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(C) To complete a vertical circle of radius $R$,the minimum speed at the lowest point $A$ must be $v_A = \sqrt{5gR}$.Given that the diameter of the ci

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$\frac{5}{4}D$

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(C) To complete a vertical circle of radius $R$,the minimum speed at the lowest point $A$ must be $v_A = \sqrt{5gR}$.Given that the diameter of the circle is $D$,the radius is $R = \frac{D}{2}$.Substituting $R$ into the velocity equation,we get $v_A = \sqrt{5g \left(\frac{D}{2}\right)} = \sqrt{\frac{5gD}{2}}$.Using the principle of conservation of mechanical energy,the potential energy at height $h$ is converted into kinetic energy at point $A$:$mgh = \frac{1}{2}mv_A^2$$gh = \frac{1}{2}v_A^2$Substituting the value of $v_A^2 = \frac{5gD}{2}$:$gh = \frac{1}{2} \left(\frac{5gD}{2}\right)$$h = \frac{5D}{4}$

$A$ body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) just completes a vertical circle of diameter $AB = D$. The height $h$ is equal to

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