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

The correct option is B

2E

Let the extension in the spring be x. Let's draw the free body drawaing of one end of the spring A (Right End)

$K_A$x

F = $K_A$x,x = $\frac{F}{K_A}$

Now we know that the potential energy stored in a streched spring is given by $\frac{1}{2}$k $x^2$

$\Rightarrow$ ($\frac{1}{2}$ $K_A$${\frac{F}{K_A})}^2$

$\Rightarrow$ $\frac{1}{2}$ $\frac{F^2}{\left(K_A\right)}$

Given Energy stored in A= E

$\Rightarrow$ $\frac{1}{2}$ $\frac{F^2}{\left(K_A\right)}$ = E ..(1)

$k_B$x

X = $\frac{F}{k_B}$

Energy = $\frac{1}{2}$ $k_B$

Similarly , for spring B

X = $\frac{F}{k_B}$

Energy = $\frac{1}{2}$ $K_A$${\frac{F}{K_A}}^2$ = $\frac{1}{2}$ $\frac{F^2}{\left(K_B\right)}$

Given $k_B$ = $\frac{k_A}{2)}$

$\Rightarrow$ Energy = $\frac{1}{2}$ $\frac{F^2}{\left(K_A\right)}$ 2

Energy B = 2E .. (ii)