$(A) \large\sqrt \frac{MgL \tan\theta}{4m} \quad$ $(B) \large\sqrt \frac{Mg L}{m \tan \theta} \quad$$(C) \large\sqrt \frac{M L\tan\theta}{4mg} \quad$$ (D) \large\sqrt \frac{mL \tan\theta}{4Mg}$

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The free body diagram for the problem is as follows:

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Since the point A is at equilibrium $\rightarrow M \vec{g} + \vec{T} + \vec{T_1} =0$

Along the horizontal axis (left to right) $\rightarrow T - T_1\cos \theta = 0 \rightarrow T_1 \cos \theta = T$

Along the vertical axis $\rightarrow -Mg + T_1\sin \theta = 0 \rightarrow T_1 \sin \theta = Mg$

Therefore, dividing the two equations $\rightarrow \large\frac{T_1 \sin \theta}{T_1 \cos \theta}$$ = \large\frac{Mg}{T} $$\rightarrow T = \large\frac{Mg}{\tan \theta}$

Velocity of the wave traveling on the string, $v = \large\sqrt \frac{T}{\mu}$, where $T$ is the tension and $\mu = \large\frac{m}{L}$, the mass of the string divided by its length.

$\Rightarrow v = \large\sqrt \frac{Mg L}{m \tan \theta}$

Now, the distance traveled by the wave from point A to point B $d = \large\frac{L}{2}$

$\Rightarrow $ Time $\; t = \large\frac{d}{v}$$ = \Large\frac{\frac{L}{2}}{\sqrt \frac{MgL}{m\tan\theta}}$$ = \large\sqrt \frac{mL \tan\theta}{4Mg}$

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