A) 300 %
B) 200 %
C) 100 %
D) 50 %
Correct Answer: A
Solution :
If suppose initial length \[{{l}_{1}}=100\] then \[{{l}_{2}}=100+100=200\] \[\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}={{\left( \frac{100}{200} \right)}^{2}}\] Þ \[{{R}_{2}}=4{{R}_{1}}\] \[\frac{\Delta R}{R}\times 100=\frac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}}\times 100=\frac{4{{R}_{1}}-{{R}_{1}}}{{{R}_{1}}}\times 100\]\[=300%.\]
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