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

Answer

C

Explanation

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

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