If a particle starts from A along the curved circular p

If a particle starts from A along the curved circular path shown in figure with tangential acceleration ‘a’. Then acceleration at B in magnitude is:

$a π \sqrt{1 + π^{2}}$

$a \sqrt{π^{2} - 1}$

$a \sqrt{1 + π^{2}}$

$2 a \sqrt{1 + π^{2}}$

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$V_{B}^{2} = V_{0}^{2} + 2 a \times \frac{π R}{2}$

$V_{B}^{2} = a π R$

Normal acceleration at $B = \frac{V_{B}^{2}}{R} = a π$

Net acceleration at $B = \sqrt{a^{2} + a^{2} π^{2}} = a \sqrt{1 + π^{2}}$