Question

A lead bullet at ${27}^{\circ }C$ just melts when stopped by an obstacle. Assuming that $25$% of its heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking is
(Melting point of lead $={327}^{\circ }C$ specific heat of lead $=0.03cal/g{m}^{\circ }C$, latent heat of fusion of lead$=6cal/gm$ and $J=4.2Joule/cal)$.

• A. $410m/sec$
• B. $1230m/sec$
• C. $307.5m/sec$
• D. None of these

Answer 1

The correct option is A $410m/sec$

Let m be the mass of lead bullet and v be the velocity.

Let kinetic energy of the bullet before striking will be $\frac{1}{2}{mv}^2$

After striking, the kinetic energy of the bullet is converted into heat.

Thus heat produced is

Q=$\frac{1}{2}{mv}^2$

$\therefore 25$ heat is absorbed by the obstacle

$Q^1=\frac{75}{100}Q=\frac{75}{100}.\frac{1}{2}{mv}^2\to \left(1\right)$

this heat $Q^1$ melts the bullet

Temperature of bullet rises from $27℃$ to $327℃$ and then it melts.

If c is specific heat of lead and L be the latent heat of lead, we have,

$Q^1=(m\times c\times \Delta t)+(m\times L)\to \left(2\right)$

From ($1$) and ($2$), we have

$\frac{75}{100}.\frac{1}{2}{mv}^2=(m\times c\times \Delta t)+(m\times L)\frac{75}{100}.\frac{1}{2}{mv}^2=m(c\times \Delta t+L)v^2=(c\times \Delta t+L)\frac{200}{75}{v}^2=(0.03\times (327-27)+6)\times 2.667v^2=40.005kcal{kg}^{-1}{v}^2=40.005\times 4.2\times {10}^{3}J{kg}^{-1}{v}^2=168021m^2{s}^{-2}v=410m/s$