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: Use the outer half instead:
The outer half of the rope has mass $\frac M4$. The force $F$ pulls it forward and the tension $T$ pulls it back: $F - T = \frac M4 a$.

Step 2: Solve:
With $a = \frac{2F}{3M}$: $F - T = \frac M4 \cdot \frac{2F}{3M} = \frac F6$, so $T = \frac{5F}{6}$.

Final Answer:
$T = \frac{5F}{6}$, option (D). \[ \boxed{\frac{5F}{6}} \]