# Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.4

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

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

**(Assume Ï€=22/7)**

**Solution:**

We know that, Surface area of sphere (SA) = 4Ï€rÂ²

(i)Radius of sphere, r = 10.5 cmSA = 4Ã—(22/7)Ã—10.5

^{2}= 1386Therefore, Surface area of sphere is 1386 cmÂ²

(ii)Radius of sphere, r = 5.6cmSA = 4Ã—(22/ 7)Ã—5.6

^{2}= 394.24Therefore, Surface area of sphere is 394.24 cmÂ²

(iii)Radius of sphere, r = 14cmSA = 4Ã—(22/7)Ã—(14)

^{2}= 2464

Therefore, Surface area of sphere is 2464 cmÂ²

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

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

**(Assume Ï€ = 22/7)**

**Solution:**

(i)Radius of sphere, r = diameter/2 = 14/2 cm = 7 cmWe know that, Surface area of sphere = 4Ï€rÂ²= 4Ã—(22/7)Ã—7

^{2}= 616Surface area of a sphere is 616 cmÂ²

(ii)Radius of sphere,r= diameter/2=21/2 = 10.5 cmWe know that, Surface area of sphere = 4Ï€rÂ²

= 4Ã—(22/7)Ã—10.5

^{2}= 1386Surface area of a sphere is 1386 cmÂ²

Therefore, the surface area of a sphere having diameter 21cm is 1386 cmÂ²

(iii)Radius(r) of sphere = 3.5/2 = 1.75 cmWe know that, Surface area of sphere = 4Ï€rÂ²

= 4Ã—(22/7)Ã—1.75

^{2}= 38.5Surface area of a sphere is 38.5 cmÂ²

**Question 3. Find the total surface area of a hemisphere of radius 10 cm. [Use Ï€=3.14]**

**Solution:**

Given, Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3Ï€rÂ²

= 3Ã—3.14Ã—10

^{2}= 942Therefore, total surface area of given hemisphere is 942 cmÂ².

**Question 4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.**

**Solution:**

We assume that r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So,

r1 = 7cm

r2 = 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4(r1)Â²/4(r2)Â²

= (r1/r2)Â²

= (7/14)Â² = (1/2)Â² = Â¼

Therefore, the ratio between the surface areas is 1:4.

**Question 5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cmÂ². (Assume **Ï€** = 22/7)**

**Solution:**

Given, Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

We know, Formula for Surface area of hemispherical bowl = 2Ï€rÂ²

= 2Ã—(22/7)Ã—(5.25)

^{2}= 173.25cmÂ²Cost of tin-plating 100 cmÂ² area = Rs 16

So, Cost of tin-plating 1 cmÂ² area = Rs 16 /100

Cost of tin-plating 173.25 cmÂ² area = Rs. (16Ã—173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cmÂ² is Rs 27.72.

**Question 6. Find the radius of a sphere whose surface area is 154 cmÂ². (Assume** Ï€** = 22/7)**

**Solution:**

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

So,

4Ï€rÂ² = 154

rÂ² = (154Ã—7)/(4Ã—22) = (49/4)

r = (7/2) = 3.5cm

Therefore, 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 the diameter of earth be d, then the diameter of moon will be d/4 (as per given statement)

Radius of earth = d/2

So, Radius of moon = Â½Ã—d/4 = d/8

Surface area of moon = 4Ï€(d/8)Â²

Surface area of earth = 4Ï€(d/2)Â²

The ratio between their surface areas is 1:16.

**Question 8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume **Ï€** =22/7)**

**Solution:**

Given:

Inner radius of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm

We know, formula for outer CSA of hemispherical bowl = 2Ï€rÂ², where r is radius of hemisphere

= 2Ã—(22/7)Ã—(5.25)Â²= 173.25

Therefore, the outer curved surface area of the bowl is 173.25 cmÂ².

**Question 9. A right circular cylinder just encloses a sphere of radius r (see fig.). 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)Surface area of sphere = 4Ï€rÂ², where r is the radius of sphere

(ii)As Height of cylinder, h = r+r =2rAnd Radius of cylinder = r

CSA of cylinder formula = 2Ï€rh = 2Ï€r(2r) (using value of h)

= 4Ï€rÂ²

(iii)Ratio between areas = (Surface area of sphere)/CSA of Cylinder)= 4r

^{2}/4r^{2}= 1/1Therefore, Ratio of the areas obtained in (i) and (ii) is 1:1.

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