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PG Exam 2016 : Physics (Banaras Hindu University (BHU), Varanasi )

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Brijmohan
Banaras Hindu University (BHU), Varanasi
PhD Condensed matter Physics

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fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES CLASSICAL MECHANICS SOLUTIONS NET/JRF (JUNE-2011) Q1. A particle of unit mass moves in a potential V x ax 2 b , where a and b are positive x2 constants. The angular frequency of small oscillations about the minimum of the potential is (a) Ans: 8b 8a (b) (c) 8a / b (d) 8b / a (b) 1 V 2b b b 4 Solution: V x ax 2 2 0 2ax 3 0 ax 4 b 0 x0 . x x x a 2V k Since , m 1 and k 2 where xo is stable equilibrium point. x x x m 0 1 2V 6b 6b b 4 Hence k 2 2a 4 2a 8a at x x0 . b x x0 a a Thus 8a . Q2. The acceleration due to gravity (g) on the surface of Earth is approximately 2.6 times that on the surface of Mars. Given that the radius of Mars is about one half the radius of Earth, the ratio of the escape velocity on Earth to that on Mars is approximately (a) 1.1 Ans: (b) 1.3 (c) 2.3 (d) 5.2 (c) Solution: Escape velocity = 2gR ge R e Escape velocity of Earth R g 2.3 where e 2 and e 2.6. Rm gm Escape velocity of Mass gm R m Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 1 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q3. The Hamiltonian of a system with n degrees of freedom is given by H q1 ,..... q n ; p1 ,....... p n ; t , with an explicit dependence on the time t. Which of the following is correct? (a) Different phase trajectories cannot intersect each other. (b) H always represents the total energy of the system and is a constant of the motion. i H / pi , p i H / qi are not valid since H has explicit time (c) The equations q dependence. (d) Any initial volume element in phase space remains unchanged in magnitude under time evolution. Ans: (a) Q4. The Lagrangian of a particle of charge e and mass m in applied electric and magnetic 1 fields is given by L mv 2 eA v e , where A and are the vector and scalar 2 potentials corresponding to the magnetic and electric fields, respectively. Which of the following statements is correct? (a) The canonically conjugate momentum of the particle is given by p mv p2 e A p e (b) The Hamiltonian of the particle is given by H 2m m (c) L remains unchanged under a gauge transformation of the potentials (d) Under a gauge transformation of the potentials, L changes by the total time derivative Ans: (d) Solution: 2 V Q5. L t 0 Consider the decay process in the rest frame of the -. The masses of the , and are M , M and zero respectively. A. The energy of is (a) M 2 2 2 M c 2M (b) M 2 2 M c2 2M (c) M M c 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com (d) M M c 2 Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 2 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (b) Solution: From conservation of energy M c 2 E E . 2 2 2 4 2 2 2 E2 and is same. p c M c and E p c since momentum of 2 M c2 E E , M 2c4 E 2 E E E 4 M2 c M c2 2 2 M2 M c2 M c 2 . and E E M c E E E 2M M B. The velocity of is (a) Ans: M 2 M c 2 2 M M 2 M 2 M c 2 2 M M (b) 2 (c) M c M (d) M c M (a) Solution: Velocity of E M 2 2 M c2 2M M c 2 1 v2 c2 2 v2 4M M 2 1 2 2 2 c M 2 M 2 2 4 2 2 M 2 M 4M M 2 v2 2M 2 M 4M M 2 v 2 M 4 M 1 v c . 