Two springs of force constants $300\, N/m$ (Spring $A$) 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 $\frac{{{E_A}}}{{{E_B}}}$. Then $\frac{{{E_A}}}{{{E_B}}}$ is equal to
JEE MAIN 2013, Diffcult
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$\mathrm{k}_{\mathrm{A}}=300 \mathrm{N} / \mathrm{m}, \quad \mathrm{k}_{\mathrm{B}}=400 \mathrm{N} / \mathrm{m}$
Let when the combination of springs is compressed by force $\mathrm{F}$. Spring $A$ is compressed by $x$. Therefore compression in spring $\mathrm{B}$
$x_{B}=(8.75-x) c m$
$\mathrm{F}=300 \times x=400(8.75-x)$
Solving we get, $x=5 \mathrm{cm}$
$x_{B}=8.75-5=3.75 \mathrm{cm}$
$\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A}\left(x_{A}\right)^{2}}{\frac{1}{2} k_{B}\left(x_{B}\right)^{2}}=\frac{300 \times(5)^{2}}{400 \times(3.75)^{2}}=\frac{4}{3}$
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