Q65 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
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 :
- A$\frac{4}{3} E$
- B4E
- Correct answerC3E
- D$2 E$
Solution
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: $U_B = 4E \cdot \frac{3}{4} = 3E$.
$x_A = \frac{m_A g}{k_A} = \frac{100g}{k_A}$ 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: $U_B = 4E \cdot \frac{3}{4} = 3E$.
Hence, the correct answer is option: 'C'.