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) gt 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

Options

A. `T_(B)-T_(A)`
B. `(T_(A)+T_(B))//5`
C. `(4T_(A)+T_(B))//5`
D. `(T_(A)-T_(B))//5`

Answer

C

Explanation

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

Applying Newtons second law on sections AB and section AC or rope we get
`T_(B)-T_(A)=Ma` and `T_(C)-T_(A)=M/5a`
Solving `T_(C)=(T_(B)+4T_(A))/5`