JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Elasticity & SHM

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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}

, is:

(A)4E
(B)2E
(C)3E✓ Correct
(D) $\frac{4}{3} E$

Answer: (C)

Solution

Each block hangs in equilibrium with extension

x = \frac{m g}{k}

, storing energy

U = \frac{1}{2} k x^{2} = \frac{m ^{2} g ^{2}}{2 k}

, so

U \propto \frac{m ^{2}}{k}

. Therefore

\frac{U _{A}}{U _{B}} = (\frac{m _{A}}{m _{B}})^{2} \frac{k _{B}}{k _{A}} = (\frac{100}{200})^{2} \frac{k _{B}}{k _{A}}

. With

4 k_{A} = 3 k_{B}

\frac{k _{B}}{k _{A}} = \frac{4}{3}

, so

\frac{U _{A}}{U _{B}} = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}

. The energy stored in B is

3 E

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Elasticity & SHM

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