MHT CET 2023 11th May Evening Shift | Rotational Motion Question 49 | Physics

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

MHT CET 2023 11th May Morning Shift

MCQ (Single Correct Answer)

-0

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is $$\mathrm{K}_{\mathrm{c}}$$. Radius of gyration of the same disc about a diameter of the disc is $$K_d$$. The ratio $$K_c: K_d$$ is

$$\sqrt{2}: 1$$

$$1: \sqrt{2}$$

$$2: 1$$

$$1: 4$$

MHT CET 2023 11th May Morning Shift

MCQ (Single Correct Answer)

-0

A disc has mass $$M$$ and radius $$R$$. How much tangential force should be applied to the rim of the disc, so as to rotate with angular velocity '$$\omega$$' in time $$\mathrm{t}$$ ?

$$\frac{M R \omega}{4 t}$$

$$\frac{\mathrm{MR} \omega}{2 \mathrm{t}}$$

$$\frac{\mathrm{MR} \omega}{\mathrm{t}}$$

$$\mathrm{MR} \omega \mathrm{t}$$

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