A man holding a rod of mass m as shown in figure. Find weight of rod experienced by him.

The correct option is mg/4.

Step-by-step Explanation:

Let us analyze the forces acting on the rod to find the force (weight) experienced by the man holding it. From the provided image, the rod of mass $m$ makes an angle of $60°$ with the horizontal ground, and it is resting on the man's shoulder at its upper end.

Let the length of the rod be $L$. The key forces acting on the rod are:
1. The weight of the rod, $mg$, acting vertically downwards at its center of gravity, which is at a distance of $\frac{L}{2}$ from the bottom end.
2. The normal force, $N$, exerted by the man's shoulder at the upper end of the rod, acting perpendicular to the rod.
3. The reaction forces exerted by the ground at the bottom end of the rod.

To find the normal force $N$ (which represents the weight of the rod experienced by the man), we can apply the condition for rotational equilibrium. By taking the torque about the point of contact of the rod with the ground (bottom end), we eliminate the unknown ground reaction forces:

$\sum \tau =0$

The torque due to the weight of the rod about the bottom end is:
${\tau }_g=mg\cdot \frac{L}{2}\cdot \cos (60°)$

The torque due to the normal force $N$ from the man's shoulder about the bottom end is:
${\tau }_N=N\cdot L$

Equating the magnitudes of these opposing torques for rotational equilibrium:
$N\cdot L=mg\cdot \frac{L}{2}\cdot \cos (60°)$

We can simplify the equation by dividing both sides by the length of the rod $L$:
$N=\frac{mg}{2}\cdot \cos (60°)$

Substitute the value of $\cos (60°)=\frac{1}{2}$:
$N=\frac{mg}{2}\cdot \frac{1}{2}$

$N=\frac{mg}{4}$

Thus, the weight of the rod experienced by the man is $\frac{mg}{4}$.