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(D) The centripetal acceleration '$a$' of a particle moving in a circular path of radius '$r$' with speed '$v$' is given by the formula: $a = \frac{v^

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$4$:$1$

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(D) The centripetal acceleration '$a$' of a particle moving in a circular path of radius '$r$' with speed '$v$' is given by the formula: $a = \frac{v^2}{r}$.From this relation,it is clear that the centripetal acceleration is directly proportional to the square of the speed: $a \propto v^2$.Let the initial speed be '$v_1 = v$' and the initial acceleration be '$a_1 = a$'.Let the final speed be '$v_2 = 2v$' and the final acceleration be '$a_2$'.Taking the ratio of the final acceleration to the initial acceleration:$\frac{a_2}{a_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{2v}{v}\right)^2 = (2)^2 = 4$.Therefore,the ratio of the acceleration after and before the change is $4:1$.

$A$ particle is moving along a circular path with constant speed and centripetal acceleration '$a$'. If the speed is doubled,the ratio of its acceleration after and before the change is:

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