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Surface Areas and Volumes : Exercise 13.8 (Mathematics NCERT Class 9th)


( Assume \pi = {{22} \over 7}, unless stated otherwise. )
Q.1     Find the volume of a sphere whose radius is

              (i) 7 cm               (ii) 0.63 cm
Sol.

(i) We have : r = radius of the sphere = 7 cm
Therefore, Volume of the sphere  = {4 \over 3}\pi {r^3}
                                                         = \left( {{4 \over 3} \times {{22} \over 7} \times 7 \times 7 \times 7} \right)c{m^3}
                                                       
 = {{4312} \over 3}c{m^3} = 1437{1 \over 3}c{m^3}
(ii) We have : r = radius of the sphere = 0.63 m

Therefore, Volume of the sphere  = {4 \over 3}\pi {r^3}
                                                         = \left( {{4 \over 3} \times {{22} \over 7} \times 0.63 \times 0.63 \times 0.63} \right){m^3}
                                                       
 = 1.05\,\,{m^3}\left( {approx} \right)


Q.2      Find the amount of water displaced by a solid spherical ball of diameter.
               (i) 28 cm                  (ii) 0.21 m
Sol.

(i) Diameter of the spherical ball = 28 cm
Therefore, Radius  = \left( {{{28} \over 2}} \right)cm = 14\,cm
Amount of water displaced by the spherical ball
 = Its\,volume\, = {4 \over 3}\pi {r^3}
                            = \left( {{4 \over 3} \times {{22} \over 7} \times 14 \times 14 \times 14} \right)c{m^3}
                            = {{34496} \over 3}c{m^3} = 11498{2 \over 3}c{m^3}

(ii) Diameter of the spherical ball = 0.21 m
Therefore, Radius  = \left( {{{0.21} \over 2}} \right)m = 0.105\,\,m
Amount of water displaced by the spherical ball
 = \,Its\,volume\, = {4 \over 3}\pi {r^3}
                             = \left( {{4 \over 3} \times {{22} \over 7} \times 0.105 \times 0.105 \times 0.105} \right){m^3}
                           
 = 0.004851\,\,{m^3}


Q.3     The diameter of a metallic ball is 4.2 cm. What is mass of the ball, if the density of the metal is 8.9 g per c{m^3}?
Sol.

Diameter of the ball = 4.2 cm
Therefore ,     Radius  = \left( {{{4.2} \over 2}} \right)cm = 2.1\,cm
Volume of the ball  = {4 \over 3}\pi {r^3}
                                 = \left( {{4 \over 3} \times {{22} \over 7} \times 2.1 \times 2.1 \times 2.1} \right)c{m^3}
                               
 = 38.808\,c{m^3}
Density of the metal is 8.9g per c{m^3}
Therefore Mass of the ball = (38.808 × 8.9) g = 345.3912 g


Q.4      The diameter of the moon is approximately one - fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Sol.

Let the diameter of the moon be r. Then, the radius of the moon  = {r \over 2}
According to the question, diameter of the earth is 4r, so its radius  = {{4r} \over 2} = 2r.
{V_1} = The volume of the moon  = {4 \over 3}\pi {\left( {{r \over 2}} \right)^3}
                                                                     = {4 \over 3}\pi {r^3} \times {1 \over 8}
 \Rightarrow 8{V_1} = {4 \over 3}\pi {r^3} ... (1)
and, {V_2} = The volume of the earth  = {4 \over 3}\pi {\left( {2r} \right)^3}
                              = {4 \over 3}\pi {r^3} \times 8
 \Rightarrow {{{V_2}} \over 8} = {4 \over 3}\pi {r^3} ... (2)
From (1) and (2) , we have
8{V_1} = {{{V_2}} \over 8}\,\,\,  \Rightarrow {V_1} = {1 \over {64}}\,{V_2}\,\,
Hence, the volume of the moon is {1 \over {64}} of the volume of the earth.


Q.5      How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Sol.

Diameter of a hemispherical bowl = 10.5 cm
Therefore , its radius  = \left( {{{10.5} \over 2}} \right)cm = 5.25\,cm
Volume of the bowl  = {2 \over 3}\pi {r^3}
                                 
 = \left( {{2 \over 3} \times {{22} \over 7} \times 5.25 \times 5.25 \times 5.25} \right)c{m^3}
                                 
 = 303.1875\,c{m^3}
Hence, the hemispherical bowl can hold 303 l (approx.) of milk.


