Q 11-10-107JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
A bullet of mass $5\ \text{g}$, travelling with a speed of $210\ \text{m s}^{-1}$, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet. The rise of temperature of the bullet if the specific heat of its material is $0.030\ \text{cal g}^{-1}\,^\circ\text{C}^{-1}$ ($1\ \text{calorie} = 4.2\times10^{7}$ ergs) is close to:
Answer: (A) $87.5^\circ\text{C}$
KE $= \tfrac12\times0.005\times210^{2} = 110.25$ J. Half of it, $55.1$ J $= \dfrac{55.1}{4.2} \approx 13.1$ cal, heats the bullet.
$$\Delta T = \frac{13.1}{5\times0.030} \approx 87.5^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics