A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block. The tension in the middle of the rope will be

Options

A. F
B. `2F//3`
C. `3 F//5`
D. `5 F//6`

Answer

D

Explanation

The acceleration of block-rope system is `a=(F)/((M+m))`
where M is the mass of block and m is the mass of rope.
So the tension in the middle of the rope will be
`T={M+(m//2)}a=(M+(m//2)F)/(M+m)`
Given that `m=M//2`
`:. T=[(M+(M//4))/(M+(M//2))]=F=(5F)/(6)`.