Surface Area and Volume : Exercise - 13.3 (Mathematics NCERT Class 10th)



Take \pi = {{22} \over 7}, unless stated otherwise.
Q.1      A metallic sphere  of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height  of the cylinder.
Sol. 

Volume of the sphere
 = {4 \over 3}\pi {r^3} = {4 \over 3} \times \pi \times {\left( {4.2} \right)^3}c{m^3}
If h is the  height of a cylinder of radius  6 cm. Then its volume,
 = \pi {\left( 6 \right)^2}h\,c{m^3} = 36\,\pi h\,c{m^3}
Since, the volume of metal in the form of sphere  and cylinder remains the same , we have
36\,\pi h = {4 \over 3} \times \pi \times 4.2 \times 4.2 \times 4.2
 \Rightarrow     h = {1 \over {36}} \times {4 \over 3} \times 4.2 \times 4.2 \times 4.2
 \Rightarrow     h = 2.744


Q.2      Metallic  spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find  the radius of the resulting sphere.
Sol.

Sum of the volumes of 3 gives spheres.
 = {4 \over 3}\pi \left( {{r_1}^3 + {r_2}^3 + {r_3}^3} \right)
 = {4 \over 3}\pi \left( {{6^3} + {8^3} + {{10}^3}} \right)c{m^3}
 = {4 \over 3}\pi \left( {216 + 512 + 1000} \right)c{m^3}
 = {4 \over 3}\pi \left( {1728} \right)c{m^3}
Let R be  the radius of the new spheres whose volume is the sum of the  volumes of 3 given spheres.
Therefore       = {4 \over 3}\pi {R^3} = {4 \over 3}\pi \left( {1728} \right)
 \Rightarrow     {R^3} = 1728
 \Rightarrow     {R^3} = {\left( {12} \right)^3}
 \Rightarrow     R = 12
Hence, the  radius of the  resulting sphere is 12 cm.

Therefore 1728 = 23 × 63 = (2 × 6)3 = 12)3


Q.3      A 20 cm deep well with  diameter 7 cm is dug and the  earth  from  digging is evenly spread out to form a platform 22 m by 14 m. Find the  height of the platform.
Sol. 

Let h m be the  required height  of the platform.
The shape of the platform will be like the shape of a cuboid 22 m × 14 m × h with a hole  in the  shape of cylinder of radius  3.5 m and depth h m.
The volume  of the platform will be equal to the  volume  of the  earth  dug out from  the well.
Now, the  volume  of the earth = Volume  of the cylindrical well
 = \pi {r^2}h
 = {{22} \over 7} \times 12.25 \times 20\,{m^3}
 = 770\,{m^3}
Also, the  volume of the platform = 22 × 14 × h {m^3}
But volume of the platform = Volume of the  well
i.e.,      22 × 14 × h = 770
h = {{770} \over {22 \times 14}} = 2.5
Therefore  Height of the platform = 2.5 m


Q.4     A well of diameter 3 m is dug 14 m deep. The earth taken out of it has  been spread evenly all around it in the shape of a circular  ring  of width 4 m to form an embankment. Find the height of the embankment.
Sol.

Let h be the required height of the embankment.
The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and  external radius (4 + 1.5)m = 5.5 m (see figure).

The  volume of the embankment will be equal  to the volume of earth  dug out from the  well.
Now,  the volume of the earth = Volume  of the cylindrical well
 = \pi \times {\left( {1.5} \right)^2} \times 14\,{m^3} = 31.5\,\pi \,{m^3}
Also  the volume of the embankment
 = \pi \left( {{{5.5}^2} - {{1.5}^2}} \right)h\,{m^3}
 = \pi \left( {5.5 + 1.5} \right)\left( {5.5 - 1.5} \right)h\,{m^3}
 = \pi \times 7 \times 4h\,{m^3} = 28\,\pi h\,{m^3}
Hence, we have
28\,\pi h = 31.5\,\pi               
 \Rightarrow     h = {{31.5} \over {28}} = 1.125
Hence, the required height  of the  embankment = 1.125 m


Q.5     A container shaped like a right  circular  cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be  filled into cones of height12 cm and diameter 6 cm, having  hemispherical shape on the top. Find the number of such  cones which  can be  filled with ice cream.
Sol.

