Rd Sharma 2018 Solutions for Class 9 Math Chapter 21 Surface Area And Volume Of A Sphere are provided here with simple step-by-step explanations. These solutions for Surface Area And Volume Of A Sphere are extremely popular among Class 9 students for Math Surface Area And Volume Of A Sphere Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2018 Book of Class 9 Math Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 21.20:

Question 1:

Find the volume of a sphere whose radius is:

(i) 2 cm

(ii) 3.5 cm

(iii) 10.5 cm

Answer:

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

Page No 21.20:

Question 2:

Find the volume of a sphere whose diameter is:

(i) 14 cm

(ii) 3.5 dm

(iii) 2.1 m

Answer:

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.

Page No 21.20:

Question 3:

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

Answer:

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.

Page No 21.20:

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:

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

Page No 21.20:

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:

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.

Page No 21.20:

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:

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.

Page No 21.20:

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 32 cm and 2 cm, find the diameter of the third ball.

Answer:

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 1st small ball (d1) = cm

Diameter of 2nd small ball (d2) = cm

Let the diameter of the 3rd 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.

Page No 21.20:

Question 8:

A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 53 cm. Find the radius of the cylinder.

Answer:

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 (rs) = 5 cm

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

So, let us take the radius of the cylinder (rc) = 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.

Page No 21.20:

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:

In the given problem,

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

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

So, volume of the 1st sphere (V1) =

Volume of the 2nd sphere (V2) =

Now, the ratio of the volumes of two balls =

Therefore, the ratio of the volumes is

Page No 21.20:

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:

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 (rs) = 3.5 cm

Radius of the cylinder (rc) = 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 .

Page No 21.20:

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:

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.

Page No 21.20:

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:

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 (rs) = 6 cm

Radius of the cylinder (rc) = 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 No 21.21:

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:

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

Page No 21.21:

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:

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

Page No 21.21:

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 223 cm. Find the diameter of the cylinder.

Answer:

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.

Page No 21.21:

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:

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

Here,

Radius of the hemisphere (rh) = 7 cm

Height of the cone (h) = 49 cm

So, let the base radius of the cone (rc) = 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 .

Page No 21.21:

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:

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 (rc) = 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

Page No 21.21:

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:

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

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

Radius of the small cone (rc) = 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.

Page No 21.21:

Question 19:

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

 

Answer:

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.

Page No 21.21:

Question 20:

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

Answer:

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.

Page No 21.21:

Question 21:

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

Answer:

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.

Page No 21.21:

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:

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

Page No 21.21:

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:

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

Page No 21.21:

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:

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 (Ve) =

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 (Vm) =

So, fraction of the volumes =

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

Page No 21.21:

Question 25:

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

Answer:

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.

Page No 21.21:

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:

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 (rc) = 16 cm

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

So, let us take the radius of the sphere (rs) = 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.

Page No 21.21:

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 π =22/7.

Answer:

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 (rc) = 12 cm

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

So, let us take the radius of the spherical ball (rs) = 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.

Page No 21.21:

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:

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 (rs) = 1.5 cm

The radius of the cylinder (rc) = 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.

Page No 21.21:

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:

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 (rs) = 1 cm

The diameter of the measuring cylinder = 10 cm

So, radius of the measuring cylinder (rc) = 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.

Page No 21.21:

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:

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 No 21.22:

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:

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 (rc) = 12 cm

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

So, let us take the radius of the spherical ball (rs) = 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.

Page No 21.22:

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:

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.



Page No 21.25:

Question 1:

Find the surface area of a sphere of radius 14 cm.

Answer:

In the given problem, we have to find the surface area of a sphere of a given radii.

Radius of the sphere (r) = 14 cm

So, surface area of the sphere =

Therefore, the surface area of the given sphere of radius 14 cm is.

Page No 21.25:

Question 2:

Find the total surface area of a hemisphere of radius 10 cm.

Answer:

In the given problem, we have to find the total surface area of a hemisphere of a given radii.

Radius of the hemisphere (r) = 10 cm

So, total surface area of the hemisphere =

Therefore, the total surface area of the given hemisphere of radius 10 cm is.

Page No 21.25:

Question 3:

Find the radius of a sphere whose surface area is 154 cm2.

Answer:

In the given problem, we have to find the radius of a sphere whose surface area is given.

Surface area of the sphere (S) = 154 cm2

Let the radius of the sphere be r cm

Now, we know that surface area of the sphere =

So,

Further, solving for r

Therefore, the radius of the given sphere is.

