## 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 = 4πr^{2}

4 x 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 = 4πr^{2}

4 x 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 = 4πr^{2}

4 x 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 = 4πr^{2}

4 x 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 = 4πr^{2}

4 x 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 = 4πr^{2}

4 x 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 = 3πr^{2}

= 3 x 3.14 x 10 x 10 cm^{2} = 942 cm^{2}

**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 = 4πr^{2} = 4 x x (7) cm^{2}

= 4 x 22 x 7 cm^{2} = 616 cm^{2}

Case II: Radius of balloon (R^{2}) = 14 cm^{2}

Surface area of balloon= 4πr_{2}^{2}=4 x x (14)^{2} cm^{2}

= 4 x 22 x 14 x 2 cm^{2} = 2464 cm^{2}

∴ Ratio of surface area = = 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 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 = 4πr^{2}

∴ 4πr^{2} = 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 =

Surface area of a sphere = 4πr^{2}

Since, the earth as well as the moon is considered to be sphere.

Surface area of the earth = 4πr^{2}

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 = 2πR^{2}

= 2 x x 5.25 x 5.25 = 2 x 22 x 0.75 x 5.25 = 173.25 cm^{2}

**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 = 4πR^{2}

(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) = 4πr^{2}