To solve this problem, we need to analyze the equilibrium of the rod supported by two knife edges.$\\$Given:$\\$• Weight of the rod $W. \\$• Distance between knife edges A and B is $d. \\$• Distance from A to the center of mass of the rod is $x. \\$We need to find the normal reaction on knife edge A.$\\$In equilibrium, the sum of vertical forces and the sum of moments about any point must be zero.$\\$Let's denote:$\\$• $R_A$ as the normal reaction at A.$\\$• $R_B$ as the normal reaction at B.$\\$From the equilibrium of vertical forces:$\\ R_A + R_B = W \quad (1) \\$Taking moments about point A (to eliminate $R_A$):$\\ R_B \cdot d = W \cdot x \\$Solve for $R_B: \\ R_B = \frac{W \cdot x}{d} \quad (2) \\$Substitute equation (2) into equation (1):$\\ R_A + \frac{W \cdot x}{d} = W \\$Rearrange to solve for $R_A: \\ R_A = W - \frac{W \cdot x}{d} \\ R_A = \frac{W \cdot d}{d} - \frac{W \cdot x}{d} \\ R_A = \frac{W \cdot (d - x)}{d} \\$Therefore, the normal reaction on A is $\frac{W (d - x)}{d}. \\$This corresponds to Option 3.