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(D) When a particle moves along a vertical circle, the minimum speed required at the highest point to complete the loop is given by the formula:$v_{\m

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$\sqrt{2} v$

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(D) When a particle moves along a vertical circle, the minimum speed required at the highest point to complete the loop is given by the formula:$v_{\min} = \sqrt{gR}$From this expression, we can see that the minimum speed is directly proportional to the square root of the radius:$v_{\min} \propto \sqrt{R}$Let the initial radius be $R_1 = R$ and the initial minimum speed be $(v_{\min})_1 = v$.Let the new radius be $R_2 = 2R$ and the new minimum speed be $(v_{\min})_2$.Using the proportionality relation:$\frac{(v_{\min})_1}{(v_{\min})_2} = \sqrt{\frac{R_1}{R_2}}$Substituting the known values:$\frac{v}{(v_{\min})_2} = \sqrt{\frac{R}{2R}} = \frac{1}{\sqrt{2}}$Solving for the new minimum speed:$(v_{\min})_2 = \sqrt{2}v$

Consider a particle moving with a minimum speed $v$ at the highest point of a vertical circle of radius $R$. If the radius of the circle is doubled, the corresponding minimum speed will be:

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