A uniform beam HK of length 10m and weight 200N is supported at H and K. A man of weight 1000N stands on the beam at a distance x from H. If the reactions at H and K are 800N and 400N respectively, fi

• Length of the beam HK, $L = 10$ m.
• Weight of the uniform beam, $W_b = 200$ N. This acts at the center of the beam, which is $L/2 = 10/2 = 5$ m from H.
• Weight of the man, $W_m = 1000$ N. This acts at point P, at a distance $x$ from H (HP = $x$).
• Reaction force at H, $R_H = 800$ N.
• Reaction force at K, $R_K = 400$ N.

For the beam to be in rotational equilibrium, the sum of clockwise moments about any point must equal the sum of anticlockwise moments about the same point. Let's take moments about point H.

Clockwise moments about H:

• Moment due to the weight of the beam: $W_b \times (\text{distance from H to center})$
• Moment due to the weight of the man: $W_m \times (\text{distance from H to P})$

Anticlockwise moments about H:

• Moment due to the reaction force at K: $R_K \times (\text{distance from H to K})$

The distance HP is 3 m.

The final answer is $\text{C. 3m}$.