A uniform meter scale of mass 0.24 kg is made of steel. It is kept on two wedges, W1 and W2, in a horizontal position. W1 is at a distance of 0.2 m from one of its ends, while W2 is at distance of 0.4 m from the other end. If the force on the scale is N1 due to W1 and N2 due to W2, then : (take g = 10.0 m s-2)

To find the forces N1 and N2, we apply the principles of static equilibrium for a uniform meter scale of length 1 m and mass 0.24 kg.

• Total Weight (W): W = mg = 0.24 kg × 10 m/s² = 2.4 N. This acts at the center of mass (0.5 m mark).
• Positions: Wedge W1 is at 0.2 m. Wedge W2 is at 0.6 m (1.0 m - 0.4 m).
• Translational Equilibrium: N1 + N2 = W = 2.4 N.
• Rotational Equilibrium: Taking torque about W1:
  N2 × (0.6 - 0.2) = W × (0.5 - 0.2)
  0.4 N2 = 2.4 × 0.3
  0.4 N2 = 0.72
  N2 = 1.8 N.
• Solving for N1: N1 = 2.4 - 1.8 = 0.6 N.

Thus, N1 = 0.6 N and N2 = 1.8 N, which matches Option C.