JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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
$4 E$
B
$2 E$
C
$3 E$
D
$\frac{4}{3} E$

Solution

At equilibrium, $k x = m g \Rightarrow x = \frac{m g}{k}$ . The stored elastic energy is:

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

Taking the ratio:

$\frac{U _{A}}{U _{B}} = (\frac{m _{A}}{m _{B}})^{2} \cdot \frac{k _{B}}{k _{A}} = (\frac{1}{2})^{2} \cdot \frac{4}{3} = \frac{1}{3}$

$∴ U_{B} = 3 U_{A} = 3 E$ .

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