NEET Uniform Circular Motion MCQs & PYQs

Q: A wheel having diameter of 3 m starts from rest and accelerates uniformly to an angular velocity of 210 rpm in 5 seconds. Angular acceleration of the wheel is :

ω = ω + α . t ⇒(210 × 2π)/60 = 0 + α × 5 ⇒ α = 1.4 π rad/sec ²

Q: Find ratio of Normal reaction of blocks. At point A & point B respectively if their masses are same and velocities are 2 m/s & 4 m/s respectively :

Centrifugal Force act radially outwards, So, N A = mg - mv²/R = m(10) - m(2)²/2= 8m So, N B = mg + mv²/R = m(10) +m(4)²/2= 18m N A : N B = 4 : 9

Q: Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is :

m(nω)²×r = 2m(ω)² ×(r/2) so, n =1

Practice NEET Uniform Circular Motion Questions

Question 1:

easy

A wheel having diameter of 3 m starts from rest and accelerates uniformly to an angular velocity of 210 rpm in 5 seconds. Angular acceleration of the wheel is :

ω = ω + α . t

⇒(210 × 2π)/60 = 0 + α × 5

⇒ α = 1.4 π rad/sec ²

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Question 2:

easy

The angular velocity of earth about its axis of rotation is :

Angular speed ω = 2π / T = 2π / (60×60×24) rad /sec

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Question 3:

easy

Two bodies of masses 10 kg and 5 kg moving on concentric orbits of radii R and r such that their period of revolution are same. The ratio of their centripetal acceleration is :

As time period of revolution is same for both the particles angular speed will be equal for both.

Centripetal Acceleration is given by ω²r.

so a₁ / a₂ = R/r

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Question 4:

easy

A stone of mass m is tied to a string of length l and rotated in a circle with a constant speed v. If the string is released, the stone flies :

In Circular motion velocity of the particle is always directed along tangent. So when string is released object moves tangentially.

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Question 5:

easy

A stone of mass 1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10/π revolution per second. Calculate the tension of the string ?

Tension force in the string will provide required centripetal force

T = mω²r= 1× (10/π ×2π)²×1= 400 N

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Question 6:

moderate

Find ratio of Normal reaction of blocks. At point A & point B respectively if their masses are same and velocities are 2 m/s & 4 m/s respectively :

Centrifugal Force act radially outwards,

So, N _A = mg - mv²/R = m(10) - m(2)²/2= 8m

So, N _B = mg + mv²/R = m(10) +m(4)²/2= 18m

N _A : N _B = 4 : 9

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Question 7:

moderate

Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is :

m(nω)²×r = 2m(ω)² ×(r/2)

so, n =1

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Question 8:

easy

A particle moves in a circle of radius 5 m with constant speed and time period 2π s. The acceleration of the particle is :

As speed of the particle is constant only centripetal acceleration will act on the object.

Centripetal Acceleration = ω²R

a _c = (2π/2π)× 5= 5 m/s²

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Question 9:

easy

Assertion: In uniform circular motion tangential acceleration of particle is zero.

Reason: In uniform circular motion net force on particle is always directed towards centre of circular path.

Tangential acceleration is zero because speed is constant. The net force is centripetal, which is directed towards the center. Thus, both are true but the Reason is not the explanation of the Assertion.

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Question 10:

easy

A particle moves in a circle of radius \(10\text{ cm}\) with constant speed and time period \(\pi\text{ s}\). The acceleration of the particle is :

The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}\). The centripetal acceleration is \(a = \omega^2 R = 2^2 \times 10 = 40\text{ cm/s}^2\).

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