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Chapter 7 Lagrangian Mechanics Previous Year Questions with Solutions

Pravegaa PYQ Series · Volume 2: Classical Mechanics & Special Relativity

CM07: Lagrangian Mechanics

46 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 1
  • GATE: 8
  • CSIR NET: 14
  • JEST: 15
  • TIFR GS: 8
CM07.1IIT JAM 2026, Q12MCQ | 2 marks

A particle of mass \(m\) moves in a potential given by \[ V(x,y,z)=-k\,\frac{y}{(x^2+y^2+z^2)} \] where \(k\) is a constant. Which of the following statements is correct?

  1. The force corresponding to the potential is central

  2. Angular momentum of the system is not conserved

  3. Linear momentum along the \(y\)-direction is conserved

  4. Energy of the system is not conserved

Show solution

In spherical polar coordinates \(y=r\sin\theta\sin\phi\), so \[ V=-\frac{k\sin\theta\sin\phi}{r}, \] which depends on the direction \((\theta,\phi)\) and not on \(r\) alone. The force is therefore not central (A is false), and the torque \(\vec{r}\times\vec{F}=-\vec{r}\times\nabla V\neq 0\), so angular momentum is not conserved (B is true).

\(\partial V/\partial y\neq0\), so \(F_y\neq 0\) and \(p_y\) is not conserved (C is false). \(V\) has no explicit time dependence, so energy is conserved (D is false). \[ \boxed{\text{Angular momentum is not conserved}} \]

Answer: (B) Angular momentum of the system is not conserved

CM07.2GATE 2026, Q28NAT | 1 mark

A projectile of mass \(m\) is launched from the ground with the initial speed \(v_0\) at an angle \(30^\circ\) from the horizontal. Take the ground to be horizontal. Ignoring the drag, the magnitude of Hamilton's action \(\int L\,dt\) for the particle from the beginning till it hits the ground is \(f\times\left(\dfrac{mv_0^3}{g}\right)\). The value of \(f\) (rounded off to two decimal places) is

Show solution

\(v_x=\frac{\sqrt3}{2}v_0\), \(v_y=\frac12v_0-gt\), \(y=\frac12v_0t-\frac12gt^2\); the time of flight is \(\tau=v_0/g\). Work in units \(m=v_0=g=1\) (so \(\tau=1\)): \begin{gather*} \int_0^1\tfrac12(v_x^2+v_y^2)\,dt=\tfrac12\cdot\tfrac34+\tfrac12\int_0^1\left(\tfrac12-t\right)^2dt=\tfrac38+\tfrac1{24}=\tfrac5{12},\\ \int_0^1 y\,dt=\int_0^1\left(\tfrac t2-\tfrac{t^2}{2}\right)dt=\tfrac14-\tfrac16=\tfrac1{12}. \end{gather*} \[ S=\int(T-V)\,dt=\frac5{12}-\frac1{12}=\frac13\;\Rightarrow\;\boxed{f=\tfrac13\approx0.33} \]

Answer: 0.33 (range 0.32 to 0.34)

CM07.3GATE 2026, Q46MCQ | 2 marks

The Lagrangian \(L_0=\frac12m\dot q^2-\frac12m\omega^2q^2\)

with the generalized coordinate \(q\) is transformed to \(L=L_0+\alpha\dfrac{df(q)}{dt}\). Consider the following statements:

(i) Expression for the canonical momentum does not change.

(ii) The equation of the motion does not change.

Which of the following options is correct for the above statements?

  1. Both (i) and (ii) are correct.

  2. Both (i) and (ii) are not correct.

  3. (i) is correct and (ii) is not correct.

  4. (i) is not correct and (ii) is correct.

