A) 410 m/sec
B) 1230 m/sec
C) 307.5 m/sec
D) None of the above
Correct Answer: A
Solution :
If mass of the bullet is m gm, then total heat required for bullet to just melt down Q1 = m c Dq + m L = m ´ 0.03 (327 ? 27) + m ´ 6 = 15 m cal \[=(15m\times 4.2)J\] Now when bullet is stopped by the obstacle, the loss in its mechanical energy \[=\frac{1}{2}(m\times {{10}^{-3}}){{v}^{2}}J\] (As \[m\ gm=m\times {{10}^{-3}}kg\]) As 25% of this energy is absorbed by the obstacle, The energy absorbed by the bullet \[{{Q}_{2}}=\frac{75}{100}\times \frac{1}{2}m{{v}^{2}}\times {{10}^{-3}}=\frac{3}{8}m{{v}^{2}}\times {{10}^{-3}}J\] Now the bullet will melt if \[{{Q}_{2}}\ge {{Q}_{1}}\] i.e. \[\frac{3}{8}m{{v}^{2}}\times {{10}^{-3}}\ge 15m\times 4.2\]Þ \[{{v}_{\min }}=410\ m/s\]
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