Question

A lead bullet just melts when stopped by an obstacle. Assuming that $5$ percent of the heat is absorbed by the obstacle, find the speed of the bullet if its initial temperature is ${27}^o$C. (Melting point of lead $={327}^o$C, specific heat of lead $=0.03$ cal/gm/${}^{o}C$, latent heat of fusion of lead $=6$ calories/gm, J$=4.2$joules/calorie)

• A. $610$ m/s.
• B. $410$ m/s.
• C. $510$ m/s.
• D. $710$ m/s.

Answer 1

The correct option is B $410$ m/s.

solution:

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 $\left({27}^o\right)$ to $\left({327}^o\right)$and then it melts.

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

,$Q^1$ =$\left(mc\Delta t\right)+\left(mL\right)\to \left(2\right)$

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

$\left(\frac{75}{100}\right)\frac{1}{2}{mv}^2=\left(mc\Delta t\right)+\left(mL\right)$

$\frac{75}{100}\frac{1}{2}{mv}^2=m(c\Delta t+L)$

$v^2=(c\Delta t+L)\frac{200}{75}$

$v^2=0.03$ x $(327-27)+6)$ x $2.667$

$v^2=40.005kcal{kg}^{-1}$

$v^2=40.005$ x $4.2$ x ${10}^{3}J{kg}^{-1}$

$v^2=168021m^2{s}^{-2}$

$v=410m/s$

hence the correct option: B