A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel `=460 K//kg^(@)//C,g=10 m//s^(2)`)

Options

A. `0.01^@C`
B. `0.1^@C`
C. `1^@C`
D. `1.1^@C`

Answer

B

Explanation

Loss in energy `=mg(h-h')`
`=0.1xx10xx(10-5.4)`
`=4.6J`
Now, `4.6J=ms Delta theta`
`=0.1xx460xxDelta theta`
`Delta theta=0.1^@C`