Show solution

\(\alpha\dfrac{df}{dt}=\alpha f'(q)\dot q\), so \[p=\frac{\partial L}{\partial\dot q}=m\dot q+\alpha f'(q)\neq m\dot q\quad\Rightarrow\quad\text{(i) is false.}\] A total time derivative does not change the variation of the action: the extra Euler–Lagrange terms \(\dfrac{d}{dt}(\alpha f')-\alpha f''\dot q=0\) cancel identically. The equation of motion \(m\ddot q=-m\omega^2q\) is unchanged, so (ii) is true. \[\boxed{\text{(i) not correct, (ii) correct}}\]

Answer: (D)

CM07.4GATE 2022, Q41MCQ | 2 marks

A particle of mass 1 kg is released from a height of 1 m above the ground. When it reaches the ground, what is the value of Hamilton's action for this motion in J s? (\(g\) is the acceleration due to gravity; take gravitation potential to be zero on the ground)

  1. \(-\dfrac23\sqrt{2g}\)

  2. \(\dfrac53\sqrt{2g}\)

  3. \(3\sqrt{2g}\)

  4. \(-\dfrac13\sqrt{2g}\)

Show solution

Measuring the height \(z\) from the ground, \(L=\tfrac12m\dot z^2-mgz\) with \(m=1\) kg. Released from rest at \(z=1\) m, \[ z(t)=1-\tfrac12gt^2,\qquad \dot z=-gt, \] and the ground is reached at \(T=\sqrt{2/g}\). Along the path, \[ L=\tfrac12g^2t^2-g\left(1-\tfrac12gt^2\right)=g^2t^2-g . \] Hence \[ S=\int_0^{T}L\,dt=\left[\frac{g^2t^3}{3}-gt\right]_0^{\sqrt{2/g}} =\frac{g^2}{3}\left(\frac2g\right)^{3/2}-g\sqrt{\frac2g}=\frac23\sqrt{2g}-\sqrt{2g}=-\frac13\sqrt{2g}. \] \[\boxed{S=-\dfrac13\sqrt{2g}\ \text{J s}}\]

Answer: (D) \(-\dfrac13\sqrt{2g}\)

CM07.5GATE 2021, Q30NAT | 1 mark

A hoop of mass \(M\) and radius \(R\) rolls without slipping along a straight line on a horizontal surface as shown in the figure. A point mass \(m\) slides without friction along thee inner surface of the hoop, performing small oscillations about the mean position. The number of degrees of freedom of the system (in integer) is .

Figure

Show solution

The hoop moves in a plane along a straight line. Its centre position \(x\) and rotation angle \(\phi\) are related by the rolling condition \(\dot x=R\dot\phi\), so the hoop has one degree of freedom. The point mass is constrained to the inner surface of the hoop, so its position relative to the hoop's centre is fixed by one angle \(\theta\). Hence the system has two independent generalized coordinates, \((x,\theta)\): \[\boxed{2}\]

Answer: 2 (range 2 to 2)

CM07.6GATE 2021, Q54MSQ | 2 marks

A uniform block of mass \(M\) slides on a smooth horizontal bar. Another mass \(m\) is connected to it by an inextensible string of length \(l\) of negligible mass, and is constrained to oscillate in the \(X\)-\(Y\) plane only. Neglect the sizes of the masses. The number of degrees of freedom of the system is two and the generalized coordinates are chosen as \(x\) and \(\theta\), as shown in the figure.

Figure

If \(p_x\) and \(p_\theta\) are the generalized momenta corresponding to \(x\) and \(\theta\), respectively, then the correct option(s) is (are)

  1. \(p_x=(m+M)\dot x+ml\cos\theta\,\dot\theta\)

  2. \(p_\theta=ml^2\dot\theta-ml\cos\theta\,\dot x\)

  3. \(p_x\) is conserved

  4. \(p_\theta\) is conserved

Show solution

With \(Y\) measured downward, the bob is at \((x+l\sin\theta,\ l\cos\theta)\), so its velocity is \((\dot x+l\cos\theta\,\dot\theta,\ -l\sin\theta\,\dot\theta)\). The Lagrangian is \[L=\frac12M\dot x^2+\frac12m\left(\dot x^2+l^2\dot\theta^2+2l\dot x\dot\theta\cos\theta\right)+mgl\cos\theta.\] Then \[p_x=\frac{\partial L}{\partial\dot x}=(m+M)\dot x+ml\cos\theta\,\dot\theta\quad\text{(A true)},\] \[p_\theta=\frac{\partial L}{\partial\dot\theta}=ml^2\dot\theta+ml\cos\theta\,\dot x\quad\text{(B false: wrong sign)}.\] \(x\) is cyclic (\(\partial L/\partial x=0\)), so \(p_x\) is conserved (C true). \(\partial L/\partial\theta=-ml\dot x\dot\theta\sin\theta-mgl\sin\theta\neq0\), so \(p_\theta\) is not conserved (D false). \[\boxed{\text{(A), (C)}}\]

