MHT CET 2024 2nd May Morning Shift | Circular Motion Question 15 | Physics

A particle is performing uniform circular motion along the circumference of the circle of diameter 1 m with frequency 4 Hz . The acceleration of the particle in $\mathrm{m} / \mathrm{s}^2$ is

$8 \pi^2$

$16 \pi^2$

$24 \pi^2$

$32 \pi^2$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

A particle moves around a circular path of radius '$$r$$' with uniform speed '$$V$$'. After moving half the circle, the average acceleration of the particle is

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

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

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

$$\frac{\mathrm{V}^2}{\pi \mathrm{r}}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

On dry road, the maximum speed of a vehicle along a circular path is '$$V$$'. When the road becomes wet, maximum speed becomes $$\frac{\mathrm{V}}{2}$$. If coefficient of friction of dry road is '$$\mu$$' then that of wet road is

$$\frac{2 \mu}{3}$$

$$\frac{\mu}{4}$$

$$\frac{\mu}{3}$$

$$\frac{3 \mu}{4}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

A string of length '$$L$$' fixed at one end carries a body of mass '$$\mathrm{m}$$' at the other end. The mass is revolved in a circle in the horizontal plane about a vertical axis passing through the fixed end of the string. The string makes angle '$$\theta$$' with the vertical. The angular frequency of the body is '$$\omega$$'. The tension in the string is

$$\mathrm{mL}^2 \omega$$

$$\mathrm{mL} \omega^2$$

$$\frac{\omega^2}{\mathrm{~mL}}$$

$$\frac{\mathrm{m} \omega^2}{\mathrm{~L}}$$

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