary produces female gametes called ova (eggs). In human beings, a single matured egg is released into the oviduct by one of the ovaries every month. Uterus is the part where development of the; baby takes place. Like sperm, an egg is also a single cell.

The first step in the process of reproduction is the fusion of a sperm and an ovum. Millions of sperms from the male are transferred into the female body. The sperms swim in the oviduct with the help of their tails to reach the egg. When they come in contact with the egg, one of the sperms may fuse with the egg.

Such fusion of the egg and the sperm is called fertilization. During fertilization, the nucleus of the sperm and the egg fuse to form a single nucleus. This results in the formation of a fertilized egg or zygote. The new individual inherits some characteristics from the mother and some characteristics from the father.

Class 8 Science Chapter 9 MP Board Question 3.
Choose the most appropriate answer:
(a) Internal fertilization occurs:
(i) in female body.
(ii) outside female body.
(iii) in male body.
(iv) outside male body.
Answer:
(i) in female body.

(b) A tadpole develops into an adult frog by the process of:
(i) fertization
(ii) metamorphosis
(iii) embedding
(iv) budding.
Answer:
(ii) metamorphosis

(c) The number of nuclei present in a zygote is:
(i) none
(ii) one
(iii) two
(iv) four.
Answer:
(iii) two

MP Board Class 8th Science Chapter 9 Question 4.
Indicate whether the following statements are True (T) or False (F):

1. Oviparous animals give birth to young ones.
2. Each sperm is a single cell.
3. External fertilization takes place in frog.
4. A new human individual develops from a cell called gamete.
5. Egg laid after fertilization is made up of a single cell.
6. Amoeba reproduces by budding.
7. Fertilization is necessary even in asexual reproduction.
8. Binary fission is a method of asexual reproduction.
9. A zygote is formed as a result of fertilization.
10. An embryo is made up of a single cell.

Answer:

1. False
2. True
3. True
4. False
5. True
6. False
7. False
8. True
9. False
10. False

Class 8 MP Board Science Chapter 9 Question 5.
Give three differences between a zygote and a foetus.
Answer:
The three differences between a zygote and a foetus are:

1. Zygote is made up of a single cell while foetus is made up of many Cells.
2. Zygote does not have well developed limbs, foetus has well developed and identifiable limbs.
3. Zygote is formed by the fertilization of sperms and ovum, foetus is formed by the repeated divisions of the zygote.

Class 8 Science Chapter 9 Question Answer Question 6.
Define asexual reproduction. Describe the two methods of asexual reproduction in animals.
Answer:
Asexual reproduction is defined as the type of reproduction in which only a single parent is involved. Budding and binary fission are the two main methods of asexual reproduction.

(i) Budding is the method of asexual reproduction that takes place in hydra. Each hydra has two or three bulges which are called buds. These buds develop into new individuals. Since new individuals in hydra are formed from buds, this type of asexual reproduction is known as budding.

(ii) Binary fission: Amoeba is a single celled animal. It begins reproduction by the division of its nucleus into two nuclei. Thus, two amoeba are formed from one amoeba. Since an animal reproduces by division into two individuals, this type of asexual reproduction is called binary fission.

Science Chapter 9 Class 8 MP Board Question 7.
In which female reproductive organ does the embryo get embedded?
Answer:
The embedding of the embryo takes place in the wall of the uterus of the female for further development.

Class 8 Science Chapter 9 Question 8.
What is metamorphosis? Give examples.
Answer:
Metamorphosis. It is the drastic change which takes place during the development of an animal into an adult. Frogs and silkworms exhibit metamorphism.
Egg of frog → Tadpole larva → Adult
Tadpoles look different from the adult. But they soon transform into an adult capable of jumping and swimming. Similarly, caterpillar or the pupa of silkworm looks quite different from adult. Also beautiful moth emerges out of the cocoon.

Class 8th Science Chapter 9 Question Answer Question 9.
Differentiate between internal fertilization and external fertilization.
Answer:
(i) Internal fertilization – The fertilization which takes place inside the body of the female is called internal fertilization. It occurs in humans and other animals such as cows, dogs and hens.

(ii) External fertilization – The fertilization which takes place outside the body of the female is called external fertilization. It is common in aquatic animals such as frogs, fish and starfish.

Class 8th Science Chapter 9  Question 10.
Complete the cross-word puzzle using’ the hints given below:
Across
1. The process of the fusion of the gametes.
6. The type of fertilization in hen.
7. The term used for bulges observed on the sides of the body of Hydra.
8. Eggs are produced here.

Down
2. Sperms are produced in these male reproductive organs.
3. Another term for the fertilized egg.
4. These animals lay eggs.
5. A type of fission in amoeba.

MP Board Class 8th Science Reproduction in Animals NCERT Extended Learning – Activities and Projects

Class 8 Science Chapter 6 Reproduction In Animals Question 1.
Visit a poultry farm. Talk to the manager of the farm and try to find out the answers to the following:
(a) What are layers and broilers in a poultry farm?
(b) Do hens lay unfertilized eggs?
(c) How can you obtain fertilized and unfertilized eggs?
(d) Are the eggs that we get in the stores fertilized or unfertilized?
(e) Can you consume fertilized eggs?
(f) Is there any difference in the nutritional value of the fertilized and undertilized eggs?
Answer:
Do yourself.

