Q. The force constant of two springs are $K_{1}$ and $K_{2}.$ Both are stretched till their elastic energies are equal. If the stretching forces are $F_{1}$ and $F_{2},$ then $F_{1}:F_{2}$ is
A) $K_{1}\ :K_{2}$
B) $K_{2}\ :K_{1}$
C) $\sqrt{K_{1}}:\sqrt{K_{2}}$
D) $K_{1}^{2}:K_{2}^{2}$
Explanation
(c) Given elastic energies are equal $i.e.,\frac{1}{2}k_{1}x_{1}^{2} = \frac{1}{2}k_{2}x_{2}^{2}$
$\Rightarrow \frac{k_{1}}{k_{2}} = \left( \frac{x_{2}}{x_{1}} \right)^{2}\ $and using$\ F = kx$
$\Rightarrow \frac{F_{1}}{F_{2}} = \frac{k_{1}x_{1}}{k_{2}x_{2}} = \frac{k_{1}}{k_{2}} \times \sqrt{\frac{k_{2}}{k_{1}}} = \sqrt{\frac{k_{1}}{k_{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.