A lead bullet at 27^{circ}C just melts when s | vedclass

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(A) Let the mass of the bullet be $m \, g$. The total heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta	he

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(A) Let the mass of the bullet be $m \, g$. The total heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta	heta + mL$.Substituting the values: $Q_1 = m 	imes 0.03 	imes (327 - 27) + m 	imes 6 = m 	imes 0.03 	imes 300 + 6m = 9m + 6m = 15m \, cal$.Converting this to Joules: $Q_1 = 15m 	imes 4.2 = 63m \, J$.The kinetic energy of the bullet is $K.E. = \frac{1}{2} M v^2$,where $M$ is the mass in kg $(M = m 	imes 10^{-3} \, kg)$.Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet to melt it.Thus,$0.75 	imes (\frac{1}{2} 	imes m 	imes 10^{-3} 	imes v^2) = 63m$.Simplifying: $\frac{0.75}{2} 	imes 10^{-3} 	imes v^2 = 63$.$0.375 	imes 10^{-3} 	imes v^2 = 63$.$v^2 = \frac{63}{0.375 	imes 10^{-3}} = \frac{63000}{0.375} = 168000$.$v = \sqrt{168000} \approx 409.87 \, m/s \approx 410 \, m/s$.

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