Class 8 Ch 9 Science Question 2.
Observe live hydra yourself and learn how they reproduce by doing the following activity:
During the summer months collect water weeds from ponds or ditches along with the pond water and put them in a glass jar. After a day or so you may see several hydra clinging to the sides of the jar. Hydra is transparent, jelly-like and with tentacles. It clings to the jar with the base of its body. If the jar is shaken, the hydra will contract instantly into a small blob, at the same time drawing its tentacles in. Now take out few hydras from the jar and put them on a watch glass. Using a hand lens or a binocular or dissection microscope, observe the changes that are taking place in their body. Note down your observations.
Answer:
Do yourself.

Class 8 Science Chapter 9 Question Answer In English Question 3.
The eggs we get from the market are generally the unfertilized ones. In case you wish to observe a developing chick embryo, get a fertilized egg from the poultry or hatchery which has been incubated for 36 hours or more. You may then be able to see a white disc-like structure on the yolk. This is the developing embryo. Sometimes if the heart and blood vessels have developed you may even see a red spot.
Answer:
Do yourself.

Reproduction In Animals Class 8 Pdf Questions And Answers Question 4.
Talk to a doctor. Find out how twinning occurs. Look for any twins in your neighbourhood, or among your friends. Find out if the twins are identical or non-identical. Also find out why identical twins are always of the same sex? If you know of any story about twins, write it in your own words.
Answer:
Do yourself.

MP Board Class 8th Science Reproduction in Animals NCERT Intext Activities and Projects

Activity 9.1
Visit some ponds or soil-flowing streams during spring or rainy season. Look out for clusters of frog’s eggs floating in water. Write down the colour and size of the eggs.
Answer:
The colour of the eggs is dull white and size is less than a centimetre to a few centimetres.

Activity 9.2
Try to collect eggs of the following organism-frog, lizard, butterfly or moth, hen and crow or any other bird. Were you able to collect eggs of all of them? Make drawings of the eggs that you have collected.
Answer:
Yes, I collected all of these eggs.

MP Board Class 8th Science Reproduction in Animals NCERT Additional Important Questions

A. Short Answer Type Questions

Ch 9 Class 8 Science Question 1.
Name two organisms which reproduce by splitting their bodies into two parts.
Answer:
Two organisms which reproduce by splitting their bodies into two parts are amoeba and Paramecium.

Reproduction In Animals Class 8 Activities Question 2.
What is reproduction? What’s the need for this process?
Answer:

• The process of producing individuals of their own kind is called reproduction.
• It is an essential for the continuation of similar units of individuals, generation after generations.

8th Science 9th Lesson Question Answer Question 3.
What is fertilization?
Answer:
The process of fusion of male and female gamete in sexual reproduction is called fertilization.

Science Lesson 9 Class 8 Question 4.
Name the male and female sex organ of animals.
Answer:
The male sex organ is called testes and female sex organ is called ovary.

Class 8 Science Chapter 9question Answer Question 5.
Give the life cycle of a frog.
Answer:
Egg → early tadpole → late tadpole → adult frog.

B. Long Answer Type Questions

Science Ch 9 Class 8 Question 6.
What are the names of male reproductory organs in human beings?
Answer:
In human beings male reproductory organs are a pair of testes, sperm ducts and a penis. Testes are responsible for producing male gametes called sperms. They are transported to the sperm ducts and through the penis, they are ejaculated in female body.

Science Class 8 Chapter 9 Question Answer Question 7.
Explain the reproductory organs of female in human beings.
Answer:
In human beings the female reproductory organs are a pair of ovaries, oviducts or fallopian tubes and the uterus. Ovaries are responsible to produce female gamete called ova, which fertilize with male sperm in female fallopian tube and travels to uterus for its development into embryo and the foetus.
Exercise for Practice

Class 8 Chapter 6 Reproduction In Animal Question 8.
Explain the process of fertilization.
Answer:
Sperm is a tiny cell and is quite active for its movement, a thread like structure, the flagellum is found which is called as the tail. Ovum, the female gamete is a large, spherical cell which remains static. When the male and female gamete fuse; a new cell called zygote is formed. This process of fusion is known as fertilization. This zygote undergoes many changes and forms a new individual.

MP Board Class 6th Sanskrit Solutions Surbhi Chapter 1 स्वराभ्यासः (संस्कृत शब्द परिचयः)

MP Board Class 6th Sanskrit Chapter 1 अभ्यासः

Class 6 Sanskrit Chapter 1 MP Board प्रश्न 1.
चित्राणि दृष्ट्वा नामानि लिखत (चित्र देखकर नाम लिखो)

उत्तर:
(क) कमलम् (कमल का फूल)
(ख) खड्गः (तलवार)
(ग) घटः (घड़ा)
(घ) छत्रम् (छाता)
(ङ) ऊष्ट्रः (ऊँट)
(च) एणः (बारहसिंगा)
(छ) धनुषम् (धनुष)
(ज) आम्रम् (आम का फल)
(झ) ढौलः (ढोलक)।

MP Board Class 6 Sanskrit Chapter 1 प्रश्न 2.
कोष्ठात् उचितंशब्दं चित्वा चित्रानुसारं लिखत- (कोष्ठक से उचित शब्द चुनकर चित्र के अनुसार लिखो)
(गणेशः, ओष्ठौ, मयूरः, फणी, ऋषिः, बकः, झरः, रथः)

उत्तर:
(क) ऋषिः (ऋषि)
(ख) गणेशः (गणेश)
(ग) औष्ठौ (ऊपर और नीचे के दोनों ओंठ)
(घ) झरः (झरना)
(ङ) मयूरः (मोर)
(च) बकः (बगुला)
(घ) फणी (साँप या सर्प)
(ज) रथः (रथ)।

