A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g=10 m/s²)
Official Solution
Correct Option:
(1)
Step 1: is at the bottom: .
Step 2: is at the top: .
Step 3: By energy conservation: .
Step 4: .
Step 5: Given .
Step 6: .
Step 7: m/s.
02
PYQ 2022
medium
physicsID: jee-main
A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass m moves against the wall with a speed v. Which of the following curve represents the correct relation between the normal reaction on the block by the wall (N) and speed of the block (v)?
1
2
3
4
Official Solution
Correct Option:
(1)
Therefore, the graph given in option A suits the best for the above relation.
So, the correct option is (A).
03
PYQ 2022
easy
physicsID: jee-main
For a particle in uniform circular motion, the acceleration a at any point P(R, θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed):
1
2
3
4
Official Solution
Correct Option:
(1)
The correct option is (C): .
As the particle in uniform circular motion experiences only centripetal acceleration of magnitude ω2R or
directed towards centre so from diagram.
04
PYQ 2022
hard
physicsID: jee-main
A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60 m, then the magnitude of displacement would be (Given cos135°= –0.7)
1
42 m
2
47 m
3
19 m
4
40 m
Official Solution
Correct Option:
(2)
Distance travelled = 60 m ⇒ Angle covered = 135° d=Rθ 60= R= = m Displacement = ⇒ = ≈47 m
05
PYQ 2022
medium
physicsID: jee-main
Motion of a particle in x–y plane is described by a set of following
and y = 4sin(ωt) m. The path of the particle will be
1
Circular
2
Helical
3
Parabolic
4
Elliptical
Official Solution
Correct Option:
(1)
The correct answer is (A) : Circular
⇒ The particle is moving in a circular motion with radius of 4 m.
06
PYQ 2023
medium
physicsID: jee-main
A stone tied to long string at its end is making revolutions in horizontal circle in every minute The magnitude of acceleration of stone is The value of _____ ( Take )
Official Solution
Correct Option:
(1)
The correct answer is 125 a=ω2R=(6028×2π)2×1.8 =(6056×722)2×1.8
=225(44)2×1.8 =2251936×1.8
x=125
07
PYQ 2023
medium
physicsID: jee-main
A car is moving with a constant speed of in a circular horizontal track of radius A bob is suspended from the roof of the car by a massless string The angle made by the string with the vertical will be : (Take )
A body of mass 200 g is tied to a spring of spring constant 12.5 N/m, while the other end of the spring is fixed at point . If the body moves about in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s, then the ratio of extension in the spring to its natural length will be:
1
2
3
4
Official Solution
Correct Option:
(2)
Mass of the body,
Spring constant, , Angular speed,
Let the natural length of the spring be , and the extension in the spring be .
The restoring force due to the spring is:
Substituting the given values:
Simplifying:
Thus, the ratio of extension to the natural length is:
09
PYQ 2023
easy
physicsID: jee-main
A small block of mass 100 g is tied to a spring of spring constant 7.5 N/m and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s about point A, then tension in the spring is-
Two concentric circular coils with radii 1 cm and 1000 cm, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _____× 10-8 H. (Take, π2 = 10)
Official Solution
Correct Option:
(1)
11
PYQ 2023
medium
physicsID: jee-main
A car is moving on a circular path of radius such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of is . The value of is _____ .
Official Solution
Correct Option:
(1)
Step 1: Relating Tangential and Centripetal Acceleration
Given:
The tangential acceleration is , and the centripetal acceleration is . The problem states that:
Step 2: Differential Equation for Velocity
Rearrange the equation:
Integrate both sides:
The integration gives:
Simplify:
Step 3: Solve for Velocity as a Function of Time
Rearrange to find :
Thus:
Step 4: Angular Displacement for Quarter Revolution
The angular displacement for circular motion is related to velocity by:
Substitute :
Integrate from to for the quarter revolution:
Solve the integral:
For , substitute and solve:
Thus, the time for the first quarter revolution is .
