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