| Two similar springs P and Q have spring constants \[{{K}_{P}}\] and \[{{K}_{Q}},\] such that \[{{K}_{P}}>{{K}_{Q}}\]. |
| They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs \[{{W}_{P}}\] and \[{{W}_{Q}}\] are related as, in case and case , respectively [NEET 2015 ] |
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 , 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 , 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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