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(B) The centripetal acceleration $a$ of a particle moving in a circular path of radius $R$ with speed $v$ is given by the formula: $a = \frac{v^2}{R}$

Vedclass · Accepted Answer

$4$:$1$

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(B) The centripetal acceleration $a$ of a particle moving in a circular path of radius $R$ with speed $v$ is given by the formula: $a = \frac{v^2}{R}$.Since the radius $R$ of the circular path remains constant,the 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 $a_1 = a = \frac{v^2}{R}$.When the speed is doubled,the new speed is $v_2 = 2v$.The new acceleration $a_2$ is given by: $a_2 = \frac{(2v)^2}{R} = \frac{4v^2}{R} = 4a$.Therefore,the ratio of the acceleration after the change to the acceleration before the change is: $\frac{a_2}{a_1} = \frac{4a}{a} = \frac{4}{1}$.Thus,the ratio 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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