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A hammer of mass 1 Kg having speed of 50 m/sec hits an iron nail of mass 200g. If the specific heat of iron is 0.105 $c a l / g^{0} C$ and half the energy is converted into heat, the raise in the temperature of nail is

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$7 . 1^{o} C$

$9 . 2^{o} C$

$10 . 5^{o} C$

$12 . 1^{o} C$

answer is A.

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Detailed Solution

$\begin{array}{l} W = J Q \Rightarrow \frac{1}{2} (\frac{1}{2} M v^{2}) = J . m . c Δ θ ∆ θ = \frac{M v^{2}}{4 J m c} \\ \Rightarrow Δ θ = {7.1}^{0} C \end{array}$

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