Answer: (A), (C)

CM07.7GATE 2021, Q56MSQ | 2 marks

The time derivative of a differentiable function \(g(q_i,t)\) is added to a Lagrangian \(L(q_i,\dot q_i,t)\) such that \[L'=L(q_i,\dot q_i,t)+\frac{d}{dt}g(q_i,t)\] where \(q_i,\dot q_i,t\) are the generalized coordinates, generalized velocities and time, respectively. Let \(p_i\) be the generalized momentum and \(H\) the Hamiltonian associated with \(L(q_i,\dot q_i,t)\). If \(p_i'\) and \(H'\) are those associated with \(L'\), then the correct option(s) is(are)

(A) Both \(L\) and \(L'\) satisfy the Euler-Lagrange's equations of motion

(B) \(p_i'=p_i+\dfrac{\partial}{\partial q_i}g(q_i,t)\)

(C) If \(p_i\) is conserved, then \(p_i'\) is necessarily conserved

(D) \(H'=H+\dfrac{d}{dt}g(q_i,t)\)

Show solution

Here \(\dfrac{dg}{dt}=\sum_j\dfrac{\partial g}{\partial q_j}\dot q_j+\dfrac{\partial g}{\partial t}\).

(A) \(\dfrac{\partial}{\partial\dot q_i}\dfrac{dg}{dt}=\dfrac{\partial g}{\partial q_i}\), so \(\dfrac{d}{dt}\dfrac{\partial}{\partial\dot q_i}\dfrac{dg}{dt}=\dfrac{d}{dt}\dfrac{\partial g}{\partial q_i}=\dfrac{\partial}{\partial q_i}\dfrac{dg}{dt}\). The Euler–Lagrange expression of \(dg/dt\) vanishes identically, so \(L\) and \(L'\) give the same equations of motion. True.

(B) \(p_i'=\dfrac{\partial L'}{\partial\dot q_i}=p_i+\dfrac{\partial g}{\partial q_i}\). True.

(C) \(\dot p_i'=\dot p_i+\dfrac{d}{dt}\dfrac{\partial g}{\partial q_i}\). Even if \(\dot p_i=0\), the second term need not vanish. False.

(D) \(H'=\sum_ip_i'\dot q_i-L'=\sum_ip_i\dot q_i+\sum_i\dfrac{\partial g}{\partial q_i}\dot q_i-L-\sum_i\dfrac{\partial g}{\partial q_i}\dot q_i-\dfrac{\partial g}{\partial t}=H-\dfrac{\partial g}{\partial t}\), which is not \(H+\dfrac{dg}{dt}\). False. \[\boxed{\text{(A), (B)}}\]

Answer: (A), (B)

CM07.8GATE 2020, Q35NAT | 1 mark

If a particle is moving along a sinusoidal curve, the number of degree of freedom of the particle is

Show solution

Take the curve as \(y=A\sin x\), \(z=0\). A free particle has 3 coordinates, and there are two equations of constraint (\(y-A\sin x=0\) and \(z=0\)). Therefore \[ \text{degrees of freedom}=3-2=\boxed{1} \] A single parameter (for example \(x\), or the arc length) fixes the position.

Answer: 1

CM07.9GATE 2019, Q55NAT | 2 marks

A projectile of mass \(1\,kg\) is launched at an angle of \(30^\circ\) from the horizontal direction at \(t=0\) and takes time \(T\) before hitting the ground. If its initial speed is \(10\,ms^{-1}\), the value of the action integral for the entire flight in the units of \(kgm^2s^{-1}\) (round off to one decimal place) is . [Take \(g=10\,ms^{-2}\)]

Show solution

Time of flight: \(T=\dfrac{2v\sin\theta}{g}=\dfrac{2\times10\times\frac12}{10}=1\,s\).

