ExamBro
ExamBro
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.
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{{rg}}\)

Step-by-step Solution

Detailed explanation

Conical pendulum

\({r}=\frac{\ell}{\sqrt{2}}\)

\(\sin \theta=\frac{{r}}{\ell}=\frac{1}{\sqrt{2}}\)

\(\theta=45^{\circ}\)

\({T} \sin \theta=\frac{{mv}^{2}}{{r}}\)

\({T} \cos \theta={mg}\)

\(\tan \theta=\frac{{v}^{2}}{{rg}} \Rightarrow {v}=\sqrt{{rg}}\)

Same subject
Explore more questions on app