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

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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

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3 F / 5

2F/3

5 F / 6

answer is D.

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Detailed Solution

The acceleration of block-rope system is $a = \frac{F}{(M + m)}$ ,
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 $T = {M + (m / 2)} a = \frac{M + (m / 2) F}{M + m}$
Given that $m = M / 2 ∴ T = [\frac{M + (M / 4)}{M + (M / 2)}] F = \frac{5 F}{6}$