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A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, $N \times 100 m / s$ is the initial speed of the bullet. The initial temperature of the bullet is $27 ° C$ and its melting point is $327 ° C .$ Latent heat of fusion of lead $= 2.5 \times 10^{4} J / k g$ and specific heat capacity of lead $= 125 J / k g - K .$ Then find value of N

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Detailed Solution

Let the mass of the bullet = m.

Heat required to take the bullet from $27 ° C t o 327 ° C = m \times (125 J / k g - K) (300 K) = m \times (3.75 \times 10^{4} J / k g)$

Heat required to melt the bullet $= m \times (2.5 \times 10^{4} J / k g)$

If the initial speed be v, the kinetic energy is $\frac{1}{2} m v^{2}$ and hence the heat developed is $\frac{1}{2} (\frac{1}{2} m v^{2}) = \frac{1}{4} m v^{2} .$ Thus, $\frac{1}{4} m v^{2} = m (3.75 + 2.5) \times 10^{4} J / k g$ or, $v = 500 m / s$

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