NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.2

# NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.2

NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.2 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 11 Surface Areas and Volumes Ex 11.2.

## NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.2

Ex 11.2 Class 9 Maths Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm

Solution:
(i) Here, r = 10.5 cm
Surface area of a sphere = 4Ï€r2

(ii) Here, r = 5.6 cm
Surface area of a sphere = 4Ï€r2

(iii) Here, r = 14 cm
Surface area of a sphere = 4Ï€r2

Ex 11.2 Class 9 Maths Question 2.
Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m

Solution:
(i) Here, diameter = 14 cm

(ii) Here, diameter = 21 cm

(iii) Here, diameter = 3.5 m

Ex 11.2 Class 9 Maths Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use Ï€ = 3.14)

Solution:
Here, radius (r) = 10 cm
Total surface area of a hemisphere = 3Ï€r2
= 3 × 3.14 × 10 × 10 cm2 = 942 cm2

Ex 11.2 Class 9 Maths 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.

Solution:
Case I: When radius (r1) = 7 cm
Surface area = 4Ï€r12 = 4 × 22/7
× (7)2 cm2
= 4 × 22 × 7 cm2 = 616 cm2

Case II: When radius (r2) = 14 cm2
Surface area = 4Ï€r2= 4 × 22/7
× (14)2 cm2
= 4 × 22 × 14 × 2 cm2 = 2464 cm2
The required ratio = 616/2464 = 1/4 or 1 : 4

Ex 11.2 Class 9 Maths 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 ₹16 per 100 cm².

Solution:
Inner diameter of the hemispherical bowl = 10.5 cm

Ex 11.2 Class 9 Maths Question 6.
Find the radius of a sphere whose surface area is 154 cm².

Solution:
Let the radius of the sphere be r cm.
Surface area of a sphere = 4Ï€r2
4Ï€r2 = 154

Thus, the required radius of the sphere is 3.5 cm.

Ex 11.2 Class 9 Maths Question 7.

The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas.

Solution:
Let the radius of the earth be r.
Radius of the moon = r/4
Surface area of a sphere = 4Ï€r2

Since, the earth as well as the moon is considered to be sphere.
Surface area of the earth = 4Ï€r2

Thus, the required ratio = 1 : 16.

Ex 11.2 Class 9 Maths 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.

Solution:
Inner radius (r) = 5 cm
Thickness = 0.25 cm

Outer radius (R) = 5.00 + 0.25 = 5.25 cm
Outer curved surface area of the bowl = 2Ï€R2

Ex 11.2 Class 9 Maths 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).

Solution:
(i) For the sphere, radius = r
Surface area of the sphere = 4Ï€r2

(ii) For the right circular cylinder,
∴ Radius of the cylinder = r
Height of the cylinder = Diameter of the sphere
Height of the cylinder (h) = 2r
Since, curved surface area of a cylinder = 2Ï€rh
= 2Ï€r(2r) = 4Ï€r2