fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
CLASSICAL MECHANICS SOLUTIONS GATE- 2010 Q1.
For the set of all Lorentz transformations with velocities along the x-axis consider the two statements given below: P: If L is a Lorentz transformation then, L-1 is also a Lorentz transformation. Q: If L1 and L2 are Lorentz transformations then, L1L2 is necessarily a Lorentz transformation. Choose the correct option (A) P is true and Q is false
(B) Both P and Q are true
(C) Both P and Q are false
(D) P is false and Q is true
Ans:
(b)
Q2.
A particle is placed in a region with the potential V (x ) =
1 2 λ 3 kx − x , where k, λ > 0. 2 3
Then, (A) x = 0 and x =
k
λ
are points of stable equilibrium
(B) x = 0 is a point of stable equilibrium and x = (C) x = 0 and x =
k
λ
k
λ
is a point of unstable equilibrium
are points of unstable equilibrium
(D) There are no points of stable or unstable equilibrium Ans:
(b)
Solution: V = ⇒
k ∂V 1 2 λx 3 ⇒ = kx − λx 2 = 0 ⇒ x = 0, x = . kx − λ ∂x 2 3
∂ 2V = k − 2λx ∂x 2
⇒ At x = 0,
∂ 2V k ∂ 2V = + ve (Stable) and ⇒ = , = −ve (unstable) At x λ ∂x 2 ∂x 2
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fiziks Q3.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 0 A π meson at rest decays into two photons, which move along the x-axis. They are both
detected simultaneously after a time, t = 10s. In an inertial frame moving with a velocity V = 0.6c in the direction of one of the photons, the time interval between the two
detections is (A) 15 s Ans:
(B) 0 s
(C) 10 s
(D) 20 s
(a)
Solution: t1 = t 0
= 10
v c v 1− c
1+
1 + 0.6 = 10 × 2 = 20sec , t 2 = t 0 = 10 1 − 0.6
v c v 1+ c 1−
1 1 − 0.6 = 10 × = 5sec 2 1 + 0.6
⇒ t1 − t 2 = 15sec Statement for Linked Answer Questions 4 and 5:
The Lagrangian for a simple pendulum is given by L = Q4.
Hamilton’s equations are then given by (A) pθ = −mgl sin θ ; θ = (C) pθ = −mθ ; θ =
Ans:
1 2 2 ml θ − mgl (1 − cos θ ) 2
pθ ml 2
pθ m
(B) pθ = mgl sin θ ; θ =
pθ ml 2
⎛g⎞ (D) pθ = −⎜ ⎟θ ; ⎝l⎠
pθ ml
θ=
(b)
Solution: H =
P ∂H ∂H Pθ2 = −Pθ ⇒ Pθ = mglsin θ; + mgl (1 − cos θ ) ⇒ = θ ⇒ θ = θ 2 .Q5. The 2 2ml ∂θ ∂Pθ ml
Poisson bracket between θ and θ is
{ }
{ }
1 ml 2
{ }
g l
(A) θ ,θ = 1
(B) θ ,θ =
{ }
(D) θ ,θ =
(C) θ ,θ =
1 m
Ans: (b) Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES P 1 1 1 ⎛ ∂θ ∂θ ∂θ ∂Pθ ⎞ ⎧ P ⎫ ⎟⎟ = 1 ⋅ 2 − 0 = 2 . θ ,θ = ⎨θ , θ 2 ⎬ where θ = θ 2 . ⇒ 2 ⎜⎜ − ml ml ml ml ⎝ ∂θ ∂Pθ ∂Pθ ∂θ ⎠ ⎩ ml ⎭
{ }
GATE- 2011
Q6.
A particle is moving under the action of a generalized potential V (q, q ) =
1+ q . The q2
magnitude of the generalized force is (A) Ans:
2(1 + q ) q3
2(1 − q ) q3
(C)
2 q3
(D)
q q3
(c) d ⎛ ∂V ⎞ ∂V 2 = Fq ⇒ Fq = 3 . ⎜ ⎟− dt ⎝ ∂q ⎠ ∂q q
Solution: Q7.
(B)
Two bodies of mass m and 2m are connected by a spring constant k. The frequency of the normal mode is (A) 3k / 2m
Ans:
(C)
k/m
2k / 3m
(D)
k / 2m
(a)
Solution: m
Q8.
