Heat And Thermodynamics Ques 8
- A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is $27^{\circ} \mathrm{C}$.
(Melting point of lead $=327^{\circ} \mathrm{C}$, specific heat of lead $=0.03 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$, latent heat of fusion of lead $=6 \mathrm{cal} / \mathrm{g}$, $J=4.2 \mathrm{~J} / \mathrm{cal}$.)
(1981, 3M)
(a) $\frac{p}{\alpha K}$
(b) $\frac{3 \alpha}{p K}$
(c) $3 p K \alpha$
(d) $\frac{p}{3 \alpha K}$
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Answer:
Correct Answer: 8.( 409.8 m/s )
Solution:
Formula:
- $75 %$ heat is retained by bullet
$ \frac{3}{4}\left[\frac{1}{2} m v^2\right]=m s \Delta \theta+m L \text { or } v=\sqrt{\frac{(8 s \Delta \theta+8 L)}{3}} $
Substituting the values, we have
$ \begin{aligned} v & =\sqrt{\frac{(8 \times 0.03 \times 4.2 \times 300)+(8 \times 6 \times 4.2)}{3 \times 10^{-3}}} \\ & =409.8 \mathrm{~m} / \mathrm{s} \end{aligned} $
BETA
