JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2026
- Medium
- MCQ
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 :

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 3E
Step-by-step Solution
Detailed explanation
For equilibrium \( kx=mg \)
\( U=\frac{1}{2}kx^{2}=\frac{1}{2}\frac{m^{2}g^{2}}{k} \)
\( U\propto\frac{m^{2}}{k} \)
\( \frac{U_{A}}{U_{B}}=(\frac{m_{A}}{m_{B}})^{2}\frac{k_{B}}{k_{A}}=(\frac{1}{2})^{2}(\frac{4}{3})=\frac{1}{3} \)
\( \frac{E}{U_{B}}=\frac{1}{3}\Rightarrow U_{B}=3E \)
\( U=\frac{1}{2}kx^{2}=\frac{1}{2}\frac{m^{2}g^{2}}{k} \)
\( U\propto\frac{m^{2}}{k} \)
\( \frac{U_{A}}{U_{B}}=(\frac{m_{A}}{m_{B}})^{2}\frac{k_{B}}{k_{A}}=(\frac{1}{2})^{2}(\frac{4}{3})=\frac{1}{3} \)
\( \frac{E}{U_{B}}=\frac{1}{3}\Rightarrow U_{B}=3E \)
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