MHT CET 2024 3rd May Morning Shift | Rotational Motion Question 29 | Physics

$\overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{r}} \times \vec{\omega}, \overrightarrow{\boldsymbol{\alpha}}=\overrightarrow{\mathbf{r}} \times \vec{a}, \vec{L}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}, \vec{\tau}=\overrightarrow{\mathrm{f}} \times \overrightarrow{\mathrm{r}}$

$\overrightarrow{\mathrm{v}}=\vec{\omega} \times \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}}, \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}}$

$\overrightarrow{\mathrm{v}}=\vec{\omega} \times \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}, \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}}$

$\overrightarrow{\mathrm{v}}=\vec{\omega} \times \overrightarrow{\mathrm{r}}, \vec{\alpha}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}, \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{f}}$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

Ratio of radius of gyration of a circular disc to that of circular ring each of same mass and radius around their respective axes is

$\sqrt{2}: 1$

$\sqrt{2}: \sqrt{3}$

$\sqrt{3} : \sqrt{2}$

$1: \sqrt{2}$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

Two circular loops P and Q of radii ' r ' and ' nr ' are made respectively from a uniform wire. Moment of inertia of loop Q about its axis is four times that of loop P about its axis. The value of ' $n$ ' is

$(2)^{1 / 3}$

$(2)^{2 / 3}$

$(2)^{3 / 4}$

$(2)^{1 / 4}$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

A solid sphere of mass ' $m$ ', radius ' $R$ ', having moment of inertia about an axis passing through center of mass as 'I' is recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through the rim (edge) \& perpendicular to plane remains 'I'. Then the radius of disc is

$\frac{2 \mathrm{R}}{\sqrt{15}}$

$\left(\sqrt{\frac{2}{15}}\right) \mathrm{R}$

$\frac{4 \mathrm{R}}{\sqrt{15}}$

$\frac{\mathrm{R}}{4}$

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