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

Q: 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

a $\mathrm{k}_{\mathrm{A}}=300 \mathrm{N} / \mathrm{m}, \quad \mathrm{k}_{\mathrm{B}}=400 \mathrm{N} / \mathrm{m}$ 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, $x=5 \mathrm{cm}$ $x_{B}=8.75-5=3.75 \mathrm{cm}$ $\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A}\left(x_{A}\right)^{2}}{\frac{1}{2} k_{B}\left(x_{B}\right)^{2}}=\frac{300 \times(5)^{2}}{400 \times(3.75)^{2}}=\frac{4}{3}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$\frac{4}{3}$
B$\frac{16}{9}$
C$\frac{3}{4}$
D$\frac{9}{16}$

JEE MAIN 2013, Diffcult

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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