JEE Advanced 2016 Paper 1

The free-body diagram of the given situation is shown in the following figure. The balancing force in $x$ and $y$ directions is given by $N+N\cos 60^{\circ}=m g \;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(1)$ That is, $N\sin 60^{\circ}=f \;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(2)$ From Eq. (1), we get $N=\frac{16}{1+(1/2)}=\frac{32}{3}$ From Eq. (2), we get the frictional force at the bottom of the stick as follows: $\left(\frac{32}{3}\right)\left(\frac{\sqrt{3}}{2}\right)=f \;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(3)$ $\Rightarrow f=\frac{16\sqrt{3}}{3} N$ From triangle $ABC$, we get $\frac{h}{x} \;\; =\sin 60^{\circ}$ $\Rightarrow x \;\; =\frac{h(2)}{\sqrt{3}} \;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(4)$ Now, the balancing torque about the bottom of the stick is $N x-\left(\frac{l}{2}\right) m g \sin 30^{\circ}=0$ $\Rightarrow \left(\frac{32}{3}\right)\left[\frac{h(2)}{\sqrt{3}}\right]=\frac{l}{2}(16)\left(\frac{1}{2}\right)$ $\Rightarrow \frac{h}{l}=\frac{3\sqrt{3}}{16}$ Hence, option (D) is the correct choice.