2 2 2 2 2 2 c M M M 2 M 2 c M 2 M 2 Q6. The Hamiltonian of a particle of unit mass moving in the xy-plane is given to be: H xp x yp y 1 2 1 2 x y in suitable units. The initial values are given to be 2 2 x 0 , y 0 1,1 1 1 and p x 0 , p y 0 , . During the motion, the curves traced out 2 2 by the particles in the xy-plane and the p x p y plane are (a) both straight lines (b) a straight line and a hyperbola respectively (c) a hyperbola and an ellipse, respectively (d) both hyperbolas Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 3 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (d) Solution: H xp x yp y 1 2 1 2 x y 2 2 . Solving Hamiltonion equation of motion H H x and y . y p y y p x p x x p p p y x H H x x y y . and x y p y p x After solving these four differential equation and eliminating time t and using boundary condition one will get x 1 1 1 and p x 2 py y NET/JRF (DEC-2011) Q7. A double pendulum consists of two point masses m attached by strings of length l as shown in the figure: The kinetic energy of the pendulum is (a) (b) 1 2 2 2 ml 2 1 2 2 1 2 1 2 cos 2 (c) 1 2 2 2 2 ml 1 2 1 2 2 1 2 cos 2 (d) Ans: 1 2 2 2 ml 1 2 2 1 2 2 2 ml 2 1 2 2 1 2 1 2 cos 2 m 2 1 l l (c) m Solution: Let co-ordinate x1 , y1 and x 2 , y 2 . K .E. x1 l sin 1 , y1 l cos 1 1 1 2 2 12 y 12 m x 2 2 mx y 2 2 1 , y 1 l sin 1 1 l cos 1 x 1 x2 l sin 1 l sin 2 , y 2 l cos 1 l cos 2 2 l cos 2 l cos 1 x 1 2 2 , y 2 l sin 1 1 l sin 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 4 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 , y 1 , x 2 , y 2 in K.E equation, one will get Put the value of x T Q8. 1 2 2 1 2 cos 1 2 . m 1 2 2 2 A constant force F is applied to a relativistic particle of rest mass m. If the particle starts from rest at t = 0, its speed after a time t is Ans: (d) Fct F t m 2c 2 2 2 (d) Solution: dp F p Ft c . dt At t=0, p=0 so c=0 p Ft Q9. (c) c 1 e Ft / mc Ft (b) c tanh mc (a) Ft / m mu u2 1 2 c Ft u F t m Ft 1 mc 2 . The potential of a diatomic molecule as a function of the distance r between the atoms is given by V r a b 12 . The value of the potential at equilibrium separation between 6 r r the atoms is: (a) 4a 2 / b Ans: (b) 2a 2 / b (c) a 2 / 2b (d) a 2 / 4b (d) Solution: V r 1 12 b V a b a 12b 0 6 7 13 0 7 6a 6 0 12 , for equilibrium 6 r r r r r r r 1 1 12b 12 b 6 2b 6 6a 6 0 r r r 6a a 1 6 a2 a2 a2 2b a b . V r a 2b 4b 4b 2b 2b 2 a a Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 5 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q10. Two particles of identical mass move in circular orbits under a central potential V r 1 2 kr . Let l1 and l2 be the angular momenta and r1, r2 be the radii of the 2 orbits respectively. If l1/l2 = 2, the value of r1 / r2 is: (a) Ans: 2 (b) 1 / 2 (c) 2 (d) 1 / 2 (a) Solution: Veff J2 1 kr 2 where J is angular momentum. 2 2mr 2 Condition for circular orbit J2 Veff 0 3 kr 0 J 2 r 4 J r 2 . mr r 2 J r r J r J Thus 1 1 1 1 1 2 since 1 2 . J2 J 2 r2 r2 J2 r2 Q11. A particle of mass m moves inside a bowl. If the surface of the bowl is given by the equation z 1 a x 2 y 2 , where a is a constant, the Lagrangian of the particle is 2 (a) (c) Ans: 1 2 gar 2 2 r 2 mr 2 1 2 r 2 sin 2 2 gar 2 2 r 2 mr 2 (b) 1 2 2 r 2 m 1 a2r 2 r 2 (d) 1 2 gar 2 2 r 2 m 1 a2r 2 r 2 (d) Solution: L 1 2 y 2 z 2 mgz mx 2 where z 1 a x2 y2 . 2 1 It has cylindrical symmetry. Thus x r cos , y r sin , z a r 2 . 2 , y and z cos r sin sin r cos r r . x a rr 1 2 gar 2 . 