Exercise 12.2

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 π.

Solution:

Given that,

Radius of cone and the hemisphere = cm

Height of the cone = Radius of the cone = cm

To find,

The volume of the solid in terms of π.

Volume of the given solid = Volume of the solid + Volume of the hemispherical solid

= 13πr2h14πr2h + 23πr332πr3

= 13π12() + π13

= π + π

= π

= π3π

∴ The volume of the solid in terms of is π 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:

Given that,

Diameter of the cylindrical part = cm

Length of the cylindrical part = cm

Height of the conical part = cm

To find,

The volume of the air contained in the model

It can be observed from the figure that the,

Height of each conical part (h1) = cm

Height of cylindrical part (h2) = 12-12 − 2-2 × Height of the conical part

= 12 − (2 × 23)

= 12 −

= cm

Diameter of the cylindrical part = cm

Radius of the cylindrical part = Radius of the conical part

= 3223

Volume of the air in the model = Volume of cylindrical part + 2(Volume of conescircle)

Volume of the cylinder = π322(8-8)

= π × 94−94 × 8

= π

Volume of 2 cones = 2 × 13πr2hr3h

= 2 × 13π × 322−323 × 2

= 23π × 9494 × (2)

= 3-3π

Volume of the air in cuboid = 18-18π + 3π

= π

= 21(22722)

= 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 below figure).

Solution:

Given that,

The length of the gulab jamun = cm

Diameter of the gulab jamun = cm

The volume of syrup in gulab jamun is % to its volume.

To find,

The volume of syrup in gulab jamun.

The diagram of gulab jamun shaped like a cylinder with two hemispherical ends is shown in the following diagram.

From the diagram,

Radius of the cylindricalhemispherical part(r1) = Radius of hemisphericalcylindrical part (r2)

= 2.822.83

= cm

The length of the hemispherical part is the same as that of the of the hemispherical part.

Length of each hemispherical part = cm

Height of the cylindrical part = − 2-2 × Length of hemispherical part

= 5 − 2()

= 5 −

= cm

Volume of one gulab jamun = Volume of cylindrical part + 2(Volume of hemispherical part)

Volume of cylindrical part = πr2hr3h

= π1.42×

= 227−227 1.42 × 2.2

= 13.55213.252

2(Volume of hemispherical part) = 2 × 23−23πr3r2

= 43−43πr3

= 43 × 227−227 × 1.431.42

= 11.49811.898

Volume of one gulab jamun = 13.55213.25 + 11.498

= 25.0526 cm3

Volume of 45 gulab jamuns = × 25.05

= 1127.251126.25 cm3

volume of sugar syrup = % of volume

= 3010030200 × 1127.25

= 338.17339.17 cm3

∴ The volume of sugar syrup found in 45 gulab jamuns is approximately 338 cm3.

Continue

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 Fig. 12.16).

Solution:

Length of the cuboid = cm

Breadth of the cuboid = cm

Height of the cuboid = cm

Radius of conical depression = cm

Height of conical depression = cm

To find,

The volume of wood in entire stand

Volume of the wood = Volume of cuboidwood - 4 × Volume of conescuboid

Volume of cuboid = lbhlb

= × ×

= cm3

4 × Volume of cones = 4 × 1323πr2r3h

= 4383 × 227 × 122123 × 1.4-1.4

= 1.471.37 cm3

Volume of the wood = − 1.47

= cm3 (Upto two decimal places)

∴ The volume of the wood in the entire stand is 523.53 cm3.

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its 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:

Given that,

Height of the conical vessel(h) = cm

Radius of conical vessel(r1) = cm

Radius of lead shots (r2) = cm

One-fourth of water flows out from the vessel.

To find,

The number of lead shots dropped in the vessel.

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

Volume of water flows out = Volume of lead shots dropped in the vessel

14−14 × Volume of the cone = n × 43−43πr23r22

14 × 1314πr12r23h = n × 43πr23

r12h = 1617nr23

Substituting the values we known, we obtain,

52() = n(16 × 0.53)

n = 20016×0.53

n =

∴ The number of lead shots dropped in the vessel is 100.

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 8g mass. (Use π = 3.14)

Solution:

Let there be two cylinders, one is of larger and the other is smaller

Given that,

Height of the larger cylinder h1 = cm

Diameter of the larger cylinder d1 = cm

Height of the smaller cylinderh2 = cm

Radius of the smaller cylinder r2 = cm

To find,

The mass of iron pole.

Radius of larger cylinder (r1) = 242−242

= cm

Total volume of pole = Volume of the larger cylinder + Volume of the smaller cylinder

Volume of the larger cylinder = πr12r13 h1

= π() ×

= π

Volume of the smaller cylinder = πr22 h2

= π × 8292 ×

= × ×

= π

Volume of the iron pole = Volume of the larger cylinder + Volume of the smaller cylinder.

= π + π

= π

= 35520()

= cm3 (Upto one decimal place)

Mass of 1 cm3 iron = g

Mass of 111532.8 cm3 iron = 111532.8 ×

= g (Upto one decimal place)

We know that g = 1 kg

892262.4 g = 892262.4 ÷ g

= kg (Upto three decimal places)

∴ The mass of the iron is 892.262 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:

Given that,

A solid with a right circular cone and a hemisphere.

Height of the conical part of the cylinder = cm

Radius of the conical part of the cylinder = cm

Radius of hemispherical part of the cylinder = cm

Radius of the outer cylinder = cm

Height of the outer cylinder = cm

To find,

The volume of the water left in the cylinder.

Volume of the water left = Volume of – Volume of the

Volume of the cylinder = πr2h

= π(602612)()

= π × × 180

= π cm3

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

= πr2r3h + πr3r2

= 13π(602603) + 23π603602

= 13π() + 23π()

= π

Volume of the water left = π − π

= π

= 360000 ×

= cm3 (Upto two decimal places)

= m3 (Upto three decimal places)

∴ The volume of the water left in the cylinder is 1.131 m3.

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 π = 3.14.

Solution:

Given that,

Height of the cylindrical part = cm

Diameter of cylindrical neck = cm

Diameter of spherical glass vessel = cm

Volume of the water that the vessel holds = cm3

To find it the above given volume is correct.

Volume of the vessel = Volume of + Volume of cylinder

Volume of sphere = π r3r2

= × × 8.5238.522

= cm3 (Upto three decimal places)

Volume of cylinder = πr2hπr3h

= × 1222 ×

= cm3 (Upto two decimal places)

Therefore, volume of the vessel = +

= cm3 (Upto two decimal places)

∴ The volume of the vessel is 346.51 cm3 and hence the child is wrong.