KVPY SX SB Exam 20141 Question Paper

Centripetal force at point A :
${\mathit{T}}_1-\mathrm{mg}=\frac{\mathit{m}{\mathit{v}}^2}{\mathit{l}}$….(1)
At point B :
${\mathit{T}}_2=\mathit{m}\mathit{g}\cos \mathit{\theta }$ ….(2)
According to question
${\mathit{T}}_1=4{\mathit{T}}_2$ ….(3)
⇒ $\mathit{m}\mathit{g}+\frac{\mathit{m}{\mathit{v}}^2}{\mathit{l}}=4\mathit{m}\mathit{g}\cos \mathit{\theta }$ [from equation (1) & (2)]
⇒ $\mathit{m}\mathit{g}\left(4\cos \mathit{\theta }-1\right)=\frac{\mathit{m}{\mathit{v}}^2}{\mathit{l}}$ ….(4)
According to conservation of energy between point A and B
Also $\frac{1}{2}\mathit{m}{\mathit{v}}^2+0=0+\mathit{m}\mathit{g}\mathit{l}\left(1-\cos \mathit{\theta }\right)$
$\mathit{m}{\mathit{v}}^2=2\mathit{m}\mathit{g}\mathit{l}\left(1-\cos \mathit{\theta }\right)$
$\frac{\mathit{m}{\mathit{v}}^2}{\mathit{l}}=2\mathit{m}\mathit{g}\left(1-\cos \mathit{\theta }\right)$ ….(5)
From equation (4) & (5)
$\mathit{m}\mathit{g}\left(4\cos \mathit{\theta }-1\right)=2\mathit{m}\mathit{g}\left(1-\cos \mathit{\theta }\right)$
⇒ $4\cos \mathit{\theta }\mathit{–}1=2-2\cos \mathit{\theta }$
⇒ $6\cos \mathit{\theta }=3$
⇒ $\cos \mathit{\theta }=\frac{1}{2}$
⇒ $\mathit{\theta }=60\mathit{º}$