Classical Mechanics — Part 2 MCQs with Explanations
Practice high-yield Physics & Applied Physics MCQs on Classical Mechanics — Part 2 with worked academic solutions for MDCAT, CSS, PMS, PPSC, FPSC.
Questions
25
Difficulty Level
Exam-Level
Target Exams
MDCAT, CSS, PMS...
Estimated Time
30 mins
Exam-Level MDCAT CSS PMS PPSC FPSC
Verified for Commission Syllabus • Oct 14, 2026
25
Verified Practice Questions
Interactive test mode with detailed academic worked solutions
Live Progress|0/25 answered
✓0 correct
✗0 wrong
0% score
🎉 Set complete! All 25 questions answered.
Question 1 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
Consider two particles with position vectors $$\overrightarrow {{{\bf{r}}_1}} $$ and $$\overrightarrow {{{\bf{r}}_2}} $$ . The force exerted by particle 2 on particle 1 is $$\overrightarrow {\bf{F}} \left( {\overrightarrow {{{\bf{r}}_1}} ,\,\overrightarrow {{{\bf{r}}_2}} } \right) = \left( {{{{\bf{\dot r}}}_2} - {{{\bf{\dot r}}}_1}} \right)\left( {{r_2} - {r_1}} \right).$$ The force is
Correct Answer: Option D — non-central and non-conservative
Detailed Academic Explanation:
non-central and non-conservative is the correct answer to “Consider two particles with position vectors $$\overrightarrow {{{\bf{r}}_1}} $$ and $$\overrightarrow {{{\bf{r}}_2}} $$ . The force exerted by particle 2 on particle 1 is $$\overrightarrow {\bf{F}} \left( {\overrightarrow {{{\bf{r}}_1}} ,\,\overrightarrow {{{\bf{r}}_2}} } \right) = \left( {{{{\bf{\dot r}}}_2} - {{{\bf{\dot r}}}_1}} \right)\left( {{r_2} - {r_1}} \right).$$ The force is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 2 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A particle constrained to move along the X-axis in a potential V = kx 2 , is subjected to an external time dependent force $$\overrightarrow {\bf{F}} \left( t \right).$$ Here, k is a constant, x, the distance from the origin and t is the time. At some time T, when the particle has zero velocity at x = 0, the external force is removed. The particle will
Correct Answer: Option D — remain at rest
Detailed Academic Explanation:
remain at rest is the correct answer to “A particle constrained to move along the X-axis in a potential V = kx 2 , is subjected to an external time dependent force $$\overrightarrow {\bf{F}} \left( t \right).$$ Here, k is a constant, x, the distance from the origin and t is the time. At some time T, when the particle has zero velocity at x = 0, the external force is removed. The particle will”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 3 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A system of four particles is in X-Y plane of these two particles of masses m are located at (1, 1) and (-1, 1). The remaining two particles each of mass 2m are located at (1, 1) and (-1, -1). The xy component of moment of inertia tensor of the system of particles is
Correct Answer: Option C — 2m
Detailed Academic Explanation:
2m is the correct answer to “A system of four particles is in X-Y plane of these two particles of masses m are located at (1, 1) and (-1, 1). The remaining two particles each of mass 2m are located at (1, 1) and (-1, -1). The xy component of moment of inertia tensor of the system of particles is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 4 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A particle of mass m is constrained to move on the plane curve xy = c (c > 0) under gravity (Y-axis vertical). The Lagrangian of the particle is given by
$$\frac{1}{2}m{\dot x^2}\left( {1 + \frac{{{c^2}}}{{{x^4}}}} \right) - \frac{{mgc}}{x}$$ is the correct answer to “A particle of mass m is constrained to move on the plane curve xy = c (c > 0) under gravity (Y-axis vertical). The Lagrangian of the particle is given by”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 5 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
An electron gains energy so that its mass becomes 2m 0 . Its speed is
Correct Answer: Option A — $$\frac{{\sqrt 3 }}{2}c$$
Detailed Academic Explanation:
$$\frac{{\sqrt 3 }}{2}c$$ is the correct answer to “An electron gains energy so that its mass becomes 2m 0 . Its speed is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 6 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
Let (p, q) and (P, Q) be two pairs of canonical variables. The transformation Q = q α cos βp, P = q α sin βp is canonical for
α = $$\frac{1}{2}$$, β = 2 is the correct answer to “Let (p, q) and (P, Q) be two pairs of canonical variables. The transformation Q = q α cos βp, P = q α sin βp is canonical for”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 7 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
Two bodies of masses m and 2m are connected by spring constant, the frequency of normal mode is
Correct Answer: Option A — $$\sqrt {\frac{{3k}}{{2m}}} $$
Detailed Academic Explanation:
$$\sqrt {\frac{{3k}}{{2m}}} $$ is the correct answer to “Two bodies of masses m and 2m are connected by spring constant, the frequency of normal mode is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 8 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A heavy symmetrical top is rotating about its own axis of symmetry (Z-axis). If $${I_1},\,{I_2}$$ and $${I_3}$$ are the principal moments of inertia along X, Y and Z axes respectively then
