Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Use p = 22/7, unless otherwise indicated
1. Find the volume of a cylinder whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Given,
r = 3.5 cm
h = 40 cm
By using the formula,
Volume of a cylinder = pr
2
h
= 22/7 × 3.5 × 3.5 × 40
= 1540 cm
3
(ii) Given,
r = 2.8 m
h =15 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 2.8 × 2.8 × 15
= 369.6 m
3
2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h)
are:
(i) d = 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m
Solution:
(i) Given,
d = 21cm
r = d/2 = 21/2cm
h = 10 cm.
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 21/2 × 21/2 × 10
= 3465 cm
3
(ii) Given,
Page 2
Use p = 22/7, unless otherwise indicated
1. Find the volume of a cylinder whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Given,
r = 3.5 cm
h = 40 cm
By using the formula,
Volume of a cylinder = pr
2
h
= 22/7 × 3.5 × 3.5 × 40
= 1540 cm
3
(ii) Given,
r = 2.8 m
h =15 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 2.8 × 2.8 × 15
= 369.6 m
3
2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h)
are:
(i) d = 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m
Solution:
(i) Given,
d = 21cm
r = d/2 = 21/2cm
h = 10 cm.
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 21/2 × 21/2 × 10
= 3465 cm
3
(ii) Given,
d = 7 m
r = d/2 = 7/2m
h = 24 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 7/2 × 7/2 × 24
= 924 m
3
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 25 cm.
Find the volume of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 25 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So,
pr
2
= 616
22/7 × r
2
= 616
r
2
= 616 × 7/22
= 196
r = v196
= 14cm
Volume of cylinder = Area of base of right circular cylinder × height
= 616 × 25
= 15400 cm
3
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
the volume of the cylinder.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
Let ‘r’ be the radius of the cylinder.
By using the formula,
Circumference of base of cylinder = 2pr
So,
Page 3
Use p = 22/7, unless otherwise indicated
1. Find the volume of a cylinder whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Given,
r = 3.5 cm
h = 40 cm
By using the formula,
Volume of a cylinder = pr
2
h
= 22/7 × 3.5 × 3.5 × 40
= 1540 cm
3
(ii) Given,
r = 2.8 m
h =15 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 2.8 × 2.8 × 15
= 369.6 m
3
2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h)
are:
(i) d = 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m
Solution:
(i) Given,
d = 21cm
r = d/2 = 21/2cm
h = 10 cm.
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 21/2 × 21/2 × 10
= 3465 cm
3
(ii) Given,
d = 7 m
r = d/2 = 7/2m
h = 24 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 7/2 × 7/2 × 24
= 924 m
3
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 25 cm.
Find the volume of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 25 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So,
pr
2
= 616
22/7 × r
2
= 616
r
2
= 616 × 7/22
= 196
r = v196
= 14cm
Volume of cylinder = Area of base of right circular cylinder × height
= 616 × 25
= 15400 cm
3
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
the volume of the cylinder.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
Let ‘r’ be the radius of the cylinder.
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88 × 7 / 2 × 22
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Volume of cylinder = pr
2
h
= 22/7 × 14 × 14 × 15
= 9240 cm
3
5. A hollow cylindrical pipe is 21 dm long. Its outré and inner diameters are 10 cm
and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
We have,
Length of cylinder = 21 dm = 210 cm
Outer diameter = 10 cm
Outer radius, R = 10/2 = 5cm
Inner diameter = 6 cm
Inner radius, r = 6/2 = 3cm
? Volume of copper used in making the pipe = p (R
2
– r
2
)h
= 22/7 (5
2
- 3
2
) 210
= 22/7 (25-9) 210
= 10560 cm
3
6. Find the (i) curved surface area (ii) total surface area and (iii) volume of a right
circular cylinder whose height is 15 cm and the radius of the base is 7 cm.
Solution:
We have,
Height of cylinder = 15 cm
Radius of base = 7 cm
(i) Curved surface area = 2prh
= 2 × 22/7 × 7 × 15
= 660 cm
2
(ii) Total surface area = 2pr(h+r)
= 2 × 22/7 × 7 (15+7)
= 968 cm
2
(iii) Volume of cylinder = pr
2
h
Page 4
Use p = 22/7, unless otherwise indicated
1. Find the volume of a cylinder whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Given,
r = 3.5 cm
h = 40 cm
By using the formula,
Volume of a cylinder = pr
2
h
= 22/7 × 3.5 × 3.5 × 40
= 1540 cm
3
(ii) Given,
r = 2.8 m
h =15 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 2.8 × 2.8 × 15
= 369.6 m
3
2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h)
are:
(i) d = 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m
Solution:
(i) Given,
d = 21cm
r = d/2 = 21/2cm
h = 10 cm.
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 21/2 × 21/2 × 10
= 3465 cm
3
(ii) Given,
d = 7 m
r = d/2 = 7/2m
h = 24 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 7/2 × 7/2 × 24
= 924 m
3
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 25 cm.
