A small ball of mass `m` is dropped from an aeroplane moving at `50 m` height above the ground with a speed of `25 sqrt(2) "metre"//sec` if half of the mechanical energy of ball with respect to ground is lost as a thermal energy due to air friction. The change in the temperature of the ball as it lands on the ground is. specific heat capacity of ball is `56.25 J//kg`. `(g = 10 m//sec^(2))`

Options

A. `5^(@)C`
B. `10^(@)C`
C. `20^(@)C`
D. `500//56.25^(@)C`

Answer

B

Explanation

`1/2(Mxxgxxg+1/2MV^(2)) = Ms Delta T`
`1/2(10xx50+1/2xx(25sqrt(2))^(2)) = 46.25xxDelta T`
`Delta T = (500+625)/(56.25xx2) = 10^(@)C`.