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Surface Areas and Volumes : Exercise 13.6 (Mathematics NCERT Class 9th)


Assume \pi = {{22} \over 7}, unless stated otherwise.
Q.1     The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000\,c{m^3} = 1\ell )

Sol.

Let r cm be the radius of the base and h cm be the height of the cylinder.
Circumference of the base = 132 cm
 \Rightarrow 2\pi r = 132
 \Rightarrow 2 \times {{22} \over 7} \times r = 132
 \Rightarrow r = \left( {{{132 \times 7} \over {2 \times 22}}} \right)cm = 21\,cm
Volume of the cylinder = \pi {r^2}h\,c{m^3}
                                        = \left( {{{22} \over 7} \times 21 \times 21 \times 25} \right)c{m^3}
                                        = 34650\,c{m^3}
Therefore Vessel can hold  = \left( {{{34650} \over {1000}}} \right)litres
i.e., 34.65 litres of water.


Q.2    The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1\,c{m^3} of wood has a mass of 0.6 g.
Sol.

We have h = Height of the cylindrical pipe = 35 cm
R = External radius  = \left( {{{28} \over 2}} \right)cm = 14\,cm
r = Internal radius  = \left( {{{24} \over 2}} \right)cm = 12\,cm
Volume of the wood used in making the pipe = Volume the external cylinder – Volume of the internal cylinder
\pi {R^2}h - \pi {r^2}h = \pi \left( {{R^2} - {r^2}} \right)h
                               = {{22} \over 7} \times \left( {{{14}^2} - {{12}^2}} \right) \times 35\,\,c{m^3}
                               = {{22} \over 7} \times 26 \times 2 \times 35\,\,c{m^3}
                               = 5720\,\,c{m^3}
Weight of 1c{m^3} = 0.6\,g
Therefore Weight of 5720 c{m^3} = \left( {5720 \times 0.6} \right)\,g
                                                            = \left( {{{5720 \times 0.6} \over {1000}}} \right)kg     [1 kg = 1000 g]
                                                           = 3.432\,kg


Q.3     A soft drink is available in two packs - (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much ?
Sol.

(i) Capacity of tin can  = \ell bh\,c{m^3}
                                       = \left( {5 \times 4 \times 15} \right)c{m^3}
                                       = 300\,c{m^3}
(ii) Capacity of plastic cylinder  = \pi {r^2}h\,c{m^3}

                                                        = \left( {{{22} \over 7} \times {7 \over 2} \times {7 \over 2} \times 10} \right)cm
                                                      
 = 385\,c{m^3}
Thus , the plastic cylinder has greater capacity by (385 - 300) = 85 cm3


Q.4     If the lateral surface of a cylinder is 94.2\,c{m^2} and its height is 5 cm , then find (i) radius of its base (ii) its volume (Use \pi = 3.14)
Sol.

(i) Let r be the radius of the base and h be the height of the cylinder. Then ,
Lateral surface  = 94.2\,c{m^2}
 \Rightarrow 2\pi rh = 94.2\,
 \Rightarrow 2 × 3.14 × r × 5 = 94.2
 \Rightarrow r = {{94.2} \over {2 \times 3.14 \times 5}} = 3
Thus , the radius of its base = 3 cm

(ii) Volume of the cylinder  = \pi {r^2}h
                                                = \left( {3.14 \times {3^2} \times 5} \right)c{m^3}
                                                = 141.3\,c{m^3}


Q.5     It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per {m^2}, Find.
              (i) Inner curved surface area of the vessel,
              (ii) Radius of the base,
              (iii) Capacity of the vessel.
Sol.

(i) Inner curved surface area of the vessel
 = {{Total\,\cos t\,\,of\,\,pa{\mathop{\rm int}} ing} \over {Rate\,of\,pa{\mathop{\rm int}} ing}}
 = \left( {{{2200} \over {20}}} \right){m^2} = 110\,{m^2}

(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.
Therefore 2\pi rh = 110
 \Rightarrow 2 \times {{22} \over 7} \times r \times 10 = 110
 \Rightarrow r \times {{110 \times 7} \over {2 \times 22 \times 10}} = {7 \over 4} = 1.75
Thus, the radius of the base = 1.75 m

(iii) Capacity of the vessel = \pi {r^2}h
                                               = \left( {{{22} \over 7} \times {7 \over 4} \times {7 \over 4} \times 10} \right){m^3}
                                             
 = 96.25\,{m^3}


Q.6     The capacity of a closed cylindrical vessel height 1 m is 15.4 litres. How many square metres metal sheet would be needed to make it?
Sol.

Capacity of a closed cylindrical vessel = 15. 4 litres
                                                                   = \left( {15.4 \times {1 \over {1000}}} \right){m^3} = 0.0154\,{m^3} [1 m3= 1000 liters]
Let r be the radius of the base and h be the height of the vessel. Then ,
Volume\,\, = \pi {r^2}h = \pi {r^2} \times 1 = \pi {r^2}                       [Since h = 1m]
Therefore \pi {r^2} = 0.0154
 \Rightarrow {{22} \over 7} \times {r^2} = 0.0154
 \Rightarrow {r^2} = {{0.0154 \times 7} \over {22}} = 0.0049
 \Rightarrow r = \sqrt {0.0049} = 0.07
Thus, the radius of the base of vessel = 0.07 m
Metal sheet needed to make the vessel = Total surface area of the vessel
                                                                   = 2\pi rh + 2\pi {r^2} = 2\pi r\left( {h + r} \right)
                                                                   = 2 \times {{22} \over 7} \times 0.07 \times \left( {1 + 0.07} \right){m^2}
                                                                 
 = 44 \times 0.01 \times 1.07\,{m^2} = 0.4708\,{m^2}


Q.7      A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Sol.

Diameter of the graphite cylinder = 1 mm = {1 \over {10}} cm
Therefore Radius\, = {1 \over {20}}cm
Length of the graphite = 14 cm
Volume of the graphite cylinder = \pi {r^2}h
                                                        = \left( {{{22} \over 7} \times {1 \over {20}} \times {1 \over {20}} \times 14} \right)c{m^3} = 0.11\,c{m^3}

Diameter of the pencil = 7mm = {7 \over {10}}cm = 0.11\,c{m^3}
Therefore Radius of the pencil = {7 \over {20}}cm
and , length of the pencil = 14 cm
Therefore Volume of the pencil = \pi {r^2}h
                                                        = \left( {{{22} \over 7} \times {7 \over {20}} \times {7 \over {20}} \times 14} \right)c{m^3}
                                                      
 = 5.39\,c{m^3}
Volume of wood = Volume of the pencil – Volume of the graphite
                              = \left( {5.39 - 0.11} \right)c{m^3}
                              = 5.28c{m^3}


Q.8     A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Sol.

Diameter of the cylindrical bowl = 7 cm
Therefore Radius = {7 \over 2}cm
Height of serving bowl = 4 cm
Therefore Soup saved in on serving = Volume of the bowl
                                                             = \pi {r^2}h
                                          = \left( {{{22} \over 7} \times {7 \over 2} \times {7 \over 2} \times 4} \right)c{m^3}
                                                            
 = 1.54\,c{m^3}
Soup served to 250 patients = \left( {250 \times 1.54} \right)c{m^3}
                                                 = 38500\,c{m^3}\,i.e.,\,38.5\ell
Hence , the hospital has to prepare 38.5 l soup daily to serve 250 patients.