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Work Energy Power

JEE Main level, single correct, medium

130 of 183

Question 130 of 183

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

Show answer and solution

Answer: (A) $\frac{4}{3}$

Given :

\mathrm{k}_{\mathrm{A}}=300 \mathrm{~N} / \mathrm{m}, \mathrm{k}_{\mathrm{B}}=400 \mathrm{~N} / \mathrm{m}

Let when the combination of springs is compressed by force F. Spring A is compressed by

x

. Therefore compression in spring

\mathrm{B}

\begin{aligned} & x_B=(8.75-x) \mathrm{cm} \\ & \mathrm{F}=300 \times \mathrm{x}=400(8.75-\mathrm{x}) \end{aligned}

Solving we get,

x=5 \mathrm{~cm}

\mathrm{x}_{\mathrm{B}}=8.75-5=3.75 \mathrm{~cm}

\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{E}_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{k}_{\mathrm{A}}\left(\mathrm{x}_{\mathrm{A}}\right)^2}{\frac{1}{2} \mathrm{k}_{\mathrm{B}}\left(\mathrm{x}_{\mathrm{B}}\right)^2}=\frac{300 \times(5)^2}{400 \times(3.75)^2}=\frac{4}{3}

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JEE Main 2025 Apr 2 Shift 1 question paper

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