Page No 21.25:

Question 4:

The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.

Answer:

In the given problem, the area available for the motorcyclist for riding will be equal to the surface area of the hollow sphere. So here, we have to find the surface area of a hollow sphere of a given diameter.

Diameter of the sphere (d) = 7 m

So, surface area of the sphere =

Therefore, the area available for the motorcyclist for riding is.

Page No 21.25:

Question 5:

Find the volume of a sphere whose surface area is 154 cm2.

Answer:

In the given problem, we have to find the volume of a sphere whose surface area is given.

So, let us first find the radius of the given sphere.

Surface area of the sphere (S) = 154 cm2

Let the radius of the sphere be r cm

Now, we know that surface area of the sphere =

So,

Further, solving for r

Therefore, the radius of the given sphere is 3.5 cm.

Now, the volume of the sphere =

Therefore, the volume of the given sphere is.

Page No 21.25:

Question 6:

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?

Answer:

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

Here, edge of the cube (s) = 44 cm

Diameter of the small spherical bullets (d) = 4 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.



Page No 21.26:

Question 7:

If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.

Answer:

In the given problem, we are given a cone and a sphere which have equal volumes. The dimensions of the two are;

Radius of the cone (rc) = r

Radius of the sphere (rs) = 2r

Now, let the height of the cone = h

Here, Volume of the sphere = volume of the cone

Further, solving for h

Therefore, the height of the cone is.

Page No 21.26:

Question 8:

If a hollow sphere of internal and external diameters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.

Answer:

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

Internal diameter of the sphere (d) = 4 cm

External diameter of the sphere (D) = 8 cm

Now, the given sphere is molded into a cone,

Diameter of the base of cone (dc) = 8 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 cone = the volume of hollow sphere

Further, solving for h,

So, height of the cone is

Page No 21.26:

Question 9:

The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height of the cone.

Answer:

In the given problem, we are given a sphere and a cone of the following dimensions:

Radius of the sphere (rs) = 5 cm

So, surface area of the sphere =

Also, radius of the cone base (rc) = 4 cm

So, curved surface area of the cone =

Now, it is given that the surface area of the sphere is 5 times the curved surface are of the cone. So, we get

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

So, let us take the height of the cone as h,

We get,

Squaring both sides,

Further, solving for h

Therefore, height of the cone is.

Page No 21.26:

Question 10:

If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.

Answer:

In the given problem, we are given a sphere inscribed in a cube. So, here we need to find the ratio between the volume of a cube and volume of sphere. This means that the diameter of the sphere will be equal to the side of the cube. Let us take the diameter as d.

Here,

Volume of a cube (V1) =

Volume of a sphere (V2) =

Now, the ratio of the volume of sphere to the volume of the cube =

So, the ratio of the volume of cube to the volume of the sphere is.

Page No 21.26:

Question 1:

Mark the correct alternative in each of the following:

In a sphere the number of faces is

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

A sphere has only a single face. Since there are no sides of a sphere, it has a single continuous face.

Therefore, the correct option is (a)

Page No 21.26:

Question 2:

The total surface area of a hemisphere of radius r is

(a) πr2

(b) 2πr2

(c) 3πr2

(d) 4πr2

Answer:

The curved surface area of a hemisphere of radius r is. So, the total surface area of a hemisphere will be the sum of the curved surface area and the area of the base.

Total surface area of a hemisphere of radius r =

Therefore, the correct option is (c)

Page No 21.26:

Question 3:

The ratio of the total surface area of a sphere and a hemisphere of same radius is

(a) 2 : 1

(b) 3 : 2

(c) 4 : 1

(d) 4 : 3

Answer:

In the given question,

The total surface area of a sphere (S1) =

The total surface area of a hemisphere (S2) =

So the ratio of the total surface area of a sphere and a hemisphere will be,

Therefore, the ratio of the surface areas is. So, the correct option is (d)

Page No 21.26:

Question 4:

A sphere and a cube are of the same height. The ratio of their volumes is

(a) 3 : 4

(b) 21 : 11

(c) 4 : 3

(d) 11 : 21

Answer:

In the given problem, we have a sphere and a cube of equal heights. So, let the diameter of the sphere and side of the cube be x units.

So, volume of the sphere (V1) =

Volume of the cube (V2) =

So, to find the ratio of the volumes,

Therefore, the ratio of the volumes of sphere and cube of equal heights is . So, the correct option is (d).

