NEET Circular Motion MCQs & PYQs

Q: A stone is moved round a horizontal circle with a 20 cm long string tied to If centripetal acceleration is 9.8 m/s2, then its angular velocity will be :

Centripetal Acceleration is given by a= ω²R ⇒ 9.8 = ω²× 1/5 ⇒ ω² =49 ⇒ ω = 7 rad/sec

Q: A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle θ1; in the next 4 sec, it rotates through an additional angle θ2. The ratio of θ2/θ1 is :

For Circular motion angle traversed is θ = ω t + ½ α t² so, θ1 =½α(2)² = 2α and θ1 + θ2 = ½α(6)² = 18 α ⇒ θ2= 16 α so θ2 /θ1= 8

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: A car has to move on a level turn of radius (R = 50m). If the coefficient of static friction between tyre and road is µ = 0.2. Find the maximum speed the car can take without skidding is given by :

For safe turning on a horizontal road. μ = v²/ rg ⇒ v= (μ rg ) 1/2 v = (0.2 × 50 × 10 ) 1/2 = 10 m/s

Q: A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the If the angular velocity of the turntable is doubled, it will just slip at a distance of :

F= mω²r As the distance will increase centrifugal force acting on the object will also increase. As slipping starts at 4 cm object will not slip at radius smaller than 4cm.

Practice NEET Circular Motion Questions

Question 1:

moderate

A point on the periphery of rotating disc has its acceleration vector making on angle 30° with velocity vector then the ratio of magnitude of centripetal acceleration to tangential acceleration is :

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Question 2:

easy

A stone is moved round a horizontal circle with a 20 cm long string tied to If centripetal acceleration is 9.8 m/s², then its angular velocity will be :

Centripetal Acceleration is given by a= ω²R

⇒ 9.8 = ω²× 1/5

⇒ ω² =49

⇒ ω = 7 rad/sec

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Question 3:

moderate

A block on a stationary horizontal table with increasing speed in a circle is seen from an inertial frame of reference. The angle between net force on the block and velocity vector is :

When net acceleration makes acute angle with velocity, speed of the particle will increase. As tangential accleration is positive, speed will increase.

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Question 4:

moderate

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle θ₁; in the next 4 sec, it rotates through an additional angle θ₂. The ratio of θ₂/θ₁ is :

For Circular motion angle traversed is θ = ω t + ½ α t²

so, θ₁ =½α(2)² = 2α

and θ₁ + θ₂ = ½α(6)² = 18 α ⇒ θ₂= 16 α

so θ₂ /θ₁= 8

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Question 5:

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 6:

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 7:

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 8:

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 9:

easy

A car has to move on a level turn of radius (R = 50m). If the coefficient of static friction between tyre and road is µ = 0.2. Find the maximum speed the car can take without skidding is given by :

For safe turning on a horizontal road.

μ = v²/ rg ⇒ v= (μ rg ) ^1/2

v = (0.2 × 50 × 10 )^1/2 = 10 m/s

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Question 10:

easy

A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the If the angular velocity of the turntable is doubled, it will just slip at a distance of :

F= mω²r

As the distance will increase centrifugal force acting on the object will also increase. As slipping starts at 4 cm object will not slip at radius smaller than 4cm.

💬 Discuss this Question