1. Model the iron bar as a uniform rod

   The iron bar is heavy and uniform, so its total weight $W$ acts at its centre of mass, i.e. at the midpoint of the bar.

   One end is on the ground and the other end rests on the person's shoulder. The bar is in static equilibrium.

2. Forces acting on the bar

   There are three main vertical forces:

   • Weight of the bar: $W$, acting downward at the midpoint.
   • Reaction from the ground at the lower end: let it be $R_g$ upward.
   • Reaction from the person's shoulder at the upper end: let it be $R_p$ upward on the bar.

   The "weight experienced by the person" is the force exerted by the bar on the person, whose magnitude equals $R_p$.

3. Take moments about the lower end

   Let the length of the bar be $L$.

   • The upper end is at distance $L$ from the lower end.
   • The centre of mass is at distance $L/2$ from the lower end.

   Since the rod makes angle $\theta$ with the horizontal, the horizontal distances are:

   • For the shoulder force at the upper end: $L\cos\theta$
   • For the weight at the midpoint: $\dfrac{L}{2}\cos\theta$

   For rotational equilibrium about the lower end:

   $R_p (L\cos\theta) = W\left(\frac{L}{2}\cos\theta\right)$

4. Solve for $R_p$

   $R_p = \frac{W}{2}$

5. Interpretation

   The person experiences the downward force exerted by the bar, equal in magnitude to $R_p$.

   Hence, the weight experienced by the person is

   $\boxed{\frac{W}{2}}$

6. Check options

   • A: $W\sin\theta$ ❌
   • B: $W$ ❌
   • C: $\dfrac{W}{2}$ ✅
   • D: $W\cos\theta$ ❌

Therefore, the correct option is C.