RD Sharma 2020 solution class 9 chapter 21 Surface Area and Volume of a Sphere Exercise 21.2

Exercise 21.2

Page-21.19

Question 1:

Find the volume of a sphere whose radius is:

(i) 2 cm

(ii) 3.5 cm

(iii) 10.5 cm

Answer 1:

In the given problem, we have to find the volume of the sphere of given radii.

(i) Radius of the sphere = 2 cm

So, volume of the sphere =

Therefore, the volume of the sphere of radius 2 cm is

(ii) Radius of the sphere = 3.5 cm

So, volume of the sphere =

Therefore, the volume of the sphere of radius 3.5 cm is

(iii) Radius of the sphere = 10.5 cm

So, volume of the sphere =

Therefore, the volume of the sphere of radius 10.5 cm is

Question 2:

Find the volume of a sphere whose diameter is:

(i) 14 cm

(ii) 3.5 dm

(iii) 2.1 m

Answer 2:

In the given problem, we have to find the volume of the spheres of different diameter.

(i) Diameter of the sphere = 14 cm

So, volume of the sphere =

So, the volume of the sphere of diameter 14 cm is.

(ii) Diameter of the sphere = 3.5 dm

So, volume of the sphere =

So, the volume of the sphere of diameter 3.5 dm is.

(iii) Diameter of the sphere = 2.1 m

So, volume of the sphere =

So, the volume of the sphere of diameter is.

Question 3:

A hemispherical tank has inner radius of 2.8 m. Find its capacity in litres.

Answer 3:

In the given problem, we have a hemispherical tank of the following dimensions:

Inner radius of the tank (r) = 2.8 m

So for the capacity of the tank, we need to find the volume of the hemispherical tank.

Volume of the tank =

Now, as we know,

So,

Therefore, the capacity of the tank is.

Question 4:

A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. find the volume of steel used in making the bowl.

Answer 4:

In the given problem, we have a hemispherical bowl of the following dimensions:

Inner radius of the bowl (r) = 5 cm

Thickness of the iron sheet = 0.25 cm

So, the outer radius of the tank (R) = inner radius + thickness of the sheet

Now, we need to find the volume of steel used to make the bowl. For that we can use the formula for calculating the volume of a hemispherical shell.

Volume of a hemispherical shell =

Therefore, the volume of the iron used to make the tank is

Question 5:

How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?

Answer 5:

In the given problem, we have a lead cube which is remolded into small spherical bullets.

Here, edge of the cube (s) = 22 cm

Diameter of the small spherical bullets (d) = 2 cm

Now, let us take the number of small bullets be x

So, the total volume of x spherical bullets is equal to the volume of the lead cube.

Therefore, we get,

Volume of the x bullets = volume of the cube

Therefore, small bullets can be made from the given lead cube.

Question 6:

A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made.

Answer 6:

In the given problem, Let the number of smaller ladoos which can be made from a bigger ladoo be x.

Here,

The radius of the bigger ladoo (r₁) = 5 cm

The radius of the smaller ladoo (r₂) = 2.5 cm

So, according to the question, volume of the big ladoo is equal to the volume of x small ladoos.

Volume of the big ladoo = volume of the x small ladoos

Therefore, can be made from one ladoo of radius 5 cm.

Question 7:

A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be $\frac{3}{2}$ cm and 2 cm, find the diameter of the third ball.

Answer 7:

In the given problem, we have a spherical ball of lead which is remolded into 3 smaller balls.

Here, Diameter of the bigger ball (d) = 3 cm

Diameter of 1^st small ball (d₁) = cm

Diameter of 2^nd small ball (d₂) = cm

Let the diameter of the 3^rd ball = x cm.

So now,

Volume of the bigger ball = sum of the volumes of the smaller balls

Further, solving for x, we get

Further,

Therefore, the diameter of the third ball is.

Question 8:

A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises $\frac{5}{3}$ cm. Find the radius of the cylinder.

Answer 8:

In the given problem, a sphere is immersed in a water filled cylinder and this leads to a rise in the water level by 5/3 cm. Here, we need to find the radius of the cylinder.

Given here,

Radius of the sphere (r_s) = 5 cm

Rise in the level of water in cylinder (h) = 5/3 cm

So, let us take the radius of the cylinder (r_c) = x cm

Now, according to the problem, the volume of the sphere will be equal to the increase in the volume of the cylinder.

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore, the radius of the cylinder is.

Question 9:

If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?

