Two springs of force constants 300 N/m (springs A) and 400 N/m (spring B ) are joined together in series. The combination is compressed by 8.75 cm. The ratio of energy stored in A and B is $\frac{{\mathrm{E}}_{\mathrm{A}}}{{\mathrm{E}}_{\mathrm{B}}}.$ The $\frac{{\mathrm{E}}_{\mathrm{A}}}{{\mathrm{E}}_{\mathrm{B}}}$ is equal to

Given,

The displacement of spring A, ${\mathrm{d}}_{\mathrm{a}}=\mathrm{kx}=300{\mathrm{x}}_{\mathrm{a}}$

The displacement of spring  B, ${\mathrm{d}}_{\mathrm{b}}=400{\mathrm{x}}_{\mathrm{b}}$

${\mathrm{x}}_{\mathrm{a}}+{\mathrm{x}}_{\mathrm{b}}=8.75\mathrm{cm}=0.0875\mathrm{m\_\_\_\_\_\_\_}\left(1\right)$

The load borne by springs connected in series is the same as the load given to the series as a whole.

$300{\mathrm{x}}_{\mathrm{a}}=400{\mathrm{x}}_{\mathrm{b}}\mathrm{\_\_\_\_\_}\left(2\right)$

solving equations 1 and 2,

$x_a=0.05m$

$x_b=0.0375m$

now the energy of the first spring,

$E_a=\frac{1}{2}\times 300\times 0.{05}^2\_\_\_\_\_\left(3\right)$

$E_b=\frac{1}{2}\times 400\times 0.{0375}^2\_\_\_\_\_\left(4\right)$

dividing equation 3 by 4,

$\frac{E_a}{E_b}=\frac{4}{3}$

Hence the correct answer is 4:3.