A lead bullet at 27,^oC just melts when stopped by an obstacle. Assuming that 25% of heat is absorbe

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10-1.Thermometry, Thermal Expansion and Calorimetry

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A lead bullet at $27\,^oC$ 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 ........ $m/s$ ( $M. P.$ of lead $= 327\,^oC,$ specific heat of lead $= 0.03\,cal/g\,^oC,$ latent heat of fusion of lead $= 6\,cal/g$ and $J = 4.2\,joule/cal$ )

A

$410$

B

$1230$

C

$307.5$

D

None of the above

Solution

$\frac{75}{100} \times \frac{1}{2} \times m \times v^{2}=m \times 0.03 \times 4.2 \times 1000 \times 300+m \times 6 \times 4.2 \times 1000$

$\Rightarrow 0.75 \times 0.5 \times v^{2}=4.2 \times 1000 \times 15$

$\Rightarrow v^{2}=168000$

$\Rightarrow v=409.87 \approx 410 m / s$

Standard 11

Physics

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