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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 $T_{B} (T_{B} > T_{A})$ . 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 $\frac{L}{5}$ from end A is

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$\frac{(T_{A} - T_{B})}{5}$

$\frac{(T_{A} + T_{B})}{5}$

$\frac{(4 T_{A} + T_{B})}{5}$

$T_{B} - T_{A}$

answer is C.

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Detailed Solution

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

Applying Newton's second law on section AB and section AC of rope we get

$T_{B} - T_{A} = Ma and T_{C} - T_{A} = \frac{M}{5} a$

Solving $T_{C} = \frac{T_{B} + 4 T_{A}}{5}$

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