2 r 2 So L m 1 a 2r 2 r 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 6 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q12. A planet of mass m moves in the inverse square central force field of the Sun of mass M . If the semi-major and semi-minor axes of the orbit are a and b , respectively, the total energy of the planet is: (a) (c) Ans: 1 1 (b) GMm a b a b (d) GMm a b 2 GMm a b GMm 1 1 a b a (a) Solution: Assume Sun is at the centre of elliptical orbit. Conservation of energy 1 2 GMm 1 2 GMm mv1 mv2 2 a 2 b v2 Conservation of momentum L mv1a mv2b b s a v2 v1 b a v1 1 2 1 2 GMm GMm 1 a2 b a m v12 v12 2 GMm mv1 mv2 2 2 a b 2 b ab 1 2 b2 a 2 1 1 b a b mv1 GMm mv12 GMm 2 ab 2 2 b a b a E Q13. 1 2 GMm b 1 GMm mv1 GMm 2 a a b a a GMm b b a GMm b GMm 1 a a b a b a b a An annulus of mass M made of a material of uniform density has inner and outer radii a and b respectively. Its principle moment of inertia along the axis of symmetry perpendicular to the plane of the annulus is: (b) 1 M b 2 a 2 2 (d) 1 M b2 a2 2 (a) 1 b4 a4 M 2 2 b a2 (c) 1 M b2 a2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 7 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (d) Q14. The trajectory on the zpz-plane (phase-space trajectory) of a ball bouncing perfectly elastically off a hard surface at z = 0 is given by approximately by (neglect friction): (a) (b) PZ PZ z z PZ PZ (c) d) z z Ans: (a) Solution: H P2 Pz2 mgz and E z mgz . 2m 2m Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 8 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (JUNE-2012) Q15. The bob of a simple pendulum, which undergoes small oscillations, is immersed in water. Which of the following figures best represents the phase space diagram for the pendulum? (a) (b) p p x (c) x (d) p p x x . Ans: (d) Solution: When simple pendulum oscillates in water it is damped oscillation so amplitude continuously decrease and finally it stops. Q16. Two events separated by a (spatial) distance 9 109m, are simultaneous in one inertial frame.The time interval between these two events in a frame moving with a constant speed 0.8 c (where the speed of light c = 3 108 m/s) is (a) 60 s Ans: (b) 40 s (c) 20 s (d) 0 s (b) ' ' x1' 9 10 9 m and t 2 t1' 0 . Then Solution: x 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 9 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES t ' v x' t1 v x' ' ' ' t2 t1' x1' v x 2 x1' v x 2 2 c2 2 1 c2 1 . t t t 2 t1 2 1 2 2 2 2 2 c2 c2 v v v v v 1 1 1 2 1 2 1 2 2 2 c c c c c t 2 t1 40 sec Put v 0.8c Q17. If the Lagrangian of a particle moving in one dimensions is given by L 2 x V x the 2x Hamiltonian is 2 x V x (b) 2x 1 (a) xp 2 V x 2 Ans: p2 V x (d) 2x 1 2 V x (c) x 2 (a) L x px x . px px x x x L and Solution: Since H p x x 2 px x 2 px x x H px x V x H px px x V x . V x H 2x 2 2x 2 Q18. A horizontal circular platform mutes with a constant angular velocity directed vertically upwards. A person seated at the centre shoots a bullet of mass m horizontally with speed v. The acceleration of the bullet, in the reference frame of the shooter, is (a) 2v to his right (c) v to his right Ans: (b) 2v to his left (d) v to his left (a) Solution: Velocity of bullet j, = v Angular . velocity= k There will be coriolis force F 2m v . a 2v i . F 2m vi Q19. The Poisson bracket r , p has the value (a) r p Ans: p (b) r (c) 3 (d) 1 (b) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 10 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1/ 2 , r , yj zk py Solution: r xi x 2 y 2 z 2 , p p xi j pz k 1/ 2 2 2 p px py pz2 r