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# A lead bullet at 27oC 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 = 327oC, specific heat of lead :0.03 cal/gmoC, latent heat of fusion of lead - 6 cal/gm and J = 4.2J/cal)

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a
410 m/s
b
1230 m/s
c
307.5 m/s
d
None of the above
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detailed solution

Correct option is A

If mass of the bullet is m gm, then total heat required for bullet to just melt downQ1=mcΔθ+mL=m×0.03(327−27)+m×6     =15mcal=(15m×4.2)JNow when bullet is stopped by the obstacle, the loss in its mechanical energy =12m×10−3v2J(As m gm=m×10−3kg)As 25% of this energy is absorbed by the obstacle,The energy absorbed by the bulletQ2=75100×12mv2×10−3=38mv2×10−3JNow the bullet will melt if Q2≥Q1 i.e.  38mv2×10−3≥15m×4.2⇒vmin=410m/s

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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