Page No 21.26:

Question 5:

The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be

(a) 27π cm3

(b) 36π cm3

(c) 108π cm3

(d) 12π cm3

Answer:

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 = 6 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 = 6 cm

Radius of the sphere = 3 cm

Now, the volume of the sphere =

Therefore, the volume of the largest sphere inside the given cube is . So, the correct option is (b).

Page No 21.26:

Question 6:

A cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be

(a) 4

(b) 3

(c) 6

(d) 8

Answer:

In the given problem, we have a cylindrical rod of the given dimensions:

Radius of the base (rc) = x units

Height of the cylinder (h) = 8x units

So, the volume of the cylinder (Vc) =

Now, this cylinder is remolded into spherical balls of same radius. So let us take the number of balls be y.

Total volume of y spheres (Vs) =

So, the volume of the cylinder will be equal to the total volume of y number of balls.

We get,

Therefore, the number of balls that will be made is. So, the correct option is (c)

Page No 21.26:

Question 7:

If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is

(a) 1 : 2

(b) 1 : 4

(c) 1 : 8

(d) 1 : 16

Answer:

Here, we are given that the ratio of the two spheres of ratio 1:8

Let us take,

The radius of 1st sphere = r1

The radius of 1st sphere = r2

So,

Volume of 1st sphere (V1) =

Volume of 2nd sphere (V2) =

Now,

Now, let us find the surface areas of the two spheres

Surface area of 1st sphere (S1) =

Surface area of 2nd sphere (S2) =

So, Ratio of the surface areas,

Using (1), we get,

Therefore, the ratio of the spheres is. So, the correct option is (b)

Page No 21.26:

Question 8:

If the surface area of a sphere is 144π m2, then its volume (in m3) is
 
(a) 288 π

(b) 316 π

(c) 300 π

(d) 188 π

Answer:

In the given problem,

Surface area of a sphere = m2

So,

Now, using the formula volume of the sphere, we get

Therefore, volume of the sphere is. Therefore the correct option is (a)

Page No 21.26:

Question 9:

If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq.cm) is

(a) 100 π

(b) 75 π

(c) 60 π

(d) 50 π

Answer:

In the given problem, Let the radius of smaller spherical balls which can be made from a bigger ball be x units.

Here,

The radius of the bigger ball (r1) = 10 cm

The radius of the smaller ball (r2) = x cm

The number of smaller balls = 8

So, volume of the big ball is equal to the volume of 8 small balls.

Volume of the big balls = volume of the 8 small balls

Further, solving for x, we get,

Now, surface area of a small ball of radius 5 cm =

Therefore, the surface area of the small spherical ball is. So, the correct option is (a).



Page No 21.27:

Question 10:

If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is

(a) π : 2

(b) π : 3

(c) π : 4

(d) π : 6

Answer:

In the given problem, we are given a sphere inscribed in a cube. So, here we need to find the ratio between the volume of a sphere and volume of a cube. This means that the diameter of the sphere will be equal to the side of the cube. Let us take the diameter as d.

Here,

Volume of a sphere (V1) =

Volume of a cube (V2) =

Now, the ratio of the volume of sphere to the volume of the cube =

So, the ratio of the volume of sphere to the volume of the cube is . Therefore, the correct option is (d)

Page No 21.27:

Question 11:

If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is

(a) 2r

(b) 3r

(c) r

(d) 4r

Answer:

In the given problem, we have a solid sphere which is remolded into a solid cone such that the radius of the sphere is equal to the height of the cone. We need to find the radius of the base of the cone.

Here, radius of the solid sphere (rs) = r cm

Height of the solid cone (h) = r cm

Let the radius of the base of cone (rc) = x cm

So, the volume of cone will be equal to the volume of the solid sphere.

Therefore, we get,

Therefore, radius of the base of the cone is. So, the correct option is (a).

Page No 21.27:

Question 12:

A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is

(a) 4πr3

(b) 83πr3

(c) 2πr3

(d) 8πr3

Answer:

In the given problem, we have a sphere inscribed in a cylinder such that it touches the top, base and the lateral surface of the cylinder. This means that the height and the diameter of the cylinder are equal to the diameter of the sphere.

So, if the radius of the sphere = r

The radius of the cylinder (rc)= r

The height of the cylinder (h) = 2r

Therefore, Volume of the cylinder =

So, the volume of the cylinder is. Therefore, the correct option is (c).

