NCERT Solutions Maths Class 10 Exercise 13.2

# NCERT Solutions Maths Class 10 Exercise 13.2

## Maths Class 10 Exercise 13.2

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of Ï€.

Solution: Radius of the hemisphere (r) = 1 cm

Volume of the hemisphere = 2/3 × Ï€r3

= 2/3 × Ï€(1)3

= 2Ï€/3 cm3

Radius of the base of the cone (r) = 1 cm

Height of the cone = 1 cm

Volume of the cone = 1/3 × Ï€r2h

= 1/3 × Ï€(1)2 × 1

= Ï€/3 cm3

Volume of the solid = Volume of the hemisphere + Volume of the cone

= 2Ï€/3 + Ï€/3

= Ï€ cm3

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution: For upper and lower conical portions, radius of the base (r) = 1.5 cm

Height of the cones (h1) = 2 cm

Volume of both the cones = 2 × 1/3 × Ï€r2h1

= 2/3 × Ï€(1.5)2 × 2

= 3Ï€ cm3

Radius of the base of the cylinder (r) = 1.5 cm

Height of the cylinder (h2) = 12 – (2 + 2) = 8 cm

Volume of the cylinder = Ï€r2h2 = Ï€(1.5)2 × 8 = 18Ï€ cm3

Volume of the model = Volume of both the cones + Volume of cylinder = 3Ï€ + 18Ï€

= 21Ï€ cm3

= 21 × 22/7

= 66 cm3

3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends, with length 5 cm and diameter 2.8 cm (see figure).

Solution: Volume of a gulab jamun = Volume of two hemispheres + Volume of a cylinder

= 2 × 2/3 × Ï€r3 + Ï€r2h

= 4/3 × Ï€(1.4)3 + Ï€(1.4)2 × 2.2

= Ï€(1.4)2[4 × 1.4/3 + 2.2]

= 1.96Ï€(5.6 + 6.6)/3

= 1.96Ï€(12.2)/3 cm3

Volume of 45 gulab jamuns = 45 × 1.96Ï€(12.2)/3 cm3

= 15 × 1.96Ï€(12.2) cm3

= 15 × 1.96 × 12.2 × 22/7 cm3

= 1127.28 cm3

Volume of the syrup in 45 gulab jamuns = 1127.28 × 30/100

= 338.184 cm3

= 338 cm3 (approx.)

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Solution: The dimensions of the cuboid = 15 cm by 10 cm by 3.5 cm

Volume of the cuboid = l × b × h

= 15 × 10 × 3.5

= 525 cm3

Volume of the conical depression = 1/3 × Ï€r2h

= 1/3 × 22/7 × 0.5 × 0.5 × 1.4

= 11/30 cm3

Volume of the four conical depressions = 4 × 11/30 = 1.47 cm3

Volume of the wood in the entire stand = 525 – 1.47 = 523.53 cm3

5. A vessel is in the form of inverted cone. Its height is 8 cm and the radius of the top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution: Radius of the top of the conical vessel (r) = 5 cm and its height (h) = 8 cm

Volume of the conical vessel = 1/3 × Ï€r2h

= 1/3 × Ï€(5)2 × 8

= 200Ï€/3 cm3

Volume of spherical lead shot = 4/3 × Ï€R3

= 4/3 × Ï€(0.5)3

= Ï€/6 cm3

Volume of water that flows out = ¼ × Volume of the cone

= ¼ × 200Ï€/3 cm3

= 50Ï€/3 cm3

Let the number of lead shots dropped in the vessel be n.

Therefore, n × Ï€/6 = 50Ï€/3

n = 50Ï€/3 × 6/Ï€

n = 100

Hence, 100 lead shots dropped in the vessel.

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use Ï€ = 3.14)

Solution: Base radius of lower cylinder (r) = 24/2 = 12 cm

Height of the lower cylinder (h) = 220 cm

Volume of the lower cylinder = Ï€r2h

= Ï€(12)2 × 220

= 31680Ï€ cm3

Base radius of upper cylinder (R) = 8 cm

Height of upper cylinder (H) = 60 cm

Volume of upper cylinder = Ï€R2H

= Ï€(8)2 × 60

= 3840Ï€ cm3

Volume of the solid iron pole = Volume of lower cylinder + Volume of upper cylinder

= 31680Ï€ + 3840Ï€

= 35520Ï€

= 35520 × 3.14

= 111532.8 cm3

Since the mass of 1 cm3 of iron = 8 g

Therefore, the mass of 111532.8 cm3 of iron = 111532.8 × 8 g

= 892262.4 g

= 892.26 kg

Hence, the mass of the pole is 892.26 kg.

7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution: Radius of the base of the right circular cone (r) = 60 cm

And height of the right circular cone (h1) = 120 cm

Volume of the right circular cone = 1/3 × Ï€r2h1

= 1/3 × Ï€(60)2 × 120

= 144000Ï€ cm3

Radius of the base of the hemisphere (r) = 60 cm

Volume of the hemisphere = 2/3 × Ï€r3

= 2/3 × Ï€(60)3

= 144000Ï€ cm3

Radius of the base of the right circular cylinder (r) = 60 cm

And height of the right circular cylinder (h2) = 180 cm

Volume of the right circular cylinder = Ï€r2h2

= Ï€(60)2 × 180

= 648000Ï€ cm3

Now, volume of water left in the cylinder = Volume of right circular cylinder – (Volume of right circular cone + Volume of hemisphere)

= 648000Ï€ – (144000Ï€ + 144000Ï€)

= 648000Ï€ – 288000Ï€

= 360000Ï€ cm3

= 0.36Ï€ m3

= 0.36 × 22/7 m3

= 1.131 m3 (approx.)

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and Ï€ = 3.14.

Solution:

The amount of water the vessel holds = Volume of spherical part + Volume of cylindrical part

= (4/3) × Ï€r3 + Ï€r2h

= (4/3) × Ï€(8.5/2)3 + Ï€(1)2 × 8

= 4/3 × 3.14 × 4.25 × 4.25 × 4.25 + 3.14 × 8

= 321.39 + 25.12

= 346.51 cm3

Hence, the child is not correct. The correct volume is 346.51 cm3.