1. Understand the setup

A uniform heavy iron bar has:

• one end on the ground,
• the other end on the man's shoulder,
• inclination with horizontal $=60^\circ$,
• weight of bar $=12\,\text{kg-wt}$.

We need the load supported by the man.

2. Forces acting on the bar

Since the bar is uniform, its weight acts at its midpoint.

Let:

• lower end on ground be point $A$,
• upper end on shoulder be point $B$,
• length of rod be $L$.

For vertical equilibrium:

• upward reaction from ground at $A$ = $R_A$,
• upward force from man's shoulder at $B$ = $R_B$,
• downward weight at midpoint = $12$ kg-wt.

So, $R_A + R_B = 12$

3. Take moments about the ground end $A$

This eliminates $R_A$.

• Weight acts at midpoint, whose horizontal distance from $A$ is $\frac{L}{2}\cos 60^\circ$
• Force by man acts at $B$, whose horizontal distance from $A$ is $L\cos 60^\circ$

Both forces are vertical, so their moment arms are these horizontal distances.

Balancing moments about $A$: $R_B\,(L\cos 60^\circ)=12\left(\frac{L}{2}\cos 60^\circ\right)$

Cancel $L\cos 60^\circ$ from both sides: $R_B=12\cdot \frac{1}{2}=6$

4. Interpretation

Thus the man supports a load of $6\,\text{kg-wt}$

5. Check options

• A: $3$ kg ✗
• B: $6$ kg ✓
• C: $6\sqrt{3}$ kg ✗
• D: $12$ kg ✗

Therefore, the correct option is B.