A) \[{{W}_{2}}/{{W}_{1}}=8\]
B) \[{{W}_{2}}/{{W}_{1}}=1/8\]
C) \[{{W}_{2}}/{{W}_{1}}=4\]
D) \[{{W}_{2}}/{{W}_{1}}=1/4\]
Correct Answer: C
Solution :
Given, \[\frac{d{{l}_{1}}}{dt}=\frac{d{{l}_{2}}}{dt}\] \[\Rightarrow \] \[{{L}_{1}}=8\,mH\] and \[{{L}_{2}}=2\,mH\] \[\Rightarrow \] \[{{p}_{1}}={{p}_{2}}\] \[\frac{dW}{dt}=\frac{dw}{dt}\] \[{{L}_{1}}{{l}_{1}}\frac{d{{l}_{1}}}{dt}={{L}_{2}}{{l}_{2}}\frac{d{{l}_{2}}}{dt}\] \[\left[ \because \frac{d{{l}_{1}}}{dt}=\frac{d{{l}_{2}}}{dt},Given \right]\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{L}_{2}}}{{{L}_{1}}}\] \[\Rightarrow \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{2\,mH}{8\,mH}\] \[\Rightarrow \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{4}\] \[\Rightarrow \] We know\[w=\frac{1}{2}L{{l}^{2}}\] \[\Rightarrow \]So, \[\frac{{{w}_{1}}}{{{w}_{2}}}=\frac{{{L}_{1}}l_{1}^{2}}{{{L}_{2}}l_{2}^{2}}=\frac{{{L}_{1}}}{{{L}_{2}}}{{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}\] \[=\left( \frac{8}{2} \right){{\left( \frac{1}{4} \right)}^{2}}=4\times \frac{1}{16}=\frac{1}{4}\] \[\frac{{{w}_{2}}}{{{w}_{1}}}=4\]You need to login to perform this action.
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