Volume of the cylinder
 = \pi {r^2}h
 = \pi {\left( {{{12} \over 2}} \right)^2} \times 15
 = \pi \times {6^2} \times 15

Volume  of a cone having  hemispherical shape on the top
 = {1 \over 3}\pi {r^2}h + {2 \over 3}\pi {r^3} = {1 \over 3}\pi {r^2}\left( {h + 2r} \right)
 = {1 \over 3}\pi {\left( {{6 \over 2}} \right)^2}\left( {12 + 2 \times {6 \over 2}} \right)
 = {1 \over 3}\pi \times {3^2} \times 18
Let the number of cone that can be filled with ice cream be h.
Then {1 \over 3}\pi \times {3^2} \times 18 \times n = \pi \times {6^2} \times 15
n = {{\pi \times 6 \times 6 \times 15} \over {\pi \times 3 \times 3 \times 18}} \times 3 = 10
Q.6 How many silver  coins, 1.75 cm in diameter and of  thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Sol. The shape of the coin will be like the shape of a cylinder of radius {{1.75} \over 2}cm
= 0.875 cm and of height 2mm  = {2 \over {10}}cm = .2\,cm
Its volume  
 = \pi {r^2}h = {{22} \over 7} \times 0.875 \times 0.875 \times .2\,c{m^3}
 = 0.48125\,c{m^3}
Volume  of the cuboid  = 5.5 \times 10 \times 3.5\,c{m^3} = 192.5\,c{m^3}
Number  of coins  required to form the cuboid
 = {{Volume\,of\,the\,cuboid} \over {Volume\,of\,the\,coin}} = {{192.5} \over {0.48125}} = 400
Hence, 400  coins must be melted to form a cuboid.


Q.6     How many silver  coins, 1.75 cm in diameter and of  thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Sol.

The shape of the coin will be like the shape of a cylinder of radius {{1.75} \over 2}cm
= 0.875 cm and of height 2mm  = {2 \over {10}}cm = .2\,cm
Its volume  
 = \pi {r^2}h = {{22} \over 7} \times 0.875 \times 0.875 \times .2\,c{m^3}
 = 0.48125\,c{m^3}
Volume  of the cuboid  = 5.5 \times 10 \times 3.5\,c{m^3} = 192.5\,c{m^3}
Number  of coins  required to form the cuboid
 = {{Volume\,of\,the\,cuboid} \over {Volume\,of\,the\,coin}} = {{192.5} \over {0.48125}} = 400
Hence, 400  coins must be melted to form a cuboid.


 

Q.7     A cylindrical  bucket, 32 m high  and with  radius of base 18 cm, is filled with sand. The bucket  is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius  and slant height  of the heap.
Sol.

Volume of the  sand = Volume of the cylindrical bucket
 = \pi {r^2}h = \pi \times 18 \times 18 \times 32\,c{m^3}

Volume of the conical heap
 = {1 \over 3}\pi {r^2}h, where  r = ? , h = 24 cm
 = {1 \over 3}\pi {r^2} \times 24\,c{m^3} = 8\,\pi {r^2}
The volume  of the conical  heap will be equal  to that  of sand.
Therefore      8\pi {r^2} = \pi \times 18 \times 18 \times 32
 \Rightarrow     {r^2} = 18 \times 18 \times 4 = {18^2} \times {2^2}
 \Rightarrow     r = 18 × 2 = 36
Here, slant height     \ell = \sqrt {{r^2} + {h^2}}
 \Rightarrow     \ell = \sqrt {{{36}^2} + {{24}^2}}
 \Rightarrow      = \sqrt {12 \times 12\left( {3 \times 3 + 2 \times 2} \right)}
 = 12\sqrt {9 + 4} = 12\sqrt {13}
Hence, the radius of the conical heap is 36 cm and its slant height is 12\sqrt {13} \,cm


Q.8     Water in a canal 6 m wid  and 1.5 m deep, is flowing with a speed of 10 km/h. How  much  area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Sol.

Width  of the  canal = 6 m
Depth of the canal = 1.5 m
Length of water column per hour = 10 km
Length of water  column in 30 minutes or {1 \over 2} hour
 = {1 \over 2} \times 10\,km = 5000\,m
Volume of water flown in 30 minutes
= 1.5 × 6 × 5000 {m^3} = 45000\,{m^3}
Since 8m = {8 \over {100}}m
i.e., 0.08 m standing water is desired
Therefore  Area irrigated in 30 minutes
 = {{Volume\,} \over {Height}} = {{45000} \over {0.08}}
 = 562500\,{m^2}.\,or\,56.25\,hectares


Q.9 A farmer connects a pipe of internal diameter 20 cm from  a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at the  rate of 3 km/h, in how much time  will the tank be filled?
Sol.

Diameter of the pipe = 20 cm
 \Rightarrow     Radius of the pipe = 10 cm
Length of water column per hour
= 3 km = 3 × 1000 × 100 cm
Volume of water flown in one hour
 = \pi \times 100 \times 300000\,c{m^3}
Tank to be  filled = Volume  of cylinder (with  r = 5m = 500 cm and h = 2m = 200 cm)
 = \pi \times 500 \times 500 \times 200\,c{m^3}
Time required to fill  the tank
 = {{Volume{\mkern 1mu} \,of{\mkern 1mu} \tan k} \over {Volume{\mkern 1mu} of{\mkern 1mu} water{\mkern 1mu} flown}}
 = {{\pi \times 500 \times 500 \times 200} \over {\pi \times 100 \times 300000}}hours
 = {5 \over 3}hours = 1{2 \over 3}hours
= 1 hour 40 minutes
= 60 + 40 minutes = 100 minutes.




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