Class 6 Sanskrit Surbhi Chapter 1 प्रश्न 3.
अधोलिखितानां शब्दानां वर्णान् पृथक्-पृथक लिखत (नीचे लिखे हुए शब्दों के वर्गों को अलग-अलग लिखो)
(क) चटका
(ख) नरेशः
(ग) पत्रम्
(घ) दर्पणः
(ङ) घटः।
उत्तर:
(क) चटका = च् + अ + ट् + अ + क् + आ।
(ख) नरेशः = न् + अ + र् + ए + श् + अ + :।
(ग) पत्रम् = प् + अ + त्र् + अ + म्।
(घ) दर्पणः = द + अ + र् + प् + अ + ण् + अ + :।
(ङ) घटः = घ् + अ + ट् + अ + :।

MP Board Class 6th Sanskrit Chapter 1 प्रश्न 4.
रिक्तस्थानानि पूरयत (खाली स्थान भरो)
(क) ज …………… यानम्,
(ख) शश………….
(ग) ग …………… शः
(घ) ………….. क्रम्
(ङ) इ …………….. का।
उत्तर:
(क) जलयानम्
(ख) शशकः
(ग) गणेश:
(घ) चक्रम्
(ङ) इष्टिका।

Class 6 Sanskrit Chapter 1 प्रश्न 5.
वर्णमालानुसार क्रमेण स्थापयत (वर्णमाला के अनुसार क्रम से रखिए)
ऊर्णाः, बकः, औषधम्, हलम्, सरः, मयूरः, रथः, आम्रम्, घटः, कमलम्।
उत्तर:
आम्रम्, ऊर्णा, औषधम्, कमलम्, घटः, बकः, मयूरः, रथः, सरः, हलम्।

योग्यताविस्तारः
MP Board Class 6 Sanskrit प्रश्न 1.
पशूनां,शाकानां, फलानां च पञ्च-पञ्च नामानि लिखत। (पशुओं, शाक (सब्जी) और फलों के पाँच-पाँच नाम लिखो)।
उत्तर:

1. पशुनाम् नामानि :
   अश्वः, ऊष्ट्रः, एणः, भल्लूकः, शशकः (घोड़ा, ऊँट, बारहसिंगा, भालू, खरगोश)।
2. शाकानाम् नामानि :
   लशुनम्, आलु, पलाण्डुम्, कूष्माण्डं, शिम्बां (लहसुन, आलू, प्याज, कुम्हड़ा, सेम)।
3. फलानाम् नामानि :
   आम्रम्, नारिकेलम्, जम्बीरम्, जम्बूफलम्, नारङ्गफलम् (आम, नारियल, नीबू, जामुन, नारंगी)।

स्वराभ्यासः शब्दार्थाः

स्वराः
अश्वः = घोड़ा। आम्रम् = आम्र। इष्टिका = ईंट। ईशः = ईश्वर। ऊष्ट्रः = ऊँट। ऊर्णा = ऊन। ऋषिः = ऋषि। एण: = बारहसिंगा। ऐन्द्रजालिकः = जादूगर। ओष्ठौ = ऊपर और नीचे के दोनों ओंठ। औषधम् = दवाई।

व्यञ्जनानि :
कमलम् = कमल। खड्गः = तलवार। गणेशः = गणेश। घटः = घड़ा। गङ्गा = गंगा नदी। चक्रम् = पहिया। छत्रम् = छाता। जलयानम् = पानी का जहाज। झरः = झरना। चञ्चु : = चोंच। टङ्कणयंत्रम् = टाइप करने की मशीन। ठठकः = ठठेरा। डमरूः = डमरू। ढौलः = ढोलक। बाणः = तीर। तरुः = पेड़, वृक्ष। थः = मरुद्यान। दर्पणः = शीशा। धनुः = धनुष। नरेशः = राजा। पर्णम् = पत्ता। फणी = साँप, सर्प। बकः = बगुला। भल्लूकः = भालू। मयूरः = मोर। यज्ञः = हवन। रथः = रथ। लड्डुः = लड्डू। वटः = बरगद। शशकः = खरगोश। षट्पदः = भौंरा। सरः = तालाब। हलम् = हल।

संयुक्त व्यञ्जनानि :
क्षत्रियः = क्षत्रिय। त्रसम् = बगीचा, या वृक्षों का झुरमुट। ज्ञानी = ज्ञानी (जानकार)।

MP Board Class 6th Sanskrit Solutions

MP Board Class 9th Maths Solutions Chapter 1 संख्या पद्धति Ex 1.5

प्रश्न 1.
बताइए नीचे दी गई संख्याओं में कौन-कौन परिमेय हैं और कौन-कौन अपरिमेय हैं:

उत्तर:
परिमेय संख्याएँ : (ii) और (iii)
अपरिमेय संख्याएँ : (i), (iv) और (v).

प्रश्न 2.
निम्नलिखित व्यंजकों में से प्रत्येक व्यंजक को सरल कीजिए:
(i) (3 + √3) (2 + √2) (2019)
(ii) (3 + √3) (3 – √3)
(iii) (√5 + √2)²
(iv) (√5 – √2)(√5 + √2)
हल:
(i) (3+ √3) (2 + √2) = 3 x 2 + 3√2 + 2√3 + √3 x √2
= 6 + 3√2 + 2√3 + √6.

