NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.4

Class 8 Chapter 11 Mensuration questions are provided here with detailed solutions, which are handy for revising the concepts. Class 8 CBSE Mathematics Chapter 11 Exercise 11.4 questions are solved by subject experts using a step-by-step problem-solving approach. These NCERT Solutions will help in boosting the confidence level among students. Download free NCERT Solutions and practise offline to score more in the exams.



NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.4

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Access Answers to NCERT Class 8 Maths Chapter 11 Mensuration Exercise 11.4 Page number 191

Exercise 11.1 Solutions: 5 Questions (Long answers)

Exercise 11.2 Solutions: 11 Questions (Long answers)

Exercise 11.3 Solutions: 10 Questions (2 Short answers, 8 Long answers)

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.4

1. Given a cylindrical tank, in which situation will you find the surface area, and in which situation will you find the volume

(a) To find how much it can hold.

(b) Number of cement bags required to plaster it.

(c) To find the number of smaller tanks that can be filled with water from it.

Solution: We find the area when a region is covered by a boundary, such as the outer and the inner surface area of a cylinder, a cone, a sphere and the surface of a wall or floor.

When the amount of space is occupied by an object, such as water, milk, coffee, tea, etc., then we have to find out the volume of the object.

(a) Volume (b) Surface area (c) Volume

2. Diameter of cylinder A is 7 cm, and the height is 14 cm. The diameter of cylinder B is 14 cm, and the height is 7 cm. Without doing any calculations, can you suggest whose volume is greater? Verify it by finding the volume of both cylinders. Check whether the cylinder with greater volume also has a greater surface area.

Ncert solution class 8 chapter 11-24

Solution: Yes, we can say that volume of cylinder B is greater since the radius of cylinder B is greater than that of cylinder A.

Find the Volume for cylinders A and B

The diameter of cylinder A = 7 cm

The radius of cylinder A = 7/2  cm

And the height of cylinder A = 14 cm

The volume of cylinder A = πr2h

=  (22/7 )×(7/2)×(7/2)×14 = 539

The volume of cylinder A is 539  cm3

Now, the diameter of cylinder B = 14 cm

The radius of cylinder B =  14/2 = 7 cm

And the height of cylinder B = 7 cm

The volume of cylinder B = πr2h

= (22/7)×7×7×7 = 1078

The volume of  cylinder B is 1078 cm3

Find the surface area for cylinders A and B

The surface area of cylinder A =  2πr(r+h )

= 2 x 22/7 x 7/2 x (7/2 + 14) = 385

The surface area of cylinder A is385 cm2

The surface area of cylinder B =  2πr(r+h)

= 2×(22/7)×7(7+7) = 616

The surface area of cylinder B is 616 cm2

Yes, a cylinder with greater volume also has a greater surface area.

3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?

Solution: Given, the base area of the cuboid = 180 cm2 and the volume of the cuboid = 900 cm3

We know that the volume of the cuboid = lbh

900 = 180×h (substituting the values)

h= 900/180 = 5

Hence, the height of the cuboid is 5 cm.

4. A cuboid is of dimensions 60 cm×54 cm×30 cm. How many small cubes with sides 6 cm can be placed in the given cuboid?

Solution:  Given, the length of the cuboid, l = 60 cm, the breadth of the cuboid, b = 54 cm and

The height of the cuboid, h = 30 cm

We know that, the volume of the cuboid = lbh = 60 ×54×30  = 97200 cm3

And the volume of the cube = (Side)3

= 6×6×6 = 216 cm3

Also, the number of small cubes =  the volume of the cuboid / the volume of the cube

= 97200/216

= 450

Hence, the required cubes are 450.

5. Find the height of the cylinder whose volume is 1.54 m3 and the diameter of the base is 140 cm.

Solution:

Given: The volume of the cylinder = 1.54 m3 and the diameter of the cylinder = 140 cm

Radius  ( r )=  d/2 = 140/2  = 70 cm

The volume of the cylinder = πr2h

1.54 = (22/7)×0.7×0.7×h

After simplifying, we get the value of h, which is

h = (1.54×7)/(22×0.7×0.7)

h = 1

Hence, the height of the cylinder is 1 m.

6. A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank.

Ncert solution class 8 chapter 11-25

Solution: Given, the radius of the cylindrical tank, r = 1.5 m and the height of the cylindrical tank, h  = 7 m

The volume of the cylindrical tank, V =  πr2h

= (22/7)×1.5×1.5 ×7

= 49.5  cm3

= 49.5×1000 liters = 49500 liters

[∵ 1 m3= 1000 litres]

Hence, the required quantity of milk is 49500 litres.

7. If each edge of a cube is doubled,

(i) how many times will its surface area increase?

(ii) how many times will its volume increase?

Solution:(i) Let the edge of the cube be “ l” .

The formula for the surface area of the cube, A =  6 l2

When the edge of the cube is doubled, then

The surface area of the cube, say A’ = 6(2l)2 = 6×4l2 = 4(6 l2)

A’ = 4A

Hence, the surface area will increase by four times.

(ii) The volume of the cube, V = l3

When the edge of the cube is doubled, then

The volume of cube, say V’ = (2l)3 = 8( l3)

V’ = 8×V

Hence, the volume will increase 8 times.

8. Water is poured into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m^3, find the number of hours it will take to fill the reservoir.

Solution:

Given, the volume of the reservoir = 108 m3

The rate of pouring water into the cuboidal reservoir = 60 litres/minute

=  60/1000  m3per minute

Since 1 litre = (1/1000 )m3

=  (60×60)/1000 m3 per hour

Therefore, (60×60)/1000 m3 water filled in reservoir will take = 1 hour

Therefore, 1 m3 water filled in reservoir will take =  1000/(60×60)  hours

Therefore, 108 m3 water filled in reservoir will take = (108×1000)/(60×60)  hours = 30 hours

Answer: It will take 30 hours to fill the reservoir.




Chapter 11 Mensuration Exercise 11.4 deals with the amount of space occupied by a three-dimensional object. Students will learn how to find the volume of a Cube, Cuboid, and Cylinder and the difference between Volume and Capacity. Download and practise NCERT Class 8 Maths Solutions offline and sharpen your solving skills.

Also, explore – 

NCERT Solutions for Class 8 Maths Chapter 11

NCERT Solutions for Class 8 

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