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A lead bullet of 10g travelling at $300 m s^{- 1}$ strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (specific heat of lead is 150J/kg . k)

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$125^{\circ} C$

$100^{\circ} C$

$150^{\circ} C$

$200^{\circ} C$

answer is C.

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Detailed Solution

$W = J Q$

$\frac{1}{2} (\frac{1}{2} m v^{2}) = m s Δ θ$

$ΔT = \frac{(0.5) (0.5) (300^{2})}{150} = 180^{\circ} C$

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