(ii) (3 + √3) (3 – √3) = (√3)² – (√3)²
= 9 – 3 = 6.

(iii) (5 + √2) = (√5)² + 2(√5) (√2) + (√2)²
= 5+ 2√10 + 2 = 7 + 2√10.

(iv) (√5 – √2) (√5 + √2) = (√5)² – (√2)² = 5 – 2 = 3.

प्रश्न 3.
अपको याद होगा कि π को एक वृत्त की परिधि (मान लीजिए c) और उसके व्यास (मान लीजिए d) के अनुपात से परिभाषित किया जाता है, अर्थात् π = \(\frac { c }{ d }\) है। यह इस तथ्य का अन्तर्विरोध करता हुआ प्रतीत होता है कि अपरिमेय है। इस अन्तर्विरोध का निराकरण आप किस प्रकार करेंगे?
उत्तर:
जब हम किसी भी युक्ति द्वारा किसी लम्बाई का मापन करते हैं तब हमको केवल सन्निकट परिमेय मान प्राप्त होता है। अत: हम यह अनुभव नहीं कर पाते हैं कि c या d परिमेय है या अपरिमेय इसलिए कोई अन्तर्विरोध नहीं है।

प्रश्न 4.
संख्या रेखा \(\sqrt { 9.3 }\) को निरूपित कीजिए।
हल:

अतः बिन्दु E संख्या रेखा पर अभीष्ट \(\sqrt { 9.3 }\) बिन्दु है। यहाँ AB संख्या रेखा है तथा B शून्य बिन्दु है।

प्रश्न 5.
निम्नलिखित के हरों का परिमेयीकरण कीजिए :

हल:

MP Board Class 9th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 6 वर्ग और वर्गमूल Ex 6.3

Class 8 Maths Exercise 6.3 Solutions In Hindi प्रश्न 1.
निम्नलिखित संख्याओं के वर्गमूल ज्ञात करने में इकाई अंक की क्या सम्भावना है –

1. 9801
2. 99856
3. 998001
4. 657666025

हल:

संख्याओं के वर्गमूल में इकाई अंक की सम्भावना –

1. 9801 में 1 या 9
2. 99856 में 4 या 6
3. 998001 में 1 या 9
4. 657666025 में 5

Mp Board Class 8 Maths Chapter 6 Exercise 6.3 प्रश्न 2.
बिना गणना किए वह संख्या बताएँ जो वास्तव में पूर्ण वर्ग नहीं है –

1. 153
2. 257
3. 408
4. 441.

हल:
संख्या (i) 153, (ii) 257, तथा (iii) 408 पूर्ण वर्ग नहीं है।

Class 8 Maths Chapter 6 Exercise 6.3 Solutions In Hindi प्रश्न 3.
बार-बार घटाने की विधि से 100 और 169 का वर्गमूल ज्ञात कीजिए।
हल:
1. 100 से 1 से प्रारम्भ करके क्रमागत विषम संख्याओं के घटाने पर,

• 100 – 1 = 99
• 99 – 3 = 96
• 96 – 5 = 91
• 91 – 7 = 84
• 84 – 9 = 75
• 75 – 11 = 64
• 64 – 13 = 51
• 51 – 15 = 36
• 36 – 17 = 19
• 19 – 19 = 0

संख्या 100 से क्रमागत विषम संख्याओं को घटाने पर 10वाँ पद शून्य (0) प्राप्त होता है।
अतः \(\sqrt { 100 } \) = 10

2. 169 से 1 से प्रारम्भ करके क्रमागत विषम संख्याओं को घटाने पर,

• 169 – 1 = 168
• 168 – 3 = 165
• 165 – 5 = 160
• – 160 – 7 = 153
• 153 – 9 = 144
• 144 – 11 = 133
• 133 – 13 = 120
• 120 – 15 = 105
• 105 – 17 = 88
• 88 – 19 = 69
• 69 – 21 = 48
• 48 – 23 = 25
• 25 – 25 = 0.

संख्या 169 से क्रमागत विषम संख्याओं को घटाने पर 13वाँ पद शून्य (0) प्राप्त होता है। अतः \(\sqrt { 169 } \) = 13

Varg Aur Vargmul Class 8 प्रश्न 4.
अभाज्य गुणनखण्ड विधि से निम्न संख्याओं का वर्गमूल ज्ञात कीजिए –

1. 729
2. 400
3. 1764
4. 4096
5. 7744
6. 9604
7. 5929
8. 9216
9. 529
10. 8100

हल:
1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Class 8 Maths Chapter 6.3 Hindi Medium प्रश्न 5.
निम्नलिखित संख्याओं में प्रत्येक के लिए वह सबसे छोटी पूर्ण संख्या ज्ञात कीजिए जिससे इस संख्या को गुणा करने पर यह एक पूर्ण वर्ग संख्या बन जाए। इस पूर्ण वर्ग संख्या का वर्गमूल भी ज्ञात कीजिए –

1. 252
2. 180
3. 1008
4. 2028
5. 1458
6. 768

हल:
1.

हम जानते हैं कि पूर्ण वर्ग संख्या में प्रत्येक पद के जोड़े बनते हैं।
परन्तु यहाँ 7 का जोड़ा नहीं है। अतः यदि 252 में 7 का गुणा कर दें तो संख्या पूर्ण वर्ग बन जाएगी।
∴ सबसे छोटी पूर्ण संख्या 7 होगी।
गुणा करने पर प्राप्त संख्या = 252 x 7 = 1764

2.

