# Surface Areas and Volumes : Exercise 13.2 (Mathematics NCERT Class 9th) Assume $\pi = {{22} \over 7}$, unless stated otherwise.
Q.1     The curved surface area of a right circular cylinder of height 14 cm is 88 $c{m^2}$. Find the diameter of the base of the cylinder.
Sol.

Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then,
Curved surface area of cylinder $= 2\,\pi rh$
$\Rightarrow$ $88 = 2 \times {{22} \over 7} \times r \times 14$
$\Rightarrow$ $r = {{88 \times 7} \over {2 \times 22 \times 14}} = 1$
Therefore Diameter of the base = 2r $= 2 \times 1 = 2\,cm$

Q.2    It is required to make a closed cylindrical tank of height 1m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Sol.

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm, radius of base = ${{140}\over 2}$ = 70 cm = .70 m
Height = 1m
Metal sheet required to make a closed cylindrical tank
= Its total surface area
$= \,2\pi r\left( {h + r} \right)$ $= 2 \times {{22} \over 7} \times 0.7\left( {1 + 0.70} \right){m^2}$
[Because h = 1 m, r = ${{140} \over 2}$ cm = 70 cm = 0.70 cm]
$= 2 \times 22 \times 0.1 \times 1.70\,{m^2}$
$= 7.48\,{m^2}$
Hence, the sheet required $= 7.48\,{m^2}$

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Q.3      A metal pipe is 77 cm long. The inner diameter of a cross- section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its.

(i) Inner curved surface area,

(ii) Outer curved surface area,

(iii) Total surface area. Sol.

We have, R = external radius $= {{4.4} \over 2}cm = 2.2\,cm$
r = internal radius $= {4 \over 2}cm = 2cm$
h = length of the pipe = 77 cm
(i)
Inner curved surface $= 2\pi rh\,c{m^2}$
$= 2 \times {{22} \over 7} \times 2 \times 77\,c{m^2}$
$= 968\,c{m^2}$

(ii) Outer curved surface $= 2\pi Rh\,c{m^2}$
$= 2 \times {{22} \over 7} \times 2.2 \times 77\,c{m^2}$

$= 1064.8\,c{m^2}$

(iii) Total surface area of a pipe
= Inner curved surface area + outer curved surface area + areas of two bases
= $2\pi rh + 2\pi Rh + 2\pi \left( {{R^2} - {r^2}} \right)$
= $\left[ {968 + 1064.8 + 2 \times {{22} \over 7}\left( {4.84 - 4} \right)} \right]$ cm2

= $\left( {2032.8 + {{44} \over 7} \times 0.84} \right)c{m^2}$
=$\left( {2032.8 + 5.28} \right)c{m^2} = 2038.08\,c{m^2}$

Q.4     The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in ${m^2}$.
Sol.

The length of the roller is 120 cm i.e., h = 1.2 m and ,
radius of the cylinder (i.e., roller) $= {{84} \over 2}cm = 42cm = 0.42\,m$.
Distance covered by roller in one revolution
= Its curved surface area = $2\pi rh$
= $\left( {2 \times {{22} \over 7} \times 0.42 \times 1.2} \right){m^2}$
= $3.168\,{m^2}$
Area of the playground = Distance covered by roller in 500 revolution.
= $\left( {500 \times 3.168} \right){m^2} = 1584\,{m^2}$
Hence , the area of playground is $1584\,{m^2}$

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Q.5     A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curve surface of the pillar at the rate of Rs 12.50 per ${m^2}$.
Sol.

Let r be the radius of the base and h be the height of the pillar.
Therefore $r = {{50} \over 2}$ cm = 25 cm = .25 m and h = 3.5 m.
Curved surface = $2\pi rh$
$= \left( {2 \times {{22} \over 7} \times 0.25 \times 3.5} \right){m^2} = 5.5\,{m^2}$
Cost of painting the curved surface @ Rs. 12. 50 per ${m^2}$
$= \left( {5.5 \times 12.5} \right)$ =  Rs. 68.75

Q.6      Curved surface area of a right circular cylinder is 4.4 ${m^2}$. If the radius of the base of the cylinder 0.7 m, Find its height.
Sol.

