**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€*r*^{3}

= 2/3
× Ï€(1)^{3}

= 2Ï€/3 cm^{3}

** **Radius of the base of the cone (

*r*) = 1 cm

Height
of the cone = 1 cm

Volume
of the cone = 1/3 × Ï€*r*^{2}*h*

=
1/3 × Ï€(1)^{2 }× 1

= Ï€/3 cm^{3}

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

= 2Ï€/3 + Ï€/3

=
Ï€ cm^{3}

**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 (*h*_{1}) = 2
cm

Volume
of both the cones = 2 × 1/3 × Ï€*r*^{2}*h*_{1}

=
2/3 × Ï€(1.5)^{2 }× 2

=
3Ï€ cm^{3}

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

Height
of the cylinder (*h*_{2}) =
12 – (2 + 2) = 8 cm

Volume
of the cylinder = Ï€*r*^{2}*h*_{2} = Ï€(1.5)^{2
}× 8 = 18Ï€ cm^{3}

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

=
21Ï€ cm^{3}

= 21
× 22/7

=
66 cm^{3}

**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€*r*^{3 }+^{ }Ï€*r*^{2}*h*

=
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 cm^{3}

Volume
of 45 gulab jamuns = 45 × 1.96Ï€(12.2)/3
cm^{3}

=
15 × 1.96Ï€(12.2) cm^{3}

=
15 × 1.96 × 12.2 ×
22/7 cm^{3}

=
1127.28 cm^{3}

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

=
338.184 cm^{3}

= 338 cm^{3} (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 cm^{3}

Volume
of the conical depression = 1/3 × Ï€*r*^{2}*h*

=
1/3 × 22/7 × 0.5 × 0.5 × 1.4

=
11/30 cm^{3}

Volume
of the four conical depressions = 4 × 11/30 = 1.47 cm^{3}

Volume
of the wood in the entire stand = 525 – 1.47 = 523.53 cm^{3}

**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€*r*^{2}*h*

=
1/3 × Ï€(5)^{2 }× 8

= 200Ï€/3 cm^{3}

Radius of the spherical lead shot (R) = 0.5 cm

Volume
of spherical lead shot = 4/3 × Ï€R^{3}

=
4/3 × Ï€(0.5)^{3}

=
Ï€/6 cm^{3}

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

=
¼ × 200Ï€/3 cm^{3}

=
50Ï€/3 cm^{3}

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 cm ^{3}**

**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€*r*^{2}*h*

=
Ï€(12)^{2}* *× 220

=
31680Ï€ cm^{3}

Base radius of upper cylinder (R) = 8 cm

Height
of upper cylinder (H) = 60 cm

Volume
of upper cylinder = Ï€*R*^{2}*H*

=
Ï€(8)^{2}* *× 60

=
3840Ï€ cm^{3}

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

= 31680Ï€ + 3840Ï€

=
35520Ï€

=
35520 × 3.14

=
111532.8 cm^{3}

Since
the mass of 1 cm^{3} of iron = 8 g

Therefore,
the mass of 111532.8 cm^{3} 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 (*h*_{1}) = 120 cm

Volume
of the right circular cone = 1/3 × Ï€*r*^{2}*h*_{1}

=
1/3 × Ï€(60)^{2}* *× 120

=
144000Ï€ cm^{3}

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

Volume
of the hemisphere = 2/3 × Ï€*r*^{3}

=
2/3 × Ï€(60)^{3}

= 144000Ï€
cm^{3}

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

And
height of the right circular cylinder (*h*_{2}) =
180 cm

Volume
of the right circular cylinder = Ï€*r*^{2}*h*_{2}

=
Ï€(60)^{2 }× 180

=
648000Ï€ cm^{3}

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€ cm^{3}

=
0.36Ï€ m^{3}

=
0.36 × 22/7 m^{3}

=
1.131 m^{3} (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
cm ^{3}. 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€r^{3} + Ï€r^{2}h

=
(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
cm^{3}

Hence,
the child is not correct. The correct volume is 346.51 cm^{3}.