अतः दी गई संख्या में 5 का जोड़ा नहीं बन रहा है। जोड़ा बनाने के लिए 5 का गुणा करना आवश्यक है।
∴ सबसे छोटी पूर्ण संख्या 5 होगी।
गुणा करने पर प्राप्त संख्या = 180 x 5 = 900

3.

अतः दी गई संख्या में 7 का जोड़ा नहीं बन रहा है। जोड़ा बनाने के लिए 7 का गुणा करना चाहिए।
∴ सबसे छोटी पूर्ण संख्या 7 होगी।
गुणा करने पर प्राप्त संख्या = 1008 x 7 = 7056

4.

अतः दी गई संख्या में 3 का जोड़ा नहीं बन रहा है। जोड़ा बनाने के लिए 3 का गुणा करेंगे।
∴ सबसे छोटी पूर्ण संख्या 3 होगी।
गुणा करने पर प्राप्त संख्या = 2028 x 3 = 6084

5.

अतः दी गई संख्या में 2 का जोड़ा नहीं बन रहा है। जोड़ा बनाने के लिए 2 का गुणा करना पड़ेगा।
∴ सबसे छोटी पूर्ण संख्या 2 होगी।
गुणा करने पर प्राप्त संख्या= 1458 x 2 = 2916

6.

अतः दी हुई संख्या में 3 का जोड़ा नहीं बन रहा है। जोड़ा बनाने के लिए 3 का गुणा करना पड़ेगा।
∴ सबसे छोटी पूर्ण संख्या 3 होगी। गुणा करने पर प्राप्त संख्या = 768 x 3

Class 8 Maths Chapter 6 Exercise 6.3 Hindi Medium प्रश्न 6.
निम्नलिखित संख्याओं में प्रत्येक के लिए सबसे छोटी पूर्ण संख्या ज्ञात कीजिए जिससे इस संख्या को भाग देने पर वह एक पूर्ण संख्या बन जाए। इस तरह ज्ञात की गई संख्या का वर्गमूल भी ज्ञात कीजिए।

1. 252
2. 2925
3. 396
4. 2645
5. 2800
6. 1620

हल:
1.

इसे पूर्ण संख्या बनाने के लिए 7 का भाग करना होगा।
∴ सबसे छोटी पूर्ण वर्ग संख्या 7 होगी।
भाग देने पर प्राप्त संख्या = 36

2.

इसे पूर्ण संख्या बनाने के लिए 13 से भाग करना होगा।
∴ सबसे छोटी पूर्ण संख्या 13 होगी।
भाग देने पर प्राप्त संख्या = 225

3.

इसे पूर्ण संख्या बनाने के लिए 11 से भाग करना होगा।
∴ सबसे छोटी पूर्ण संख्या 11 होगी।
भाग देने पर प्राप्त संख्या = 36

4.

इसे पूर्ण संख्या बनाने के लिए 5 से भाग करना होगा।
∴ सबसे छोटी पूर्ण संख्या 5 होगी।
भाग देने पर प्राप्त संख्या = 529

5.

इसे पूर्ण संख्या बनाने के लिए 7 से भाग करना होगा।
∴ सबसे छोटी पूर्ण संख्या 7 होगी।
भाग देने पर प्राप्त संख्या = 400

6.

इसे पूर्ण संख्या बनाने के लिए 5 से भाग करना होगा।
∴ सबसे छोटी पूर्ण संख्या 5 होगी।
भाग देने पर प्राप्त संख्या = 324

Class 8 Maths 6.3 In Hindi प्रश्न 7.
एक विद्यालय में कक्षा VIII के सभी विद्यार्थियों ने प्रधानमन्त्री राहत कोष में ₹ 2,401 दान में दिए। प्रत्येक विद्यार्थी ने उतने ही रुपये दान में दिए जितने कक्षा में विद्यार्थी थे। कक्षा में विद्यार्थियों की संख्या ज्ञात कीजिए।
हल:
माना कि कक्षा में विद्यार्थियों की संख्या = x है।
अतः विद्यार्थियों द्वारा दान में दी गई कुल धनराशि = ₹ x × x = ₹ x² परन्तु राहत कोष में दिए गए रुपये = ₹ 2,401

अतः कक्षा में विद्यार्थियों की संख्या = 49 है।

Class 8 Maths Chapter 6.3 प्रश्न 8.
एक बाग में 2025 पौधे इस प्रकार लगाए जाने हैं कि प्रत्येक पंक्ति में उतने ही पौधे हों, जितनी पंक्तियों की संख्या हो। पंक्तियों की संख्या और प्रत्येक पंक्ति में पौधों की संख्या ज्ञात कीजिए।
हल:
माना कि पंक्तियों की संख्या = x है।
और प्रत्येक पंक्ति में पौधों की संख्या = x
∴ कुल पौधों की संख्या =x × x = x²
लेकिन पौधों की संख्या = 2025
अब, प्रश्नानुसार,

अतः पंक्तियों की अभीष्ट संख्या = 45 तथा प्रत्येक पंक्ति में पौधों की संख्या = 45

Mp Board Class 8 Maths Solutions English Medium प्रश्न 9.
वह सबसे छोटी वर्ग संख्या ज्ञात कीजिए जो 4, 9 और 10 प्रत्येक से विभाजित हो जाए।
हल:
वह सबसे छोटी वर्ग संख्या जो 4, 9 और 10 प्रत्येक से विभाजित हो जाए, 4, 9 और 10 का ल. स. होगी।

