A lead bullet at $$27$$°C just melts when stopped by an obstacle. Assuming that $$25$$% of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking (M$$.P. of lead $$=$$ $$327$$°C, specific heat of lead $$=$$ $$0.03$$ $$cal/gm$$°C, latent heat of fusion of lead $$=$$ $$6$$ $$cal/gm$$ and J $$=$$ $$4.2$$ $$joule/cal)

Question

A lead bullet at $$27$$°C just melts when stopped by an obstacle. Assuming that $$25$$% of heat is absorbed by the obstacle, then  the  velocity of the bullet at the time of striking (M$$.P. of lead $$=$$ $$327$$°C, specific heat of lead $$=$$ $$0.03$$ $$cal/gm$$°C, latent heat of fusion of lead $$=$$ $$6$$ $$cal/gm$$ and J $$=$$ $$4.2$$ $$joule/cal)

Infinity Learn · Accepted Answer

If mass of the bullet is m gm,then total heat required for bullet to just melt $$downQ1$$ $$=$$ m c  $$+$$ m L $$=$$ m x $$0.03$$ (327$$ – $$27) $$+$$ m x $$6$$    $$=$$ $$15$$ m cal     Now when bullet is stopped by the obstacle, the loss in its mechanical energy  $$(As$$  $$)As$$ $$25$$% of this energy is absorbed by the obstacle,The energy absorbed by the bullet Now the bullet will melt if  i.e.

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