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)
N₁ = 1.6 N and N₂ = 0.8 N
N₁ = 0.8 N and N₂ = 1.6 N
N₁ = 0.6 N and N₂ = 1.8 N
N₁ = 1.8 N and N₂ = 0.6 N
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-1 – 2024
For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W₁ is at 0.2 m from one end (say, the 0m mark), and W₂ is at 0.4 m from the other end (the 1m mark), placing it at 1.0 – 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N₁ + N₂ = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W₁ (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 – 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N₂ (at 0.6m) creates an anticlockwise torque N₂ * (0.6 – 0.2) = N₂ * 0.4 Nm. Setting the sum of torques to zero: 0.72 – 0.4 N₂ = 0, which gives N₂ = 0.72 / 0.4 = 1.8 N. Substituting N₂ into the force equation: N₁ + 1.8 = 2.4, which gives N₁ = 2.4 – 1.8 = 0.6 N.