इस संख्या को पूर्ण वर्ग बनाने के लिए 5 से गुणा करना पड़ेगा।
∴ सबसे छोटी अभीष्ट वर्ग संख्या = 180 x 5 = 900

Mp Board Class 8th Maths Solutions प्रश्न 10.
वह सबसे छोटी वर्ग संख्या ज्ञात कीजिए जो प्रत्येक 8, 15 और 20 से विभाजित हो जाए।
हल:
8, 15 और 20 से विभाजित होने वाली पूर्ण वर्ग संख्या 8, 15 और 20 का ल. स. होगी।

इसे पूर्ण बर्ग बनाने के लिए 2 x 3 x 5 = 30 से गुणा करना पड़ेगा।
∴ सबसे छोटी अभीष्ट वर्ग संख्या = 120 x 30
= 3600

पाठ्य-पुस्तक पृष्ठ संख्या # 110

Class 8 Maths Chapter 6.3 In Hindi प्रश्न 1.
क्या आप हमें 5 या 6 अंकों वाली संख्या के वर्गमूल में अंकों की संख्या बता सकते हैं?
हल:
हाँ, 5 या 6 अंकों वाली संख्या के वर्गमूल में अंकों की संख्या 3 होगी।

पाठ्य-पुस्तक पृष्ठ संख्या # 111

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 6.2)

Class 8th Maths Chapter 6 Exercise 6.3 In Hindi प्रश्न 1.
क्या हम कह सकते हैं कि एक पूर्ण वर्ग संख्या में यदि n अंक हैं तो उसके वर्गमूल में \(\frac{n}{2}\) अंक होंगे। यदि n सम है या (\(\frac{n+1}{2}\)) होंगे यदि n विषम है?
हल:
हाँ, हम कह सकते हैं कि यदि n अंकों की कोई पूर्ण वर्ग संख्या है, तो उसके वर्गमूल में –

1. \(\frac{n}{2}\) अंक होंगे यदि n सम है तथा
2. अंक होंगे यदि (\(\frac{n+1}{2}\)) विषम है।

पाठ्य-पुस्तक पृष्ठ संख्या # 112

प्रयास कीजिए (क्रमांक 6.14)

Class 8 Maths Chapter 6 Exercise 6.3 प्रश्न 1.
निम्नलिखित संख्याओं के वर्गमूल में अंकों की संख्या को गणना के बिना ज्ञात कीजिए –

1. 25600
2. 100000000
3. 36864

हल:

1. क्योंकि संख्या 25600 में 5 अंक हैं, अतः इसके वर्गमूल में अंकों की संख्या \(\frac{5+1}{2}\) अर्थात् 3 अंक होगी।
2. क्योंकि संख्या 100000000 में अंकों की संख्या = 9, अतः इसके वर्गमूल में अंकों की संख्या \(\frac{9+1}{2}\) अर्थात् 5 अंक होगी।
3. क्योंकि संख्या 36864 में अंकों की संख्या = 5, अत: इसके वर्गमूल में अंकों की संख्या = \(\frac{5+1}{2}\), अर्थात् 3 अंक होगी।

पाठ्य-पुस्तक पृष्ठ संख्या # 115

प्रयास कीजिए (क्रमांक 6.15)

Class 8 Maths Exercise 6.3 Solutions प्रश्न 1.
निम्नलिखित संख्याओं के निकटतम पूर्ण संख्याओं का अनुमान लगाइए –

1. \(\sqrt { 80 } \)
2. \(\sqrt { 1000 } \)
3. \(\sqrt { 350 } \)
4. \(\sqrt { 500 } \)

हल:
1. हम जानते हैं कि –
64 < 80 < 81
8 < \(\sqrt { 80 } \) < 9
∵ 81, 80 से अधिक निकटतम है।
∴ निकटतम पूर्ण संख्या = 9

2. हम जानते हैं कि –
31² और 32² = 1024
31² \(\sqrt { 1000 } \) < 32
∵ 1000, 1024 के अधिक निकटतम है
∴ निकटतम पूर्ण संख्या = 32

3. हम जानते हैं कि –
18² = 324 और 19² = 361
18 < \(\sqrt { 1000 } \) < 19
क्योंकि 350, 324 की अपेक्षा 361 के अधिक नजदीक है।
∴ निकटतम पूर्ण संख्या = 19

4. हम जानते हैं कि –
22² = 484 और 23² = 529
22 < \(\sqrt { 500 } \) < 23
∴ 500, 529 की अपेक्षा 484 के अधिक निकट है।
∴ निकटतम पूर्ण संख्या = 22

MP Board Class 8th Maths Solutions

In this article, we will share MP Board Class 6th Sanskrit Solutions Chapter 4 सङ्ख्याबोधः Pdf, Class 6 Sanskrit Chapter 4 Question Answer, These solutions are solved subject experts from the latest edition books.

