A) l
B) 2l
C) l/2
D) 4l
Correct Answer: C
Solution :
\[l=\frac{FL}{AY}\Rightarrow l\propto \frac{L}{{{r}^{2}}}\] (F and Y are constant) \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{L}_{2}}}{{{L}_{1}}}\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=2\times {{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{2}\]\[\therefore {{l}_{2}}=\frac{{{l}_{1}}}{2}\] i.e. the change in the length of other wire is \[\frac{l}{2}\]You need to login to perform this action.
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