Laws Of Motion Question 419
Question: Two men of masses m and m/2 starts climbing up on two massless strings fixed at the ceiling with acceleration g and g/2 respectively. The ratio of tensions in the two strings will be:
Options:
A) 2 : 1
B) 4 : 1
C) 4 : 3
D) 8 : 3
Show Answer
Answer:
Correct Answer: D
Solution:
[d] FBD of man of mass(m) FBD of man of mass (m/2)
$ T _{1}=mg+mg $
$ T _{2}=\frac{m}{2}g+\frac{m}{2}.\frac{g}{2} $
$ T _{1}=2mg $
$ T _{2}=\frac{mg}{2}[ \frac{3}{2} ]=\frac{3mg}{4} $
$ \Rightarrow $ $ T _{1}:T _{2}::8:3 $
