A steel ball of mass 0.1 kg falls freely from a height of 10m 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 = 460J/kg/ $^{0}$ C, g=10ms $^{- 2}$ )

{0.01}^{0} C

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{0.1}^{0} C

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1^{0} C

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{1.1}^{0} C

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Solution

The correct option is B ${0.1}^{0} C$
Change in potential energy $=$ Energy absorbed by the ball
$m g (h_{1} - h_{2}) = m c Δ T$
$Δ T = \frac{(10) (10 - 5.4)}{460} = {0.1}^{o} C$

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