A) E
B) 2E
C) E/2
D) E/4
Correct Answer: C
Solution :
[c] Here, \[{{k}_{Q}}=\frac{{{k}_{p}}}{2}\] According to Hooke's law \[\therefore {{F}_{p}}=-{{k}_{p}}{{x}_{p}}\] \[{{F}_{Q}}=-{{k}_{Q}}{{x}_{Q}}\Rightarrow \frac{{{F}_{p}}}{{{F}_{Q}}}=\frac{{{k}_{p}}}{{{k}_{Q}}}\frac{{{x}_{p}}}{{{x}_{Q}}}\] \[{{F}_{p}}={{F}_{Q}}\] [Given] \[\therefore \,\,\frac{{{x}_{p}}}{{{x}_{Q}}}=\frac{{{k}_{Q}}}{{{k}_{p}}}\] ?. (i) Energy stored in a spring is\[U=\frac{1}{2}k{{x}^{2}}\] \[\therefore \,\,\frac{{{U}_{p}}}{{{U}_{Q}}}=\frac{{{k}_{p}}x_{p}^{2}}{{{k}_{Q}}x_{Q}^{2}}=\frac{{{k}_{p}}}{{{k}_{Q}}}\times \frac{k_{Q}^{^{2}}}{k_{^{p}}^{2}}=\frac{1}{2}\] \[\left[ \therefore {{k}_{Q}}=\frac{{{k}_{p}}}{2} \right]\] \[\Rightarrow \,\,\,{{U}_{p}}=\frac{{{U}_{Q}}}{2}=\frac{E}{2}\] \[[\therefore \,\,\,\,{{U}_{Q}}=E]\]You need to login to perform this action.
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