Question

The ring shown in figure is given a constant horizontal acceleration $\left(a_{0}=g / \sqrt{3}\right)$. The maximum deflection of the string from the vertical is $\theta_{0}$. Then

• A. $\theta_{0}=30^{\circ}$
• B. $\theta_{0}=60^{\circ}$
• C. At maximum deflection, tension in string is equal to $mg$.
• D. At maximum deflection, tension in string is equal to $\frac{2 m g}{\sqrt{3}} $

Answer 1

Answer: (A), (D)

$T \cos \theta_{0}=m g$ ...(i)
$T \sin \theta_{0}=m a_{0}$ ...(ii)
Dividing Eq. (ii) by (i), we get
$\tan \theta_{0}=\frac{a}{g}$
$\Rightarrow \theta_{0}=30^{\circ}$
$T=\frac{m g}{\cos 30^{\circ}}=\frac{2 m g}{\sqrt{3}}$