JEE Main 2022 · 25 Jun · Shift 2 · Physics Q33

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PhysicsSingle correct+4 / -1Question 33 of 90

A disc with a flat small bottom beaker placed on it at a distance $R$ from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $\omega$ . The coefficient of static friction between the bottom of the beaker and the surface of the disc is $\mu$ . The beaker will revolve with the disc if :

R\le\dfrac{\mu g}{2\omega^2}

R\le\dfrac{\mu g}{\omega^2}

R\ge\dfrac{\mu g}{2\omega^2}

R\ge\dfrac{\mu g}{\omega^2}

Answer and worked solution

Official answer: B

Friction supplies the centripetal force: $m\omega^2R\le\mu mg$ , so $R\le\dfrac{\mu g}{\omega^2}$ .

(B)

Topics:Circular motion dynamics Friction

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JEE Main 2022 · 25 Jun · Shift 2, Physics question 33