Find the volume of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 25 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So,
pr
2
= 616
22/7 × r
2
= 616
r
2
= 616 × 7/22
= 196
r = v196
= 14cm
Volume of cylinder = Area of base of right circular cylinder × height
= 616 × 25
= 15400 cm
3
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
the volume of the cylinder.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
Let ‘r’ be the radius of the cylinder.
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88 × 7 / 2 × 22
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Volume of cylinder = pr
2
h
= 22/7 × 14 × 14 × 15
= 9240 cm
3
5. A hollow cylindrical pipe is 21 dm long. Its outré and inner diameters are 10 cm
and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
We have,
Length of cylinder = 21 dm = 210 cm
Outer diameter = 10 cm
Outer radius, R = 10/2 = 5cm
Inner diameter = 6 cm
Inner radius, r = 6/2 = 3cm
? Volume of copper used in making the pipe = p (R
2
– r
2
)h
= 22/7 (5
2
- 3
2
) 210
= 22/7 (25-9) 210
= 10560 cm
3
6. Find the (i) curved surface area (ii) total surface area and (iii) volume of a right
circular cylinder whose height is 15 cm and the radius of the base is 7 cm.
Solution:
We have,
Height of cylinder = 15 cm
Radius of base = 7 cm
(i) Curved surface area = 2prh
= 2 × 22/7 × 7 × 15
= 660 cm
2
(ii) Total surface area = 2pr(h+r)
= 2 × 22/7 × 7 (15+7)
= 968 cm
2
(iii) Volume of cylinder = pr
2
h
= 22/7 × 7 × 7 × 15
= 2310 cm
3
7. The diameter of the base of a right circular cylinder is 42 cm and its height is 10
cm. Find the volume of the cylinder.
Solution:
We have,
Diameter of base of cylinder = 42 cm
Radius of base = d/2 = 42/2 = 21cm
Height = 10 cm
? Volume of cylinder = pr
2
h
= 22/7 × 21 × 21 × 10
= 13860 cm
3
8. Find the volume of cylinder, the diameter of whose base is 7 cm and height being
60 cm. Also, find the capacity of the cylinder in litres.
Solution:
We have,
Diameter of base = 7 cm
Radius of base = d/2 = 7/2 cm
Height of cylinder = 60 cm
Volume of cylinder = pr
2
h
= 22/7 × 7/2 × 7/2 × 60
= 2310 cm
3
Capacity of cylinder in litres = 2310 / 1000 = 2.31 litres.
9. A rectangular strip 25 cm× 7 cm is rotated about the longer side. Find the volume
of the solid, thus generated.
Solution:
Given,
Dimensions of rectangular strip = 25 cm × 7 cm
When it rotated about longer side it becomes,
Radius of base = 7 cm
Height of cylinder = 25 cm
Volume of cylinder = pr
2
h
= 22/7 × 7 × 7 × 25
= 3850 cm
3
10. A rectangular sheet of paper, 44 cm × 20 cm, is rolled along its length to form a
Page 5
Use p = 22/7, unless otherwise indicated
1. Find the volume of a cylinder whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Given,
r = 3.5 cm
h = 40 cm
By using the formula,
Volume of a cylinder = pr
2
h
= 22/7 × 3.5 × 3.5 × 40
= 1540 cm
3
(ii) Given,
r = 2.8 m
h =15 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 2.8 × 2.8 × 15
= 369.6 m
3
2. Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h)
are:
(i) d = 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m
Solution:
(i) Given,
d = 21cm
r = d/2 = 21/2cm
h = 10 cm.
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 21/2 × 21/2 × 10
= 3465 cm
3
(ii) Given,
d = 7 m
r = d/2 = 7/2m
h = 24 m
By using the formula,
Volume of cylinder = pr
2
h
= 22/7 × 7/2 × 7/2 × 24
= 924 m
3
3. The area of the base of a right circular cylinder is 616 cm
2
and its height is 25 cm.
Find the volume of the cylinder.
Solution:
We have,
Area of base of right circular cylinder = 616 cm
2
Height of cylinder = 25 cm
Let the radius of cylinder be ‘r’ cm
By using the formula,
Area of base of right circular cylinder = pr
2
So,
pr
2
= 616
22/7 × r
2
= 616
r
2
= 616 × 7/22
= 196
r = v196
= 14cm
Volume of cylinder = Area of base of right circular cylinder × height
= 616 × 25
= 15400 cm
3
4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find
the volume of the cylinder.
Solution:
We have,
Circumference of base of cylinder = 88 cm
Height of cylinder = 15 cm
Let ‘r’ be the radius of the cylinder.