p r p r p r p r p r p r , p = x p p x x x y p y p y y z pz pz y r p x px y p y z pz p r r p r p r p r p Q20. Consider the motion of a classical particle in a one dimensional double-well potential V x 2 1 2 x 2 . If the particle is displaced infinitesimally from the minimum 4 on the x-axis (and friction is neglected), then (a) the particle will execute simple harmonic motion in the right well with an angular frequency 2 (b) the particle will execute simple harmonic motion in the right well with an angular frequency =2 (c) the particle will switch between the right and left wells (d) the particle will approach the bottom of the right well and settle there Ans: (b) Solution: V x 1 2 x 2 2 V 2 x 2 2 2 x 0 x 0 , x 2 . 4 x 4 2V 2V 2 . At , 3 x 2 0 so V is maximum. Thus it is unstable point x 0 x 2 x 2 2V x 2 4 and it is stable equilibrium point with x 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 2V x 2 x x0 2 1. Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 11 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q21. What is proper time interval between the occurrence of two events if in one inertial frame events are separated by 7.5 108m and occur 6.5 s a part? (a) 6.50 s Ans: (b) 6.00 s (c) 5.75 s (d) 5.00 s There is not possible to calculate proper time interval Solution: x 2 x1 7.5 10 8 m and t 2 t1 6.5 s . v v v t2 2 x2 t1 2 x1 t t x x1 2 2 ' c c 2 1 c t2 t1' . v2 v2 v2 v2 1 2 1 2 1 2 1 2 c c c c In this problem v is not given so its not possible to calculate proper time interval. NET/JRF (DEC-2012) Q22. A solid cylinder of height H, radius R and density , floats vertically on the surface of a liquid of density 0 . The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The frequency of oscillation is (a) Ans: g 0H 0 (b) g H g (c) (d) 0H 0 g H (d) Solution: From Newton s law of motion ma mg 0 Agh where A is area of cross section, m AH . AHa AHg 0 Agh a 1 Q23. 0 gh 0 g H H Three particles of equal mass (m) are connected by two identical massless springs of stiffness constant (K) as shown in the figure K K m m m If x1, x2 and x3 denote the horizontal displacement of the masses from their respective equilibrium positions the potential energy of the system is Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 12 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 2 2 K x12 x2 x3 2 (c) Ans: 1 2 2 K x12 2 x2 x3 2 x2 x1 x3 2 1 2 2 K x12 x2 x3 x2 x1 x3 2 (d) 1 2 K x12 2 x2 2 x2 x1 x3 2 (c) Solution: V V Q24. (b) (a) 1 1 2 2 K x2 x1 K x3 x2 , 2 2 1 1 1 2 2 2 2 2 K x2 x12 2 x2 x1 K x3 x2 2 x3 x2 V K x12 2 x2 x3 2 x2 x1 x3 2 2 2 A planet of mass m moves in the gravitational field of the Sun (mass M). If the semimajor and semi-minor axes of the orbit are a and b respectively, the angular momentum of the planet is (a) Ans: 2GMm2 a b (b) 2GMm2 a b (c) 2GMm2 ab a b 2GMm2 ab a b (d) (d) Solution: Assume Sun is at the centre of elliptical orbit. Conservation of energy 1 2 GMm 1 2 GMm mv1 mv2 2 a 2 b v2 b Conservation of momentum L mv1a mv2b s a v2 v1 b a v1 1 a2 1 2 1 2 GMm GMm b a mv1 mv2 m v12 v12 2 GMm ab 2 2 a b 2 b 1 1 1 2 b2 a 2 b b a mv12 GMm mv1 GMm 2 a b a ab 2 2 b 1 b v1 2GM a b a 2GMab 2GMm2 ab b 1 L mv1a m 2GM a m L a b a a b b a Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 13 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q25. The Hamiltonian of a simple pendulum consisting of a mass m attached to a massless string of length l is H p 2 mgl 1 cos . If L denotes