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JEE Main 2026 (23 Jan, Shift 1) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2026 (23 Jan, Shift 1)

Q3JEE Main 2026 · 23 JAN · SHIFT 1Medium

Spring & Elastic Energy

Two blocks with masses 100 g and 200 g are attached to the ends of springs AA and BB as shown in figure. The energy stored in AA is EE. The energy stored in BB, when spring constants kA,kBk_A, k_B of AA and BB, respectively satisfy the relation 4kA=3kB4k_A = 3k_B, is:

Figure

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Answer: (d)

  1. Each spring holds its block in equilibrium: F=mgF = mg, so E=F22kE = \frac{F^2}{2k}.
  2. EA=(0.1g)22kA=EE_A = \frac{(0.1g)^2}{2k_A} = E, EB=(0.2g)22kBE_B = \frac{(0.2g)^2}{2k_B}
  3. EBEA=4⋅kAkB=4⋅34=3\frac{E_B}{E_A} = 4 \cdot \frac{k_A}{k_B} = 4 \cdot \frac{3}{4} = 3
  4. So the answer is (d) 3E3E.
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