Laws Of Motion Question 411

Question: 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 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 $

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Answer:

Correct Answer: C

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}+4T _{A}}{5} $



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