A) \[\frac{4}{3}\]
B) \[\frac{16}{9}\]
C) \[\frac{3}{4}\]
D) \[\frac{9}{16}\]
Correct Answer: C
Solution :
We have \[{{k}_{1}}=\frac{{{F}_{A}}}{{{x}_{B}}}\] As, \[x={{x}_{A}}+{{x}_{B}}\] \[\Rightarrow \] \[\frac{{{k}_{A}}+{{k}_{B}}}{{{k}_{A}}{{k}_{B}}}=\frac{1}{{{k}_{eq}}}\] or \[\frac{k}{{{k}_{eq}}}=\frac{{{k}_{A}}{{k}_{B}}}{{{k}_{A}}+{{k}_{B}}}\] \[=\frac{300\times 400}{700}=\frac{1200}{7}\] \[\therefore \] \[F=kx=\frac{1200}{7}\times 8.75\times {{10}^{-2}}=15\,N\]\[F={{k}_{A}}\,{{x}_{A}}\] \[\Rightarrow \] \[{{x}_{A}}=\frac{F}{{{k}_{A}}}=\frac{15}{300}=20\] and \[{{x}_{b}}=\frac{F}{{{k}_{B}}}=\frac{15}{400}\] \[\frac{{{x}_{A}}}{{{x}_{B}}}=\frac{4}{3}\] \[\therefore \] \[\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{k}_{A}}x_{A}^{2}}{{{k}_{B}}\times {{x}_{B}}^{2}}\] \[\left( \because \,E=\frac{1}{2}k{{x}^{2}} \right)\] \[=\frac{3}{4}\propto \,{{\left( \frac{4}{3} \right)}^{2}}=\frac{4}{3}\]
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