A lead bullet at 27°C just melts when sto | vedclass

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Question

(a) If mass of the bullet is $m \,gm,$

then total heat required for bullet to just melt down

$Q_1 = m c \Delta 	heta + m L = m 	imes 0.03 (327

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(a) If mass of the bullet is $m \,gm,$

then total heat required for bullet to just melt down

$Q_1 = m c \Delta 	heta + m L = m 	imes 0.03 (327 -27) + m 	imes 6$

     $= 15\, m\, cal $$ = (15m 	imes 4.2)J$

Now when bullet is stopped by the obstacle, the loss in its mechanical energy $ = \frac{1}{2}(m 	imes {10^{ - 3}}){v^2}J$

(As $m\;gm = m 	imes {10^{ - 3}}kg$)

As $25\%$ of this energy is absorbed by the obstacle,

The energy absorbed by the bullet

${Q_2} = \frac{{75}}{{100}} 	imes \frac{1}{2}m{v^2} 	imes {10^{ - 3}} = \frac{3}{8}m{v^2} 	imes {10^{ - 3}}J$

Now the bullet will melt if ${Q_2} \ge {Q_1}$

i.e. $\frac{3}{8}m{v^2} 	imes {10^{ - 3}} \ge 15m 	imes 4.2$==> ${v_{\min }} = 410\;m/s$

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