The ring shown in figure is given a constant horizontal acceleration (a0 = g/3).The maximum deflection of the string from the vertical is θ0, then

Tcos⁡θ0=mg ……..(i)Tsin⁡θ0=ma0 …….(ii)Dividing Eq. (ii) by (i), we gettan⁡θ0=ag⇒θ0=30∘T=mgcos⁡30∘=2mg3

The ring shown in figure is given a constant horizontal acceleration (a0 = g/3).The maximum deflection of the string from the vertical is θ0, then

Tcos⁡θ0=mg ……..(i)Tsin⁡θ0=ma0 …….(ii)Dividing Eq. (ii) by (i), we gettan⁡θ0=ag⇒θ0=30∘T=mgcos⁡30∘=2mg3

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The ring shown in figure is given a constant horizontal acceleration (a₀ = g/ $\sqrt{3}$ ).The maximum deflection of the string from the vertical is $θ_{0}$ , then

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$θ_{0} = 30^{\circ}$

$θ_{0} = 60^{\circ}$

At maximum deflection, tension in string is equal to mg.

At maximum deflection, tension in string is equal to $\frac{2 mg}{\sqrt{3}}$ .

answer is A, D.

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Detailed Solution

$Tcos θ_{0} = mg$ ……..(i)

$Tsin θ_{0} = {ma}_{0}$ …….(ii)

Dividing Eq. (ii) by (i), we get

$\begin{array}{l} \tan θ_{0} = \frac{a}{g} \Rightarrow θ_{0} = 30^{\circ} \\ T = \frac{mg}{\cos 30^{\circ}} = \frac{2 mg}{\sqrt{3}} \end{array}$

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The ring shown in figure is given a constant horizontal acceleration (a0 = g/3).The maximum deflection of the string from the vertical is θ0, then

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The ring shown in figure is given a constant horizontal acceleration (a₀ = g/ $\sqrt{3}$ ).The maximum deflection of the string from the vertical is $θ_{0}$ , then