$(A)$ Given: Radius $r = 1 \,m$, Centrifugal force $F = 4 \times 10^{-12} \,N$, Mass $m = 1.6 \times 10^{-27} \,kg$.We know that the centrifugal force

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$(A)$ Given: Radius $r = 1 \,m$, Centrifugal force $F = 4 \times 10^{-12} \,N$, Mass $m = 1.6 \times 10^{-27} \,kg$.We know that the centrifugal force is given by the formula $F = m \omega^2 r$.Rearranging the formula to solve for angular velocity $\omega$, we get $\omega^2 = \frac{F}{mr}$.Substituting the given values: $\omega^2 = \frac{4 \times 10^{-12}}{1.6 \times 10^{-27} \times 1}$.$\omega^2 = \frac{4}{1.6} \times 10^{-12 - (-27)} = 2.5 \times 10^{15} = 25 \times 10^{14}$.Taking the square root of both sides: $\omega = \sqrt{25 \times 10^{14}} = 5 \times 10^7 \,rad/s$.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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Assume a proton is rotating along a circular path of radius $1 \,m$ under a centrifugal force of $4 \times 10^{-12} \,N$. If the mass of the proton is $1.6 \times 10^{-27} \,kg$, then its angular velocity of rotation is

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A
$5 \times 10^7 \,rad/s$
B
$10^{15} \,rad/s$
C
$2.5 \times 10^7 \,rad/s$
D
$5 \times 10^{14} \,rad/s$

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3-2.Motion in Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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Assume a proton is rotating along a circular path of radius $1 \,m$ under a centrifugal force of $4 \times 10^{-12} \,N$. If the mass of the proton is $1.6 \times 10^{-27} \,kg$, then its angular velocity of rotation is

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