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A small sphere of mass $m$ suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and it is released, then

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Thread tension at the moment when the vertical component of the sphere's velocity is maximum will be $m g$

The angle $θ$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is along horizontal is $\cos^{- 1} \frac{1}{\sqrt{3}}$

Thread tension as a function of $θ$ is $T = 3 m g \cos θ$

Total acceleration of sphere as a function of $θ$ (with the vertical) is $g \sqrt{1 + 3 \cos^{2} θ}$

answer is A, B, C.

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Detailed Solution

Between $A$ and $B$ $m g L \cos θ = \frac{1}{2} m {V_{B}}^{2}$

$∴ {V_{B}}^{2} = 2 g L \cos θ$

Now $a_{r} = \frac{{V_{B}}^{2}}{L} = 2 g \cos θ$

and $a_{t} = g \sin θ$

$∴ a = \sqrt{a_{t}^{2} + a_{r}^{2}} = g \sqrt{1 + 3 \cos^{2} θ}$

Now, at $B$ $T_{B} - m g \cos θ = \frac{m V_{B}^{2}}{L}$

$T_{B} = 3 m g \cos θ$

When total acceleration vector directed horizontally

$\tan (90 - θ) = \frac{a_{t}}{a_{r}} = \frac{g \sin θ}{2 g \cos θ} = \frac{1}{2} \tan θ$

on solving $θ = \cos^{- 1} \frac{1}{\sqrt{3}}$

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