Well-designed AP 8th Class Maths Textbook Solutions Chapter 11 Mensuration Exercise 11.4 offers step-by-step explanations to help students understand problem-solving strategies.

## Mensuration Class 8 Exercise 11.4 Solutions – 8th Class Maths 11.4 Exercise Solutions

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.

Solution:

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 finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area ?

Solution:

Diameter of cyliner ‘A’ = 7 cm

Height of cylinder A’ = 14 cm

Diameter of cyliner ‘B’ = 14 cm

Height of cylinder ’B’ = 7 cm

Volume of cylinder ‘B’ is highest because it was more radius.

Now volume of cylinder A = πr²h

= 11 × 7 × 7 = 539 cm³

Volume of cylinder ‘B’ = πr²h

∴ Volume of cylinder ‘B’ is more.

Now Surface area of cylinder A = 2πrh + 2πr² (or) 2πr(h + r)

Surface area of cylinder ‘B’ = 2πr(r + h)

= 44 × 14 = 616 cm²

Result : Cylinder with greater volume also has greater surface area.

Question 3.

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

Solution:

Given volume (v) of cuboid = v

= 900 cm³

base area (A) = lb =x180 cm³

then height = \(\frac{lbh}{lb}\)

= \(\frac{900 \mathrm{~cm}^2}{180 \mathrm{~cm}^2}\) = 5 cm

∴ height of 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 ?

Solution:

Dimensions of cuboid = 60 cm × 54 cm × 30 cm

∴ V_{1} = volume of cuboid

= 60 x 54 x 30 cm^{3} …………(1)

Side of cube (s) = 6 cm

volume of cube = V, = s3 = 6 × 6 × 6 cm^{3} ………….(2)

then no.of cubes that can be placed in cuboid

= \(\frac{\text { Volume of cuboid }}{\text { Volume of cube }}\) = \(\frac{V_1}{V_2}\)

So 450 cubes can be placed in that cuboid.

Question 5.

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

Solution:

Volume of cylinder V = πr²h = 1.54 m3

diameter of base of cylinder = d = 140cm.

∴ radius of base of cylinder = r

= \(\frac{d}{2}\) = \(\frac{140}{2}\) = 70 cm = \(\frac{70}{100}\) = 0.7 cm

∴ V = \(\frac{22}{7}\) × 0.7 × 0.7 × h = 1.54 m³

⇒ h = \(\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}\) = 1m

height of given cylinder = h = 1 meter.

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 ?

Solution:

Shape of milk tank = cylinder

Radius of tank = r = 1.5 m

Length of tank = h = 7 m

then quality of milk that can be stored in tank = Volume of tank in litres

= πr²h

= latex]\frac{22}{7}[/latex] × 1.5 × 1.5 × m³ = 49.5 m³

1m³ = 1000 L

49.5 m3 equals to 49.5 × 1000 L = 49,500 L

49,500 L milk can be stored in it.

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 ?

Solution:

Let

Side of cube (i) = s

then its surface area = 6s^{2};

and its volume = s^{3}

now let side of cube is doubled = 2s

then its surface area = 6(2s)^{2}

= 6(4s^{2}) = 24s^{2}

So surface area increased to

\(\left(\frac{24 s^2}{6 s^2}\right)\) = 4 times ………….(1)

Volume of increased cube

= 2s × 2s × 2s = 8 s³

and volume is increased to \(\frac{8 s^3}{s^3}\) = 8 times …………..(2)

Question 8.

Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.

Solution:

Volume of reservoir = 108 m³

= 108 × 1000 L(∵ m³ = 1000 L)

Volume of water poured in 1 minute = 60 L

then let the reservoir filled in’t’ minutes

Volume of water poured in’t’ minutes = Volume of reservoir

= (60 × t) = 108 × 1000

⇒ t = \(\frac{108 × 1000}{60}\) = 1800 minutes

So t = 1800 minutes, converting in to hours \(\frac{10800}{30}\) = 30 hours

It will take 30 hours to fill the reservoir.