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JEE Mains · Physics · STD 11 - 13. oscillations
Two springs of force constants \(300\, N/m\) (Spring \(A\)) and \(400\, N/m\) (Spring \(B\)) are joined together in series . The combination is compressed by \(8.75\, cm\). The ratio of energy stored in \(A\) and \(B\) is \(\frac{{{E_A}}}{{{E_B}}}\). Then \(\frac{{{E_A}}}{{{E_B}}}\) is equal to
- A \(\frac{4}{3}\)
- B \(\frac{16}{9}\)
- C \(\frac{3}{4}\)
- D \(\frac{9}{16}\)
Answer & Solution
Correct Answer
(A) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}_{\mathrm{A}}=300 \mathrm{N} / \mathrm{m}, \quad \mathrm{k}_{\mathrm{B}}=400 \mathrm{N} / \mathrm{m}\) Let when the combination of springs is compressed by force \(\mathrm{F}\). Spring \(A\) is compressed by \(x\). Therefore compression in spring \(\mathrm{B}\)…
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