Question

A smooth wire bent in the form of circle of radius $a$, a pearl moves slowly on the circumference, this circle is rotated about diameter $AB$ with speed $\omega$, when pearl reaches to point $P$ it stops sliding then value of ${\omega }^2$ will be

• A. $\frac{2g}{a}$
• B. $\frac{2g}{a\sqrt{3}}$
• C. $\frac{a\sqrt{3}}{g}$
• D. $\frac{2g}{g\sqrt{3}}$

Answer 1

The correct option is B $\frac{2g}{a\sqrt{3}}$

$\begin{array}{c}mg\cos \theta =\frac{m{w}^{2}a\sin \theta }{2}\\ \Rightarrow w^2=\frac{2g\cos \theta }{a}\\ =\frac{2g}{a\sqrt{3}}\\ Hence,\\ optionBiscorrectanswer.\end{array}$