(C) Given: Radius $R = 180 \, cm = 1.8 \, m$. Frequency $f = \frac{28}{60} \, Hz$. Angular velocity $\omega = 2 \pi f = 2 \times \frac{22}{7} \times \

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(C) Given: Radius $R = 180 \, cm = 1.8 \, m$. Frequency $f = \frac{28}{60} \, Hz$. Angular velocity $\omega = 2 \pi f = 2 \times \frac{22}{7} \times \frac{28}{60} = \frac{44}{15} \, rad/s$. The centripetal acceleration is $a = \omega^2 R$. Substituting the values: $a = \left(\frac{44}{15}\right)^2 \times 1.8 = \frac{1936}{225} \times 1.8$. Simplifying: $a = \frac{1936 \times 1.8}{225} = \frac{1936}{125} \, m s^{-2}$. Comparing this with $\frac{1936}{x} \, m s^{-2}$,we get $x = 125$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ stone tied to a $180 \, cm$ long string at its end is making $28$ revolutions in a horizontal circle in every minute. The magnitude of the acceleration of the stone is $\frac{1936}{x} \, m s^{-2}$. The value of $x$ is: (Take $\pi = \frac{22}{7}$)

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A
$124$
B
$123$
C
$125$
D
$122$

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3-2.Motion in Plane

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Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ stone tied to a $180 \, cm$ long string at its end is making $28$ revolutions in a horizontal circle in every minute. The magnitude of the acceleration of the stone is $\frac{1936}{x} \, m s^{-2}$. The value of $x$ is: (Take $\pi = \frac{22}{7}$)

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