A) same
B) half
C) double
D) four times
Correct Answer: B
Solution :
Here \[{{l}_{1}}={{l}_{2}};{{n}_{1}}=n,\,\,{{n}_{2}}=\frac{n}{2}\] Using the relation, \[nlBA=CQ\] \[l=\frac{CQ}{nBA}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{Q}_{1}}}{{{Q}_{2}}}\times \frac{{{n}_{2}}}{{{n}_{1}}}\] \[\Rightarrow \] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}\times \frac{{{n}_{2}}}{{{n}_{1}}}=1\] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{n}{n/2}=2\] \[\left( \begin{matrix} \because \,\,{{l}_{1}}={{l}_{2}} \\ \therefore \,\,\frac{{{l}_{1}}}{{{l}_{2}}}=1 \\ \end{matrix} \right)\] \[{{Q}_{2}}=\frac{{{Q}_{1}}}{2}\]
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