question_answer

A perfectly straight portion of a uniform rope has mass M and length L. At end A of the segment, the tension in the rope is \[{{T}_{A}}\] and at end B it is \[{{T}_{B}}({{T}_{B}}>{{T}_{A}})\]. Neglect effect of gravity and no contact force acts on the rope in between points A and B. The tension in the rope at a distance L/5 from end A is

A) \[{{T}_{B}}-{{T}_{A}}\]

B) \[({{T}_{A}}+{{T}_{B}})/5\]

C) \[(4{{T}_{A}}+{{T}_{B}})/5\]

D) \[({{T}_{A}}-{{T}_{B}})/5\]

Correct Answer: C

Solution :

The F.B.D. of section of rope between A and B having acceleration a towards left is

Applying Newton?s second law on section AB and section AC of rope we get

\[{{T}_{B}}-{{T}_{A}}=Ma\] and \[{{T}_{C}}-{{T}_{A}}=\frac{M}{5}a\]

Solving \[{{T}_{C}}=\frac{{{T}_{B}}+4{{T}_{A}}}{5}\]