A perfectly straight protion 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)gtT_(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`