Answer 9:

In the given problem,

Let us take the radius of 1^st sphere (r₁) = x cm

So, the radius of 2^nd sphere (r₂) = 2x cm

So, volume of the 1^st sphere (V₁) =

Volume of the 2^nd sphere (V₂) =

Now, the ratio of the volumes of two balls =

Therefore, the ratio of the volumes is

Question 10:

A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.

Answer 10:

In the given problem, we need to find the height of water in a given cylinder when a hemispherical bowl full of water is emptied into it.

Here,

Radius of the hemispherical bowl (r_s) = 3.5 cm

Radius of the cylinder (r_c) = 7 cm

So, according to the question, the volume of the water in the hemispherical bowl will be equal to the volume of the water in cylinder.

Let us take the height of the water level in cylinder as h cm

So,

Volume of hemispherical bowl = volume of the water in cylinder

Therefore, the height of the water in the cylinder is equal to .

Question 11:

A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.

Answer 11:

In the given problem, we have a cylinder of the following dimensions:

Let the diameter of the cylinder be d.

So, the height of the cylinder (h) =

Now, the volume of the cylinder (V₁) =

Also, volume of the cylinder is equal to the volume of a sphere of radius 4 cm.

So, volume of the sphere =

Now, it is given that the volume of the sphere is equal to the volume of the cylinder.

So,

So, the diameter of the base of cylinder is 8 cm

Therefore, radius of the base of cylinder is.

Question 12:

A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.

Answer 12:

In the given problem, we need to find the height of water in a given cylinder when a hemispherical bowl full of water is emptied into it.

Here,

Radius of the hemispherical bowl (r_s) = 6 cm

Radius of the cylinder (r_c) = 4 cm

So, according to the question, the volume of the water in the hemispherical bowl will be equal to the volume of the water in cylinder.

Let us take the height of the water level in cylinder as h cm

So,

Volume of hemispherical bowl = volume of the water in cylinder

Therefore, the height of the water in the cylinder is equal to .

Page-21.20

Question 13:

The diameter of a coper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

Answer 13:

In the given problem, the first step is to find the volume of the sphere.

Here,

The diameter of the sphere = 18 cm

So, the radius of the sphere =

Therefore, the volume of the sphere =

Now, the sphere is melted and drawn into a long wire of uniform cross section. So, the volume of both the sphere and the wire will be equal.

For the wire,

Length of the wire = 108 m

= 10800 cm

Let the radius of the wire = x cm

So the volume of the wire =

According to the question,

So, the radius of the wire is 0.3 cm.

Therefore, the diameter of the wire is

Question 14:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.

Answer 14:

In the given problem, we are given a sphere which is melted and drawn into a wire. So, here, the first step is to find the volume of the sphere.

Here,

The diameter of the sphere = 6 cm

So, the radius of the sphere =

Therefore, the volume of the sphere =

Now, the sphere is melted and drawn into a long wire of uniform cross section. So, the volume of both the sphere and the wire will be equal.

For the wire,

Diameter of the wire = 0.2 cm

Radius of the wire = 0.1 cm

Let the length of the wire = x cm

So, the volume of the wire =

Now, volume of sphere = volume of wire,

So, the length of the wire is

Question 15:

The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5cm respectively. If it is melted and recast into a solid cylinder of height $2\frac{2}{3}$cm. Find the diameter of the cylinder.

Answer 15:

In the given problem, we have a spherical shell which is remolded into a solid cylinder. So here, we first find the volume of the spherical shell.

We are given that,

Radius of internal surface of the spherical shell = 3 cm

Radius of external surface of the spherical shell = 5 cm

So,

The volume of the spherical shell =

Where, R = external radius

r = internal radius

So,

Volume of the shell =

Next we find the volume of the cylinder.

Height of the cylinder =

Let us take the radius of the cylinder as r cm.

So,

Volume of the cylinder =

Now, according to the problem, the volume of shell will be equal to the volume of the solid cylinder. So, we get

Volume of spherical shell = Volume of cylinder

Since, the radius is 7 cm; the diameter of the cylinder will be 14 cm.

Therefore, the diameter of the cylinder is.

Question 16:

A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

Answer 16:

In the given problem, we have a lead hemisphere which is remolded into a right circular cone.

Here,

Radius of the hemisphere (r_h) = 7 cm

Height of the cone (h) = 49 cm

So, let the base radius of the cone (r_c) = x cm

Now, the volume of the hemisphere will be equal to volume of the volume of the cone.

