These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.4

Question 1.

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

Answer:

(a) Volume

(b) Surface area

(c) volume

Question 2.

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

Answer:

Volume of cylinder B will be greater than the volume of cylinder A.

For cylinder A

r = \(\frac { 7 }{ 2 }\) cm, h = 14 cm

Volume of the cylinder A = πr^{2} h cm units

= \(\frac{22}{7}\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 14 cm^{3}

= 11 × 7 × 7 cm^{3} = 539 cm^{3}

Surface Area of cylinder A = 2π r(h + r)

= 2 × \(\frac{22}{7}\) × \(\frac { 7 }{ 2 }\) 14 + \(\frac { 7 }{ 2 }\)

= 22 × \(\frac { 35 }{ 2 }\) = 385 cm^{2}

For cylinder B

Here r = \(\frac { 14 }{ 2 }\) = 7 cm , h = 7 cm

Volume of the cylinder B = π r^{2} h

= \(\frac{22}{7}\) × 7 × 7 × 7 = 1078 cm^{3}

Surface area of cylinder = 2 π r (h + r)

= 2 × \(\frac{22}{7}\) × 7(7 + 7) = 44 × 14 = 616 cm^{2}

Thus, cylinder B has greater volume than cylinder A.

Thus, the cylinder of great capacity has greater surface area.

Question 3.

Find the height of a cuboid whose base area is 180 cm^{2} and volume is 900 cm^{3}?

Answer:

Base area of the cuboid = 180 cm^{2}

∴ lb = 180

Volume of the cuboid = 900 cm^{3}

Base Area × height = 900

lb × h = 900

180 × h = 900

h= \(\frac{900}{180}\) = 5

∴ Height of the cuboid = 5 cm

Question 4.

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

Answer:

Here l = 60 cm, b = 54 cm, h = 30 cm

Volume of the cuboid = l × b × h cm^{3} = 60 × 54 × 30 cm^{3}

Side of a cube = 6 cm

Volume of cube = l^{3} cm^{3} = 6 × 6 × 6 cm^{3}

Required number of small cubes

= \(\frac{\text { Volume of the cuboid }}{\text { Volume of the small cube }}\)

= \(\frac{60 \times 54 \times 30}{6 \times 6 \times 6}\) = 10 × 9 × 5 = 450

Number of small cubes = 450.

Question 5.

Find the height of the cylinder whose volume is 1.54 m^{3} and diameter of the base is 140 cm?

Answer:

Radius of the cylinder = \(\frac { 140 }{ 2 }\) = 70 cm

Let the height of the cylinder be “h”

Volume of the cylinder = 1.54 m^{3}

= 1.54 × 100 × 100 × 100 cm^{3}

πr^{2} h = 154 × 100 × 100 cm^{3}

h = \(\frac{22}{7}\) × 70 × 70 × h = 154 × 100 × 100

= 100 cm = 1 m

∴ height of the cylinder = 1m

Question 6.

A milk tank is in the form of 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?

Answer:

Radius of the tank = 1.5 m = \(\frac{15}{10}\) m

Length of the tank = 7 m

Volume of the tank = πr^{2} h cu. m

= \(\frac{22}{7}\) × \(\frac{15}{10}\) × \(\frac{15}{10}\) × 7m^{3}

= \(\frac{22 \times 15 \times 15}{10 \times 10}\) × 1000 litres

\(\frac{495}{10}\) × 1000 litres = 49500 litres

Quantity of milk in the tank = 49500 litres

Question 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?

Answer:

Let the original edge be ‘x’

∴ increased edge 2x

∴ original surface area = 6x^{2}

increased surface area = 6(2x)^{2} = 24x^{2}

= \(\frac{\text { Increase surface area }}{\text { Original surface area }}\) = \(\frac{24 x^{2}}{6 x^{2}}\)

Surface area is increased 4 times.

(ii) Original volume = x^{3}

Increased volume = (2x)^{3}

= \(\frac{\text { Increased volume }}{\text { Original volume }}\) = \(\frac{8 x^{3}}{x^{3}}\)

Volume is increased by 8 times

Question 8.

Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m^{3}. Find the number of hours it will take to fill the reservoir.

Answer:

Volume of the reservoir = 108 m^{3}

= 108 × 1000 litres = 108000 litres

Amount of water poured in one minute = 60 litres

Time taken to fill the reservoir = \(\frac{108000}{60}\)

= 1800 minutes \(\frac{1800}{60}\) hours = 30 hours

The required number of hours = 30.