Classical Mechanics MCQ Quiz - Objective Question with Answer for Classical Mechanics - Download Free PDF

Last updated on Jun 28, 2025

Latest Classical Mechanics MCQ Objective Questions

Classical Mechanics Question 1:

For the transformation
                                 Q=ln (1+𝑞1/2 cos 𝑝) ,𝑃 = 2𝑞 1/2 (1+𝑞1/2cos 𝑝) sin 𝑝 
the generating function is

Answer (Detailed Solution Below)

Option 4 :

Classical Mechanics Question 1 Detailed Solution

Solution:

F3 = F3(p, Q, t)

∂F3/∂p = -q,    ∂F3/∂Q = -P

⇒ Q = log(1 + q1/2 cos p) ⇒ eQ = 1 + q1/2 cos p

⇒ (eQ - 1) / cos p = q1/2 ⇒ q = ((eQ - 1)2) / cos2 p

⇒ P = 2(1 + q1/2 cos p) q1/2 sin p ⇒ P = 2eQ q1/2 sin p ⇒ 2eQ(eQ - 1) tan p

⇒ ∂F3/∂p = -q = -((eQ - 1)2 / cos2 p) ⇒ F3 = -∫(eQ - 1)2 sec2p dp

⇒ F3 = - (eQ - 1)2 tan p + f1(Q)   ........A

∂F3/∂Q = -P = -2(e2Q - eQ) tan p

F3 = -2 ∫ (e2Q - eQ) tan p + f2(p)

= - (eQ - 1)2 tan p + tan p + f2(p)   ....(B)

Equating A and B: f1(Q) = 0,   f2(p) = -tan p

So F3 = - (eQ - 1)2 tan p

Classical Mechanics Question 2:

A non-relativistic particle of mass 𝑚 and charge 𝑞 is moving in a magnetic field 𝐵⃗(𝑥,𝑦,𝑧). If 𝑣⃗denotes its velocity and {…}P.B. denotes the Poisson Bracket, then   is equal to 

Answer (Detailed Solution Below)

Option 3 :

Classical Mechanics Question 2 Detailed Solution

Solution:

H = ( p – qA )2 / 2m + qφ

⇒ H = |p|2 / 2m + q2 |A|2 / 2m – (q / m) p · A

Use p = pxi + pyj + pzk

and A = Axi + Ayj + Azk

∂H/∂px = (px – qAx) / m = ẋ

∂H/∂py = (py – qAy) / m = ẏ

∂H/∂pz = (pz – qAz) / m = ż

Let us find {x, ẏ}:

{x, ẏ} = ∂x/∂x ∂ẏ/∂px – ∂x/∂px ∂ẏ/∂x + ∂x/∂y ∂ẏ/∂py – ∂x/∂py ∂ẏ/∂y

0 = – (1/m)(q) ( ∂Ay/∂x – ∂Ax/∂y ) = (q/m2)( ∇ × A )z = (q/m2) Bk

⇒ {vi, vj} = (q/m2) Bk εij ⇒ εijk {vi, vj} = (q/m2) Bl εijk εij

= (q/m2) Blkl = 2qBk / m2

Classical Mechanics Question 3:

The Lagrangian of a system is 
                       
Which one of the following is conserved?

  1. 12𝑥˙+3𝑦˙ 
  2. 12𝑥˙−3y
  3. 3𝑥˙−12𝑦˙ 
  4.  3𝑥˙+3𝑦˙

Answer (Detailed Solution Below)

Option 1 : 12𝑥˙+3𝑦˙ 

Classical Mechanics Question 3 Detailed Solution

Solution:

L = (15/2) m x2 + 6mx + 3my2 − mg(x + 2y)

⇒  (∂L/∂) − ∂L/∂x = 0 ⇔ 15m + 6mÿ + mg = 0 ........(1)

⇒  (∂L/∂) − ∂L/∂y = 0 ⇔ 6m + 6mÿ + 2mg = 0 .......(2)

Use operation 2(1) − (2)

24m + 6 = 0 ⇒ d/dt (4x + y) = 0 ⇒ 4 + = 0 ⇒ 12 + 3 = c

Classical Mechanics Question 4:

A frictionless track is defined by 𝑧 = 𝑧o - ,as shown in the figure. 

