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

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θ0=30∘

θ0=60∘

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

At maximum deflection, tension in string is equal to 2mg3.

answer is A.

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Detailed Solution

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

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