Two springs $A$ and $B(k_{A}=2 k_{B})$ are stretched by a applying forces of equal magnitudes at the four ends. If the energy stored in $A$ is $E$, that in $B$ is

Given:

$\begin{array}{l} k_{A}=2 k_{B} . \\ f_{A}=F_{B} \end{array}$

so;

$\begin{aligned} k_{A} x_{A}=k_{B} x_{B} & \Rightarrow 2 k_{B} x_{A}=k_{B} x_{B} \\ & \Rightarrow 2 x_{A}=x_{B} \end{aligned}$

also $\frac{1}{2} k_{A} x_{A}^{2}=E$

$\text { so } \begin{aligned} \frac{1}{2} k_{B} x_{B}^{2}=\frac{1}{2} \frac{k_{A}}{2} 4 x_{A}^{2} & =2\left(\frac{1}{2} k_{A} x_{A}^{2}\right) \\ & =2 E . \end{aligned}$

so energy stored $=2 E$

so (B) is correct.