12
PYQ 2024
medium
physicsID: jee-main
A cyclist starts from the point P of a circular ground of radius 2 km and travels along its circumference to the point S. The displacement of a cyclist is :
1
6 km
2
√8 km
3
4 km
4
8 km
Official Solution
Correct Option:
(2)
Since the cyclist travels along the circumference from point P to point S, which are opposite ends of the diameter of the circle, we can visualize the displacement as the straight-line distance between P and S.
1. Determine the Displacement:
The radius R of the circular ground is given as 2 km.
Points P and S form a diameter of the circle.
The displacement from P to S is the length of the diagonal of the square formed by the radii OP and OS.
Using the Pythagorean theorem, we find:
Answer:
13
PYQ 2024
medium
physicsID: jee-main
A car of 800 kg is taking a turn on a banked road of radius 300 m and angle of banking 30°. If the coefficient of static friction is 0.2, then the maximum speed with which the car can negotiate the turn safely: (Given , ).
1
70.4 m/s
2
51.4 m/s
3
264 m/s
4
102.8 m/s
Official Solution
Correct Option:
(2)
The problem asks for the maximum safe speed with which a car can navigate a banked circular turn, considering the forces of gravity, normal reaction, and static friction.
Concept Used:
When a car is taking a turn on a banked road, the necessary centripetal force is provided by the horizontal components of the normal force and the frictional force. For the car to move at the maximum safe speed ( ), it has a tendency to slide up the incline. Therefore, the force of static friction ( ) acts down the incline.
By balancing the forces in the vertical direction and equating the net horizontal force to the required centripetal force ( ), we can derive the formula for the maximum safe speed:
where:
is the acceleration due to gravity.
is the radius of the turn.
is the coefficient of static friction.
is the angle of banking.
Step-by-Step Solution:
Step 1: List the given parameters.
Mass of the car, (Note: the mass is not needed for the final calculation as it cancels out).
Radius of the turn, .
Angle of banking, .
Coefficient of static friction, .
Acceleration due to gravity, .
Given value, .
Step 2: Calculate the value of .
Step 3: Substitute the given values into the formula for maximum speed.
Step 4: Perform the numerical calculation step-by-step.
First, calculate the value of the expression in the parenthesis:
Now, substitute this back into the main equation:
Step 5: Calculate the final value of the maximum speed.
Rounding to one decimal place, we get 51.4 m/s.
The maximum speed with which the car can negotiate the turn safely is approximately 51.4 m/s.
14
PYQ 2024
medium
physicsID: jee-main
A particle connected with light thread is performing vertical circular motion. Speed at point B (Lowermost point) is sufficient, so that it is able to complete its circular motion. Ignoring air friction, find the ratio of kinetic energy at A to that at B. (A being top-most point)
1
2
3
4
Official Solution
Correct Option:
(1)
The correct option is (A):
15
PYQ 2025
medium
physicsID: jee-main
A uniform disc of radius is rotating about an axis passing through its diameter with angular speed 800 rpm. A torque of magnitude is applied on the disc for 40 sec. If the final angular speed of the disc is 2100 rpm, find the diameter of the disc if its mass is 1 kg.
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Use the equation for angular velocity.
We are given that the initial angular speed is 800 rpm and the final angular speed is 2100 rpm. The equation for angular velocity is given by:
where is the final angular velocity, is the initial angular velocity, is the angular acceleration, and is the time. Converting rpm to rad/s:
Now, substitute the values into the equation:
Solving for :
Step 2: Use the equation for torque.
We are given a torque of . The torque is related to the angular acceleration and moment of inertia by:
For a disc rotating about its diameter, the moment of inertia is:
Substitute the values into the torque equation:
Simplifying:
Since , this simplifies to:
Solving for :
Step 3: Calculate the Diameter.
The diameter is twice the radius:
Thus, the correct answer is .
16
PYQ 2026
medium
physicsID: jee-main
Find speed given to particle at lowest point so that tension in string at point A becomes zero.
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Condition for zero tension at point A. At point A, when tension becomes zero, the centripetal force is provided only by the component of weight.
Step 2: Calculating speed at point A.
