A) 4 times
B) unchanged
C) \[\frac{1}{2}\]times
D) 2 times
Correct Answer: A
Solution :
Volume remains unchanged \[{{A}_{1}}{{l}_{1}}={{A}_{2}}{{l}_{2}}\] \[{{A}_{1}}\times l={{A}_{2}}\times 2l\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}=\frac{l}{2l}\times \frac{1}{2}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{4}\] \[\Rightarrow \] \[{{R}_{2}}=4{{R}_{1}}\] Resistance becomes 4 times the initial resistance.
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