Q: A lead bullet strikes against a steel plate with a velocity 200 m/s. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and target, then the rise in temperature of the bullet is (specific heat capacity of lead = 125 Jkg^{-1} K^{-1})

(a) 80 °C

(b) 60 °C

(c) 160 °C

(d) 40 °C

Ans: (a)

Sol: $\large \frac{1}{2}mv^2 \times \frac{1}{2} = m s \Delta T$

$\large \frac{v^2}{4 s } = \Delta T $

$\large \Delta T = \frac{(200)^2}{4 \times 125}$

Δ T = 80°C