NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise

NCERT solutions for class 9 Maths

Surface Areas and Volumes

Chapter 13 Exercise 13.4

EX 13.4 QUESTION 1.

Find the surface area of a sphere of radius
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
(i) Radius of sphere (r)= 10.5 cm
Surface area of a sphere = 4πr²
4 x $\frac { 22 }{ 7 }$ x 10.5 x 10.5 = 4 x 22 x 1.5 x 1.5 = 1386.00 cm²

(ii) Radius of sphere (r) = 5.6 cm
Surface area of a sphere = 4πr²
4 x $\frac { 22 }{ 7 }$ x 5.6 x 5.6 = 4 x 22 x 0.8 x 5.6 = 394.24 cm²

(iii) Radius of sphere (r) = 14 cm
Surface area of a sphere = 4πr²
4 x $\frac { 22 }{ 7 }$ x 14 x 14 = 4 x 22 x 2 x 14 = 2464 cm²

Ex 13.4 Question 2.

Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Solution:
(i) Radius of sphere (r) = 14 /2 = 7 cm
Surface area of a sphere = 4πr²
4 x $\frac { 22 }{ 7 }$ x 7 x 7 = 4 x 22 x 7 = 616 cm²

(ii) Radius of sphere (r) = 21/2 = 10.5 cm
Surface area of a sphere = 4πr²
4 x $\frac { 22 }{ 7 }$ x 10.5 x 10.5 = 4 x 22 x 4.5 x 10.5 = 1386 cm²

(iii) Radius of sphere (r) = 3.5/2 = 1.75 cm
Surface area of a sphere = 4πr²
4 x $\frac { 22 }{ 7 }$ x 1.75 x 1.75 = 4 x 22 x 0.25 x 1.75 = 38.50 cm²

Ex 13.4 Question 3.

Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
Radius of hemisphere (r) = 10 cm
Surface area of hemisphere = 3πr²
= 3 x 3.14 x 10 x 10 cm² = 942 cm²

Ex 13.4 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:
Radius of balloon (r) = 7 cm
Surface area of balloon = 4πr² = 4 x $\frac { 22 }{ 7 }$ x (7) cm²
= 4 x 22 x 7 cm² = 616 cm²

Case II: Radius of balloon (R²) = 14 cm²
Surface area of balloon= 4πr₂²=4 x $\frac { 22 }{ 7 }$ x (14)² cm²
= 4 x 22 x 14 x 2 cm² = 2464 cm²
∴ Ratio of surface area = $\frac { 616 }{ 2464 }$ = $\frac { 1 }{ 4 }$ or 1 : 4

Ex 13.4 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 ₹16 per 100 cm².
Solution:
Internal radius of the hemispherical bowl (r) = 10.5 /2 = 5.25 cm
2 x $\frac { 22 }{ 7 }$ x 5.25 x 5.25 = 2 x 22 x 0.75 x 5.25 = 173.25 cm².

Cost of tin – plating at the rate of ₹16 per 100 cm² = ₹ 173.25 x 16 / 100 = ₹ 27.72

Ex 13.4 Question 6.

Find the radius of a sphere whose surface area is 154 cm².
Solution:
Let the radius of the sphere = r cm.
Surface area of a sphere = 4πr²
∴ 4πr² = 154

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

Ex 13.4 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 = r.
∴ Radius of the moon = $\frac { r }{ 4 }$
Surface area of a sphere = 4πr²
Since, the earth as well as the moon is considered to be sphere.
Surface area of the earth = 4πr²

Thus, the required ratio = 1 : 16.

Ex 13.4 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 + 025 cm = 5.25 cm
∴ Outer curved surface area of the bowl = 2πR²
= 2 x $\frac { 22 }{ 7 }$ x 5.25 x 5.25 = 2 x 22 x 0.75 x 5.25 = 173.25 cm²

Ex 13.4 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) Radius of the sphere = Radius of the cylinder
∴ Surface area of the sphere = 4πR²

(ii) For the right circular cylinder,
Radius of the cylinder = r
Height of the cylinder = Diameter of the sphere = 2r
Curved surface area of a cylinder = 2πrh= 2πr(2r) = 4πr²

NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.4)

Published by admin on August 30, 2020August 30, 2020