JEE Main Physics question
Two masses of 3 kg and 5 kg are suspended from the ends of an unstretchable massless cord passing over a frictionless pulley. When the masses are released, the pressure on the pulley is (g = 10 ms⁻²)
Answer: (B)
Solution
T = 2m₁m₂g/(m₁+m₂) = (2×3×5×10)/8 = 37.5 N. Total downward thrust on the pulley = 2T = 75 N = 7.5 kgf.