(B)
k
2m ω =
k
μ
=
2mm 2m k 3k = . where reduce mass μ = = 2m 2m + m 3 2m 3
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 (A) α = 2, β = 1/2 Ans:
(B) α = 2, β =2
(C) α = 1, β = 1
(D) α = 1/2, β = 2
(d)
Solution:
∂Q ∂P ∂Q ∂P ⋅ − ⋅ =1 ∂q ∂p ∂p ∂q
⇒ αq α −1 cos(βp ) × q α β cos(βp ) − q α β (− sin (βp )) × αq α −1 sin (βp ) = 1
αq 2α −1 β (cos 2 βp + sin 2 βp ) = 1 ⇒ αβq 2α −1 = 1 ⇒ α = , β = 2 . 1 2
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fiziks Q9.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Two particles each of rest mass m collide head-on and stick together. Before collision, the
speed of each mass was 0.6 times the speed of light in free space. The mass of the final entity is (A) 5m / 4 Ans:
(B) 2m
(C) 5m / 2
(D) 25 m / 8
(c)
Solution: From conservation of energy mc 2
1−
2
v c2
+
mc 2
1−
2
v c2
= m1c 2 ⇒
2mc 2 1−
2
v c2
= m1c 2
Since v = 0.6c ⇒ m1 = 5m / 2
GATE- 2012
Q10.
In a central force field, the trajectory of a particle of mass m and angular momentum L in plane polar coordinates is given by, 1 m (1 + ε cos θ ) = r L2
where, ε is the eccentricity of the particle’s motion. Which one of the following choice for ε gives rise to a parabolic trajectory? (a) ε = 0 Ans:
(c) 0 < ε < 1
(d) ε > 1
(b)
Solution: Q11.
(b) ε = 1
l m = (1 + ε cos θ ) for parabolic trajectory ε = 1 . r l2
A particle of unit mass moves along the x-axis under the influence of a potential,
V ( x ) = x( x − 2) . The particle is found to be in stable equilibrium at the point x = 2. The 2
time period of oscillation of the particle is (a) Ans:
π 2
(b) π
(c)
3π 2
(d) 2π
(b)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ∂V 2 2 2 V ( x ) = x( x − 2 ) ⇒ = ( x − 2 ) + 2 x( x − 2 ) = 0 ⇒ x = 2, x = ∂x 3 ∂ 2V ∂ 2V 2 ( ) ( ) = 2 x − 2 + 2 x − 2 + 2 x ⇒ = 2× 2 = 4 ∂x 2 ∂x 2
∂ 2V ∂x 2
⇒ω = Q12.
⇒ω = x=2
2π =2 T
⇒T =π .
A rod of proper length l0 oriented parallel to the x-axis moves with speed 2c/3 along the x-axis in the S-frame, where c is the speed of the light in free space. The observer is also moving along the x-axis with speed c/2 with respect to the S-frame. The length of the rod as measured by the observer is (b) 0.48l0
(a) 0.35l0 Ans:
(c) 0.87l0
(d) 0.97l0
(d)
Solution: l = l0 1 −
u2x = 0.97 l0 c2
Q13. A particle of mass m is attached to fixed point O by a weightless inextensible string of
length a. It is rotating under the gravity as shown in the figure. The
z
Lagrangian of the particle is
(
θ
)
1 L(θ , φ ) = ma 2 θ 2 + sin 2 θφ 2 − mga cos θ where θ and φ are the 2
polar angles. The Hamiltonian of the particles is (a) H =
1 2ma 2
1 H= 2ma 2 (c) H = Ans:
aθ
m
⎛ 2 pφ2 ⎞ ⎟ − mga cos θ ⎜ pθ + 2 ⎟ ⎜ sin θ ⎠ ⎝
(b)
g
⎛ 2 pφ2 ⎞ ⎜ pθ + ⎟ + mga cosθ 2 ⎜ ⎟ sin θ ⎝ ⎠
1 ( pθ2 + pφ2 ) − mga cos θ 2 2ma
(d) H =
1 ( pθ2 + pφ2 ) + mga cos θ 2 2ma
(b)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 Solution: H = Pθθ + Pφφ − L = Pθ θ + Pφ φ − ma 2 θ 2 + sin 2 θφ 2 + mga cos θ 2
(
)
Pφ ∂L P ∂L = ma 2 sin 2 θφ ⇒ φ = = Pθ ⇒ ma 2 θ = Pθ ⇒ θ = θ 2 and Pφ = ∂φ ma 2 sin 2 θ ma ∂θ
Put the value of θ and φ 2 2 ⎛ Pφ Pφ ⎞ ⎞⎟ Pθ 1 2 ⎜ ⎛ Pθ ⎞ 2 ⎛ ⎟ + mga cosθ H = Pθ × + Pφ × − − ma ⎜ ⎟ + sin θ ⎜⎜ ⎜ ⎝ ma 2 ⎠ 2 ma 2 ma 2 sin 2 θ ma 2 sin 2 θ ⎟⎠ ⎟ ⎝ ⎠ ⎝
Pθ2 Pφ2 Pφ2 Pθ2 H= − + − + mga cosθ ma 2 2ma 2 ma 2 sin 2 θ 2ma 2 sin 2 θ 1 H= 2ma 2
⎛ 2 Pθ2 ⎜ Pθ + ⎜ sin 2 θ ⎝
⎞ ⎟ + mga cos θ ⎟ ⎠
Statement for Linked Answer Questions 14 and 15:
Q14.
