MHT CET 2023 12th May Morning Shift | Rotational Motion Question 47 | Physics

A thin uniform circular disc of mass '$$\mathrm{M}$$' and radius '$$R$$' is rotating with angular velocity '$$\omega$$', in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same radius but of mass $$\left(\frac{M}{2}\right)$$ is placed gently on the first disc co-axially. The new angular velocity will be

$$\frac{2}{3} \omega$$

$$\frac{4}{5} \omega$$

$$\frac{5}{4} \omega$$

$$\frac{3}{2} \omega$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio

$$1: 2$$

$$\sqrt{2}: 1$$

$$2: 1$$

$$1: \sqrt{2}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

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From a disc of mass '$$M$$' and radius '$$R$$', a circular hole of diameter '$$R$$' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

$$\frac{13 \mathrm{MR}^2}{32}$$

$$\frac{11 \mathrm{MR}^2}{32}$$

$$\frac{9 \mathrm{MR}^2}{32}$$

$$\frac{7 \mathrm{MR}^2}{32}$$

MHT CET 2023 11th May Morning Shift

MCQ (Single Correct Answer)

-0

A particle moves along a circular path with decreasing speed. Hence

its resultant acceleration is towards the centre.

it moves in a spiral path with decreasing radius.

the direction of angular momentum remains constant.

its angular momentum remains constant

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