Classical Mechanics MCQ Quiz in मल्याळम - Objective Question with Answer for Classical Mechanics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Classical Mechanics ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Classical Mechanics MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Classical Mechanics MCQ Objective Questions

Top Classical Mechanics MCQ Objective Questions

Classical Mechanics Question 1:

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

  1. 1+12αβ=0
  2. 112αβ=0
  3. 1 + 2αβ = 0
  4. 1 - 2αβ = 0

Answer (Detailed Solution Below)

Option 3 : 1 + 2αβ = 0

Classical Mechanics Question 1 Detailed Solution

Explanation:

  • Given the transformations X=αpx and P=βx², 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: X,P=(Xx)(Pp)(Xp)(Px)
  • Substituting X and P from the problem statement, we get: X,P=((αp/x)x)((βx²)p)((αp/x)p)((βx²)x)
  • Computing partial derivatives, we get:X,P=(αpx²)(0)(αx)(2βx)=2αβ
  • Setting the above equation to 1: 2αβ=1
  • Finally, 1+2αβ=0

Classical Mechanics Question 2:

Earth may be assumed to be an axially symmetric freely rotating rigid body. The ratio of the principal moments of inertia about the axis of symmetry and an axis perpendicular to it is 33:32. If T0 is the time taken by earth to make one rotation around its axis of symmetry, then the time period of precession is closest to

  1. 33 T0
  2. 33 T0/2
  3. 32 T0
  4. 16 T0

Answer (Detailed Solution Below)

Option 3 : 32 T0

Classical Mechanics Question 2 Detailed Solution

Explanation:

In this case, we use the formula for the precession of an axially symmetric freely rotating rigid body, caused by the torque resulting from external forces (such as gravity) trying to alter the tilt of the rotation axis. This forms what is known as a gyroscope.

The formula for the precession rate (angular frequency) of a freely spinning body is given by:

 ωprecession=ω0×(IaxisIperpendicular1)

where

  • ω0 is the rotational angular frequency of the earth about its axis of symmetry,
  • Iperpendicular is the moment of inertia of the earth about an axis perpendicular to the axis of symmetry,
  • Iaxis is the moment of inertia of the earth about the axis of symmetry.

Solving for the time period of precession, which is the reciprocal of the precession rate, we get: Tprecession=T0×1(IaxisIperpendicular1)

It is given that the ratio of the moment of inertia about the axis of symmetry Iaxis to the moment of inertia about an axis perpendicular to this Iperpendicular is 33:32, which makes: IaxisIperpendicular=3332

Substituting this into the precession period formula, we get: Tprecession=T0×1(33321) 

Tprecession=32T0

Classical Mechanics Question 3:

A particle of mass m moves under the influence of a force F(x) = -kx^3, where x is the displacement of the particle from its equilibrium position and k is a positive constant. If the particle is initially at rest at x = a, then what is the velocity of the particle when it passes through its equilibrium position?

  1. 2kma2 
  2. k2ma2 
  3. 3kma2 
  4. 2k3ma2 

Answer (Detailed Solution Below)

Option 2 : k2ma2 

Classical Mechanics Question 3 Detailed Solution

Explanation:

We can use Newton's second law of motion to solve this problem.

The equation of motion for the particle is given by:

md2xdt2=kx3 

We can rewrite this equation as:

d2xdt2=kmx3  

We can multiply both sides of this equation by dxdt  to obtain:

ddt(12(dxdt)2)=kmx3dxdt .

We can integrate both sides of this equation with respect to time to obtain:

12(dxdt)2=k4m(a4x4) 

At the equilibrium position x = 0, we have:

12(dxdt)2=k4ma4 

Therefore, the velocity of the particle when it passes through its equilibrium position is:

|dxdt|=k2ma2. 

Therefore, the correct option is (2).

