JEE Mains · Physics · STD 11 - 13. oscillations
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle \((\theta)\) of thread deflection in the extreme position will be :
- A \(\tan ^{-1}(\sqrt{2})\)
- B \(2 \tan ^{-1}\left(\frac{1}{2}\right)\)
- C \(\tan ^{-1}\left(\frac{1}{2}\right)\)
- D \(2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)\)
Answer & Solution
Correct Answer
(B) \(2 \tan ^{-1}\left(\frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
Loss in kinetic energy \(=\) Gain in potential energy \( \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta)\) \( \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta)\) Acceleration at lowest point \(=\frac{\mathrm{v}^2}{\ell}\) Acceleration at…
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