A particle of mass m slides under the gravity without friction along the parabolic path
y = ax 2 axis shown in the figure. Here a is a constant.
y
m x The Lagrangian for this particle is given by
Ans:
(a) L =
1 2 mx − mgax 2 2
(b) L =
1 m(1 + 4a 2 x 2 )x 2 − mgax 2 2
(c) L =
1 2 mx + mgax 2 2
(d) L =
1 m 1 + 4a 2 x 2 x 2 + mgax 2 2
(
)
(d)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 Solution: Equation of constrain is given by y = ax 2 , K.E T = m ( x 2 + y 2 ) 2 y = 2axx ⇒ T =
1 1 m ( x 2 + 4ax 2 x 2 ) = mx 2 (1 + 4ax 2 ) 2 2
V = − mgy = − mgax 2 . Since particle is moving downward direction so potential V is
negative. ∵ L = T −V ⇒ L =
(
)
1 m 1 + 4a 2 x 2 x 2 + mgax 2 2
Q15. The Lagrange’s equation of motion of the particle for above question is given by (b) m (1 + 4a 2 x 2 ) x = −2mgax − 4ma 2 xx 2
(a) x = 2 gax
(
)
(c) m 1 + 4a 2 x 2 x = 2mgax + 4ma 2 xx 2 Ans:
(d) x = −2 gax
(c)
Solution:
d ⎛ dL ⎞ dL ⇒ m(1 + 4a 2 x 2 ) x = 4ma 2 xx 2 + 2mgax ⎜ ⎟= dt ⎝ dx ⎠ dx
GATE- 2013
Q16.
In the most general case, which one of the following quantities is NOT a second order tensor? (a) Stress
(b) Strain
(c) Moment of inertia
(d) Pressure
Ans: (b) Solution: Strain is not a tensor. Q17.
An electron is moving with a velocity of 0.85c in the same direction as that of a moving photon. The relative velocity of the electron with respect to photon is
Ans:
(a) c
(b) − c
(c) 0.15c
(d) − 0.15c
(b)
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fiziks Q18.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The Lagrangian of a system with one degree of freedom q is given by L = αq 2 + βq 2 ,
where α and β are non-zero constants. If p q denotes the canonical momentum conjugate to q then which one of the following statements is CORRECT? (a) p q = 2 β q and it is a conserved quantity. (b) p q = 2 β q and it is not a conserved quantity. (c) p q = 2α q and it is a conserved quantity. (d) p q = 2αq and it is not a conserved quantity. Ans: (d) Solution: Q19.