By using the formula,
Circumference of base of cylinder = 2pr
So,
2pr = 88
2 × 22/7 × r = 88
r = 88 × 7 / 2 × 22
= 616/44
= 14cm
Radius of cylinder = 14 cm
? Volume of cylinder = pr
2
h
= 22/7 × 14 × 14 × 15
= 9240 cm
3
5. A hollow cylindrical pipe is 21 dm long. Its outré and inner diameters are 10 cm
and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
We have,
Length of cylinder = 21 dm = 210 cm
Outer diameter = 10 cm
Outer radius, R = 10/2 = 5cm
Inner diameter = 6 cm
Inner radius, r = 6/2 = 3cm
? Volume of copper used in making the pipe = p (R
2
– r
2
)h
= 22/7 (5
2
- 3
2
) 210
= 22/7 (25-9) 210
= 10560 cm
3
6. Find the (i) curved surface area (ii) total surface area and (iii) volume of a right
circular cylinder whose height is 15 cm and the radius of the base is 7 cm.
Solution:
We have,
Height of cylinder = 15 cm
Radius of base = 7 cm
(i) Curved surface area = 2prh
= 2 × 22/7 × 7 × 15
= 660 cm
2
(ii) Total surface area = 2pr(h+r)
= 2 × 22/7 × 7 (15+7)
= 968 cm
2
(iii) Volume of cylinder = pr
2
h
= 22/7 × 7 × 7 × 15
= 2310 cm
3
7. The diameter of the base of a right circular cylinder is 42 cm and its height is 10
cm. Find the volume of the cylinder.
Solution:
We have,
Diameter of base of cylinder = 42 cm
Radius of base = d/2 = 42/2 = 21cm
Height = 10 cm
? Volume of cylinder = pr
2
h
= 22/7 × 21 × 21 × 10
= 13860 cm
3
8. Find the volume of cylinder, the diameter of whose base is 7 cm and height being
60 cm. Also, find the capacity of the cylinder in litres.
Solution:
We have,
Diameter of base = 7 cm
Radius of base = d/2 = 7/2 cm
Height of cylinder = 60 cm
Volume of cylinder = pr
2
h
= 22/7 × 7/2 × 7/2 × 60
= 2310 cm
3
Capacity of cylinder in litres = 2310 / 1000 = 2.31 litres.
9. A rectangular strip 25 cm× 7 cm is rotated about the longer side. Find the volume
of the solid, thus generated.
Solution:
Given,
Dimensions of rectangular strip = 25 cm × 7 cm
When it rotated about longer side it becomes,
Radius of base = 7 cm
Height of cylinder = 25 cm
Volume of cylinder = pr
2
h
= 22/7 × 7 × 7 × 25
= 3850 cm
3
10. A rectangular sheet of paper, 44 cm × 20 cm, is rolled along its length to form a
cylinder. Find the volume of the cylinder so formed.
Solution:
We have,
Dimensions of rectangular sheet = 44cm × 20cm
When it rolled along its length it becomes,
Radius of base = length/2p
= 44×7 / 2×22
= 7cm
Height of cylinder = 20 cm
? Volume of cylinder = pr
2
h
= 22/7 × 7 × 7 × 20
= 3080 cm
3
11. The volume and the curved surface area of cylinder are 1650 cm
3
and 660
cm
2
respectively. Find the radius and height of the cylinder.
Solution:
We have,
Volume of cylinder = 1650 cm
3
Curved surface area = 660 cm
2
Volume of cylinder/curved surface area = 1650/660
pr
2
h/ 2prh = 1650/660
r/ 2 = 5/2
r = 5cm
Surface area = 660 cm
2
2prh = 660
2 × 22/7 × 5 × h = 660
h = 660×7 / 2×22×5
= 4620/220
= 21cm
? Radius = 5cm and height = 21cm
12. The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio
5:3. Calculate the ratio of their volumes.
Solution:
We have,
Ratio of radii of two cylinder = 2:3
Radius of cylinder 1 = r
1
Radius of cylinder 2 = r
2
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.81/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 have been prepared according to the Class 8 exam syllabus. Information about Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 covers topics
like and Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2.
Introduction of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2
In this doc you can find the meaning of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 defined & explained in the simplest way possible. Besides explaining types of
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 theory, EduRev gives you an ample number of questions to practice Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
MCQs
,Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 | Mathematics (Maths) Class 8
,past year papers
,Viva Questions
,study material
,Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 | Mathematics (Maths) Class 8
,Previous Year Questions with Solutions
,video lectures
,Free
,Exam
,Summary
,Extra Questions
,mock tests for examination
,Semester Notes
,Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 | Mathematics (Maths) Class 8
,Important questions
,Sample Paper
,shortcuts and tricks
,practice quizzes
,ppt
,Objective type Questions
;
Additional Information about Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 for Class 8 Preparation
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Free PDF Download
The Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 is an invaluable resource that delves deep into the core of the Class 8 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 now and kickstart your journey towards success in the Class 8 exam.
Importance of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2
The importance of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Notes
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Notes on EduRev are your ultimate resource for success.
Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Class 8 Questions
The "Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 on the App
Students of Class 8 can study Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Mensuration – III (Surface Area and Volume of a Right Circular Cylinder) Exercise 22.2 is prepared as per the latest Class 8 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup