Questions

A lead bullet at 27^{o}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^{o}C, specific heat of lead :0.03 cal/gm^{o}C, 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

NEW

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 40^{o}C. When m gram of ice at -10^{o}C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^{o}C. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta}}{500}\right)\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$where $\mathrm{\theta}$ is temperature in ^{o}C. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice}}=0.5\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water}}=1.0\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$ 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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