RD Sharma 2020 solution class 9 chapter 21 Surface Area and Volume of a Sphere Exercise 21.1

Exercise 21.1

Page-21.7

Question 1:

Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Answer 1:

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

Question 2:

Find the surface area of a sphere of diameter.

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

Answer 2:

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.

Question 3:

Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm (Use $\mathrm{\pi }=3.14$)

Answer 3:

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

Question 4:

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

Answer 4:

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

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 5:

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

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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 6:

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

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 7:

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

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 8:

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

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 9:

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 (S_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 surface area of the moon (S_m) =

Ratio of surface areas =

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

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 10:

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

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 11:

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

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 12:

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

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 13:

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