Properties Of Solids And Liquids Question 56
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) [RPMT 1996]
Options:
A) $ \frac{2V^{2}}{JS} $
B) $ \frac{V^{2}}{4JS} $
C) $ \frac{V^{2}}{J} $
D) $ \frac{V^{2}S}{2J} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ W=JQ $
therefore $ \frac{1}{2}( \frac{1}{2}mV^{2} )=J\times mS\Delta \theta $
therefore $ \Delta \theta =\frac{V^{2}}{4JS} $
