JEE Main Physics question
To the free end of a spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is d. If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is (force constant k)
Answer: (B)
Solution
Slow lowering: equilibrium d = mg/k. Sudden release (free fall onto the spring,
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account