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(C) The centripetal force $F$ acting on a particle of mass $m$ moving with angular velocity $\omega$ in a circular path of radius $R$ is given by $F =

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$R/4$

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(C) The centripetal force $F$ acting on a particle of mass $m$ moving with angular velocity $\omega$ in a circular path of radius $R$ is given by $F = m\omega^2 R$.Since the centripetal force $F$ and the mass $m$ are kept constant,we have $R = \frac{F}{m\omega^2}$.This implies $R \propto \frac{1}{\omega^2}$.Let the original radius be $R_1 = R$ and original angular velocity be $\omega_1 = \omega$.Let the new radius be $R_2$ and the new angular velocity be $\omega_2 = 2\omega$.Using the proportionality $R_1 \omega_1^2 = R_2 \omega_2^2$,we get $R \cdot \omega^2 = R_2 \cdot (2\omega)^2$.$R \cdot \omega^2 = R_2 \cdot 4\omega^2$.$R_2 = \frac{R}{4}$.

$A$ particle moves with constant angular velocity in a circular path of a certain radius and is acted upon by a certain centripetal force $F$. If the centripetal force $F$ is kept constant but the angular velocity is doubled,the new radius of the path (original radius $R$) will be:

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