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Oscillations

JEE Main level, single correct, medium

73 of 194

Question 73 of 194

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 $4 k_{A}=3 k_{B^{\prime}}$ , is :

Show answer and solution

Answer: (C) 3E

For blocks hanging in equilibrium, spring extensions are:

x_A = \frac{m_A g}{k_A} = \frac{100g}{k_A}

and

x_B = \frac{m_B g}{k_B} = \frac{200g}{k_B}

.
Energy stored:

E = \frac{1}{2}k_A x_A^2 = \frac{5000g^2}{k_A}

and

U_B = \frac{1}{2}k_B x_B^2 = \frac{20000g^2}{k_B}

.
Taking ratio:

\frac{U_B}{E} = 4 \cdot \frac{k_A}{k_B}

.
Given

4k_A = 3k_B

, we have

\frac{k_A}{k_B} = \frac{3}{4}

.
Therefore:

U_B = 4E \cdot \frac{3}{4} = 3E

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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