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}


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.
surface

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}


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.


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}


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. 2 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)