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(B) For a particle to complete a vertical circular motion,the minimum velocity at the top point $A$ is $v_A = \sqrt{g\ell}$.Using the law of conservat

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$2g$

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(B) For a particle to complete a vertical circular motion,the minimum velocity at the top point $A$ is $v_A = \sqrt{g\ell}$.Using the law of conservation of energy between point $A$ and point $C$:$\frac{1}{2}mv_A^2 + mg(2\ell) = \frac{1}{2}mv_C^2 + mg\ell$$\frac{1}{2}m(g\ell) + 2mg\ell = \frac{1}{2}mv_C^2 + mg\ell$$\frac{1}{2}g\ell + mg\ell = \frac{1}{2}mv_C^2$$\frac{3}{2}g\ell = \frac{1}{2}v_C^2 \implies v_C^2 = 3g\ell$.The centripetal acceleration at point $A$ is $a_A = \frac{v_A^2}{\ell} = \frac{g\ell}{\ell} = g$.The centripetal acceleration at point $C$ is $a_C = \frac{v_C^2}{\ell} = \frac{3g\ell}{\ell} = 3g$.The increase in centripetal acceleration is $\Delta a = a_C - a_A = 3g - g = 2g$.

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