Surface Areas and Volumes : Exercise 13.4 (Mathematics NCERT Class 9th)


Q.1     Find the surface area of a sphere of radius :
            (i) 10. 5 cm       (ii) 5.6 cm       (iii) 14 cm
Sol.

(i) We have :
r = radius of the sphere = 10.5 cm

Surface area  = 4\pi {r^2}
                   
 = \left( {4 \times {{22} \over 7} \times 10.5 \times 10.5} \right)c{m^2}
                            = 1386\,c{m^2}

(ii) We have :
r = radius of the sphere = 5.6 cm

Surface area  = 4\pi {r^2}  = \left( {4 \times {{22} \over 7} \times 5.6 \times 5.6} \right)c{m^2}
                                                                   = 394.24\,c{m^2}

(iii) We have :
r = radius of the sphere = 14 cm

Surface area  = \left( {4 \times {{22} \over 7} \times 14 \times 14} \right)c{m^2}
                     = 2464\,c{m^2}


Q.2     Find the surface area of a sphere of diameter :
           (i) 14 cm              (ii) 21 cm                        (iii) 3.5 m
Sol.

(i) Here , r = \left( {{{14} \over 2}} \right)cm = 7cm
Surface area = 4\pi {r^2}  = \left( {4 \times {{22} \over 7} \times 7 \times 7} \right)c{m^2}
                                                 = 616\,c{m^2}

(ii) Here , r = \left( {{{21} \over 2}} \right)cm = 10.5\,cm
Therefore Surface area = 4\pi {r^2}  = \left( {4 \times {{22} \over 7} \times 10.5 \times 10.5} \right)c{m^2}
                                                                 = 1386\,c{m^2}

(iii) Here , r = \left( {{{3.5} \over 2}} \right)cm = 1.75\,m
                               = \left( {4 \times {{22} \over 7} \times 1.75 \times 1.75} \right){m^2}
                       = 38.5\,{m^2}


Q.3     Find the total surface area of a hemisphere of radius 10 cm (Use \pi = 3.14)
Sol.

Here , r = 10 cm  
Total surface area of hemisphere  = 3\pi {r^2}  = \left( {3 \times 3.14 \times 10 \times 10} \right)c{m^2}
                                                                                                        = 942\,c{m^2}


Q.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.
Sol.

Let {r_1}\,\,and\,\,{r_2} be the radius of balloons in the two cases.
Here , {r_1} = 7\,cm\,\,and\,\,{r_2} = 14\,cm
Therefore , ratio of their surface area  = {{4\pi {r_1}^2} \over {4\pi {r_2}^2}} = {{r_1^2} \over {r_2^2}}
                                                                            = {{7 \times 7} \over {14 \times 14}} = {1 \over 4}
Thus, the required ratio of their surface areas = 1 : 4


Q.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 Rs 16 per 100 c{m^2}
Sol.

Here r = \left( {{{10.5} \over 2}} \right)cm = 5.25\,cm
Curved surface area of the hemisphere  = 2\pi {r^2} = \left( {2 \times {{22} \over 7} \times 5.25 \times 5.25} \right)c{m^2}
                                                          
 = 173.25\,c{m^2}
Rate of tin - plating is Rs 16 per 100 c{m^2}.
Therefore Cost of tin - plating the hemisphere  = \left( {173.25 \times {{16} \over {100}}} \right)
                                                                       = Rs. 27.72


Q.6      Find the radius of a sphere whose surface area is 154\,c{m^2}
Sol.

Let r be the radius of the sphere.
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
Thus, the radius of the sphere is 3.5 cm


Q.7     The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol.

Let the diameter of earth be R and that of the moon will be {R \over 4}
The radii of moon and earth are {R \over 8}and\,{R \over 2} respectively.
Ratio of their surface area  = {{4\pi {{\left( {{R \over 8}} \right)}^2}} \over {4\pi {{\left( {{R \over 2}} \right)}^2}}} = {{{1 \over {64}}} \over {{1 \over 4}}}
                                                      = {1 \over {64}} \times {4 \over 1} = {1 \over {16}}i.e.,\,1\,:16


Q.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.
Sol.

Inner radius r = 5 cm
Thickness of steel = 0.25 cm
Therefore , outer radius R = (r + 0.25) cm
                                        = (5 + 0.25) cm  = 5.25

Therefore Outer curved surface = 2\pi {R^2}
                                                                  = \left( {2 \times {{22} \over 7} \times 5.25 \times 5.25} \right)c{m^2}
                                                                   = 173.25\,c{m^2}


Q.9 A right circular cylinder just encloses a sphere of radius r (see figure). Find

3
(i) Surface area of the sphere,
(ii) Curved surface area of the cylinder,
(iii) Ratio of the areas obtained in (i) and (ii).
Sol.

(i) The radius of the sphere r, so its surface area = 4\pi {r^2}

(ii) Since the right circular cylinder just encloses a sphere of radius r. So the radius of cylinder = r and its height = 2r
Curved surface of cylinder ={\rm{2}}\pi {\rm{rh}}
2\pi r\left( {2r} \right)         [Since r = r , h = r]

= 4\pi {r^2}

(iii) Ratio of area = 4\pi {r^2}:\,4\pi {r^2} = 1:1



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