A) \[{{W}_{P}}={{W}_{Q}};{{W}_{P}}>{{W}_{Q}}\]
B) \[{{W}_{P}}={{W}_{Q}};{{W}_{P}}={{W}_{Q}}\]
C) \[{{W}_{P}}>{{W}_{Q}};{{W}_{Q}}>{{W}_{P}}\]
D) \[{{W}_{P}}<{{W}_{Q}};{{W}_{Q}}<{{W}_{P}}\]
Correct Answer: C
Solution :
Given, \[{{K}_{P}}>{{K}_{Q}}\] In case (a), the elongation is same i.e. \[{{x}_{1}}={{x}_{2}}=x\] So, \[{{W}_{P}}=\frac{1}{2}{{K}_{P}}{{x}^{2}}\] and \[{{W}_{Q}}=\frac{1}{2}{{K}_{Q}}{{x}^{2}}\] \[\therefore \] \[\frac{{{W}_{P}}}{{{W}_{Q}}}=\frac{{{K}_{P}}}{{{K}_{Q}}}>1\] \[\Rightarrow \,\,\,\,\,{{W}_{P}}>{{W}_{Q}}\] In case (b), the spring force is same i.e. \[{{F}_{1}}={{F}_{2}}=F\] So, \[{{x}_{1}}=\frac{F}{{{K}_{P}}},{{x}_{2}}\frac{F}{{{K}_{Q}}}\] \[\therefore \] \[{{W}_{P}}=\frac{1}{2}{{K}_{P}}x_{1}^{2}=\frac{1}{2}{{K}_{p}}\frac{{{F}^{2}}}{K_{P}^{2}}=\frac{1}{2}\frac{{{F}^{2}}}{{{K}_{P}}}\] and \[{{W}_{Q}}=\frac{1}{2}{{K}_{Q}}x_{2}^{2}=\frac{1}{2}{{K}_{Q}}.\frac{{{F}^{2}}}{K_{Q}^{2}}=\frac{1}{2}\frac{{{F}^{2}}}{{{K}_{Q}}}\] \[\therefore \] \[\frac{{{W}_{P}}}{{{W}_{Q}}}=\frac{{{K}_{Q}}}{{{K}_{P}}}<1\] \[\Rightarrow \,\,\,\,\,\,\,\,{{W}_{P}}<{{W}_{Q}}\]
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