Composite & Cut-Out Bodies question from JEE Main 2016 (10 Apr, Shift 1)

Answer: (b)

1. Take B as origin, BC along $x$. Let $AB = l_1$, $BC = l_2$; mass $\propto$ length.
2. Centre of AB at $x = \frac{l_1}{2}\cos 60^\circ = \frac{l_1}{4}$; centre of BC at $x = \frac{l_2}{2}$. Point A is at $x = \frac{l_1}{2}$.
3. $\frac{l_1\cdot\frac{l_1}{4} + l_2\cdot\frac{l_2}{2}}{l_1 + l_2} = \frac{l_1}{2} \Rightarrow 2l_2^2 - 2l_1 l_2 - l_1^2 = 0$
4. With $r = \frac{l_2}{l_1}$: $2r^2 - 2r - 1 = 0 \Rightarrow r = \frac{1+\sqrt{3}}{2} \approx 1.37$
5. So the answer is (b) $1.37$.