A) l
B) 2l
C) \[\frac{l}{2}\]
D) \[\frac{l}{4}\]
Correct Answer: A
Solution :
\[l=\frac{FL}{AY}=\frac{FL}{\pi {{r}^{2}}Y}\therefore l\propto \frac{FL}{{{r}^{2}}}\] (Y = constant) \\[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{F}_{2}}}{{{F}_{1}}}\times \frac{{{L}_{2}}}{{{L}_{1}}}{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=2\times 2\times {{\left( \frac{1}{2} \right)}^{2}}=1\] \ \[{{l}_{2}}={{l}_{1}}\] i.e. increment in its length will be l.
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