Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in figure. the energy stored in A is E. The energy stored in B, when spring constants $$K_{A},K_{B}$$ of A and B, respectively satisfy the relation $$4K_{A}=3K_{B}$$ is:

Let extension in spring A = $$x_A$$ and extension in spring B = $$x_B$$

Energy stored in spring A = $$E_A$$ = E and energy stored in spring B = $$E_B$$

From Force balance in spring A,

$$K_Ax_A\ =\ mg\ =\ 0.1\ \times\ 10\ =\ 1N$$

$$\therefore\ x_A\ =\ \frac{1}{K_A}$$

$$E_A\ =\ \frac{1}{2}K_Ax_A^2\ =\ \frac{1}{2}K_A\times\ \frac{1}{K_A^2}=\frac{0.5}{K_A}$$

Similarly, from spring B,

$$K_Bx_B\ =\ 0.2\ \times\ 10\ =\ 2$$

$$\therefore\ x_B\ =\ \frac{2}{K_B}$$

$$E_B\ =\ \frac{1}{2}K_Bx_B^2\ =\ \frac{1}{2}K_B\times\ \frac{4}{K_B^2}\ =\ \frac{2}{K_B}$$

Since, $$K_B\ =\ \frac{4}{3}K_A$$

$$\therefore\ E_B\ =\ \frac{2}{\frac{4}{3}K_A}=\frac{1.5}{K_A}=3\ \times\ \frac{0.5}{K_A}=3E$$