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A lead bullet penetrates into a solid object and melts. Assuming that 40 \%  of its kinetic energy is used to heat it, the initial speed of bullet is :
(Given, initial temperature of the bullet =127^{\circ} \mathrm{C},
Melting point of the bullet =327^{\circ} \mathrm{C},
Latent heat of fusion of lead =2.5 \times 10^{4} \mathrm{~J} \mathrm{~kg}^{-1},
Specific heat capacity of lead =125 \mathrm{~J} / \mathrm{kg} \mathrm{K} )

Option: 1

125 \mathrm{~ms}^{-1}


Option: 2

500 \mathrm{~ms}^{-1}


Option: 3

250 \mathrm{~ms}^{-1}


Option: 4

600 \mathrm{~ms}^{-1}


Answers (1)

best_answer

Let initial speed \mathrm{= u}
\underset{127^{\circ}C}{\mathrm{bullet}}\: \overset{Q_{1}}{\rightarrow}\underset{327^{\circ}C}{\mathrm{bullet}}\: \overset{Q_{2}}{\rightarrow}\,\underset{327^{\circ}C}{\mathrm{bullet\, melts}}

total heat required.
\mathrm{Q =Q_{1}+Q_{2}}
      \mathrm{=m s \Delta T+m L_{f} }
       \mathrm{=m\left[125 \times 200+2.5 \times 10^{4}\right] }

Given,
\mathrm{40 \% \, K E =Q }
\mathrm{\frac{40}{100} \times \frac{1}{2} m u^{2} =m \times\left[250 \times 100+2.5 \times 10^{4}\right] }
\mathrm{\frac{2}{5} \times \frac{u^{2}}{2} =5 \times 10^{4} }
\mathrm{u^{2} =25 \times 10^{4} }

\mathrm{u =5 \times 10^{2} \mathrm{m/s} }

The correct answer is (2)

Posted by

sudhir.kumar

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