# 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,
(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.

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