JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2021
- Hard
- MCQ
A particle of mass \({m}\) is suspended from a ceiling through a string of length \(L\). The particle moves in a horizontal circle of radius \(r\) such that \({r}=\frac{{L}}{\sqrt{2}}\). The speed of particle will be:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\sqrt{{rg}}\)
Step-by-step Solution
Detailed explanation
Conical pendulum
\({r}=\frac{\ell}{\sqrt{2}}\)
\(\sin \theta=\frac{{r}}{\ell}=\frac{1}{\sqrt{2}}\)
\(\theta=45^{\circ}\)
\({T} \sin \theta=\frac{{mv}^{2}}{{r}}\)
\({T} \cos \theta={mg}\)
\(\tan \theta=\frac{{v}^{2}}{{rg}} \Rightarrow {v}=\sqrt{{rg}}\)
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