# NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

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## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

NCERT TEXTBOOK EXERCISES

Question 1. Find the surface area of a sphere of radius.

(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm

Solution:

(i) We have, r = 105 cm

Question 2. Find the surface area of a sphere of diameter

(i) 14 cm
(ii) 21 cm
(iii) 3.5 m

Solution:

Question 3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Solution:

We have, r = 10 cm
Total surface area of a hemisphere = 3πr2
= 3 x 3.14 x (10)2
= 9.42 x 100
= 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.

Solution:

Let initial radius, r1 = 7 cm
After increases, r2 = 14 cm
Surface area for initial balloon = 4πr12 = 4 x 22/7 x 7 x 7 = 88 x 7
A1 = 616 cm2
Surface area for increasing balloon = 4πr22 = 4x 22/7 x 14 x 14 = 88 x 28
A2 = 2464 cm2
∴ Required ratio = A1 : A2 = 616 : 2464 = 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 ₹16 per 100 cm2.

Solution:

We have, inner diameter = 10.5 cm
Inner radius = 10.5/2 cm = 5.25 cm
Curved surface area of hemispherical bowl of inner side = 2πr2

Question 6. Find the radius of a sphere whose surface area is 154 cm2.

Solution:

Surface area of a sphere = 154 cm2

Hence, the radius of the sphere 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 areas.

Solution:

Let diameter of the Earth = d1

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:

= ( 5 + 0.25) cm = 5.25 cm

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:

The radius of the sphere = r
Height of the cylinder = Diameter = 2r
(i) Surface area of the sphere A1 = 4πr2
(ii) Curved surface area of the cylinder = 2πrh
A2 = 2π x r x 2r
A2 = 4πr2
(iii) Required ratio = A1 :A2 = 4πr: 4πr2 = 1 : 1

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 PDF

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