Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 11 Mensuration Ex 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?

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 cyclinder 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?

Solution:

In volume of a cylinder, the radius is multiplied by squaring it.

∴ Volume of cylinder B will be more than that of A.

Volume of cylinder A

Radius (r) = cm. Height (h) = 14 cm

∴ Volume of a cylinder A = πr^{2}h

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

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

= 11 × 7 × 7 cm^{3} = 539 cm^{3}

Volume of cyclinder B

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

∴ Volume of the cylinder B = πr^{2}h

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

= 22 × 7 × 7 = 1078 cm^{3}

Thus, cylinder B has greater volume.

Now,

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

= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × [\(\frac{7}{2}\) + 14] cm^{2}

= 22 × \(\frac{35}{2}\) cm^{2} = 11 × 35 cm^{2} = 385 cm^{2}

Surface area of cylinder B = 2πrh(r + h)

= 2 × \(\frac{22}{7}\) × 7 × [7 + 7] cm^{2}

= 2 × 22 × [14] cm^{2} = 616 cm^{2}

Thus, the cylinder of greater capacity has greater (more) surface area.

Question 3.

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

Solution:

Let the height of the cuboid = h cm

∴ Base area × Height = Volume

or 180 × h = 900

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

Hence, the required 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 he placed in the given cuboid?

Solution:

Volume of the cuboid

= 60 cm × 54 cm × 30 cm

= (60 × 54 × 30) cm^{3}

Volume of the small cube = (6 × 6 × 6) cm^{3}

= 10 × 9 × 5 = 450

Thus, 450 small cubes can be placed in the given 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:

Diameter = 140 cm

⇒ Radius (r) = \(\frac{140}{2}\) = 70 cm

Let height of the cylinder be h m.

∴ Volume = πr^{2}h = \(\frac{22}{7}\) × (70)^{2} × h

= \(\frac{22}{7}\) × 70 × 70 × h

Since, volume of the cylinder is 1.54 m^{3}.

∴ 22 × 10 × 70 × h = 1.54 × 1000000

= 1540000

or h = \(\frac{1540000}{22×10×70}\) = 100 cm or 1.

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 lit tank?

Solution:

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

Height = 7m

∴ Volume of the tank = πr^{2}h

= \(\frac{22}{7}\) × (\(\frac{15}{10}\))^{2} × 7 m^{3}

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

= \(\frac{11×3×3}{2}\) m^{3} = \(\frac{99}{2}\) m^{3} = 49.5 m^{3}

∵ 1 m^{3} = 1000 litres

∴ Quantity of milk in the tank

= 49.5 × 1000 litres

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

Question 7.

If each of a cube is doubled,

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

(ii) How many times will its volume increase?

Solutoion:

Let the original edge be x

∴ Increased edge = 2x

(i) ∴ Original S.A. = 6x^{2}

Increased SA. = 6(2x)^{2} = 24x^{2}

Since, \(\frac{24 x^{2}}{6 x^{2}}\) = 4

∴ Surface area is increased by 4 times

(ii) Original volume = x^{3}

Increased volume = (2x)^{3} = 8x^{3}

∵ \(\frac{8 x^{3}}{x^{3}}\) = 8

∴ 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?

Solution:

Volume of the reservoir

= 108 m^{3
}

∵ 1 m^{3} = 1000 litres

∴ Capacity of the reservoir

108 × 1000 litres = 108000 litres

Amount of water poured in 1 minute

= 60 litres

∴ Amount of water to be poured in 1 hour

= 60 × 60 litres

Thus, number of hours required to fill the reservoir = \(\frac{108000}{60×60}\) = 30

∴ The required number of hours = 30