Correct Answer: Option C — $${I_1} = {I_2},\,{I_1} \ne {I_3}$$
Detailed Academic Explanation:
$${I_1} = {I_2},\,{I_1} \ne {I_3}$$ is the correct answer to “A heavy symmetrical top is rotating about its own axis of symmetry (Z-axis). If $${I_1},\,{I_2}$$ and $${I_3}$$ are the principal moments of inertia along X, Y and Z axes respectively then”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 9 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A hoop is pivoted at a point on circumference. The period of small oscillations in the plane of loop is
Correct Answer: Option A — $$2\pi \sqrt {\frac{{2R}}{g}} $$
Detailed Academic Explanation:
$$2\pi \sqrt {\frac{{2R}}{g}} $$ is the correct answer to “A hoop is pivoted at a point on circumference. The period of small oscillations in the plane of loop is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 10 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
The Lagrangian of a system is given by $$L = \frac{1}{2}{\dot q^2} + q\dot q - \frac{1}{2}{q^2}$$ It describes motion of a
Correct Answer: Option A — harmonic oscillator
Detailed Academic Explanation:
harmonic oscillator is the correct answer to “The Lagrangian of a system is given by $$L = \frac{1}{2}{\dot q^2} + q\dot q - \frac{1}{2}{q^2}$$ It describes motion of a”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 11 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A particle of mass M moving in a straight line with speed v collides with a stationary particle of the same mass. In the centre of mass coordinate system, the first particle is deflected by 90°. The speed of second particle after collision in laboratory system will be
Correct Answer: Option A — $$\frac{v}{{\sqrt 2 }}$$
Detailed Academic Explanation:
$$\frac{v}{{\sqrt 2 }}$$ is the correct answer to “A particle of mass M moving in a straight line with speed v collides with a stationary particle of the same mass. In the centre of mass coordinate system, the first particle is deflected by 90°. The speed of second particle after collision in laboratory system will be”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 12 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
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
Correct Answer: Option A — $$\frac{1}{7}a$$
Detailed Academic Explanation:
$$\frac{1}{7}a$$ is the correct answer to “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”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 13 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
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
Correct Answer: Option B — $$\frac{m}{2}{\left( {\frac{{dq}}{{dt}} - q} \right)^2}$$
Detailed Academic Explanation:
$$\frac{m}{2}{\left( {\frac{{dq}}{{dt}} - q} \right)^2}$$ is the correct answer to “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”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 14 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A rod of length L 0 makes an angle θ 0 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
$$\theta = {\tan ^{ - 1}}\left( {\gamma \cot {\theta _0}} \right)$$ is the correct answer to “A rod of length L 0 makes an angle θ 0 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”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 15 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A particle of mass m moves in a potential V(x) = $$\frac{1}{2}$$ mω 2 x 2 + $$\frac{1}{2}$$ mμv 2 , where x is the position coordinate, v is the speed and ω, μ are constants. The canonical momentum of the particle is
Correct Answer: Option D — p = (1 - μ)mv
Detailed Academic Explanation:
p = (1 - μ)mv is the correct answer to “A particle of mass m moves in a potential V(x) = $$\frac{1}{2}$$ mω 2 x 2 + $$\frac{1}{2}$$ mμv 2 , where x is the position coordinate, v is the speed and ω, μ are constants. The canonical momentum of the particle is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 16 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
For a particle moving in a central field,
Correct Answer: Option C — the motion is confined in a plane
Detailed Academic Explanation:
the motion is confined in a plane is the correct answer to “For a particle moving in a central field,”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 17 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
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-m 2 /s, is
Correct Answer: Option B — -80 $${\bf{\hat k}}$$
Detailed Academic Explanation:
-80 $${\bf{\hat k}}$$ is the correct answer to “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-m 2 /s, is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 18 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
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
Correct Answer: Option A — $$m\dot r{\text{ and }}m{r^2}\dot \theta $$
Detailed Academic Explanation:
$$m\dot r{\text{ and }}m{r^2}\dot \theta $$ is the correct answer to “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”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 19 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
The Lagrangian of a particle of mass m moving in a plane is given by L = $$\frac{1}{2}$$ [m(v x 2 + v y 2 )] + 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