MP Board Class 6th Sanskrit Solutions Surbhi Chapter 4 सङ्ख्याबोधः

MP Board Class 6th Sanskrit Chapter 4 अभ्यासः

MP Board Class 6 Sanskrit Chapter 4 प्रश्न 1.
उचितशब्देन रिक्तस्थानं पुरयत-
(उचित शब्द से खाली स्थान की पूर्ति करो)
(क) एकं ………… अस्ति। (बालकः/बालिका/फलम्)
(ख) तिस्त्रः …………….. सन्ति। (बालिकाः/बालकाः/फलानि)
(ग) सप्तः …………. सन्ति। (गजः/अश्वाः /अजा)
उत्तर:
क) फलम्
(ख) बालिकाः
(ग) अश्वाः

Class 6 Sanskrit Chapter 4 MP Board प्रश्न 2.
उचितपरिवर्तनेन रिक्तस्थानं पूरयत (उचित परिवर्तन से रिक्त स्थान को पूरा करो)
(लता, बाला, छात्र, गृह, बालक, नृपः, अजा, गृह, मनुष्य, गज)
(क) द्वै ………..।
(ख) तिस्त्रः ……….।
(ग) द्वौ ……….।
(घ) चत्वारि …………।
(ङ) षट् ………..।
(च) सप्त …………
(छ) नव …………।
(ज) दश……………।
(झ) पञ्च …………।
(ब) त्रयः ………….।
उत्तर:
(क) लते
(ख) बालाः
(ग) छात्रौः
(घ) गृहाणि
(ङ) बालकाः
(च) नृपाः
(छ) अजाः
(छ) गृहाणि
(झ) मनुष्याः
(ञ) गजाः।

Class 6 Sanskrit Surbhi Chapter 4 प्रश्न 3.
उचितसङ्ख्यापदेन पूरयत(उचित संख्या शब्द से पूरा करो)
(क) …………. फलम्।
(ख) …………. जनाः।
(ग) ………………. लते।
(घ) …………… मयूराः।
(ङ) …………… अश्वौ।
(च) ……………. नेत्रे।
(छ) ……………. पुष्पाणि।
(ज) …………… पर्वताः।
(झ) …………… गजाः।
(ज) ……….. माला।
उत्तर:
(क) एकं
(ख) सप्त
(ग) द्वे
(घ) पञ्च
(ङ) द्वौः
(च) द्वे
(छ) चत्वारि
(ज) अष्ट
(झ) नव
(ञ) एका।

Class 6th Sanskrit Chapter 4 प्रश्न 4.
संस्कृतसङ्ख्याः लिखत(संस्कृत संख्या को लिखो)
(क) २………… (स्त्रीलिङ्ग)
(ख) ४ ……….. (पुल्लिङ्ग)
(ग) २ ………… (नपुंसकलिङ्ग)
(घ) १………… (पुल्लिङ्ग)।
(ङ) ३…………. (स्त्रीलिङ्ग)
(च) २ ………… (नपुंसकलिङ्ग)
(छ) २…………(पुल्लिङ्ग)
(ज) १ ……….. (नपुंसकलिङ्ग)
(झ) ३ ………… (पुल्लिङ्ग)
(ञ) १ ………… (स्त्रीलिङ्ग)
उत्तर:
(क) द्वे
(ख) चत्वारः
(ग) त्रीणि
(घ) एकः
(ङ) तिस्रः
(च) द्वे
(छ) द्वौ
(ज) एकम्
(झ) त्रयः
(ञ) एका।

Sanskrit Class 6 Chapter 4 प्रश्न 5.
गणनां कृत्वां संख्यां लिखत (गिनती करके संख्या लिखो)

उत्तर:
(क) द्वे
(ख) तिस्त्रः
(ग) त्रीणि
(घ) षट्
(ङ) दश
(च) दश
(छ) पञ्च
(ज) अष्ट
(झ) चत्वारः
(ञ) नव।

योग्यताविस्तर :
सङ्ख्यां कण्ठस्थीकुरुत (सङ्ख्याओं को कण्ठाग्र करो)

सङ्ख्याबोधः हिन्दी अनुवाद :

1. एकः वृक्षः।
एक पेड़।

2. द्वौ बालको।
दो बालक।

3. त्रयः अश्वाः।
तीन घोड़े।

4. चत्वारः घटाः।
चार घड़े।

5. पञ्च कुक्कुराः।
पाँच कुत्ते।

6. षट् कपोताः।
छ: कबूतर।

7. सप्त शुकाः।
सात तोते।

8. अष्ट मूषकाः।
आठ चूहे।

9. नव मृगाः।
नौ हिरण।

10. दश मयूराः।
दस मोर।

1. एका बालिका।
एक लड़की।

2. द्वे बालिके।
दो लड़कियाँ।

3. तिस्त्रः मालाः।
तीन मालाएँ।

4. चतस्त्रः अजाः।
चार बकरियाँ।

5. पञ्च चटकाः।
पाँच चिड़ियाँ।

6. षट् घटिकाः।
छः घड़ियाँ।

7. सप्त मक्षिका
सात मक्खियाँ।

8. आष्टा छुरिका
आठ छुरियाँ।

9. नव पिपीलिका
नौ चींदियाँ।

10. दस लड़कियाँ।
दश बालिका

अनुवाद :
पुल्लिङ्ग-एक, दो, तीन, चार, पाँच।
स्त्रीलिङ्ग :
एक, दो, तीन, चार, पाँच।
नपुंसकलिङ्ग :
एक, दो, तीन, चार, पाँच।

नोट :
संस्कृत में-पुल्लिङ्ग, स्त्रीलिङ्ग तथा नपुंसकलिङ्ग में संख्यावाचक शब्दों के रूपों में भेद उपर्युक्त तालिका में देखा जा सकता है।

MP Board Class 6th Sanskrit Solutions

MP Board Class 7th General English Solutions Practice Test Paper 1

Class 7 English Practice Test 1 Question 1.
Match the columns :
(सभी प्रश्न अनिवार्य हैं।)

1. a swarm of – (a) people
2. a herd of – (b) ships
3. a fleet of – (c) cattle
4. a crowd of – (d) bees.