A particle is constrained to slide down the track under the action of gravity. The tangential acceleration at position (𝑥,𝑧) would be

Answer (Detailed Solution Below)

Option 2 :

Classical Mechanics Question 4 Detailed Solution

Calculation:

Tangential acceleration is given by:

aθ = g ⋅ sinθ

z = z0 - (x2 / 4z0)

tanθ = (dz / dx) = - (x / 2z0) ⇒ sinθ = (x / √(x2 + 4z02)) ⇒ aθ = (g ⋅ x) / √(x2 + 4z02)

Classical Mechanics Question 5:

A particle of rest mass 𝑚o and energy 𝐸 collides with another particle at rest, with the same rest mass. What is the minimum value of 𝐸 so that after the collision, there may be four particles of rest mass 𝑚o

  1. 4𝑚o𝑐2
  2. 3𝑚o𝑐2
  3. 7𝑚o𝑐2
  4. 16𝑚o𝑐2

Answer (Detailed Solution Below)

Option 3 : 7𝑚o𝑐2

Classical Mechanics Question 5 Detailed Solution

Solution: 

For the first particle, the four-vector momentum before collision is:

( E / c , p, 0, 0 )

Second particle is at rest, so its momentum vector before collision is:

( m0c2 / c , 0, 0, 0 )

Where,

p2 = (E2 / c2) - m02 c2

The quantity:

i (E2 / c2 - p2)

is invariant before and after the collision.

Therefore,

((E + m0c2) / c)2 - p2 ≥ (4m0c2)2 / c2

⇒ ((E + m0c2) / c)2 - [(E2 / c2) - m02 c2] ≥ (4m0c2)2 / c2

For minimum energy,

((E + m0c2) / c)2 - [(E2 / c2) - m02 c2] = (4m0c2)2 / c2

E = 7 m0c2

Top Classical Mechanics MCQ Objective Questions

When an object undergoes acceleration

  1. A force always acts on it
  2. It always moves down
  3. It always moves up
  4. It always falls towards the earth

Answer (Detailed Solution Below)

Option 1 : A force always acts on it

Classical Mechanics Question 6 Detailed Solution

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The correct answer is Option 1.Key Points

  • When an object undergoes acceleration, it means there is a change in its velocity. This change in velocity can occur either in terms of speed, direction, or both.
  • A force always acts on it:
    • This statement is generally true.
    • According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma).
    • So, if there's acceleration, there must be a force acting on the object.

Additional Information

  • Acceleration is a fundamental concept in physics that describes the rate of change of velocity with respect to time.
  • Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. Therefore, any change in speed, direction, or both constitutes acceleration.
  • The formula for acceleration (a) is a = F/m where a is acceleration. F is the net force acting on an object and m is the mass of the object.
  • This formula states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
  • In simpler terms, if you apply a force to an object, it will accelerate, and the acceleration will be larger if the force is stronger or if the object has less mass.
  • Acceleration can occur in various forms:
    • Linear Acceleration: Change in speed in a straight line.
    • Angular Acceleration: Change in rotational speed or direction.
    • Centripetal Acceleration: Acceleration directed towards the center of a circular path.
  • Acceleration can be positive or negative:
    • Positive Acceleration: Speeding up in the positive direction.
    • Negative Acceleration (Deceleration): Slowing down or moving in the opposite direction.
  • Gravity is a common force causing acceleration. Near the Earth's surface, objects in free fall experience acceleration due to gravity, denoted as g (approximately 9.8 m/s²).

A ball, initially at rest, is dropped from a height h above the floor bounces again and again vertically. If the coefficient of restitution between the ball and the floor is 0.5, the total distance travelled by the ball before it comes to rest is

  1. 8h/3
  2. 5h/3
  3. 3h
  4. 2h

Answer (Detailed Solution Below)

Option 2 : 5h/3

Classical Mechanics Question 7 Detailed Solution

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Concept:

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

Calculation:

v =  and v = e

0 = (ev)2 - 2gh1

h1 =  = e2h

Similarly, h2 = e4h

H = h + 2h1 + 2h2 +...∞ 

= h + 2(e2h + e4h + ... ∞)

= h + 2e2h()

= h × ()

The coefficient of restitution between the ball and the floor is 0.5.

e = 0.5

H = 5h/3

The correct answer is option (2).

A uniform circular disc on the xy-plane with its centre at the origin has a moment of inertia I0 about the x- axis. If the disc is set in rotation about the origin with an angular velocity ω = ω0(ĵ + k̂), the direction of its angular momentum is along

  1. -î + ĵ + k̂
  2. -î + ĵ + 2k̂
  3. ĵ + 2k̂
  4. ĵ + k̂

Answer (Detailed Solution Below)

Option 3 : ĵ + 2k̂

Classical Mechanics Question 8 Detailed Solution

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Concept:

We are using the angular momentum formula which is  and then using this formula for  planes.

By using matrices for values of  and  we get the value of magnitude and direction of angular momentum.

Explanation:

 A circular disc is in rotation with the origin as center in  plane.

Given,

  •   where  is angular velocity
  •  is the moment of inertia

 

We are using formula for angular momentum

For denoting angular momentum in planes we will use vector notations for angular momentum ,

Using the perpendicular axis theorem,

putting values of  and  in matrix equation, we get

By multiplication above matrices, we get

This is the magnitude of angular momentum. The direction of angular momentum is .

 

The minor axis of Earth's elliptical orbit divides the area within it into two halves. The eccentricity of the orbit is 0.0167. The difference in time spent by Earth in the two halves is closest to

  1. 3.9 days
  2. 4.8 days
  3. 12.3 days
  4. 0 days

Answer (Detailed Solution Below)

Option 1 : 3.9 days

Classical Mechanics Question 9 Detailed Solution

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Concept:

 We are using Kepler's law here which states that the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.

