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 Joule-kg^-1oC^-1, g: 10 ms^-2)

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0.01°C

1.1°C

1°C

0.1°C

answer is B.

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Detailed Solution

According to energy conservation, change in potential energy of the ball appears in the form of heat which raises the temperature of the ball.

i.e., $mg (h_{1} - h_{2}) = mcΔθ$
$\Rightarrow Δθ = \frac{g (h_{1} - h_{2})}{c} = \frac{10 (10 - 5 .4)}{460} = 0 {.1}^{\circ} C$

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