MHT CET 2025 20th April Morning Shift | Rotational Motion Question 14 | Physics

An inclined plane makes an angle $30^{\circ}$ with the horizontal. A solid sphere rolling down an inclined plane from rest without slipping has linear acceleration ( $\mathrm{g}=$ acceleration due gravity) ( $\sin 30^{\circ}=0.5$ )

$\frac{5 \mathrm{~g}}{7}$

$\frac{5 \mathrm{~g}}{14}$

$\frac{2 \mathrm{~g}}{3}$

$\frac{\mathrm{g}}{3}$

MHT CET 2025 20th April Morning Shift

MCQ (Single Correct Answer)

-0

Two spheres each of mass $M$ and radius $R$ are connected with a massless rod of length 4 R . The moment of inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod will be

MHT CET 2025 20th April Morning Shift Physics - Rotational Motion Question 10 English

$\frac{21}{5} \mathrm{MR}^2$

$\frac{84}{5} \mathrm{MR}^2$

$\frac{42}{5} \mathrm{MR}^2$

$\frac{5}{21} \mathrm{MR}^2$

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

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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

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

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

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

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

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