With \(\dot x=v\cos\theta=5\sqrt3\,ms^{-1}\), \(\dot y=v\sin\theta-gt=5-10t\) and \(y=5t-5t^2\), the Lagrangian is \[ L=\frac12m\left(\dot x^2+\dot y^2\right)-mgy=\frac12\left[75+(5-10t)^2\right]-10\left(5t-5t^2\right)=100t^2-100t+50 . \] \[ S=\int_0^TL\,dt=\int_0^1\left(100t^2-100t+50\right)dt=\frac{100}{3}-50+50=33.3\,kg\,m^2s^{-1}. \] \(\boxed{S\approx33.3}\)

Answer: 33.3 (range 33.2 to 33.4)

CM07.10CSIR NET Dec 2025, Q68Part C | 5 marks

The Lagrangian \(L=L(x,y,\dot x,\dot y)\) is invariant under the transformation \(x\to x+\epsilon y\) and \(y\to y+\epsilon x\), for any infinitesimal real parameter \(\epsilon\).

If \(P_x\), \(P_y\) denote canonically conjugate momenta corresponding to \(x\), \(y\) respectively, then the corresponding conserved quantity is

  1. \(yP_x-xP_y\)

  2. \(yP_x+xP_y\)

  3. \(xP_x+yP_y\)

  4. \(xP_x-yP_y\)

Show solution

For a symmetry \(\delta q_i=\epsilon f_i(q)\) of the Lagrangian, Noether's theorem gives the conserved quantity \(Q=\sum_ip_if_i\). Here \(f_x=y\), \(f_y=x\): \[\boxed{Q=yP_x+xP_y}\] (Check: \(\dot Q=\dot yP_x+y\dot P_x+\dot xP_y+x\dot P_y=\dot y\frac{\partial L}{\partial\dot x}+\dot x\frac{\partial L}{\partial\dot y}+y\frac{\partial L}{\partial x}+x\frac{\partial L}{\partial y}=\delta L/\epsilon=0\).)

Answer: (2) \(yP_x+xP_y\)

CM07.11CSIR NET Jun 2025, Q52Part C | 5 marks

A massless rod of length \(l\) is hinged at the extreme end of a vertical spring whose other end is fixed to the ground. A point mass \(m\) is fixed at end of the rod, as shown in the figure.FigureAssume harmonic motion of the spring given by \(h(t)=h_0(2+\cos\omega t)\), where \(h_0>l\). The equation of motion of the mass (confined to the plane of the figure) is given by

  1. \(l\ddot\theta+\omega^2h_0\sin\theta\sin\omega t-g\sin\theta=0\)

  2. \(l\ddot\theta+\omega^2h_0\sin\theta\cos\omega t-g\sin\theta=0\)

  3. \(l\ddot\theta+\omega^2h_0\sin\theta\cos\omega t+g\sin\theta=0\)

  4. \(l\ddot\theta-\omega^2h_0\sin\theta\sin\omega t+g\sin\theta=0\)

Show solution

The mass is above the pivot, with \(\theta\) measured from the upward vertical. Its coordinates are \(x=l\sin\theta\) and \(y=h(t)+l\cos\theta\), so \[ L=\frac m2\left(l^2\dot\theta^2+\dot h^2-2l\dot h\sin\theta\,\dot\theta\right)-mg\,(h+l\cos\theta) . \] Lagrange's equation: \[ \frac{d}{dt}\left(ml^2\dot\theta-ml\dot h\sin\theta\right)=-ml\dot h\cos\theta\,\dot\theta+mgl\sin\theta \;\Rightarrow\; l\ddot\theta-(g+\ddot h)\sin\theta=0 . \] The pivot's acceleration acts like extra gravity, \(g\to g+\ddot h\). With \(\ddot h=-h_0\omega^2\cos\omega t\): \[ \boxed{l\ddot\theta+\omega^2h_0\sin\theta\cos\omega t-g\sin\theta=0} \]

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