Equilibrium, Third Law & Frames question from JEE Main 2013 (9 Apr, Shift 1)

Answer: (d)

1. The string's line passes through the centre: distance from the support to the centre $= 8 + 5 = 13\text{ cm}$.
2. The centre is $5\text{ cm}$ (one radius) from the wall, so $\sin\theta = \frac{5}{13}$ and $\cos\theta = \frac{12}{13}$ ($\theta$ = angle with the wall).
3. The wall's normal force is horizontal, so vertically: $T\cos\theta = W$
4. $T = \frac{13W}{12}$
5. So the answer is (d) $13\frac{W}{12}$.