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×4.2)J

Now when bullet is stopped by the obstacle, the loss in its mechanical energy =12(m×103)v2J (As m gm=m×103kg )

As 25% of this energy is absorbed by the obstacle,

The energy absorbed by the bullet Q2=75100×12mv2×103=38mv2×103J

Now the bullet will melt if Q2Q1 i.e. 38mv2×10315m×4.2

therefore vmin=410 m/s



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