Question

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

• A. \(\frac{F}{3}\)
• B. \(\frac{3F}{4}\)
• C. \(\frac{4F}{5}\)
• D. \(\frac{5F}{6}\)

Hint:

Find the common acceleration, then apply F = ma to what lies beyond the middle point.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The block and the rope move together with the same acceleration. The tension at a point in the rope is the force that pulls everything on the block side of that point.

Step 2: Key Formula or Approach:
Total mass \(= M + \frac M2 = \frac{3M}{2}\), so \(a = \frac{F}{3M/2} = \frac{2F}{3M}\).

Step 3: Detailed Explanation:
At the middle of the rope, the part between the middle and the block is half of the rope, with mass \(\frac{M}{4}\).
The tension \(T\) at the middle pulls the block together with this half rope:
\[ T = \left(M + \frac M4\right)a = \frac{5M}{4} \times \frac{2F}{3M} = \frac{5F}{6} \]
Option A, \(\frac F3\), would be the tension on the block alone, with no allowance for the rope mass. Option B, \(\frac{3F}{4}\), does not follow from the mass split \(M + \frac M4\).

Final Answer:
The tension in the middle is \(\frac{5F}{6}\), option (D). \[ \boxed{\frac{5F}{6}} \]