A) 2 mm
B) 0.5 mm
C) 4 mm
D) 0.25 mm
Correct Answer: B
Solution :
\[l=\frac{FL}{AY}\,\,\,\Rightarrow \,\,\,l\propto \frac{L}{{{r}^{2}}}\] (F and Y are same) \[\therefore \,\,\,\,\,\,\,\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{L}_{2}}}{{{L}_{1}}}{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=2\times {{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{2}\] \[\Rightarrow \,\,\,\,\,\,\,{{l}_{2}}=\frac{{{l}_{1}}}{2}=\frac{l}{2}=0.5\,\,mm\]You need to login to perform this action.
You will be redirected in
3 sec