A) 12 mm
B) 0.75 mm
C) 15 mm
D) 6 mm
Correct Answer: A
Solution :
\[l=\frac{FL}{AY}\Rightarrow l\propto \frac{1}{{{r}^{2}}}\] (F,L and Y are constant) \[\frac{{{l}_{2}}}{{{l}_{1}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}={{(2)}^{2}}\Rightarrow {{l}_{2}}=4{{l}_{1}}=4\times 3=12mm\]
You need to login to perform this action.
You will be redirected in
3 sec
