NCERT Class 9 Maths Solutions Chapter - 13 Surface Areas and Volumes, Ex - 13.8

Ex - 13.8

Question 1.  Assume , unless stated otherwise.
Find the volume of a sphere whose radius is
(i)    7 cm        (ii)    0.63 m

Solution

(i)    Radius of sphere = 7 cm
        Volume of sphere =  

(ii)    Radius of sphere = 0.63 m
        Volume of sphere =

        m³(approximately)

Question 2.  Assume , unless stated otherwise

Find the amount of water displaced by a solid spherical ball of diameter

(i)    28 cm        (ii)    0.21 m

Solution

(i)    Radius (r) of ball = 
       Volume of ball =   
       Thus, the amount of water displaced is .

(ii)    Radius (r) of ball =  = 0.105 m
        Volume of ball =  
        Thus, the amount of water displaced is 0.004851 m³.

Question 3.  Assume , unless stated otherwise.

The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Solution

Radius (r) of metallic ball =  
    Volume of metallic ball =   

    Mass = Density  Volume = (8.9 * 38.808) g = 345.3912 g
    
    Thus, the mass of the ball is approximately 345.39 g.

Question 4.  Assume , unless stated otherwise.

The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution

Let diameter of earth be d. So, radius earth will be  .
    Then, diameter of moon will be  . So, radius of moon will be  .
    Volume of moon =    
    Volume of earth =   
      
    Thus, the volume of moon is  of volume of earth.

Question 5.  Assume , unless stated otherwise.

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Solution

Radius (r) of hemispherical bowl =   = 5.25 cm

Volume of hemispherical bowl

                                                                 
                                            = 303.1875 cm³

Capacity of the bowl 
                              = 0.3031875 litre = 0.303 litre (approximately)

Thus, the hemispherical bowl can hold 0.303 litre of milk.

Question 6.  Assume  , unless stated otherwise.

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Solution

Inner radius (r₁) of hemispherical tank  = 1 m
     Thickness of hemispherical tank       = 1 cm = 0.01 m
Outer radius (r₂) of hemispherical tank = (1 + 0.01) m = 1.01 m

Volume of iron used to make the tank  = 

                                                            

Question 7.  Assume 

Find the volume of a sphere whose surface area is 154 cm².

Solution

Let the radius of the sphere be r.
Surface area of sphere = 154 cm²
r² = 154 cm²

   
 Volume of sphere

Question 8.  Assume , unless stated otherwise.

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the

(i)  inside surface area of the dome,

(ii) volume of the air inside the dome.

Solution

(i)    Cost of white washing the dome from inside = Rs 498.96
       Cost of white washing 1 m² area = Rs 2
       CSA of inner side of dome =   = 249.48 m²

(ii)    Let inner radius of hemispherical dome be r.
        CSA of inner side of dome = 249.48 m²
        2r² = 249.48 m²
         
       
        Volume of air inside the dome = Volume of the hemispherical dome
                                                    
                                                    = 523.908 m³
      
        Thus, the volume of air inside the dome is approximately 523.9 m³.

Question 9.  Assume , unless stated otherwise.

Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the

(i)    radius r' of the new sphere,    (ii)    ratio of S and S'.

Solution

(i)    Radius of 1 solid iron sphere = r
       Volume of 1 solid iron sphere 
       Volume of 27 solid iron spheres  
       It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of

       this iron sphere will be equal to volume of 27 solid iron spheres.
       Radius of the new sphere = r'.
       Volume of new sphere  
       

(ii)    Surface area of 1 solid iron sphere of radius r = 4r²
        Surface area of iron sphere of radius r' = 4 (r')²    = 4  (3r)² = 36 r²
         

Question 10.  Assume , unless stated otherwise.

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Solution

Radius (r) of capsule  
Volume of spherical capsule
                                         
Thus, approximately 22.46 mm³ of medicine is required to fill the capsule.

Ex-13.7

Ex-13.9