1. **Problem Statement:** A light rod AB of length 10 m is supported at its ends A and B. A downward force of 500 N acts at a point 2 m from end A. We need to find the magnitudes of the reactions at supports A and B when the rod is in equilibrium. 2. **Equilibrium Conditions:** For the rod to be in equilibrium, the sum of vertical forces and the sum of moments about any point must be zero. 3. **Sum of Vertical Forces:** $$ R_A + R_B = 500 $$ where $R_A$ and $R_B$ are the reaction forces at supports A and B respectively. 4. **Sum of Moments about A:** Taking moments about point A (counterclockwise positive): $$ R_B \times 10 - 500 \times 2 = 0 $$ 5. **Calculate $R_B$:** $$ R_B \times 10 = 500 \times 2 $$ $$ R_B = \frac{1000}{10} = 100 $$ 6. **Calculate $R_A$ using vertical forces sum:** $$ R_A + 100 = 500 $$ $$ R_A = 500 - 100 = 400 $$ 7. **Answer:** The reactions at the supports are: $$ R_A = 400N, \quad R_B = 100N $$ This corresponds to option (b).