Classical Mechanics MCQs Question and Answer in Engineering Physics

51

In an electromagnetic field, which one of the following remains invariant under Lorentz transformation?

A) $$\overrightarrow {\bf{E}} \times \overrightarrow {\bf{B}} $$ B) E² - c²B² C) B² D) E²

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Answer & Solution

Answer: Option B

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52

A car is moving with constant linear acceleration a along horizontal X-axis. A solid sphere of mass M and radius R is found rolling without slipping on the horizontal floor of the car in the same direction as seen from an inertial frame outside the car. The acceleration of the sphere in the inertia frame is

A) $$\frac{1}{7}a$$ B) $$\frac{2}{7}a$$ C) $$\frac{3}{7}a$$ D) $$\frac{5}{7}a$$

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Answer: Option A

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53

The Hamiltonian of a particle is $$H = \frac{{{p^2}}}{{2m}} + pq,$$ where q is generalised coordinate and p is the corresponding canonical momentum. The Lagrangian is

A) $$\frac{m}{2}{\left( {\frac{{dq}}{{dt}} + q} \right)^2}$$ B) $$\frac{m}{2}{\left( {\frac{{dq}}{{dt}} - q} \right)^2}$$ C) $$\frac{m}{2}\left[ {{{\left( {\frac{{dq}}{{dt}}} \right)}^2} + q\frac{{dq}}{{dt}} - {q^2}} \right]$$ D) $$\frac{m}{2}\left[ {{{\left( {\frac{{dq}}{{dt}}} \right)}^2} - q\frac{{dq}}{{dt}} + {q^2}} \right]$$

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Answer & Solution

Answer: Option B

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54

A rod of length L₀ makes an angle θ₀ with the Y-axis in its rest frame while the rest frame moves to the right along the X-axis with relativistic speed v with respect to lab frame. If $$\gamma = {\left( {1 - \frac{{{v^2}}}{{{c^2}}}} \right)^{ - \frac{1}{2}}},$$ the angle in the lab frame is

A) $$\theta = {\tan ^{ - 1}}\left( {\gamma \tan {\theta _0}} \right)$$ B) $$\theta = {\tan ^{ - 1}}\left( {\gamma \cot {\theta _0}} \right)$$ C) $$\theta = {\tan ^{ - 1}}\left( {\frac{{\tan {\theta _0}}}{\gamma }} \right)$$ D) $$\theta = {\tan ^{ - 1}}\left( {\frac{{\cot {\theta _0}}}{\gamma }} \right)$$

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Answer & Solution

Answer: Option B

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55

A particle of mass m moves in a potential V(x) = $$\frac{1}{2}$$ mω²x² + $$\frac{1}{2}$$ mμv², where x is the position coordinate, v is the speed and ω, μ are constants. The canonical momentum of the particle is

A) p = m(1 + μ)v B) p = mv C) p = mμv D) p = (1 - μ)mv

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Answer & Solution

Answer: Option D

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56

For a particle moving in a central field,

A) the kinetic energy is a constant of motion B) the potential energy is velocity dependent C) the motion is confined in a plane D) the total energy is not conserved

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Answer & Solution

Answer: Option C

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57

A particle of mass 2 kg is moving such that at time t second. Its position in metre is given by $$\overrightarrow {\bf{r}} \left( t \right) = 5{\bf{\hat i}} - 2{t^2}{\bf{\hat j}}.$$ The angular momentum of the particle at t = 2 s about the origin in kg-m²/s, is

A) -40 $${\bf{\hat k}}$$ B) -80 $${\bf{\hat k}}$$ C) 80 $${\bf{\hat k}}$$ D) 40 $${\bf{\hat k}}$$

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Answer & Solution

Answer: Option B

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58

The Lagrangian of a particle moving in a plane under the influence of central potential is given by $$L = \frac{1}{2}m\left( {{{\dot r}^2} = {r^2}{{\dot \theta }^2}} \right) - V\left( r \right).$$ The generalised momenta corresponding to r and θ are given by

A) $$m\dot r{\text{ and }}m{r^2}\dot \theta $$ B) $$m\dot r{\text{ and }}mr\dot \theta $$ C) $$m{{\dot r}^2}{\text{ and }}m{r^2}\dot \theta $$ D) $$m{{\dot r}^2}{\text{ and }}m{r^2}{{\dot \theta }^2}$$

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Answer & Solution

Answer: Option A

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59

The Lagrangian of a particle of mass m moving in a plane is given by L = $$\frac{1}{2}$$ [m(v_x² + v_y²)] + a(xv_y - yv_x) where, v_x and v_y are velocity components and a is constant, The canonical momenta of the particle are given by

A) p_x = mv_x and p_y = mv_y B) p_x = mv_x + ay and p_y = mv_y + ax C) p_x = mv_x - ay and p_y = mv_y + ax D) p_x = mv_x - ay and p_y = mv_y - ax

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Answer & Solution

Answer: Option C

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60

Three particles of mass m each situated at x₁(t), x₂(t) and x₃(t) respectively are connected by two spring constants k and unstretched lengths $$l$$. The system is free to oscillate only in one-dimension along the straight line joining all the three particles. The Lagrangian of the system is

A) $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - l} \right)^2} + \frac{k}{2}\left( {{x_3} - {x_2} - l} \right)$$ B) $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_3} - l} \right)^2} + \frac{k}{2}\left( {{x_3} - {x_2} - l} \right)$$ C) $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - l} \right)^2} - \frac{k}{2}{\left( {{x_3} - {x_2} + l} \right)^2}$$ D) $$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - L} \right)^2} - \frac{k}{2}{\left( {{x_3} - {x_2} - l} \right)^2}$$

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Answer & Solution

Answer: Option A

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