Answer:

1. → (d)
2. → (c)
3. → (b)
4. → (a)

Class 7th English Practice Test 1 Question 2.
Do as directed :
(निर्देशानुसार कार्य करें)

(i) Use articles :
Sindbad set out on his voyage. He went to ……. nearest harbour and got into ……. excellent ship.
Answer:
the, an

(ii) Give plurals of : sheep, mango
(बहुवाहन बनाओ: sheep, mango)
Answer:
Sheep, mangoes.

(iii) Change the form of narration :
(कथन का रूप बदलिए)
Mahek said, “Where do you live ?”
Answer:
Mahek asked where I lived.

(iv) Change in interrrogative :
(प्रश्नवाचक में बदलिए)
Anita likes to ride an elephant.
Answer:
Does Anita like to ride an elephant?

(v) Use the correct form of the verb given :
(दी गई क्रिया का सही रूप प्रयोग कीजिए )
It ……… yesterday.
Answer:
rained.

(vi) Give one word for :
(एक शब्द दीजिए)
One who trains a horse.
Answer:
Horse trainer.

(vii) Change the voice :
(वाच्य बदलिए)
He reads the book.
Answer:
The book is read by him.

(viii) Use appropriate prepositions in the blank spaces :
(रिक्त स्थानों में उपयुक्त भरिए)

(a) The mangoes are liked ………. my wife.
(b) She always agrees ……… my decisions.
Answer:
(a) by
(b) to.

MP Board Class 7th English Question Paper Question 3.
Answer the questions given below (any five)
(नीचे दिए प्रश्नों के उत्तर दीजिए। कोई पाँच)

(i) Why did Bhola and his wife decide to sell the horse?
Answer:
Bhola went to the market to sale the horse.

(ii) What bet did the thugs want to have with Bhola?
Answer:
The thugs asked Bhola if he would sell his mangoes.

(iii) Why is Anita very excited today?
Answer:
Anita is very excited today to see Appu at her door.

(iv) How did Appu drink a bucket full of water?
Answer:
Appu drank a bucketful of water at one go.

(v) Why did Vinod decide to travel by an A/C coach?
(व्हाय डिड विनोद डिसाइड टू ट्रेवल बाइ ए/सी क्लास?)
विनोद ने ए/सी क्लास से जाने का फैसला क्यों किया?
(a) He had a lot of money. (ही हैड अलॉट ऑफ मनी।)
उसके पास बहुत पैसा था।
(b) The train was overcrowded.
(द ट्रेन वॉज़ ओवरक्रोउडेड।)
गाड़ी में ज्यादा भीड़ थी।
(c) He was sick.
(ही वॉज़ सिक।)
वह बीमार था।
Answer:
(b) The train was overcrowded.

(vi) What did Vinod and other passengers think about the young man?
Answer:
They thought that the young man did not have a valid ticket for A/C class.

MP Board Class 7th English Practice Test 1 Question 4.
(a) Write the first four lines of the poem, ‘The Rainbow’.
(‘The Rainbow’ कविता की चार पंक्तियां लीखिए)
Answer:
Students can write themselves.

(b) Who is the poet of this poem?
Answer:
Christina Georgina Rossetti.

MP Board Class 7th English Paper 2023 Question 5.
You are Anita, studying in Govt. Middle School, Bhopal. Write an application to the HeadMaster of your school requesting him to grant you leave for three days.
(आप शासकीय मिडिल स्कूल, भोपाल में अध्ययनरत अनीता हो। अपने स्कूल के प्रधानाध्यापक को तीन दिन के अवकाश हेतु एक प्रार्थना-पत्र लिखिए।)
Answer:
See Application in Grammar section.
OR
You are Ahmad/Mukesh. Write a letter to your sister, asking her to come and visit you on the coming festival.
(आप अहमद/मुकेश हैं। अपनी बहिन को आगामी पर्व पर आपके पास मिलने आने को एक पत्र लिखिए।)
Answer:
See Letters in Grammar section.

Practice Test 1 Class 7 English Question 6.
Write an essay on any one the topics given below :
(नीचे दिए शीर्षकों में से किसी एक पर निबन्ध लिखिए।)
1. My pet animal.
2. The festival of Holi.
3. A visit to a market.
4. Bal-sabha.

Class 7 English Question Paper With Answer Question 7.
Read the given story and answer the questions given below it:
(दी गई कहानी को पढ़िए और इसके नीचे लिखे गए प्रश्नों के उत्तर लिखिए।)

(A) Write answer to the following questions :
(निम्न प्रश्नों के उत्तर लिखिए)

(a) What fell from the tree ?
Answer:
A nut fell down from the tree.

(b) Did Cutie, the cock go with Henny, the hen?
Answer:
Yes, Cutie, the cock also went with Henny the hen.

(c) Where was Rinku, the rabbit standing?
Answer:
Rinku, the rabbit was standing next to a carrot.

(d) Who did they all meet?
Answer:
They all met Furry, the fox.

(e) Where did Furry, the fox take them?
Answer:
Furry, the fox took them into a cave.

(B) Match the two columns :
(दोनों स्तम्भों (कॉलम) का मिलान कीजिए)

1. Henny – (a) the rabbit
2. Furry – (b) the duck
3. Rinku – (c) the fox
4. Ducky – (d) the hen

Answer:

1. → (d)
2. → (c)
3. → (a)
4. → (b)

MP Board Class 7th English Solutions