A ball is dropped on a floor from a height of `2.0m`. After the collision it rises up to a height of `1.5m`. Assume that `40%` of the mechanical energy lost goes as thermal energy into the ball.Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is `800J K^(-1)`

Options

A. `10^(-3@)C`
B. `1.5xx10^(-3@)C`
C. `2xx10^(-3@)C`
D. `2.5xx10^(-3@)C`

Answer

D

Explanation

Loss in potential enery`xx(40)/(100)=`Heat
`mg(2-1.5)xx0.4=msDeltatheta`
`1xx10xx0.5xx0.4=800Deltatheta`
`Deltatheta=0.0025^@C=2.5xx10^(-3@)C`