A lead bullet penetrates into a solid object | vedclass

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(B) Let $m$ be the mass of the bullet and $v$ be its initial velocity.The kinetic energy of the bullet is $K.E. = \frac{1}{2} m v^2$.According to the

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$500$

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(B) Let $m$ be the mass of the bullet and $v$ be its initial velocity.The kinetic energy of the bullet is $K.E. = \frac{1}{2} m v^2$.According to the problem,$40 \%$ of this energy is used to heat the bullet to its melting point and then melt it.Heat required to raise the temperature from $127^{\circ} C$ to $327^{\circ} C$ is $Q_1 = m c \Delta T = m 	imes 125 	imes (327 - 127) = m 	imes 125 	imes 200 = 25000 m \, J$.Heat required to melt the bullet is $Q_2 = m L = m 	imes 2.5 	imes 10^{4} = 25000 m \, J$.Total heat required $Q = Q_1 + Q_2 = 25000 m + 25000 m = 50000 m \, J$.Equating $40 \%$ of kinetic energy to total heat: $0.40 	imes (\frac{1}{2} m v^2) = 50000 m$.$0.2 v^2 = 50000$.$v^2 = 250000$.$v = 500 \, m \, s^{-1}$.

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