So, we get,

Volume of the hemisphere = volume of the cone

Further, solving for x

Further,

Therefore, base radius of the cone is .

Question 17:

A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.

Answer 17:

In the given problem, we have a hollow sphere of given dimensions;

Internal radius of the sphere (r) = 2 cm

External radius of the sphere (R) = 4 cm

Now, the given sphere is molded into a cone,

So, radius of the cone (r_c) = 4 cm

Now, the volume of hollow sphere is equal to the volume of the cone.

So, let the height of cone = h cm

Therefore, we get

Volume of hollow sphere = the volume of cone

Further, solving for h,

So, height of the cone is

Now, slant height (l) of a cone is given by the formula:

So, we get,

Therefore, slant height of the cone is

Question 18:

A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.

Answer 18:

In the given problem, we have a metallic sphere which is remolded into small cones.

Here, radius of the metallic sphere (r_s) = 10.5 cm

Radius of the small cone (r_c) = 3.5 cm

Height of the small cone (h) = 3 cm

Now, let us take the number of small cones be x

So, the total volume of x cones will be equal to the volume of the metallic sphere.

Therefore, we get,

Therefore, small cones can be made from the given bigger metallic sphere.

Question 19:

A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

 

Answer 19:

In the given problem, we are given a cone and a hemisphere which have equal bases and have equal volumes. We need to show that the ratio of their heights.

So,

Let the radius of the cone and hemisphere be x cm.

Also, height of the hemisphere is equal to the radius of the hemisphere.

Now, let the height of the cone = h cm

So, the ratio of the height of hemisphere to the height of the cone =

Here, Volume of the hemisphere = volume of the cone

Therefore, the ratio of the heights is.

Question 20:

The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.

Answer 20:

In the given problem, the largest sphere is carved out of a cube and we have to find the volume of the sphere.

Side of a cube = 10.5 cm

So, for the largest sphere in a cube, the diameter of the sphere will be equal to side of the cube.

Therefore, diameter of the sphere = 10.5 cm

Radius of the sphere = 5.25 cm

Now, the volume of the sphere =

Therefore, the volume of the largest sphere inside the given cube is.

Question 21:

A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between.

Answer 21:

In the given problem, a sphere is present inside a cube such that it touches its sides.

So here,

Side of the cube (s) = 4 cm

Also,

Diameter of the sphere = 4 cm

Radius of the sphere (r) = 2 cm

Now, to find the volume of the gap, we subtract the volume of the sphere from the volume of the cube.

Volume of the gap = volume of cube − volume of sphere

Therefore, the volume of the gap between the cube and the sphere is.

Question 22:

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank.

Answer 22:

In the given problem, we have a hemispherical tank of the following dimensions:

Inner radius of the tank (r) = 1 m

Thickness of the iron sheet = 1 cm

= 0.01 m

So, the outer radius of the tank (R) = inner radius + thickness of the sheet

Now, we need to find the volume of iron used to make the tank. For that we can use the formula for calculating the volume of a hemispherical shell.

Volume of a hemispherical shell =

Therefore, the volume of the iron used to make the tank is

Question 23:

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Answer 23:

In the given problem, we have a capsule in the form of a sphere. We need to find out how much medicine is required to fill in the given capsule. So, for that we find the volume of the capsule.

Here,

Diameter of the capsule = 3.5 mm

Therefore, Radius of the capsule =

mm

Now, Volume of the sphere =

Therefore, the amount of medicine needed to fill the given capsule is

Question 24:

The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Answer 24:

In the given problem, we have two objects; moon and earth.

Let us take the diameter of the earth = d km

Therefore, the radius of the earth = km

So, the volume of the earth (V_e) =

Now, according to the problem, the diameter of moon is one-fourth of the diameter of earth.

So, the diameter of the moon = km

Therefore, the radius of the moon = km

So, the volume of the moon (V_m) =

So, fraction of the volumes =

Therefore, the fraction of the volume of moon to volume of earth is

Question 25:

A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Answer 25:

In the given problem, we are given a cone and a hemisphere which have equal bases and have equal volumes. We need to find the ratio of their heights.

So,

Let the radius of the cone and hemisphere be x cm.

Also, height of the hemisphere is equal to the radius of the hemisphere.

Now, let the height of the cone = h cm

So, the ratio of the height of hemisphere to the height of the cone =

Here, Volume of the hemisphere = volume of the cone

Therefore, the ratio of the heights is.

