MHT CET 2021 24th September Evening Shift | Rotational Motion Question 29 | Physics

The moment of inertia of a thin uniform rod of mass 'M' and length 'L' about an axis passing through a point at a distance $$\frac{L}{4}$$ from one of its ends and perpendicular to the length of the rod is

$$\frac{\mathrm{ML}^2}{48}$$

$$\frac{7 \mathrm{ML}^2}{48}$$

$$\frac{5 \mathrm{ML}^2}{48}$$

$$\frac{9 \mathrm{ML}^2}{48}$$

MHT CET 2021 24th September Morning Shift

MCQ (Single Correct Answer)

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Two identical particles each of mass '$$m$$' are separated by a distance '$$d$$'. The axis of rotation passes through the midpoint of '$$\mathrm{d}$$' and is perpendicular to the length $$\mathrm{d}$$. If '$$\mathrm{K}$$' is the average rotational kinetic energy of the system, then the angular frequency is

$$2 \mathrm{~d} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$$

$$\frac{\mathrm{d}}{2} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$$

$$\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$$

$$\frac{d}{4} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$$

MHT CET 2021 24th September Morning Shift

MCQ (Single Correct Answer)

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A body of mass '$$\mathrm{m}$$' and radius of gyration '$$\mathrm{K}$$' has an angular momentum $$\mathrm{L}$$. Its angular velocity is

$$\frac{\mathrm{K}^2}{\mathrm{~mL}}$$

$$\mathrm{mK}^2 \mathrm{~L}$$

$$\frac{\mathrm{mK}^2}{\mathrm{~L}}$$

$$\frac{\mathrm{L}}{\mathrm{mK}^2}$$

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

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The moment of inertia of a body about the given axis, rotating with angular velocity 1 rad/s is numerically equal to 'P' times its rotational kinetic energy. The value of 'P' is

$$\frac{1}{4}$$

$$\frac{1}{2}$$

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