Q.6     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.
Sol.

Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel. Then,
R = 1.01 (as thickness = 1 cm = .01 m)
and r = 1 m.
Volume of iron used = External volume – Internal volume
                                     = {2 \over 3}\pi {R^3} - {2 \over 3}\pi {r^3}
                                    = {2 \over 3}\pi \left( {{R^3} - {r^3}} \right)

                              = {2 \over 3} \times {{22} \over 7} \times \left[ {{{\left( {1.01} \right)}^3} - {{\left( 1 \right)}^3}} \right]{m^3}
                                     = {{44} \over {21}} \times \left( {1.030301 - 1} \right){m^3}
                                     = \left( {{{44} \over {21}} \times 0.030301} \right){m^3}
                                     = 0.06348\,\,{m^3}\,\left( {approx} \right)


Q.7      Find the volume of a sphere whose surface area is 154 c{m^2}.
Sol.

Let r cm be the radius of the sphere.
So, surface area = 154\,c{m^2}
 \Rightarrow 4\pi {r^2} = 154
 \Rightarrow 4 \times {{22} \over 7} \times {r^2} = 154
 \Rightarrow {r^2} = {{154 \times 7} \over {4 \times 22}} = 12.25
 \Rightarrow r = \sqrt {12.25} = 3.5\,cm
Now , Volume  = {4 \over 3}\pi {r^3}
                          = \left( {{4 \over 3} \times {{22} \over 7} \times 3.5 \times 3.5 \times 3.5} \right)c{m^3}
                          = {{539} \over 3}c{m^3} = 179{2 \over 3}c{m^3}


Q.8      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 metre, find the
               (i) Inside surface area of the dome,
               (ii) Volume of the air inside the dome.
Sol.

(i) Inside surface area of the dome  = {{Total\,\,\cos t\,\,of\,\,white\,\,washing} \over {Rate\,of\,white\,washing}}
                                                            
= \left( {{{498.96} \over {2.00}}} \right){m^2} = 249.48\,\,{m^2}

 (ii) Let r be the radius of the dome.
Therefore Surface area  = 2\pi {r^2}
 \Rightarrow 2 \times {{22} \over 7} \times {r^2} = 249.48
 \Rightarrow {r^2} = {{249.48 \times 7} \over {2 \times 22}} = 39.69
 \Rightarrow r = \sqrt {39.69} = 6.3\,m
Volume of the air inside the dome = Volume of the dome
                                                          = {2 \over 3}\pi {r^3} = {2 \over 3} \times {{22} \over 7} \times 6.3 \times 6.3 \times 6.3\,{m^3}
                                                          
 = 523.9\,{m^3}\left( {approx} \right)


Q.9     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'.
Sol.

(i) Volume of 27 solid sphere of radius r = 27 \times {4 \over 3}\pi {r^3} ... (1)
Volume of the new sphere of radius r' = {4 \over 3}\pi {r^3} ... (2)
According to the problem, we have
{4 \over 3}\pi r{'^3} = 27 \times {4 \over 3}\pi {r^3}
 \Rightarrow r{'^3} = 27{r^3} = {\left( {3r} \right)^3}
Therefore r' = 3r

(ii) Required ratio  = {S \over {S'}} = {{4\pi {r^2}} \over {4\pi r{'^2}}} = {{{r^2}} \over {{{\left( {3r} \right)}^2}}}
                                  
 = {{{r^2}} \over {9{r^2}}} = {1 \over 9} = 1:9


Q.10    A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine \left( {in\,m{m^3}} \right) is needed to fill this capsule?
Sol.

Diameter of the spherical capsule = 3.5 mm
Radius  = {{3.5} \over 2}mm
              = 1.75\,\,mm
Medicine needed for its filling = Volume of spherical capsule
                                                      = {4 \over 3}\pi {r^3}
                                                      = \left( {{4 \over 3} \times {{22} \over 7} \times 1.75 \times 1.75 \times 1.75} \right){mm^3}
                                                    
 = 22.46\,{mm^3}\,\,\left( {approx} \right)



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