MHT CET 2025 26th April Morning Shift | Rotational Motion Question 3 | Physics

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' $I_1$ '. The same rod is bent into a ring and its moment of inertia about the diameter is ' $I_2$ '. Then $I_1 / I_2$ is

$\frac{3 \pi^2}{2}$

$\frac{2 \pi^2}{3}$

$\frac{\pi^2}{3}$

$\frac{\pi^2}{9}$

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

A bob of mass ' $m$ ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' $R$ ' and mass ' $m$ '. When released from the rest, the bob starts falling vertically downwards. If the bob has covered a vertical distance ' $h$ ', then angular speed of wheel will be (There is no slipping between string and wheel, g - acceleration due to gravity)

$\frac{2}{\mathrm{R}} \sqrt{\frac{\mathrm{gh}}{3}}$

$\frac{1}{\mathrm{R}} \sqrt{\frac{2 \mathrm{gh}}{3}}$

$\mathrm{R} \sqrt{\frac{2 \mathrm{gh}}{3}}$

$2 R \sqrt{\frac{g h}{3}}$

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

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A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ' $\omega$ '. Its centre of mass rises to a maximum height of

( $\mathrm{g}=$ acceleration due gravity)

$\frac{L^2 \omega^2}{2 g}$

$\frac{\mathrm{L} \omega}{6 \mathrm{~g}}$

$\frac{\mathrm{L} \omega}{2 \mathrm{~g}}$

$\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}$

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

Using Bohr's quantisation condition, what is the rotational energy in the second orbit for a diatomic molecule?

( $I=$ moment of inertia of diatomic molecule and $\mathrm{h}=$ Planck's constant)

$\frac{\mathrm{h}}{2 \mathrm{I} \pi^2}$

$\frac{h^2}{2 I \pi^2}$

$\frac{\mathrm{h}^2}{2 \mathrm{I}^2 \pi^2}$

$\frac{\mathrm{h}}{2 \mathrm{I}^2 \pi}$

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