JEE Main Physics question
A body of mass 0.05 kg is observed to fall with an acceleration of 9.5 ms⁻². The opposing force of air on the body is (g = 9.8 ms⁻²)
Answer: (A)
Solution
mg − F = ma ⟹ F = m(g − a) = 0.05 × (9.8 − 9.5) = 0.05 × 0.3 = 0.015 N.