Classical Mechanics MCQs Question and Answer in Engineering Physics Page-3 section-1

21.
For the given transformation
1. Q = p and P = -q
2. P = q and Q = p
where p, q are canonically conjugate variables, which one of the following statements is true?

A. Both 1 and 2 are canonical

B. Only 1 is canonical

C. Only 2 is canonical

D. Neither 1 nor 2 is canonical

Answer & Solution Discuss in Board Save for Later

22.
The Lagrangian for a simple pendulum is given by $$L = \frac{1}{2}m{L^2}{{\dot \theta }^2} - mgL\left( {1 - \cos \theta } \right)$$
Hamiltonian equations are given by

A. $${{\dot p}_\theta } = - mgL\,\sin \theta ,\,\dot \theta = \frac{{{p_\theta }}}{{m{L^2}}}$$

B. $${{\dot p}_\theta } = mgL\,\sin ,\,\dot \theta = \frac{{{p_\theta }}}{{m{L^2}}}$$

C. $${p_\theta } = - m\dot \theta ,\,\dot \theta = \frac{{{p_\theta }}}{m}$$

D. $${{\dot p}_\theta } = - \left( {\frac{g}{L}} \right)\theta ,\,\dot \theta = \frac{{{p_\theta }}}{{mL}}$$

Answer & Solution Discuss in Board Save for Later

23.
Two events are separated by a distance of 6 × 10⁵ km and the first event occurs 1 s before the second event. The interval between two events is

A. time like

B. light like

C. space like

D. Cannot be determined from the information given

Answer & Solution Discuss in Board Save for Later

24.
The lagrangian of a particle of mass m moving in one dimension is $$L = \exp \left[ {\left( {\alpha t} \right)\frac{{M{{\dot x}^2}}}{2} - k{x^2}} \right]$$ where, $$\alpha $$ and k are positive constants. The equation of motion of the particle is

A. $$\ddot x + \alpha \dot x = 0$$

B. $$\ddot x + \frac{k}{m}x = 0$$

C. $$\ddot x - \alpha \dot x + \frac{k}{m}x = 0$$

D. $$\ddot x + \alpha \dot x + \frac{k}{m}x = 0$$

Answer & Solution Discuss in Board Save for Later

25.
Although mass energy equivalence of special relativity allows conversion of a photon to an electron-positron pair, such a process cannot occur in free space because

A. mass is not conserved

B. energy is not conserved

C. momentum is not conserved

D. charge is not conserved

Answer & Solution Discuss in Board Save for Later

26.
The Lagrangian forthe Kepler problem is given by $$L = \frac{1}{2}\left[ {{{\dot r}^2} + {r^2}{{\dot \theta }^2}} \right] + \frac{\mu }{r}\,\,\,\,\,\,\,\left( {\mu > 0} \right)$$
where, $$\left( {r,\,\theta } \right)$$ denotes the polar coordinates and mass of the particle is unity, then

A. $${p_\theta } = 2{r^2}\dot \theta $$

B. $${p_r} = 2\dot r$$

C. the angular momentum of the particle about the centre of attraction is a constant

D. the total energy of the particle is time dependent

Answer & Solution Discuss in Board Save for Later

27.
For a simple harmonic oscillator, the Lagrangian is $$L = \frac{1}{2}{{\dot q}^2} - \frac{1}{2}{q^2},\,{\text{if }}A\left( {p,\,q} \right) = \frac{{p + iq}}{2}$$ and H(p, q) is the Hamiltonian of the system, the Poisson bracket, {A(p, q), H(p, q)} is given by

A. iA (p, q)

B. A^∗ (p, q)

C. -iA^∗ (p, q)

D. -iA (p, q)

Answer & Solution Discuss in Board Save for Later

28.
A bead of mass m slides along a straight frictionless rigid wire rotating in a horizontal plane with a constant angular speed ω. The axis of rotation is perpendicular to the wire and passes through one end of the wire. If r is the distance of the mass from the axis of rotation and v is its speed, then the magnitude of the Coriolis force is

A. $$\frac{{m{v^2}}}{r}$$

B. $$\frac{{2m{v^2}}}{r}$$

C. mvω

D. 2mvω

Answer & Solution Discuss in Board Save for Later

29.
If for a system of N particles of different masses m₁, m₂, . . . m_N with position vectors $${\overrightarrow {\bf{r}} _1},\,{\overrightarrow {\bf{r}} _2},\,.\,.\,.\,{\overrightarrow {\bf{r}} _N}$$ and corresponding velocities $${\overrightarrow {\bf{v}} _1},\,{\overrightarrow {\bf{v}} _2},\,.\,.\,.\,{\overrightarrow {\bf{v}} _N}$$ respectively such that $$\sum\limits_i {\overrightarrow {{{\bf{v}}_i}} = 0,} $$ then

