JEE Main & Advanced
Physics
Simple Harmonic Motion
Question Bank
Critical Thinking
question_answer
A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of [IIT-JEE (Screening) 1999]
A) \[(2/3)k\]
B) \[(3/2)k\]
C) \[3k\]
D) \[6k\]
Correct Answer:
B
Solution :
Force constant \[(k)\propto \frac{1}{\text{Length}\,\text{of}\,\text{spring}}\] \[\Rightarrow \] \[\frac{K}{{{K}_{1}}}=\frac{{{l}_{1}}}{l}=\frac{\frac{2}{3}l}{l}\] \[\Rightarrow \] \[{{K}_{1}}-\frac{3}{2}K\].