Q104 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
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$\dfrac{F}{3}$
- B$\dfrac{3F}{4}$
- C$\dfrac{4F}{5}$
- Correct answerD$\dfrac{5F}{6}$
Solution
Mass of the block = $M$
Mass of the rope = $\dfrac{M}{2}$
Total mass of the system = $M + \dfrac{M}{2} = \dfrac{3M}{2}$
Acceleration of the system $a = \dfrac{F}{\dfrac{3M}{2}} = \dfrac{2F}{3M}$
The tension in the middle of the rope accelerates the block and the rear half of the rope.
Mass behind the middle point $m' = M + \dfrac{1}{2}\left(\dfrac{M}{2}\right) = M + \dfrac{M}{4} = \dfrac{5M}{4}$
Tension at the middle of the rope $T = m' a = \left(\dfrac{5M}{4}\right)\left(\dfrac{2F}{3M}\right) = \dfrac{5F}{6}$
Answer: $\dfrac{5F}{6}$
Mass of the rope = $\dfrac{M}{2}$
Total mass of the system = $M + \dfrac{M}{2} = \dfrac{3M}{2}$
Acceleration of the system $a = \dfrac{F}{\dfrac{3M}{2}} = \dfrac{2F}{3M}$
The tension in the middle of the rope accelerates the block and the rear half of the rope.
Mass behind the middle point $m' = M + \dfrac{1}{2}\left(\dfrac{M}{2}\right) = M + \dfrac{M}{4} = \dfrac{5M}{4}$
Tension at the middle of the rope $T = m' a = \left(\dfrac{5M}{4}\right)\left(\dfrac{2F}{3M}\right) = \dfrac{5F}{6}$
Answer: $\dfrac{5F}{6}$
Hence, the correct answer is option: 'D'.