Remember:When an objects which are converted from one shape to another, or when a liquid which originally filled one container of a particular shape is poured into another container of a different shape or size, their volumns remians constant.

Example: A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

Solution : Volume of cone = ${1/3}πr^2h$= $1/3$ × π × 6 × 6 × 24 $cm^3$
Volumn of Sphere:If $r$ is the radius of the sphere, then its volume is $ 4/3 πr^3$
Since, the volume of clay in the form of the cone and the sphere remains the same, therefore,

			$ 4/3 πr^3$= $1/3$ × π × 6 × 6 × 24
		    $ r^3 $ =  3 × 3 × 24 = $3^3$  × $2^3$
			 $r=6$
Therefore, the radius of the sphere is 6 cm.

Example: Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use π = 3.14)

Note: Unit of measurement for the dimension shall be same. In this case the dimension of sump is 1.57 m × 1.44 m × 95cm i.e one of the measurement is in $cm$.Convert it to $m$ before calculating the volumn

Solution : The volume of water in the overhead tank equals the volume of the water removed from the sump.
Now, the volume of water in the overhead tank (cylinder) = $πr^2h$ = 3.14 × 0.6 × 0.6 × 0.95 $m^3$
The volume of water in the sump when full = l × b × h = 1.57 × 1.44 × 0.95 $m^3$

The volume of water left in the sump after filling the tank =
= [(1.57 × 1.44 × 0.95) – (3.14 × 0.6 × 0.6 × 0.95)] $m^3$ = (1.57 × 0.6 × 0.6 × 0.95 × 2) $m^3$

So, the height of the water left in the sump = ${volume-of- water -left -in -the- sump} /{l × b}$

					=${1.57 × 0.6 × 0.6 × 0.95 × 2  }/{1.57 × 1.44 }$
					= 0.475 $m$ 
					= 47.5 $cm$
	Also,  ${Capacity-of-tank  }/{Capacity-of-sump}$ = ${3.14 × 0.6 × 0.6 × 0.95 }/{1.57 × 1.44 × 0.95}$ =  $1/2$

Therefore, the capacity of the tank is half the capacity of the sump.


Example: A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

Solution : The volume of the rod = $πr^2h$ =π ×$(1/2)^2$ × 8 =2π
The length of the new wire of the same volume = 18 m = 1800 cm
If r is the radius (in cm) of cross-section of the wire, its volume = π × $r^2$ × 1800 $cm^3$

Therefore, π ×  $r^2$  × 1800 = volume of the rod 
			π ×  $r^2$  × 1800 =  2π
i.e.,          $r^2$  = $1/900$
i.e.,          r = $1/30$
So,  the  diameter  of  the  cross  section,  i.e., the  thickness  of  the  wire  is ${1/15}$ $cm$ i.e., 0.67mm (approx.).

Example : A hemispherical tank full of water is emptied by a pipe at the rate of $3{4/7}$ litres per second. How much time will it take to empty half the tank, if it is 3m in diameter? (Take π = $22/7$)

 Solution : Radius of the hemispherical tank = $3/2$ m
 Volume of the tank =  ${2/3}× {22/7}× {3/2}$ $m^3$ =$99/14$ $m^3$
 So,  the volume of the water to be emptied  = ${1/2}×99/14$ $m^3$
 Since 1000 liters = 1 $m^3$,
									=$99/28×1000$  $liters$
									=$99000/28$ $liters$
									
Since,  $25/7$  litres of water is emptied in 1 second,$99000/28$ litres of water will be emptied 
                    in $99000/28 × 7/25$  seconds, i.e., in 16.5 minutes.
					
Time taken to empty half the tank = 16.5 minutes.