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Physics · Newton's Laws of Motion

JEE Main 2026 — 23 January, Morning Shift — Question 32

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 kA,kB\mathrm{k}_{\mathrm{A}}, \mathrm{k}_{\mathrm{B}} of A and BB, respectively satisfy the relation 4kA=3kB4 \mathrm{k}_{\mathrm{A}}=3 \mathrm{k}_{\mathrm{B}}, is :

Question figure
  1. Option A:

    4E

  2. Option B:

    2E

  3. Option C:

    3E

    Correct
  4. Option D:

    43E\frac{4}{3} \mathrm{E}

Answer: C

Step-by-step solution

For equilibrium kx=mg\mathrm{kx}=\mathrm{mg} U=12kx2=12 m2 g2k\mathrm{U}=\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \frac{\mathrm{~m}^{2} \mathrm{~g}^{2}}{\mathrm{k}} U∝m2k\mathrm{U} \propto \frac{\mathrm{m}^{2}}{\mathrm{k}} UAUB=(mAmB)2kBkA=(12)2(43)=13\frac{\mathrm{U}_{\mathrm{A}}}{\mathrm{U}_{\mathrm{B}}}=\left(\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}\right)^{2} \frac{\mathrm{k}_{\mathrm{B}}}{\mathrm{k}_{\mathrm{A}}}=\left(\frac{1}{2}\right)^{2}\left(\frac{4}{3}\right)=\frac{1}{3} EUB=13⇒UB=3E\frac{\mathrm{E}}{\mathrm{U}_{\mathrm{B}}}=\frac{1}{3} \Rightarrow \mathrm{U}_{\mathrm{B}}=3 \mathrm{E}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Spring Force and Combination of Springs