A) 8
B) 4
C) 2
D) 1
Correct Answer: A
Solution :
\[l=\frac{FL}{\pi {{r}^{2}}r}\]Þ \[l\propto \frac{F}{{{r}^{2}}}\] (Y and L are constant) \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{F}_{2}}}{{{F}_{1}}}\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=2\times {{(2)}^{2}}=8\] \ \[{{l}_{2}}=8{{l}_{1}}=8\times 1=8mm\]You need to login to perform this action.
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