the Lagrangian, the value of 2ml 2 dL is: dt (a) (b) g p cos l (c) Ans: 2g p sin l g p sin 2 l (d) lp 2 cos (a) Solution: p2 dL L where H 2 mgl 1 cos . L, H 2ml dt t 2 2 H p , L ml mgl 1 cos . H , i H p L pi q p ml 2 2 i Hence we have to calculate L, H which is only defined into phase space i.e. p and . Then L p 2 mgl 1 cos 2ml 2 L, H L H p L H 2g L dL 2g p sin and 0 p sin p l t dt l Q26. Two bodies of equal mass m are connected by a massless rigid rod of length l lying in the xy-plane with the centre of the rod at the origin. If this system is rotating about the z-axis with a frequency , its angular momentum is (a) ml 2 / 4 Ans: (b) ml 2 / 2 (c) ml 2 (d) 2ml 2 (b) Solution: Since rod is massless i.e. M = 0. Moment of inertia of the system m1 r12 m2 r22 , m1 m2 m and r1 r2 l 2 ml 2 ml 2 ml 2 ml 2 . Angular momentum, J I and J . 4 4 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 14 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q27. Which of the following set of phase-space trajectories is not possible for a particle obeying Hamilton s equations of motion? (a) (b) P P x x (c) (d) P P x Ans: x (b) Solution: Phase curve does not cut each other Q28. The muon has mass 105 MeV/c2 and mean life time 2.2 s in its rest frame. The mean distance traversed by a muon of energy 315 MeV before decaying is approximately, (a) 3 105 km Ans: (c) 6.6 m (b) 2.2 cm (d) 1.98 km (d) Solution: Since E 315MeV and m0 105 E mc 2 E m0 c 2 2 315 v 1 2 c Now, t t0 v2 1 2 c , MeV . c2 m0 c 2 2 315 v 1 2 c t 0 2.2 s t 2.2 10 6 8 1 9 105 1 2 v 0.94c . v c2 t = 6.6 s Now the distance traversed by muon is vt 0.94c 6.6 10 6 = 1.86 km . Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 15 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (JUNE-2013) Q29. The area of a disc in its rest frame S is equal to 1 (in some units). The disc will appear distorted to an observer O moving with a speed u with respect to S along the plane of the disc. The area of the disc measured in the rest frame of the observer O is ( c is the speed of light in vacuum) u2 (a) 1 c 2 Ans: 1/ 2 u2 (b) 1 c 2 1 / 2 u2 (c) 1 c 2 u2 (d) 1 c 2 1 (a) Solution: Area of disc from S frame is 1 i.e. a 2 1 or a a 1 Area of disc from S frame is a b a a 1 u2 u2 u2 1 1 1 c2 c2 c2 u2 where b a 1 2 . c Q30. A planet of mass m and an angular momentum L moves in a circular orbit in a potential, V r k / r , where k is a constant. If it is slightly perturbed radially, the angular frequency of radial oscillations is (a) mk 2 / 2 L3 Ans: (b) mk 2 / L3 (c) 2mk 2 / L3 3mk 2 / L3 (d) (b) Solution: Veff Veff L2 k L2 k 2 0 . For circular orbit 3 2 r r mr r 2mr L2 L2 k r r . Thus 0 mk mr 3 r 2 2 k d Veff dr 2 r r0 2 3L 2k 3 4 mr r r r0 k , m r 3m k 2m k m k4 2k = 3 L6 L6 L6 L2 mk 3 2 3L L2 m mk Veff 4 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 4 3 4 3 Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 16 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q31. d 2V dr 2 r r0 m mk 2 . L3 The number of degrees of freedom of a rigid body in d space-dimensions is (a) 2d (c) d d 1 / 2 (b) 6 Ans: (c) Q32. (d) d ! A system is governed by the Hamiltonian H 1 p x ay 2 1 p x bx 2 2 2 where a and b are constants and p x , p y are momenta conjugate to x and y respectively. For what values of a and b will the quantities p x 3 y and p y 2 x be conserved? (a) a 3, b 2 (b) a 3, b 2 (c) a 2, b 3 (d) a 2, b 3 Ans: (d) Solution: Poisson bracket px 3y, H 0 and p y 