A. total momentum must be zero

B. total angular momentum must be independent of the choice of the origin

C. the total force on the system must be zero

D. the total torque on the system must be zero

Answer & Solution Discuss in Board Save for Later

30.
The Lagrangian of a free particle in spherical polar coordinates is given by $$L = \frac{1}{2}m\left[ {{{\dot r}^2} + r{{\dot \theta }^2} + {r^2}{{\dot \phi }^2}{{\sin }^2}\theta } \right]$$
The quantity that conserved is

A. $$\frac{{\partial L}}{{\partial \dot r}}$$

B. $$\frac{{\partial L}}{{\partial \dot \theta }}$$

C. $$\frac{{\partial L}}{{\partial \dot \phi }}$$

D. $$\frac{{\partial L}}{{\partial \dot \phi }} + \dot r\dot \theta $$

Answer & Solution Discuss in Board Save for Later

Classical Mechanics MCQs Question and Answer in Engineering Physics

21. For the given transformation 1. Q = p and P = -q 2. P = q and Q = p where p, q are canonically conjugate variables, which one of the following statements is true?

Answer & Solution

22. The Lagrangian for a simple pendulum is given by $$L = \frac{1}{2}m{L^2}{{\dot \theta }^2} - mgL\left( {1 - \cos \theta } \right)$$ Hamiltonian equations are given by

Answer & Solution

23. Two events are separated by a distance of 6 × 105 km and the first event occurs 1 s before the second event. The interval between two events is

Answer & Solution

24. The lagrangian of a particle of mass m moving in one dimension is $$L = \exp \left[ {\left( {\alpha t} \right)\frac{{M{{\dot x}^2}}}{2} - k{x^2}} \right]$$ where, $$\alpha $$ and k are positive constants. The equation of motion of the particle is

Answer & Solution

25. Although mass energy equivalence of special relativity allows conversion of a photon to an electron-positron pair, such a process cannot occur in free space because

Answer & Solution

26. The Lagrangian forthe Kepler problem is given by $$L = \frac{1}{2}\left[ {{{\dot r}^2} + {r^2}{{\dot \theta }^2}} \right] + \frac{\mu }{r}\,\,\,\,\,\,\,\left( {\mu > 0} \right)$$ where, $$\left( {r,\,\theta } \right)$$ denotes the polar coordinates and mass of the particle is unity, then

Answer & Solution

27. For a simple harmonic oscillator, the Lagrangian is $$L = \frac{1}{2}{{\dot q}^2} - \frac{1}{2}{q^2},\,{\text{if }}A\left( {p,\,q} \right) = \frac{{p + iq}}{2}$$ and H(p, q) is the Hamiltonian of the system, the Poisson bracket, {A(p, q), H(p, q)} is given by

Answer & Solution

Answer & Solution

Answer & Solution

30. The Lagrangian of a free particle in spherical polar coordinates is given by $$L = \frac{1}{2}m\left[ {{{\dot r}^2} + r{{\dot \theta }^2} + {r^2}{{\dot \phi }^2}{{\sin }^2}\theta } \right]$$ The quantity that conserved is

Answer & Solution

21.
For the given transformation
1. Q = p and P = -q
2. P = q and Q = p
where p, q are canonically conjugate variables, which one of the following statements is true?

22.
The Lagrangian for a simple pendulum is given by $$L = \frac{1}{2}m{L^2}{{\dot \theta }^2} - mgL\left( {1 - \cos \theta } \right)$$
Hamiltonian equations are given by

23.
Two events are separated by a distance of 6 × 10⁵ km and the first event occurs 1 s before the second event. The interval between two events is

24.
The lagrangian of a particle of mass m moving in one dimension is $$L = \exp \left[ {\left( {\alpha t} \right)\frac{{M{{\dot x}^2}}}{2} - k{x^2}} \right]$$ where, $$\alpha $$ and k are positive constants. The equation of motion of the particle is

25.
Although mass energy equivalence of special relativity allows conversion of a photon to an electron-positron pair, such a process cannot occur in free space because

26.
The Lagrangian forthe Kepler problem is given by $$L = \frac{1}{2}\left[ {{{\dot r}^2} + {r^2}{{\dot \theta }^2}} \right] + \frac{\mu }{r}\,\,\,\,\,\,\,\left( {\mu > 0} \right)$$
where, $$\left( {r,\,\theta } \right)$$ denotes the polar coordinates and mass of the particle is unity, then

27.
For a simple harmonic oscillator, the Lagrangian is $$L = \frac{1}{2}{{\dot q}^2} - \frac{1}{2}{q^2},\,{\text{if }}A\left( {p,\,q} \right) = \frac{{p + iq}}{2}$$ and H(p, q) is the Hamiltonian of the system, the Poisson bracket, {A(p, q), H(p, q)} is given by

30.
The Lagrangian of a free particle in spherical polar coordinates is given by $$L = \frac{1}{2}m\left[ {{{\dot r}^2} + r{{\dot \theta }^2} + {r^2}{{\dot \phi }^2}{{\sin }^2}\theta } \right]$$
The quantity that conserved is