A 2 g bullet moving with a velocity of 200 m/s is brought to. a sudden stoppage by an obstacle. The total heat produced goes to the bullet. If the specific heat of the bullet is 0.03 cal/g-^oC, the rise in its temperature will be

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${0.1580}^{\circ} C$

${1.58}^{\circ} C$

${158.0}^{\circ} C$

${15.80}^{\circ} C$

answer is A.

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Detailed Solution

The kinetic energy of the bullet will be utilized to melt the bullet

$\begin{array}{l} \frac{1}{2} {mv}^{2} = (msΔθ) J \\ \frac{1}{2} \times 2 \times 10^{- 3} \times (200)^{2} = 2 \times 0 .03 \times Δθ \times 4 .2 \\ Δθ = 158^{\circ} C \end{array}$

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