 

Explanation:

Using Kepler's second law, 

eccentricity(e)=>

Now,

 

For the transformation x → X =  p → P = βx2 between conjugate pairs of a coordinate and its momentum, to be canonical, the constants α and β  must satisfy

  1. 1 + 2αβ = 0
  2. 1 - 2αβ = 0

Answer (Detailed Solution Below)

Option 3 : 1 + 2αβ = 0

Classical Mechanics Question 10 Detailed Solution

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Explanation:

  • Given the transformations  and , we have to check them against the requirements of canonical transformations.
  • The main requirement, among others, is that the Poisson bracket {X, P} = 1.
  • In terms of partial derivatives, the Poisson bracket can be written as: 
  • Substituting X and P from the problem statement, we get: 
  • Computing partial derivatives, we get:
  • Setting the above equation to 1: 
  • Finally, 

Which of the following terms, when added to the Lagrangian L(x, y, , ) of a system with two degrees of freedom, will not change the equations of motion?

Answer (Detailed Solution Below)

Option 2 :

Classical Mechanics Question 11 Detailed Solution

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Concept:

The Lagranges equation of motion of a system is given by

Calculation:

The Lagrangian L depends on 

L(x,y,,)

L' = L(x,y,,)

Similarly 

The correct answer is option (2).

The trajectory of a particle moving in a plane is expressed in polar coordinates (r, θ) by the equations , where the parameters r0, β  and ω are positive. Let vr and ar denote the velocity and acceleration, respectively, in the radial direction. For this trajectory

  1. a< 0 at all times irrespective of the values of the parameters
  2. ar​ > 0 at all times irrespective of the values of the parameters
  3. 0\) and ar > 0 for all choices of parameters
  4. 0\) however, ar = 0 for some choices of parameters

Answer (Detailed Solution Below)

Option 4 : 0\) however, ar = 0 for some choices of parameters

Classical Mechanics Question 12 Detailed Solution

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Explanation:

We will first write the velocity vector in polar co-ordinates and write the radial velocity and radial acceleration and by differentiating we get the desired solution.

Given,-     ----------------1

Velocity in polar co-ordinates is given by the sum of radial and transverse velocity.

Radial velocity 

Radial acceleration 

Now,  and 

 => 0\)

 and 

 

If  for some choices of parameters.

So, the correct answer is  -0\) however,  for some choices of parameters.

The Hamiltonian of a system with two degrees of freedom is H = q1p1 - q2p2 , where a > 0 is a constant. The function q1q2 + λp1P2 is a constant of motion only if λ is

  1. 0
  2. 1
  3. -a
  4. a

Answer (Detailed Solution Below)

Option 1 : 0

Classical Mechanics Question 13 Detailed Solution

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Concept:

The Hamiltonian of a system specifies its total energy—i.e., the sum of its kinetic energy (that of motion) and its potential energy (that of position)—in terms of the Lagrangian function derived in earlier studies of dynamics and of the position and momentum of each of the particles.

Calculation:

H = q1p1 - q2p2 +  , where a > 0 is a constant.

f = (q1q2 + λp1p2)

 = [f,H] + 

 = 0 ⇒  = [f,H] = 0

[f,H] = 

q2 . q1 - λ p2 (p1 + 2aq1) + q1(-q2) - λ p1(-p2) = 0

∴ λ = 0

The correct answer is option (1).

The Hamiltonian of a two particle system is H = p1p2 + q1q2, where q1 and q2 are generalized coordinates and p1 and p2 are the respective canonical momenta. The Lagrangian of this system is

Answer (Detailed Solution Below)

Option 4 :

Classical Mechanics Question 14 Detailed Solution

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Concept:

We will use the relationship of Hamiltonian and Lagrangian which is given by

  • For a two-particle system, we can write the above equation as 

 

Explanation:

Given, 

  •  (given)

substitute value of H, we get,

  • --------------------------------1
  • Using Hamilton equations, 
  •  and 
  • Put the value of  and  in equation 1, to get the value of Lagrangian
  •  

 

So, the correct answer is .

 

A system of two identical masses connected by identical springs, as shown in the figure, oscillates along the vertical direction.


The ratio of the frequencies of the normal modes is

Answer (Detailed Solution Below)

Option 1 :

Classical Mechanics Question 15 Detailed Solution

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Concept:

We will first write lagrangian for a given condition then we use the equation for normal nodes which is given by 

Explanation:

Given m are two identical masses, k is spring constant and  and are displacement of spring first and spring second respectively.

  • Using matrix we can write operators of  and  as,
  •  and 
  • Put these values in equation- to get the ratio of normal modes of frequencies,
  • Solution of above equation is 
  • ratio of both frequencies 

 

So, the correct answer is  

 

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