Question

A lead bullet, of initial temperature 27?C and speed v km/h penetrates into a solid object and melts. If 50% of the kinetic energy is used to heat it, the value of v in km/h is (for lead melting point = 600 K, latent heat of fusion $=2.5\times 104Jk{g}^{-1};$ specific heat $=125Jk{g}^{-1}{K}^{-1}$ )

• A. 3600
• B. 1800
• C. 1200
• D. 1000

Answer 1

Answer: (B)

Heat energy required to just melt the bullet $Q=Q_1+Q_2$ Here, $Q_1=ms\Delta \theta$ $=m\times 125\times (327-27)$ $=3.75\times {10}^{4}m$ $Q_2=mL=m\times 2.5\times {10}^4$ $=2.5\times {10}^{4}m$ $\therefore$ $Q=6.25\times {10}^{4}m$ If v be the speed of bullet, then 50% of $\frac{1}{2}m{v}^2$ should be equal to Q. Thus, $0.5\times \frac{1}{2}m{v}^2=6.25\times {10}^{4}m$ $\Rightarrow$ $v=\sqrt{\frac{6.25\times {10}^4\times 2}{0.5}}$ $=500m/s=1800km/h$