A steel ball of mass 0.1 , kg falls freely fr | vedclass

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(B) According to the law of conservation of energy,the loss in potential energy of the ball is converted into heat energy,which is absorbed by the bal

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$0.1$

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(B) According to the law of conservation of energy,the loss in potential energy of the ball is converted into heat energy,which is absorbed by the ball,causing a rise in its temperature.The loss in potential energy is given by:$\Delta PE = mg(h_1 - h_2)$This energy is equal to the heat absorbed by the ball:$Q = mc\Delta 	heta$Equating the two:$mg(h_1 - h_2) = mc\Delta 	heta$Solving for the rise in temperature $\Delta 	heta$:$\Delta 	heta = \frac{g(h_1 - h_2)}{c}$Given:$g = 10 \, m \, s^{-2}$$h_1 = 10 \, m$$h_2 = 5.4 \, m$$c = 460 \, J \, kg^{-1} \, ^\circ C^{-1}$$\Delta 	heta = \frac{10 	imes (10 - 5.4)}{460}$$\Delta 	heta = \frac{10 	imes 4.6}{460}$$\Delta 	heta = \frac{46}{460} = 0.1 \, ^\circ C$

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