Correct Answer: Option C — px = mvx - ay and py = mvy + ax
Detailed Academic Explanation:
px = mvx - ay and py = mvy + ax is the correct answer to “The Lagrangian of a particle of mass m moving in a plane is given by L = $$\frac{1}{2}$$ [m(v x 2 + v y 2 )] + 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”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 20 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
Three particles of mass m each situated at x 1 (t), x 2 (t) and x 3 (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
$$L = \frac{m}{2}\left[ {{{\left( {\frac{{d{x_1}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_2}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{x_3}}}{{dt}}} \right)}^2}} \right] - \frac{k}{2}{\left( {{x_1} - {x_2} - l} \right)^2} + \frac{k}{2}\left( {{x_3} - {x_2} - l} \right)$$ is the correct answer to “Three particles of mass m each situated at x 1 (t), x 2 (t) and x 3 (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”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 21 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
The Lagrangian of a particle' of mass m is $$L = \frac{m}{2}\left[ {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} \right] - \frac{V}{2}\left( {{x^2} + {y^2}} \right) + W\sin \omega t$$ where V, W and ω are constants, then conserved quantities are
Correct Answer: Option A — energy and z-component of linear momentum
Detailed Academic Explanation:
energy and z-component of linear momentum is the correct answer to “The Lagrangian of a particle' of mass m is $$L = \frac{m}{2}\left[ {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} \right] - \frac{V}{2}\left( {{x^2} + {y^2}} \right) + W\sin \omega t$$ where V, W and ω are constants, then conserved quantities are”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 22 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
The Lagrangian of a diatomic molecule is given by $$L = \frac{m}{2}\left( {\dot x_1^2 + \dot x_2^2} \right) - \frac{k}{2}{x_1}{x_2}$$ where, m is the mass of each atom and x 1 , x 2 are displacements from equilibrium position and k > 0. The normal frequencies are
Correct Answer: Option D — $$ \pm {\left( {\frac{k}{{2m}}} \right)^{\frac{1}{2}}}$$
Detailed Academic Explanation:
$$ \pm {\left( {\frac{k}{{2m}}} \right)^{\frac{1}{2}}}$$ is the correct answer to “The Lagrangian of a diatomic molecule is given by $$L = \frac{m}{2}\left( {\dot x_1^2 + \dot x_2^2} \right) - \frac{k}{2}{x_1}{x_2}$$ where, m is the mass of each atom and x 1 , x 2 are displacements from equilibrium position and k > 0. The normal frequencies are”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 23 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A particle of charge q, mass m and linear momentum $$\overrightarrow {\bf{p}} $$ enters an electromagnetic field of vector potential $$\overrightarrow {\bf{A}} $$ and scalar potential $$\phi $$. The Hamiltonian of particle is
$$\frac{1}{{2m}}{\left( {\overrightarrow {\bf{p}} - \frac{q}{c}\overrightarrow {\bf{A}} } \right)^2} + q\phi $$ is the correct answer to “A particle of charge q, mass m and linear momentum $$\overrightarrow {\bf{p}} $$ enters an electromagnetic field of vector potential $$\overrightarrow {\bf{A}} $$ and scalar potential $$\phi $$. The Hamiltonian of particle is”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 24 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
A mass m is connected on either side with a spring each of spring constants k 1 and k 2 . The free ends of springs are tied to rigid supports. The displacement of the mass is x from equilibrium position. Which one of the following is true?
Correct Answer: Option B — The angular momentum of the mass is zero about the equilibrium point and its Lagrangian $$\frac{1}{2}m{{\dot x}^2} - \frac{1}{2}\left( {{k_1} + {k_2}} \right){x^2}$$
Detailed Academic Explanation:
The angular momentum of the mass is zero about the equilibrium point and its Lagrangian $$\frac{1}{2}m{{\dot x}^2} - \frac{1}{2}\left( {{k_1} + {k_2}} \right){x^2}$$ is the correct answer to “A mass m is connected on either side with a spring each of spring constants k 1 and k 2 . The free ends of springs are tied to rigid supports. The displacement of the mass is x from equilibrium position. Which one of the following is true”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
Question 25 of 25
★ MDCAT
★ CSS
★ MDCAT
★ CSS
If a particle moves outwards in a plane along a curved trajectory described by r = aθ, θ = ωt, where a and ω are constants, then its
Correct Answer: Option B — angular momentum is conserved
Detailed Academic Explanation:
angular momentum is conserved is the correct answer to “If a particle moves outwards in a plane along a curved trajectory described by r = aθ, θ = ωt, where a and ω are constants, then its”. This question is standard in competitive examination syllabi for Classical Mechanics.
💡Key Conceptual Anchor:Classical Mechanics
✓Link copied to clipboard
Continue Your Practice
Atomic And Molecular Physics — Part 1
Master the next curriculum module with 40 exam-standard MCQs.
💡Examiner's Advisory for Physics & Applied Physics
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.