Properties Of Solids And Liquids Question 110
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) [IIT 1981]
Options:
A) 410 m/sec
B) 1230 m/sec
C) 307.5 m/sec
D) None of the above
Show Answer
Answer:
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}mv^{2}\times {{10}^{-3}}=\frac{3}{8}mv^{2}\times {{10}^{-3}}J $
Now the bullet will melt if $ Q _{2}\ge Q _{1} $ i.e. $ \frac{3}{8}mv^{2}\times {{10}^{-3}}\ge 15m\times 4.2 $
therefore $ {v _{\min }}=410\ m/s $
BETA