2 y, H 0 py (b 3) x(3b b2 ) 0 and px (a 2) y (2a a 2 ) 0 a 2, b 3 Q33. The Lagrangian of a particle of mass m moving in one dimension is given by L 1 2 bx mx 2 where b is a positive constant. The coordinate of the particle x t at time t is given by: (in following c1 and c 2 are constants) (a) b 2 t c1t c2 2m (b) c1t c 2 bt bt (c) c1 cos c 2 sin m m Ans: bt bt (d) c1 cosh c 2 sinh m m (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 17 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Equation of d L L d b 0 m b 0 0 mx x x dt x dt motion b m x b t2 d 2x b dx b c1t c2 t c1 x m 2 m dt m dt 2 NET/JRF (DEC-2013) Q34. Let A, B and C be functions of phase space variables (coordinates and momenta of a mechanical system). If , represents the Poisson bracket, the value of A, B, C A, B , C is given by (a) 0 Ans: (b) B, C , A (c) A, C , B (d) C , A , B (d) Solution: We know that Jacobi identity equation A, B, C B, C, A C, A, B 0 Now A, B, C A, B , C B, C , A C, A , B Q35. A particle moves in a potential V x 2 y 2 z2 . Which component(s) of the angular 2 momentum is/are constant(s) of motion? (a) None (b) L x , L y and L z (c) only L x and L y (d) only L z Ans: (d) Solution: A particle moves in a potential V x 2 y 2 V r , , r 2 sin 2 cos 2 r 2 sin 2 sin 2 V r , , r 2 sin 2 z2 2 r2 cos 2 2 r2 cos 2 2 Now is cyclic-co-ordinate p i.e Lz is constant of motion. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 18 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q36. The Hamiltonian of a relativistic particle of rest mass m and momentum p is given by H p 2 m 2 V x , in units in which the speed of light c 1 . The corresponding Lagrangian is 2 V x (a) L m 1 x 2 V x (c) L 1 mx Ans: 2 V x (b) L m 1 x (d) L 1 2 V x mx 2 2p p 2 m2 x (b) Solution: H p p 2 m 2 V x H 1 x p 2 p 2 m 1 2 2 1/ 2 p xm 2 1 x p 2 m2 V x p H xp p H x Now L x Put value p Q37. xm 1 x 2 2 V x L m 1 x A pendulum consists of a ring of mass M and radius R suspended by a massless rigid rod of length l attached to its rim. When the pendulum oscillates in the plane of the ring, the time period of oscillation is (a) 2 (b) (c) 2 Ans: l R g 2 R 2 2 Rl l 2 g R l (d) 2 g 2 g l 2 R2 2R 2 1/ 4 2Rl l 2 1/ 4 (c) Solution: The moment of inertia about pivotal point is given by I I c.m Md 2 MR 2 M (l R ) 2 If ring is displaced by angle then potential energy is Mg (l R) cos The Lagrangian is given by Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 19 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES L 1 2 1 2 Mg (l R) cos I V ( ) = ( MR 2 M (l R)2 ) 2 2 d L L 2 2 0 ( MR M (l R) ) Mg (l R) sin 0 dt Mg (l R) 0 For small oscillation sin ( MR 2 M (l R) 2 ) 2 R 2 2 Rl l 2 Time period is given by 2 . g R l Q38. Consider a particle of mass m attached to two identical y springs each of length l and spring constant k (see the figure). The equilibrium configuration is the one where the o springs are unstretched. There are no other external forces on the system. If the particle is given a small displacement along the x -axis, which of the following describes the equation of x o motion for small oscillations? x (a) m Ans: kx3 0 l2 kx 0 (b) m x 2kx 0 (c) m x x (d) m kx2 0 l (a) y Solution: The lagrangian of system is given by L 1 2 V ( x) mx 2 l The potential energy is given by k V ( x) x 2 l 2 2 V ( x) k x 2 l 2 1 2 1 2 2 k l x2 l 2 2 l x o 1 2 l l 2 o x 2 For small oscillation one can approximate potential by Taylor expansion Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 20 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 1 2 2 2 1 x 2 1 x 4 x 2 2 V ( x) kl 1 2 1 V ( x) kl 1 2 4 1 l 8l 2 l x2 V ( x) kl 2 l 2 2 x 4 V ( x) k 2 . 