JEE Main 2024 (Online) 5th April Evening Shift | Circular Motion Question 3 | Physics

JEE Main 2024 (Online) 5th April Evening Shift

MCQ (Single Correct Answer)

-1

A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $$9 \mathrm{~m}$$ and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $$\mathrm{m} / \mathrm{s}^2$$ ) :

$$4 \pi^2 \mathrm{~ms}^{-2}$$

$$16 \pi^2 \mathrm{~ms}^{-2}$$

$$57600 \pi^2 \mathrm{~ms}^{-2}$$

Zero

JEE Main 2024 (Online) 4th April Evening Shift

MCQ (Single Correct Answer)

-1

A cyclist starts from the point $$P$$ of a circular ground of radius $$2 \mathrm{~km}$$ and travels along its circumference to the point $$\mathrm{S}$$. The displacement of a cyclist is:

JEE Main 2024 (Online) 4th April Evening Shift Physics - Circular Motion Question 4 English

$$\sqrt8$$ km

4 km

6 km

8 km

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

A ball of mass $0.5 \mathrm{~kg}$ is attached to a string of length $50 \mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400 \mathrm{~N}$. The maximum possible value of angular velocity of the ball in $\mathrm{rad} / \mathrm{s}$ is, :

1600

1000

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

A particle moving in a circle of radius $\mathrm{R}$ with uniform speed takes time $\mathrm{T}$ to complete one revolution.

If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4 R$. The angle of projection $\theta$ is then given by :

$\sin ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}$

$\sin ^{-1}\left[\frac{\pi^2 \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}$

$\cos ^{-1}\left[\frac{\pi \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}$

$\cos ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}$

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