MHT CET 2025 25th April Evening Shift | Rotational Motion Question 5 | Physics

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)

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

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

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

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

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

The moment of inertia of a solid sphere of mass ' $m$ ' and radius ' $R$ ' about its diametric axis is ' $I$ '. Its moment of inertia about a tangent in the plane is

2.5 I

3.0 I

3.5 I

4 I

MHT CET 2025 25th April Morning Shift

MCQ (Single Correct Answer)

-0

Two discs of moment of inertia ' $\mathrm{I}_1$ ' and ' $\mathrm{I}_2$ ' and angular speeds ' $\omega_1$ ' and ' $\omega_2$ ' are rotating along the collinear axes passing through their centre of mass and perpendicular to their plane. If the two discs are made to rotate together along the same axis. The rotational kinetic energy of the system will be

$\frac{I_1 \omega_1+I_2 \omega_2}{2\left(I_1+I_2\right)^2}$

$\frac{\left(I_1 \omega_1-I_2 \omega_2\right)^2}{2\left(I_1+I_2\right)}$

$\quad \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{2\left(I_1-I_2\right)}$

$\frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{2\left(I_1+I_2\right)}$

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