MHT CET 2021 21th September Morning Shift | Circular Motion Question 20 | Physics

A particle moves in a circular orbit of radius '$$r$$' under a central attractive force, $$F=-\frac{k}{r}$$, where $$\mathrm{k}$$ is a constant. The periodic time of its motion is proportional to

$$r^{\frac{1}{2}}$$

$$\mathrm{r}^{\frac{2}{3}}$$

$$r$$

$$r^{\frac{3}{2}}$$

MHT CET 2021 20th September Evening Shift

MCQ (Single Correct Answer)

-0

A particle at rest starts moving with a constant angular acceleration of $$4 \mathrm{~rad} / \mathrm{s}^2$$ in a circular path. At what time the magnitude of its centripetal acceleration and tangential acceleration will be equal?

$$\frac{1}{4} \mathrm{~S}$$

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

$$\frac{1}{2} \mathrm{~S}$$

$$\frac{1}{3} \mathrm{~S}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

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In non-uniform circular motion, the ratio of tangential to radial acceleration is ($$r=$$ radius, $$\alpha=$$ angular acceleration and $$v=$$ linear velocity)

$$\frac{r \alpha}{v}$$

$$\frac{v^2}{r a}$$

$$\frac{r \alpha^2}{v^2}$$

$$\frac{r^2 \alpha}{v^2}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

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A particle is moving in a radius $$R$$ with constant speed $$v$$. The magnitude of average acceleration after half revolution is

$$\frac{2 \pi}{R v^2}$$

$$\frac{2 R}{\pi v}$$

$$\frac{2 v^2}{\pi R}$$

$$\frac{2 V}{\pi R^2}$$

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