Question

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. $\frac{V^{2}S}{2J}$

Answer 1

Answer: (B)

$\frac{V^{2}}{4JS}$

Answer 2

$W = JQ$ ⇒ $\frac{1}{2}\left( \frac{1}{2}mV^{2} \right) = J \times mS\Delta\theta$ ⇒ $\Delta\theta = \frac{V^{2}}{4JS}$