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Question 13:

The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is

(a) 2 : 1

(b) 1 : 1

(c) 2 : 3

(d) 1 : 2

Answer:

In the given problem, we need to find the ratio between the volume of a sphere and volume of a circumscribing right circular cylinder. This means that the diameter of the sphere and the cylinder are the same. Let us take the diameter as d.

Here,

Volume of a sphere (V1) =

As the cylinder is circumscribing the height of the cylinder will also be equal to the height of the sphere. So,

Volume of a cylinder (V2) =

Now, the ratio of the volume of sphere to the volume of the cylinder =

So, the ratio of the volume of sphere to the volume of the cylinder is . Therefore, the correct option is (c)

Page No 21.27:

Question 14:

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

(a) 1 : 2

(b) 2 : 1

(c) 4 : 1

(d) 2 : 1

Answer:

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 cone to the height of the hemisphere =

Here, Volume of the hemisphere = volume of the cone

Therefore, the ratio of the heights of the cone and the hemisphere is. So, the correct option is (b)

Page No 21.27:

Question 15:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is

(a) 1 : 2 : 3

(b) 2 : 1 : 3

(c) 2 : 3 : 1

(d) 3 : 2 : 1

Answer:

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 find the ratio of their volumes.

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)

Therefore, the ratio of the volumes of the given cone, hemisphere and the cylinder is . So, the correct option is (a).



Page No 21.8:

Question 1:

Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Answer:

Here we need to find the surface area of spheres of different radii

We solve it using the formula,

Curved surface area of the sphere

(i) Radius = 10.5cm

Therefore, the surface area of the sphere of radius 10.5cm is

(ii) Radius = 5.6cm

Therefore, the surface area of the sphere of radius 5.6cm is

(iii) Radius = 14cm

Therefore, the surface area of the sphere of radius 14cm is

Page No 21.8:

Question 2:

Find the surface area of a sphere of diameter.

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

Answer:

Here we need to find the surface area of spheres of different diameter.

We solve it using the formula,

Curved surface area of the sphere =

(i) Diameter = 14 cm

Therefore, the surface area of the sphere of diameter 14 cm is.

(ii) Diameter = 21 cm

Therefore, the surface area of the sphere of diameter 21 cm is.

(iii) Diameter = 3.5 cm

Therefore, the surface area of the sphere of diameter 3.5 cm is.

Page No 21.8:

Question 3:

Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm (Use π=3.14)

Answer:

In the given problem,

Radius of the hemisphere = 10 cm

Now, the total surface area of a hemisphere =

Therefore, the total surface area of the hemisphere is

Also,

The total surface area of the solid hemisphere =

Therefore, the total surface area of the hemisphere is

Page No 21.8:

Question 4:

The surface area of a sphere is 5544 cm2, find its diameter

Answer:

In the given problem,

Surface area of a sphere = 5544 cm2

Also, surface area of a sphere =

So according to the problem,

Now,

Therefore, diameter of the sphere = cm

= 42 cm

Therefore, the diameter of the sphere is

Page No 21.8:

Question 5:

A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 4 per 100 cm2.

Answer:

In the given problem, we have a hemispherical bowl.

Here,

Diameter of the bowl = 10.5 cm

So, Radius of the bowl = cm

So, the curved surface area of the bowl

Now, the rate of tin plating per 100 cm2 = Rs 4

The rate of tin plating per 1 cm2 = Rs

So, the cost of tin plating the bowl

= Rs 6.93

Therefore, the cost of tin plating the hemispherical bowl from inside is

Page No 21.8:

Question 6:

The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs 2 per sq. m.

Answer:

In the given problem, the dome of the building is in a form of a hemisphere.

Here,

Radius of the hemisphere = 63 dm

= 6.3 m

So, to find the cost of painting, we first find the curved surface area of the hemisphere.

Curved surface area of the sphere

Now, the rate of painting per m2 = Rs 2

So, the cost of painting the dome

Therefore, the cost of painting the dome is

Page No 21.8:

Question 7:

Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth's surface is covered by water?

Answer:

In the given problem, let us assume that the earth is a sphere and the radius of the earth is given equal to 6370 km.

The total surface area of earth can be calculated using the formula,

So, the total surface area of earth is equal to 510109600 km2.

Now, according to the question three-fourth of the earth’s surface is covered with water. This means that one-fourth of the surface is covered with land.

Therefore, we get,

The area of land covered by water

Therefore the area of earth’s surface covered by land is equal to

Page No 21.8:

Question 8:

A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.

