## 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€r

^{2}

(ii) Here, r = 5.6 cm

Surface area of a sphere = 4Ï€r^{2}

(iii) Here, r = 14 cm

Surface area of a sphere = 4Ï€r^{2}

**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€r

^{2}

= 3 × 3.14 × 10 × 10 cm

^{2}= 942 cm

^{2}

**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 (r

_{1}) = 7 cm

Surface area = 4Ï€r

_{1}

^{2}= 4 × 22/7 × (7)

^{2}cm

^{2}

= 4 × 22 × 7 cm

^{2}= 616 cm

^{2}

Case II: When radius (r_{2}) = 14 cm^{2}

Surface area = 4Ï€r_{2}^{2 }= 4 × 22/7 × (14)^{2} cm^{2}

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

∴ 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€r

^{2}

∴ 4Ï€r

^{2}= 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€r

^{2}

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

Surface area of the earth = 4Ï€r^{2}

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€R^{2}

**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€r

^{2}

(ii) For the right circular cylinder,

Radius of the cylinder = Radius of the sphere

∴ 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€r^{2}

**Related Links:**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**