Question 26:

A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?

Answer 26:

In the given problem, a sphere is immersed in a water filled cylinder and this leads to a rise in the water level by 9 cm. Here, we need to find the radius of the cylinder.

Given here,

Radius of the cylinder (r_c) = 16 cm

Rise in the level of water in cylinder (h) = 9 cm

So, let us take the radius of the sphere (r_s) = x cm

Now, according to the problem, the volume of the ball will be equal to the increase in the volume of the cylinder; as the volume of water replaced by the spherical ball will lead to rise in the level of water.

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore, the radius of the spherical ball is.

Question 27:

A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use $\mathrm{\pi } =22/7$.

Answer 27:

In the given problem, a spherical iron ball is immersed in a water filled cylinder and this leads to a rise in the water level by 6.75 cm. Here, we need to find the radius of the ball.

Given here,

Radius of the cylinder (r_c) = 12 cm

Rise in the level of water in cylinder (h) = 6.75 cm

So, let us take the radius of the spherical ball (r_s) = x cm

Now, according to the problem, the volume of the spherical ball will be equal to the increase in the volume of the cylinder as the volume of water replaced by the ball is increases the level of water in the cylinder.

So,

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore the radius of the spherical ball is.

Question 28:

A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?

Answer 28:

In the given problem, let us take the number of spheres required to raise the level of oil by 2 cm be x.

Also, it is given that,

The radius of sphere (r_s) = 1.5 cm

The radius of the cylinder (r_c) = 6 cm

So, according to the question; the volume of the iron spheres will be equal to the volume of the oil increased.

We get,

Volume of x iron spheres = volume of the oil raised

Therefore, the number of iron spheres required to raise the level of oil by 2 cm is.

Question 29:

A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?

Answer 29:

In the given problem, let us take the change in the level of water be h cm.

Also, it is given that,

The diameter of the spherical balls = 2 cm

So, radius of spherical balls (r_s) = 1 cm

The diameter of the measuring cylinder = 10 cm

So, radius of the measuring cylinder (r_c) = 5 cm

So, according to the question; the volume of the 4 spherical balls will be equal to the volume of the water increased.

We get,

Volume of 4 spheres = volume of the water raised

Therefore, the water level increases by.

Question 30:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3.

Answer 30:

In the given problem, we are given a cone, a hemisphere and a cylinder which stand on equal bases and have equal heights. We need to show that their volumes are in the ratio

1 : 2 : 3

So,

Let the radius of the cone, cylinder and hemisphere be x cm.

Now, the height of the hemisphere is equal to the radius of the hemisphere. So, the height of the cone and the cylinder will also be equal to the radius.

Therefore, the height of the cone, hemisphere and cylinder = x cm

Now, the next step is to find the volumes of each of these.

Volume of a cone (V₁) =

Volume of a hemisphere (V₂) =

Volume of a cylinder (V₃) =

So, now the ratio of their volumes = (V₁) : (V₂) : (V₃)

Hence proved, the ratio of the volumes of the given cone, hemisphere and the cylinder is .

Page-21.21

Question 31:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Answer 31:

In the given problem, a spherical form ball is immersed in a water filled cylinder and this leads to a rise in the water level by 6.75 cm. Here, we need to find the radius of the ball.

Given here,

Radius of the cylinder (r_c) = 12 cm

Rise in the level of water in cylinder (h) = 6.75 cm

So, let us take the radius of the spherical ball (r_s) = x cm

Now, according to the problem, the volume of the spherical ball will be equal to the increase in the volume of the cylinder as the volume of water replaced by the ball is increases the level of water in the cylinder.

So,

Volume of the sphere = increase in volume of the water in cylinder

Further,

Therefore the radius of the spherical ball is.

Question 32:

A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.

Answer 32:

In the given problem, we are given a cone, a sphere and a cylinder which have equal diameter. Also, the height of cylinder and cone is equal to the diameter of the sphere. We need to find the ratio of their volumes.

So,

Let the diameter of the cone, cylinder and sphere be x cm.

Now, the height of the cone and cylinder is equal to the diameter of the hemisphere. Therefore, the height of the cone and cylinder = x cm

Now, the next step is to find the volumes of each of these.

Volume of a cone (V₁) =

Volume of a sphere (V₂) =

Volume of a cylinder (V₃) =

So, now the ratio of their volumes = (V₂) : (V₃) : (V₁)

Therefore, the ratio of the volumes of the given sphere, cylinder and cone is.