A small shpere of mas ‘m’ suspended by a thread is first taken a side, so that the thread forms the right angle with the vertical and then released, then which options are correct (Here $\text{'}\theta \text{'}\Rightarrow$ is the angle of deflection)

${\text{v}}^{\text{2}}=2\text{g}l\cos \theta$

${\text{a}}_{\text{C}}=\frac{{\text{v}}^2}{l}=2\text{g}\cos \theta$

${\text{a}}_{\text{t}}=\text{g}\sin \theta$

$\therefore \text{a}=\sqrt{{\text{a}}_{\text{t}}^{\text{2}}+{\text{a}}_{\text{t}}^{\text{2}}}$

$=\sqrt{{\left(2\text{g}\cos \theta \right)}^2+{\left(\text{g}\sin \theta \right)}^2}$

$=\text{g}\sqrt{4{\cos }^2\theta +1-{\cos }^2\theta }$        

b) $\text{T}-\text{mg}\cos \theta =\frac{{\text{mv}}^2}{l}$

$\text{T}-\text{mg}\cos \theta =2\text{mg}\cos \theta$

$\text{T}=3\text{mg}\cos \theta$

${\text{v}}_{\text{y}}=\text{v}\sin \theta$

$=\sqrt{2gl\cos \theta }.\sin \theta$

${\text{v}}_{\text{y}}=\sqrt{2\text{g}l}{\cos }^{1/2}\theta \sin \theta$

$\frac{{\text{dv}}_{\text{y}}}{{\text{d}}_{\theta }}=0\text{at}$

$\frac{\text{d}}{\text{dt}}\sqrt{2\text{g}l}{\left(\cos \theta \right)}^{1/2},\sin \theta =0$

$\frac{\text{d}}{\text{dt}}\left[{\left(\cos \theta \right)}^{\text{1/2}}\right]\sin \theta =0$

${\cos }^2\theta .\cos \theta +\sin \theta .\frac{1}{2}{\left(\cos \theta \right)}^{1/2}\left(-\sin \theta \right)$

${\cos }^2\theta =\frac{1}{2}{\sin }^2\theta$

$\cot \theta =\frac{1}{\sqrt{2}}$                                       

$\therefore \text{T}=3\text{mg}\times \frac{1}{\sqrt{3}}$

$=\sqrt{3}\text{mg}$