JEE Main Physics question
Natural length of a spring is 60 cm, and its spring constant is 4000 N/m. A mass of 20 kg is hung from it. The extension produced in the spring is: (Take g = 9.8 m/s²)
Answer: (A)
Solution
At equilibrium, F = kx ⟹ x = mg/k = (20×9.8)/4000 = 196/4000 = 0.049 m = 4.9 cm.