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(B) The relation between linear speed $v$,radius $r$,and angular velocity $\omega$ is given by $v = r\omega$.Given that the linear speed becomes $v_2

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(B) The relation between linear speed $v$,radius $r$,and angular velocity $\omega$ is given by $v = r\omega$.Given that the linear speed becomes $v_2 = 4v_1$ and the angular velocity becomes $\omega_2 = 2\omega_1$.Using the relation $r = \frac{v}{\omega}$,the new radius $r_2$ is:$r_2 = \frac{v_2}{\omega_2} = \frac{4v_1}{2\omega_1} = 2 \left( \frac{v_1}{\omega_1} ight) = 2r_1$.The centripetal acceleration is given by $a_c = r\omega^2$.Therefore,the ratio of the new centripetal acceleration $a_{c2}$ to the initial centripetal acceleration $a_{c1}$ is:$\frac{a_{c2}}{a_{c1}} = \frac{r_2 \omega_2^2}{r_1 \omega_1^2} = \left( \frac{r_2}{r_1} ight) \left( \frac{\omega_2}{\omega_1} ight)^2$.Substituting the values $r_2 = 2r_1$ and $\omega_2 = 2\omega_1$:$\frac{a_{c2}}{a_{c1}} = (2) 	imes (2)^2 = 2 	imes 4 = 8$.Thus,the centripetal acceleration becomes $8$ times the initial value.

If the linear speed of a particle performing uniform circular motion is increased to $4$ times and its angular velocity is doubled,then the centripetal acceleration of the particle becomes .....

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