Connected Bodies & Pulleys question from JEE Main 2010 (AIEEE)

Answer: (d)

1. On a smooth incline at angle $\theta$ (from horizontal), acceleration $g\sin\theta$ along it; vertical component $= g\sin^2\theta$.
2. A ($60^\circ$): $a_A = g\sin^2 60^\circ = \frac{3g}{4}$. B ($30^\circ$): $a_B = g\sin^2 30^\circ = \frac{g}{4}$.
3. Relative vertical acceleration $= \frac{3g}{4} - \frac{g}{4} = \frac{g}{2} = 4.9\text{ m s}^{-2}$.
4. So the answer is (d) $4.9\text{ m s}^{-2}$ in vertical direction.