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 specific heat capacity of the ball is `800J kg^(-1) K^(-1)`

Answer

A, B, C

Explanation

Let the mass of hall be m kg
and `V_(1) = sqrt(2gh) = sqrt(40)`
`V_(2)= sqrt(2gh) = sqrt(30)`
So change in K.E.
`= (1)/(2)xx mxx 40- ((1)/(2)m)xx 30 = ((10)/(2))m`
`= 5m`
That is utilisedto increases in temperature of the ball
`((40)/(100)) xx (10)/(2)m = m xx 800 xx Delta t`
`rArr Delta t = (1)/(400) = 0.0025`
`= 2.5 xx 10^(-3)^(@)C`