# NCERT Solutions for Class 9th Mathematics

## Chapter 13. Surface Areas and Volumes

### Exercise 13.4

**Exercise 13.4**

Assume π = 22/7 , unless stated otherwise.

**Q1. Find the surface area of a sphere of radius:**

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

**Solution:**

**Q2. Find the surface area of a sphere of diameter:**

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

**Solution:**

(i) Surface area of sphere = 4πr^{2}

**Q3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)**

**Solution:**

Total surface area of hemisphere = 3πr^{2}

⇒ 3 × 3.14 × 10 × 10

⇒ 3 × 314

⇒ 942 cm^{2}

**Q4. 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:**

Surface area of sphere = 4πr^{2}

**Q5. 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 cm ^{2}.**

**Solution:**

Curved surface area of hemisphere = 2πr^{2}

**Q6. Find the radius of a sphere whose surface area is 154 cm ^{2}.**

**Solution:**

Area = 154 cm^{2}

Surface area of sphere = 4πr^{2}

**Q7. 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 diameter of earth = x

**Q8. 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 = 5 cm, width = 0.25 cm, radius = 5.25 cm

Curved surface area of hemisphere = 2πr^{2}

⇒ 44 × 0.75 × 5.25

⇒ 173.25

**Q9. A right circular cylinder just encloses a sphere of**

**radius r (see Fig. 13.22). 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:**

Radius of sphere = r, radius of cylinder = r + r = 2r

(i) surface area of sphere = 4πr^{2}

(ii) curved surface area of cylinder = 2πrh

⇒ 2πr(2r)

⇒ 4πr^{2}