NCERT Class 9 Maths Solutions Chapter - 13 Surface Areas and Volumes, Ex - 13.4

Ex - 13.4

Question 1.  Assume  , unless stated otherwise.
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 = 10.5 cm
     Surface area of sphere = 

(ii) Radius of sphere = 5.6 cm
     Surface area of sphere =  = 

(iii) Radius of sphere = 14 cm
     Surface area of sphere =  =  

Question 2.  Assume  , unless stated otherwise.

Find the surface area of a sphere of diameter:

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

Solution

(i)     Radius of sphere 
        Surface area of sphere  

(ii)    Radius of sphere 
        Surface area of sphere       

(iii)    Radius of sphere 
        Surface area of sphere =  

Question 3.  Assume , unless stated otherwise.

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

Solution

Radius of hemisphere = 10 cm
Total surface area of hemisphere 

Question 4.  Assume , unless stated otherwise.

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 spherical balloon = 7 cm
Radius of spherical balloon, when air is pumped into it = 14 cm

Question 5.  Assume , unless stated otherwise.

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 Rs 16 per 100 

Solution

Inner radius (r) of hemispherical bowl = 
Surface area of hemispherical bowl  = 

        
Cost of tin-plating 100  area = Rs 16
Cost of tin-plating 173.25  area = Rs 27.72

Thus, the cost of tin-plating the inner side of hemispherical bowl is Rs 27.72

Question 6.  Assume , unless stated otherwise.

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

Solution

Let radius of the sphere be r.
 Surface area of the sphere = 154 
    = 154 cm2    

    
Thus, the radius of the sphere is 3.5 cm.

Question 7.  Assume , unless stated otherwise.

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface area.

Solution

Let diameter of earth be d. Then, diameter of moon will be .
    Radius of earth =  
    Radius of moon = 
    Surface area of moon =  
    Surface area of earth =  
    Required ratio =  
    
Thus, the required ratio of the surface areas is 1:16.

Question 8.  Assume , unless stated otherwise.

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 of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl = 

                            
  Thus, the outer curved surface area of the bowl is 173.25 .

Question 9.  Assume , unless stated otherwise.

A right circular cylinder just encloses a sphere of radius r. 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 = 

  (ii)  Height of cylinder = r + r = 2r
        Radius of cylinder = r
        CSA of cylinder =
    
(iii)   Required ratio = 

Ex-13.3

Ex-13.5