A steel ball of mass $200 \mathrm{~g}$ falls freely from a height of 20 $\mathrm{m}$ and bounces to a height…
A steel ball of mass $200 \mathrm{~g}$ falls freely from a height of 20 $\mathrm{m}$ and bounces to a height of $10.8 \mathrm{~m}$ from the ground. If the energy lost in this process is absorbed by the ball, the rise in its temperature is ( $\mathrm{g}=10 \mathrm{~ms}^{-2}$, specific heat capacity of steel is $460 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ )
$0.1^{\circ} \mathrm{C}$
$1^{\circ} \mathrm{C}$
$0.2^{\circ} \mathrm{C}$
$2^{\circ} \mathrm{C}$
Solution
Apply the conservation of energy $\mathrm{mg}\left(\mathrm{h}_1-\mathrm{h}_2\right)=\mathrm{mc} \Delta \mathrm{T}$
$\begin{aligned}
& 10 \times(20-10.8)=460 \times \Delta \mathrm{T} \\
& \frac{92}{460}=\Delta \mathrm{T} \Rightarrow \Delta \mathrm{T}=2^{\circ} \mathrm{C}
\end{aligned}$