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A steel ball of mass 0.1 kg falls freely...

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)`)

Text Solution

Verified by Experts

Loss in gravitational PE = Heat energy ,
`0.1 xx 10(10 - 5.4) = 0.1 xx 460 xx Delta theta`
`implies Deltatheta = 0.1^(0)C` (rise in temperature)
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