Connected Bodies & Pulleys question from MHT CET 2026 (18 Apr, Shift 1)

Answer: (d)

1. Rope mass $= \frac{M}{2}$, total mass $= \frac{3M}{2}$, so $a = \dfrac{2F}{3M}$.
2. Behind the middle of the rope: block $+$ half the rope $= M + \frac{M}{4} = \frac{5M}{4}$.
3. $T = \frac{5M}{4} \times \dfrac{2F}{3M} = \dfrac{5F}{6}$
4. So the answer is (d).