A piece of lead falls from a height of 100m on a fixed non-conducting slab which brings it to rest. If the specific heat of lead is 30.6 cal/kg °C, the increase in temperature of the slab immediately after collision

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6.72°C

8.72°C

5.62°C

7.62°C

answer is B.

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Detailed Solution

$\begin{array}{l} h = 100 m; s = 30.6 cal / {kg}^{0} C \\ \begin{array}{l} m gh = J . m sΔ θ \Rightarrow Δ θ = \frac{gh}{Js} = \frac{9.8 \times 100}{4.2 \times 30.6} = \frac{980}{128.52} \\ = {7.625}^{\circ} C \end{array} \end{array}$

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