A) \[6m{{m}^{2}}\]
B) \[8m{{m}^{2}}\]
C) \[10\,m{{m}^{2}}\]
D) \[12\,m{{m}^{2}}\]
Correct Answer: B
Solution :
\[l=\frac{FL}{AY}\therefore l\propto \frac{1}{A}\] (F,L and Y are constant) \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\Rightarrow {{A}_{2}}={{A}_{1}}\left( \frac{0.1}{0.05} \right)\]= \[2{{A}_{1}}=2\times 4=8m{{m}^{2}}\]
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