Q. A lead ball moving with a velocity $V$ strikes a wall and stops. If 50% of its energy is converted into heat, then what will be the increase in temperature (Specific heat of lead is $S$)
A) $\frac{2V^{2}}{JS}$
B) $\frac{V^{2}}{4JS}$
C) $\frac{V^{2}}{J}$
D) $\frac{V^{2}S}{2J}$
Explanation
(b) $W = JQ \Rightarrow \frac{1}{2}\left( \frac{1}{2}mV^{2} \right) = J \times mS\mathrm{\Delta}\theta \Rightarrow \mathrm{\Delta}\theta = \frac{V^{2}}{4\ JS}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.