Classical Mechanics Question 4:

A ball bouncing of a rigid floor is described by the potential energy function

V(x) = {mgx for x>0 for x0

Which of the following schematic diagrams best represents the phase space plot of the ball?

  1. F1 Vinanti Teaching 03.03.23 D1
  2. F1 Vinanti Teaching 03.03.23 D2
  3. F1 Vinanti Teaching 03.03.23 D3
  4. F1 Vinanti Teaching 03.03.23 D4

Answer (Detailed Solution Below)

Option 2 : F1 Vinanti Teaching 03.03.23 D2

Classical Mechanics Question 4 Detailed Solution

Concept:

Phase - space diagram or phase plot is a plot between generalized coordinate x and generalized momentum p.

Calculation:

The Lagrangian and Hamiltonian for the system is given as follows:

L = T - V

12mx˙2mgx

H = p22m+mgx 

E = p22m + mgx

⇒ p2 = 2m(E − mgx) which is equation of parabola.

At x = 0,

p = ±2mE

At p = 0,

x = Emg

For x < 0, when the potential energy is ∞ then the kinetic energy will be 0.

The correct answer is option (2).

Classical Mechanics Question 5:

The Lagrangian of a system is given by

L = 12 ml2[θ˙2 + sin2θφ˙2] − mgl cos θ, where m, l and g are constants.

Which of the following is conserved?

  1. φ˙sin2θ
  2. φ˙sinθ
  3. φ˙sinθ
  4. φ˙sin2θ

Answer (Detailed Solution Below)

Option 1 : φ˙sin2θ

Classical Mechanics Question 5 Detailed Solution

Concept:

Lagrangian mechanics enables us to find the equations of motion when the Newtonian method is proving difficult.

In Lagrangian mechanics we start, as usual, by drawing a large, clear diagram of the system, using a ruler and a compass.

Explanation:

As φ is cyclic coordinate, so Lφ˙ = pφ =ml2sin2θϕ˙, is a constant since m, l and g are constants.

Now using the Euler Lagrangian equation for motion we get:

ddt(Lϕ˙)=0.

Thus,  ml2sin2θϕ˙ is conserved.

The correct answer is option (1).

Classical Mechanics Question 6:

The relativistic form of Newton's second law of motion is

  1. F = mcc2v2dvdt
  2. F = mc2v2cdvdt
  3. F = mc2c2v2dvdt
  4. F = mc2v2c2dvdt

Answer (Detailed Solution Below)

Option 3 : F = mc2c2v2dvdt

Classical Mechanics Question 6 Detailed Solution

Explanation:

We known that:

P = mv1v2c2 ⇒ F = dPdtmdvdt11v2c2+mv(12)1(1v2c2)3/22vc2dvdt

⇒ F = mdvdt11v2c2(1+12v2/c2(1v2c2))=mdvdt(1v2/2c2(1v2c2)3/2)

mdvdt[(1v2/c2)1/2(1v2/c2)(1v2/c2)1/2]=mc2(c2v2)dvdt

The correct option is option (3).

Classical Mechanics Question 7:

A particle of mass m slides under the gravity without friction along the parabolic path y = ax2, as shown in the figure. Here a is a constant.

F2 Arbaz Teaching 04-05-2023 D20

The Lagrange’s equation of motion of the particle for above question is given by

  1. ẍ = 2gax
  2. m(1 + 4a2x2)ẍ = −2mgax − 4ma2xẋ2
  3. m(1 + 4a2x2)ẍ = 2mgax + 4ma2xẋ2
  4. ẍ = −2gax

Answer (Detailed Solution Below)

Option 2 : m(1 + 4a2x2)ẍ = −2mgax − 4ma2xẋ2

Classical Mechanics Question 7 Detailed Solution

Concept:

Lagrange's equation of motion is a mathematical equation that is used to describe the motion of a physical system in terms of its Lagrangian function.

The Lagrangian function is a mathematical function that describes the energy of a system in terms of its position and velocity.

