MHT CET 2025 19th April Evening Shift | Rotational Motion Question 13 | Physics

A body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity V . If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of plane is

$\frac{\mathrm{V}}{\sqrt{2}}$

2 V

$\frac{\mathrm{V}}{2}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

If $\vec{F}=(5 \hat{i}-10 \hat{j})$ and $\vec{r}=(4 \hat{i}-3 \hat{j})$, then the torque acting on the object will be

$\hat{\mathrm{i}}-2 \hat{\mathrm{j}}$

$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}$

$25 \hat{\mathrm{k}}$

$-25 \hat{\mathrm{k}}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

Four particles each of mass M are placed at the corners of a square of side $L$. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

$\mathrm{L}$

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

$\frac{\mathrm{L}}{4}$

$\frac{\mathrm{L}}{\sqrt{2}}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

A thin uniform rod of mass ' $m$ ' and length ' $L$ ' is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above pivot and then released. The angular acceleration of the rod when it makes an angle ' $\theta$ ' with the vertical is [consider negligible friction at the pivot] ( $\mathrm{g}=$ acceleration due to gravity)

MHT CET 2025 19th April Morning Shift Physics - Rotational Motion Question 13 English

$\frac{3 g \sin \theta}{2 L}$

$\frac{3 g \cos \theta}{2 L}$

$\frac{2 g \sin \theta}{3 L}$

$\frac{2 g \cos \theta}{3 L}$

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