Step 3: Applying conservation of mechanical energy. At lowest point, initial speed is . Using M.E.C.:
Step 4: Substituting values.
Step 5: Final speed at lowest point.
17
PYQ 2026
medium
physicsID: jee-main
A block of mass is released from height on a smooth plane. If the normal force on the top of the circular part is , find .
1
2
3
4
Official Solution
Correct Option:
(1)
Concept: Two key physics principles are used here:
Conservation of Mechanical Energy: Since the surface is smooth, mechanical energy is conserved.
Centripetal Force at the Top of the Loop: At the top of the circular path, both weight and normal reaction act toward the center.
These equations allow us to determine the velocity at the top and then compute the required height. Step 1: Apply centripetal force condition at the top of the circular path.} At the top: Given Substitute: Step 2: Apply conservation of mechanical energy.} Initial energy (at height ): Energy at the top of the circular path (height ): Using conservation of energy: However, since the block first descends to the bottom and then climbs the circular track of radius , the initial reference height corresponds to an additional above the base. Thus, Therefore,
18
PYQ 2026
easy
physicsID: jee-main
In case of vertical circular motion of a particle by a thread of length , if the tension in the thread is zero at an angle as shown in the figure, the velocity at the bottom point (A) of the vertical circular path is ( = gravitational acceleration ).
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Apply condition of zero tension. At the given point where the string makes an angle with the horizontal, the tension in the string is zero. Hence, the centripetal force is provided only by the component of gravitational force towards the centre.
Step 2: Apply conservation of mechanical energy. Let the velocity at the given point be and velocity at the lowest point be . The vertical height difference between the two positions is Using energy conservation,
Step 3: Simplify and calculate velocity at the bottom.
19
PYQ 2026
medium
physicsID: jee-main
A block of mass is at rest w.r.t. a hollow cylinder which is rotating with angular speed , radius of the cylinder is . Find the minimum coefficient of friction between the block and the cylinder.
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Understanding the problem. The block is at rest with respect to the rotating hollow cylinder. The block will experience a centrifugal force due to the rotation of the cylinder, which can be balanced by the frictional force between the block and the cylinder. To prevent the block from slipping, we need to find the minimum coefficient of friction that ensures the block stays at rest. Step 2: Forces acting on the block. The centrifugal force on the block is given by:
This force tends to push the block outward. The frictional force is given by:
where is the coefficient of friction, and is the acceleration due to gravity. For the block to stay at rest with respect to the rotating cylinder, the frictional force must equal the centrifugal force. Therefore,
Step 3: Solving for . Cancelling from both sides:
Thus, the minimum coefficient of friction is:
20
PYQ 2026
medium
physicsID: jee-main
For the given circuit arrangement, find the charge on the capacitor in steady state.
1
2
3
4
Official Solution
Correct Option:
(2)
Step 1: Identify the steady state. In the steady state, the capacitor behaves as an open circuit because no current flows through it once it is fully charged. Therefore, the equivalent circuit simplifies to the resistances without considering the capacitor in terms of current flow. Step 2: Analyze the resistances. In the given circuit, the resistances are arranged in a combination of series and parallel. First, we need to calculate the equivalent resistance of the resistors. - The and resistors are in parallel. The equivalent resistance of these two resistors is:
- Now, the , , and resistors are in series. So, the total equivalent resistance is:
Step 3: Apply Ohm’s law. Now, we can use Ohm’s law to find the total current in the circuit. The total voltage is , and the total resistance is , so the total current is:
Step 4: Calculate the charge on the capacitor. The charge on the capacitor is given by:
where and is the voltage across the capacitor. In the steady state, the voltage across the capacitor is the same as the voltage across the capacitor branch. Using the current through the branch, we can find this voltage by multiplying the current by the total resistance in series with the capacitor:
Now, calculating the charge:
Thus, the charge on the capacitor is .
About Uniform Circular Motion - JEE-MAIN
Uniform Circular Motion is a vital chapter for JEE-MAIN aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Uniform Circular Motion PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Uniform Circular Motion carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