Explanation:

The Euler equation of motion is given as:

ddt(dLdx˙)=dLdx ⇒ mẍ(1 + 4a2x2) = −4ma2xẋ2 − 2mgax.

The correct option is option (2).

Classical Mechanics Question 8:

A particle of mass m slides under the gravity without friction along the parabolic path y = ax2, as shown in the figure. Here a is a constant.

F2 Arbaz Teaching 04-05-2023 D20

The Lagrangian for this particle is given by

  1. L = 12mẋ2 − mgax2
  2. 12m(1 + 4a2x2)ẋ2 − mgax2
  3. 12m2 + mgax2
  4. 12m(1 + 4a2x2)ẋ2 + mgax2

Answer (Detailed Solution Below)

Option 2 : 12m(1 + 4a2x2)ẋ2 − mgax2

Classical Mechanics Question 8 Detailed Solution

Explanation:

Equation of constrain is given by y = ax2, K.E., T = 12m(ẋ2 + ẏ2)

ẏ = 2axẋ ⇒ T = 12m(ẋ2 + 4a2x22) = 12mẋ2(1 + 4ax2)

V = mgy = mgax2.

∵ L = T − V ⇒ L = 12m(1 + 4a2x2)ẋ2 − mgax2   

The correct option is option (2).

Classical Mechanics Question 9:

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

  1. xx¨yy¨
  2. xy¨yx¨
  3. xy˙yx˙
  4. yx˙2+xy˙2

Answer (Detailed Solution Below)

Option 2 : xy¨yx¨

Classical Mechanics Question 9 Detailed Solution

Concept:

The Lagranges equation of motion of a system is given by

ddtLq˙Lq=0

Calculation:

The Lagrangian L depends on 

L(x,y,̇x,̇y)

ddtLx˙Lx=0

ddtLy˙Ly=0

L' = L(x,y,x˙,y˙)

ddtLxLx=ddtLxLx+y¨=0

y˙=c1

Similarly x˙=c2

The correct answer is option (2).

Classical Mechanics Question 10:

A particle of mass m moves in a potential that is V=12m(ω12x2+ω22y2+ω32z2) in the coordinates of a non-inertial frame F. The frame F is rotating with respect to an inertial frame with an angular velocity k̂Ω, where k̂ is the unit vector along their common z - axis. The motion of the particle is unstable for all angular frequencies satisfying

  1. (Ω2ω12)(Ω2ω22)>0
  2. (Ω2ω12)(Ω2ω22)<0
  3. (Ω2(ω1+ω2)2)(Ω2|ω1ω2|2)>0
  4. (Ω2(ω1+ω2)2)(Ω2|ω1ω2|2)<0

Answer (Detailed Solution Below)

Option 4 : (Ω2(ω1+ω2)2)(Ω2|ω1ω2|2)<0

Classical Mechanics Question 10 Detailed Solution

CONCEPT:

As we know,

F=V

CALCULATION:

Given: V=12m(ω12x2+ω22y2+ω32z2)

As we have;

F=V

⇒ F=m(ω12xi^+ω22yj^+ω32zk^)

Now the centrifugal force on the particle is written as;

Fcf=mω×(ω×r)

Here, r=xi^+yj^+zk^  and ω=k^Ω

so we have;

ω×r=Ω(yi^xj^)

⇒ Fcf=mω×Ω(yi^xj^)

⇒ Fcf=mΩ2[k^×(yi^xj^)]

Fcf=mΩ2(xi^+yj^)

Since, F>Fcf

m(ω12xi^+ω22yj^+ω32zk^)>mΩ2(xj^+yi^)

Now, compare the x,y, and z components we have;

ω12x>Ω2x

Ω2ω12<0

Similarly, Ω2ω22<0

and ω32>0

Combining we have;

(Ω2ω12)(Ω2ω22)<0

Hence, option 2 is the correct answer.

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