Q. The force constant of a wire is k and that of another wire of the same material is 2k. When both the wires are stretched, then work done is
A) $W_{2} = 2W_{1}^{2}$
B) $W_{2} = 2W_{1}$
C) $W_{2} = W_{1}$
D) $W_{2} = 0.5\ \ W_{1}$
Explanation
(b) Work done in stretching the wire
$W = \frac{1}{2} \times force\ constant\ \times x^{2}$
For first wire, $\ \ \ \ \ W_{1} = \frac{1}{2} \times kx^{2} = \frac{1}{2}kx^{2}$
For second wire, $W_{2} = \frac{1}{2} \times 2k \times x^{2} = kx^{2}$
Hence , $W_{2} = 2W_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.