1
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' $\omega$ '. Its centre of mass rises to a maximum height of [ $\mathrm{g}=$ gravitational acceleration]

A
$\frac{\omega^2 \mathrm{~L}^2}{12 \mathrm{~g}^2}$
B
$\frac{\omega^2 L^2 g}{6}$
C
$\frac{\omega^2 g}{12 \mathrm{~L}^2}$
D
$\frac{\omega^2 L^2}{24 \mathrm{~g}}$
2
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The moment of inertia of thin square plate PQRS of uniform thickness, about an axis passing through centre ' O ' and perpendicular to the plane of the plate is $\left(\mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3, \mathrm{I}_4\right.$ are respectively the moments of inertia about axis $1,2,3,4$ which are in the plane of the plate as shown in figure)

MHT CET 2024 9th May Morning Shift Physics - Rotational Motion Question 35 English

A
$\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$
B
$\mathrm{I}_1+\mathrm{I}_3+\mathrm{I}_4$
C
$\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3+\mathrm{I}_4$
D
$ \mathrm{I}_1+\mathrm{I}_3$
3
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A circular disc of radius ' $R$ ' and thickness $\frac{R}{8}$ has moment of inertia 'I' about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere then moment of inertia of sphere about an axis passing through diameter is

A
$I$
B
$\frac{21}{3}$
C
$\frac{\mathrm{I}}{5}$
D
$\frac{\mathrm{I}}{10}$
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Two solid spheres ( A and B ) are made of metals having densities $\rho_A$ and $\rho_B$ respectively. If there masses are equal then ratio of their moments of inertia $\left(\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}\right)$ about their respective diameter is

A
$\left(\frac{\rho_B}{\rho_A}\right)^{2 / 3}$
B
$\left(\frac{\rho_A}{\rho_B}\right)^{2 / 3}$
C
$\frac{\rho_A}{\rho_B}$
D
$\frac{\rho_B}{\rho_A}$
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