Let r be the radius of the base and h be the height of the cylinder. Then,
Curved surface area = 4.4 ${m^2}$
$\Rightarrow$ $2\pi rh = 4.4$
$\Rightarrow$ $2 \times {{22} \over 7} \times 0.7 \times h = 4.4$ [since r = 0.7]
$\Rightarrow$ $h = \left( {{{4.4 \times 7} \over {2 \times 22 \times 0.7}}} \right)\,m = 1\,m$
Thus, the height of the cylinder = 1 metre.

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Q.7     The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find.

(i) Its inner curved surface area.

(ii) The cost of plastering this curved surface at the rate of Rs 40 per ${m^2}$

Sol.

(i) Let r be the radius of the face and h be depth of the well. Then,
Curved surface = $2\pi rh$
= $\left( {2 \times {{22} \over 7} \times {{3.5} \over 2} \times 10} \right){m^2} = 110\,{m^2}$

(ii) Cost of plastering is Rs 40 per ${m^2}$
Therefore , cost of plastering the curved surface = Rs (110 × 40) = Rs 4400.

Q.8     In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Sol.

Total radiating surface in the system
= Curved surface area of the pipe = $2\pi rh$
[where $r = {5 \over 2}cm = 2.5\,cm = {{2.5} \over {100}}m = 0.025\,m\,and\,h = 28\,m$]
$= \left( {2 \times {{22} \over 7} \times 0.025 \times 28} \right){m^2} = 4.4\,{m^2}$

Q.9     Find
(i) the lateral or curved surface area of cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.

(ii) How much steel was actually used, if ${1 \over {12}}$ of the steel actually used was wasted in making the closed tank.

Sol.

(i) Here, $r = \left( {{{4.2} \over 2}} \right)m = 2.1\,m\,\,and\,\,h = 4.5\,m$
Lateral surface area $= 2\,\pi rh\,{m^2}$
$= \left( {2 \times {{22} \over 7} \times 2.1 \times 4.5} \right){m^2}$
$= 59.4\,{m^2}$

(ii) Since ${1 \over {12}}$ of the actual steel used was wasted,
the area of the steel which has gone into the tank = $\left( {1 - {1 \over {12}}} \right)$ of $x = {{11} \over {12}}of\,x.$
Steel used = Letarl surface area + 2 x Area of base
= $\left( {2\pi rh + 2\pi {r^2}} \right)c{m^2}$ = $\left( {59.4 + 2 \times {{22} \over 7} \times 2.1 \times 2.1} \right)c{m^2}$
= $\left( {59.4 + 27.72} \right)cm = 87.12\,c{m^2}$
Therefore,   actual steel used =${{11} \over {12}} \times x = 87.12$
$\Rightarrow$ $x = {{87.12 \times 12} \over {11}} = 95.04\,\,{m^2}$
Hence, the actual area of the steel used = $95.04\,{m^2}$

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Q.10    In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. Sol.

Here , $r = \left( {{{20} \over 2}} \right)$ cm = 10 cm and h = 30 cm + 2 × 2.5 cm (i.e., margin) = 35 cm
Cloth required for covering the lampshade
= Its curved surface area = $2\pi rh$
= $\left( {2 \times {{22} \over 7} \times 10 \times 35} \right)c{m^2}$
= $2200\,c{m^2}$

Q.11   The students of a Vidyalaya were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors , how much cardboard was required to be bought for the competition?
Sol.

Cardboard required by each competitor
= Curved surface area of one penholder + base area = $2\pi rh + \pi {r^2}\,$
[ $where\,\,r = 3cm\,,h = 10.5\,cm$]
= $\left[ {\left( {2 \times {{22} \over 7} \times 3 \times 10.5} \right) + {{22} \over 7} \times 9} \right]c{m^2}$
= $\left( {198 + 28.28} \right)c{m^2}$ = $226.28\,c{m^2}\left( {approx} \right)$
Cardboard required for 35 competitors = $\left( {35 \times 226.28} \right)c{m^2}$
= $7920\,c{m^2}\left( {approx} \right)$