4l x 4 1 2 k 2 So Lagrangian of system is given by L mx 2 4l The Lagranges equation of motion kx3 d L L m x 0. 0 dt x l2 x NET/JRF (JUNE-2014) Q39. The time period of a simple pendulum under the influence of the acceleration due to gravity g is T . The bob is subjected to an additional acceleration of magnitude 3 g in the horizontal direction. Assuming small oscillations, the mean position and time period of oscillation, respectively, of the bob will be (a) 0 o to the vertical and 3T (c) 60 o to the vertical and T / 2 Ans: (b) 30 o to the vertical and T / 2 (d) 0 o to the vertical and T / 3 (c) Solution: T 2 g l g 3g 2 g 2 T 4g 2 2 g l T l 1 T 2 T T 2 2g g 2 2 3g g g T cos mg , T sin 3 mg tan 3 60o Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 21 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q40. A particle of mass m and coordinate q has the Lagrangian L 1 2 2 qq mq 2 2 where is a constant. The Hamiltonian for the system is given by (a) (b) p2 2 m q (c) Ans: p 2 qp 2 2m 2m 2 p2 qp 2 2m 2 m q 2 (d) pq 2 (b) L where L Solution: H qp 1 1 2 2 qq mq 2 2 L p qq p q m q q p mq q m q p2 p2 1 p2 L H qp m q m q 2 m q 2 2 m q 2 L H qp L H qp Q41. p2 p2 m q m q 2 m q 2 p2 p2 m q 2 m q H p2 2 m q The coordinates and momenta xi , pi i 1, 2, 3 of a particle satisfy the canonical Poisson bracket relations xi , p j ij . If C1 x2 p3 x3 p2 and C 2 x1 p 2 x2 p1 are constants of motion, and if C 3 C1 , C 2 x1 p3 x3 p1 , then (a) C 2 , C3 C1 and C3 , C1 C 2 (b) C 2 , C3 C1 and C3 , C1 C 2 (c) C 2 , C3 C1 and C3 , C1 C 2 (d) C 2 , C3 C1 and C3 , C1 C 2 Ans: (d) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 22 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: C1 x2 p3 x3 p2 , C2 x1 p2 x2 p1 , C3 x1 p3 x3 p1 C2 C3 C2 C3 C2 C3 C2 C3 C2 C3 C2 C3 x1 p1 p1 x1 x2 p2 p2 x2 x3 p3 p3 x3 C2 , C3 C2 , C3 p2 x3 x2 p3 0 x1 0 0 x1 0 x1 p2 x3 x2 p3 C1 C3 C1 C3 C1 C3 C1 C3 C1 C3 C1 C3 C1 x1 p1 p1 x1 x2 p2 p2 x2 x3 p3 p3 x3 C3 , C1 C3 , C1 p3 0 x3 0 0 x3 0 p3 p1x2 x1 p2 x1 p2 x2 p1 C2 Q42. The recently-discovered Higgs boson at the LHC experiment has a decay mode into a photon and a Z boson. If the rest masses of the Higgs and Z boson are 125 GeV/c 90 GeV/c 2 2 and respectively, and the decaying Higgs particle is at rest, the energy of the photon will approximately be (a) 35 3 GeV Ans: (b) 35 GeV (c) 30 GeV (d) 15 GeV (c) H B PH Z B Solution: From conservation of momentum 0 P 1 P 2 1 P 2 P 1 P 2 P Now EH B EPH EZ B EPH EZB M HB c2 2 2 2 2 2 2 2 4 EP P 1 c 0 and EZ B P 2 c M ZB c H EZ B EPH E EZ B EPH ZB 2 4 EPH M Z c P 1 P 2 B 2 4 MZ c B M HB c 2 EPH M H B c 2 2 2 2 MZ c B M HB M z2B c 2 M HB EPH EZB EPH M HB c2 M 2 HB M z2B c 2 M HB 4 125 125 90 90 c EPH 4 30.1GeV 2 125 c Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 23 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q43. A canonical transformation relates the old coordinates q, p to the new ones Q, P by the relations Q q 2 and P p / 2q . The corresponding time independent generating function is (a) P / q 2 Ans: (b) q 2 P (d) qP 2 (c) q 2 / P (b) 2 Solution: Q q ; P p / 2q F2 F p 2 P 2q F2 q 2 P f ( P) q q F2 Q q 2 F2 q 2 P f (q ) P comparing both side f (q) f ( P) 0 F2 q 2 P NET/JRF (DEC-2014) Q44. The equation of motion of a system described by the time-dependent Lagrangian 1 2 V x is L e t mx 2 mx (a) m x mx (b) m x mx (c) m x Ans: dV 0 dx dV 0 dx (d) m x dV 0 dx dV 0 dx (a) L L V t 1 2 V x and Solution: L e t mx e t mx e x x x 2 d L L d t V t V t t mx e t 0 e mx e mxe e 0 dt x x dt x x V t V m x mx mx e 0 mx 0 x x Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 24 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q45. 