Question

A lead bullet at $27$ 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.P. of lead = $327$ specific heat of lead = $0.03cal/gm/$, latent heat of fusion of lead = $6cal/gm$ and J = $4.2Joule/cal$)

• A. 410 m/sec
• B. 1230 m/sec
• C. 307.5 m/sec
• D. $820m/s$

Answer 1

The correct option is A 410 m/sec

$\frac{75}{100}\times \frac{1}{2}\times m\times v^2=m\times 0.03\times 4.2\times 1000\times 300+m\times 6\times 4.2\times 1000\Rightarrow 0.75\times 0.5\times v^2=4.2\times 1000\times 15\Rightarrow v^2=168000\Rightarrow v=409.87\approx 410m/s$