CMC Medical
CMC-Medical Ludhiana Solved Paper-2015
question_answer
Consider a situation that a rope is hanging from a balloon with which a monkey hangs. The balloon and monkey are in the air and at the rest. If the monkey reaches the top of the rope of length 1 which masses of balloon and monkey are M and m then by what distance does the balloon descend?
A) \[\frac{mL}{m+M}\]
B) \[\frac{2mL}{3m+M}\]
C) \[\frac{mL}{2m+3M}\]
D) \[\frac{mL}{\frac{m}{2}+M}\]
Correct Answer:
A
Solution :
Consider the diagram Initially the monkey and balloon are at rest. So, CM is at rest P. When balloon decends through a length L, then CM will shift (from P) \[{{r}_{0}}=\frac{mL+M\times 0}{M+m}=\frac{mL}{M+m}\] So, the balloon descends through a distance\[\frac{mL}{M+m}\]