# NCERT solution class 9 chapter 13 Surface Areas and Volumes exercise 13.4 mathematics

## EXERCISE 13.4

#### Question 1:

Find the surface area of a sphere of radius:

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm (i) Radius (r) of sphere = 10.5 cm

Surface area of sphere = 4πr2 Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.

(ii) Radius(r) of sphere = 5.6 cm

Surface area of sphere = 4πr2 Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 cm2.

(iii) Radius (r) of sphere = 14 cm

Surface area of sphere = 4πr2 Therefore, the surface area of a sphere having radius 14 cm is 2464 cm2.

#### Question 2:

Find the surface area of a sphere of diameter:

(i) 14 cm (ii) 21 cm (iii) 3.5 m (i) Radius (r) of sphere =  Surface area of sphere = 4πr2 Therefore, the surface area of a sphere having diameter 14 cm is 616 cm2.

(ii) Radius (r) of sphere = Surface area of sphere = 4πr2 Therefore, the surface area of a sphere having diameter 21 cm is 1386 cm2.

(iii) Radius (r) of sphere = m

Surface area of sphere = 4πr2 Therefore, the surface area of the sphere having diameter 3.5 m is 38.5 m2.

#### Question 3:

Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14] Radius (r) of hemisphere = 10 cm

Total surface area of hemisphere = CSA of hemisphere + Area of circular end of hemisphere Therefore, the total surface area of such a hemisphere is 942 cm2.

#### Question 4:

The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Radius (r1) of spherical balloon = 7 cm

Radius (r2) of spherical balloon, when air is pumped into it = 14 cm Therefore, the ratio between the surface areas in these two cases is 1:4.

#### 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 16 per 100 cm2 Inner radius (r) of hemispherical bowl Surface area of hemispherical bowl = 2πr2 Cost of tin-plating 100 cm2 area = Rs 16

Cost of tin-plating 173.25 cm2 area = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72.

#### Question 6:

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

Surface area of sphere = 154

∴ 4πr= 154 cm2 Therefore, the radius of the sphere whose surface area is 154 cm2 is 3.5 cm.

#### Question 7:

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

Let the diameter of earth be d. Therefore, the diameter of moon will be .

Radius of earth = Radius of moon = Surface area of moon = Surface area of earth = Required ratio  Therefore, the ratio between their surface areas will be 1:16

#### Question 8:

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. Inner radius of hemispherical bowl = 5 cm

Thickness of the bowl = 0.25 cm

∴ Outer radius (r) of hemispherical bowl = (5 + 0.25) cm

= 5.25 cm

Outer CSA of hemispherical bowl = 2πr2 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

#### Question 9:

A right circular cylinder just encloses a sphere of radius r (see figure). Find (i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii). (i) Surface area of sphere = 4πr2

(ii) Height of cylinder = r + r = 2r

CSA of cylinder = 2πrh

= 2π(2r)

= 4πr2

(iii)  Therefore, the ratio between these two surface areas is 1:1