Answer:

In the given problem, we are given a shape in which a cylinder is placed over a hemisphere. The radius of the hemisphere and the cylinder is equal and also the height of the cylinder is equal to the radius.

So, let draw a figure representing this shape and take the radius as r cm.

So, let us find the value of r first. From the figure, we can see that,

Also, height of the sphere = 3.5 cm (as radius and height of the cylinder are equal)

Now, the surface area of the shape is equal to the surface area of the hemisphere and surface area of cylinder together.

Surface area = surface area of hemisphere + surface area of cylinder

Therefore, the surface area of the given shape is

Page No 21.8:

Question 9:

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

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 surface area of the earth (Se) =

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 surface area of the moon (Sm) =

Ratio of surface areas =

Therefore, the ratio of the surface areas of moon to surface area of earth is

Page No 21.8:

Question 10:

A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 cm, find the cost of painting it, given the cost of painting is Rs 5 per 100 cm2.

Answer:

In the given problem, a hemispherical dome of the building needs to be painted. So, we need to find the surface area of the dome.

Here, we are given the circumference of the hemispherical dome as 17.6 m and as we know that circumference of the hemisphere is given by. So, we get

So, now we find the surface area of the hemispherical dome.

So, the curved surface area of the dome is 49.28 m2

Since the rate of the painting is given in cm2, we have to convert the surface area from m2 to cm2.

So, we get

Curved surface area =

= 492800 cm2

Now, the rate of painting per 100 cm2 = Rs 5

The rate of painting per 1 cm2 =

So, the cost of painting the dome =

Therefore, the cost of painting the hemispherical dome of the building is

Page No 21.8:

Question 11:

A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at Rs 7 per 100 cm2.

Answer:

In the given problem, we have a toy in the form of a cone surmounted on a hemisphere. So, let us first draw a diagram of the toy with the given dimensions.

So here,

Height of the cone = 15 cm

Diameter of the cone base and hemisphere = 16 cm

Therefore, radius of the cone base and hemisphere =

To find the lateral surface area of the cone, we first need to find the slant height of the cone, which is given by the following formula,

So, we get

Now, let us find the surface area of the toy first,

Total Surface area = lateral surface area of the cone + surface area of the hemisphere

So, the total surface area of the toy is 829.71 cm2

Now, the rate of painting the toy = Rs 7 per 100 cm2

= Rs per cm2

So, the cost of painting the given toy = total surface area x rate of painting

Therefore, the cost of painting the toy is

Page No 21.8:

Question 12:

A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder by 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.

Answer:

In the given problem, we are given a storage tank which consists of a circular cylinder with a hemisphere adjoined on either side.

So here,

External diameter of the cylinder = 1.4 m

External radius of the cylinder = m

m

Height of the cylinder = 8 m

Now, we find the total surface area of the storage tank.

Total surface area = Surface area of cylinder + surface area of the hemisphere

Therefore, the external surface area of the storage tank is 38.28 m2 .

Now, the rate of painting the storage tank on outside = Rs 10 per m2

So, the cost of painting the storage tank on outside =

Therefore, the cost of painting the storage tank on outside is



Page No 21.9:

Question 13:

The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.

Answer:

In the given problem, we have 8 spheres mounted upon small cylindrical supports, as shown in the diagram.

Now, the spheres are to be painted silver. So, let us first find the surface area of the spheres first.

Diameter of the sphere = 21 cm

So, radius of the sphere = 10.5 cm

Now,

This is the surface area of one sphere. We have 8 such spheres. So,

Total surface area of 8 spheres =

cm2

At the junction of the sphere and the cylinder, there is some portion which would be covered. Hence, that would not be painted silver.

The covered area

So, for 8 spheres,

The covered area

Therefore, the total area to be painted

Now, the rate of silver paint per cm2 = 25 paisa

=Rs 0.25

So, the cost of silver paint for 8 spheres =

= Rs 2757.86

Now, the cylinders are to be painted black. So, let us find the surface area of the cylinders.

Radius of the cylinder = 1.5 cm

Height of the cylinder = 7 cm

Now,

This is the surface area of one cylinder. We have 8 such cylinders. So,

Total surface area of 8 cylinder =

cm2

Now, the rate of black paint per cm2 = 5 paise

= Rs 0.05

So, the cost of silver paint for 8 cylinders =

Therefore,

The total cost of the paint = cost of silver paint + cost of black paint

Therefore, the cost of painting would be



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