1 1 A particle of mass m is moving in the potential V x ax 2 bx 4 where a, b are 2 4 positive constants. The frequency of small oscillations about a point of stable equilibrium is (a) Ans: a/m 2a / m (b) (c) 3a / m (d) 6a / m (b) 1 1 Solution: V x ax 2 bx 4 2 4 1 V a 2 2 x 0 ax bx3 0 x a bx 0 ,0 x b 2V a 3bx 2 x 2 2V x 2 a 3b x b Q46. 1 a 2 At x 0, 2V x 2 m 2V a (Negative so it is unstable point) x 2 a 2a (Positive so it is stable point) b 2a m The radius of Earth is approximately 6400 km . The height h at which the acceleration due to Earth s gravity differs from g at the Earth s surface by approximately 1% is (a) 64 km Ans: (b) 48 km (c) 32 km (d) 16 km (c) g 2h g 2h g 2h 1 1 h 32 k .m. g R g R g R According to the special theory of relativity, the speed v of a free particle of mass m and Solution: Q47. total energy E is: (a) v c 1 mc 2 E mc 2 (c) v c 1 E (b) v 2 2 E mc 2 1 m E mc 2 (d) v c 1 E Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 25 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (c) 2 mc 2 v 2 mc 2 v2 m2 c 4 Solution: E 1 2 1 v c 1 c E c2 E2 v2 E 1 2 c mc 2 Q48. 2 The Hamiltonian of a classical particle moving in one dimension is H p2 q 4 where 2m is a positive constant and p and q are its momentum and position respectively. Given that its total energy E E 0 the available volume of phase space depends on E 0 as 3/ 4 (a) E 0 (c) Ans: (b) E 0 (d) is independent of E 0 E0 (a) Solution: H V q P2 q4 2m E0 Phase area p dq 1 E 4 A p dq 2mE A E 1/2 0 Q49. E 1/4 0 q p 2mE0 E0 / A E 1/4 3/4 0 E0 / 1/4 2mE0 A mechanical system is described by the Hamiltonian H q, p result of the canonical transformation generated by F q, Q p2 1 m 2 q 2 . As a 2m 2 Q , the Hamiltonian in q the new coordinate Q and momentum P becomes (a) (b) 1 2 2 m 2 2 Q P P 2m 2 (c) Ans: 1 2 2 m 2 2 Q P Q 2m 2 1 2 m 2 2 P Q 2m 2 (d) 1 2 4 m 2 2 Q P P 2m 2 (d) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 26 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q p2 1 m 2 q 2 , F F1 q, Q q 2m 2 F Q 1 p 2 p .(a) q q Solution: H F1 1 1 P P q Q q P From equation (a) and (b) p QP 2 H Q50. .(b) q 1 P p2 1 Q2 P4 1 1 1 2 4 1 m 2 q 2 m 2 2 Q P m 2 P 2 2m 2 2m 2 2 P 2m The probe Mangalyaan was sent recently to explore the planet Mars. The inter-planetary part of the trajectory is approximately a half-ellipse with the Earth (at the time of launch), Sun and Mars (at the time the probe reaches the destination) forming the major axis. Assuming that the orbits of Earth and Mars are approximately circular with radii RE and RM , respectively, the velocity (with respect Sun Earth Mars RE RM to the Sun) of the probe during its voyage when it is at a distance r RE r RM from the Sun, neglecting the effect of Earth and Mars, is (a) (c) Ans: 2GM 2GM R E RM RE rRM 2GM (d) r R E RM r (b) R E RM r r R E RM 2GM r (b) Solution: Total energy E K / 2a where 2a major axis and 2a RE RM . R RM r 1 2 GMm GMm mv v 2GM E 2 r r RE RM RE RM Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 27

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