A lead bullet (specific heat = 0.032 cal/gm oC) is completely stopped when it strikes a target with a velocity of 300 m/s. The heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be

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A lead bullet (specific heat = 0.032 cal/gm ^oC) is completely stopped when it strikes a target with a velocity of 300 m/s. The heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be

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1.67^oC

16.7^oC

167.4^oC

267.4^oC

answer is C.

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Detailed Solution

$\begin{array}{l} S = 0 .032 cal / {gm}^{\circ} C = \frac{0 .032}{10^{- 3} kg} = 32 cal / {kg}^{0} C \\ V = 300 m / s \\ Δθ = \frac{1}{4} \frac{v^{2}}{Js} \Rightarrow \frac{1}{2} \times \frac{3 \times 3 \times 10^{4}}{4 .2 \times 32} = \frac{90000}{268 .8} = 334 {.8}^{\circ} C \end{array}$
Rise in temperature of bullet $(Δθ)_{b}$
$\frac{Δθ}{2} = \frac{334 .8}{2} = 167 {.4}^{0} C$