∂L = pq but ∂q
∂L ≠0 ∂q
The relativistic form of Newton’s second law of motion is (a) F = (c) F =
mc
dv c 2 − v 2 dt
(b) F =
mc 2 dv c 2 − v 2 dt
m c 2 − v 2 dv c dt
(d) F = m
c 2 − v 2 dv dt c2
Ans: Solution: P =
mv 1−
v2 c2
⇒F=
dP dv =m ⋅ dt dt
1 −2v dv ⎛ 1⎞ + mv ⎜ − ⎟ ⋅ ⋅ 2 3/ 2 c dt ⎝ 2 ⎠ ⎛ v2 ⎞ v2 1− 2 1− 2 ⎟ ⎜ c ⎝ c ⎠ 1
⎛ ⎛ ⎞ 2 2 ⎜ v 2 ⎟ ⎜ 1− v ⎜ 1 ⎜ 1 dv dv 2c 2 c ⎟=m 1+ ⇒F =m ⎜ 2 dt dt ⎜ ⎛ v 2 ⎞ 3 2 v 2 ⎜ 2 ⎛1 − v ⎞ ⎟ ⎟ 1 − 2 ⎜⎜ ⎜ 2 ⎟⎟ ⎜ ⎜1 − c 2 ⎟ c ⎝ ⎝ c ⎠⎠ ⎠ ⎝⎝
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
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fiziks Q20.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Consider two small blocks, each of mass M, attached to two identical springs. One of the
springs is attached to the wall, as shown in the figure. The spring constant of each spring is k . The masses slide along the surface and the friction is negligible. The frequency of one of the normal modes of the system is,
Ans:
(a)
3+ 2 2
k M
(b)
3+ 3 2
k M
(c)
3+ 5 2
k M
(d)
3+ 6 2
k M
k
k M
M
(c)
Solution: T = V =
1 2 1 2 mx1 + mx 2 , 2 2
(
)
(
1 2 1 1 1 1 2 kx1 + k ( x 2 − x1 ) = kx12 + k x 22 + x12 − 2 x 2 x1 = k 2 x 2 + x 2 − 2 x 2 x1 2 2 2 2 2
⎛m 0 ⎞ ⎛ 2k T =⎜ ⎟; V = ⎜ ⎝ 0 m⎠ ⎝ −k
)
−k ⎞ ⎟ k ⎠
2k − ω 2 m −k 3+ 5 = 0 ⇒ ( 2 k − ω 2 m )( k − ω 2 m ) − k 2 = 0 ⇒ ω = 2 2 −k k −ω m
k m
GATE- 2014
Q21.
If the half-life of an elementary particle moving with speed 0.9c in the laboratory frame is
(
5 × 10 −8 s, then the proper half-life is _______________ ×10 −8 s. c = 3 × 10 8 m / s
Ans:
)
2.18
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES t0
Solution: t =
1−
Q22.
2
v c2
,
v2 t0 = t × 1 − 2 c
=
⇒ 2.18 ×10−8 s
t0 = 5 × 10 −8 × .19
Two masses m and 3m are attached to the two ends of a massless spring with force constant K . If m = 100 g and K = 0.3 N / m , then the natural angular frequency of oscillation is ________ Hz .
Ans:
0.318
Solution: f =
1 2π
Q23.
Ans:
k
μ
μ=
m1.m2 3m.m 3m = = m1 + m2 4m 4
ω=
4k = 2 = 0.318 Hz 3m
The Hamilton’s canonical equation of motion in terms of Poisson Brackets are (a) q = {q, H }; p = {p, H }
(b) q = {H , q}; p = {H , p}
(c) q = {H , p}; p = {H , p}
(d) q = {p, H }; p = {q, H }
(a)
Solution:
df ∂f ∂q ∂f ∂p ∂f = . + . + dt ∂q ∂t ∂p ∂t ∂t
∂f df df ∂f ∂H ∂f ∂H ∂f ⇒ = { f , H} + = . − . + dt ∂t dt ∂q ∂p ∂p ∂q ∂t dq dp = {q, H } and = { p, H } dt dt Q24.
A bead of mass m can slide without friction along a mass less rod kept at 45 o with the vertical as shown in the figure. The rod is rotating about the vertical axis with a constant angular speed ω . At any instant r is the distance of the bead from the origin. The zˆ momentum conjugate to r is ω (a) mr (b)
1 2
mr
1 (c) mr 2 (d) 2mr
m 45 o r
xˆ Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Ans:
(a)
Solution: L =
1 m(r 2 + r 2θ 2 + r 2 sin 2 θφ 2 ) − mgr cos θ 2
equation of constrain is θ = L=
π
and it is given φ = ω
4
1 1 1 m( r 2 + r 2 ω 2 ) − mgr 2 2 2
the momentum conjugate to r is p r = Q25.
∂L = p r = mr ∂r
A particle of mass m is in a potential given by V (r ) = −
a ar02 + r 3r 3
when a and r0 are positive constants. When disturbed slightly from its stable equilibrium position it undergoes a simple harmonic oscillation. The time period of oscillation is (a) 2π Ans:
mr03 2a
(b) 2π
m r0 a
3
2m r0 a
3
(d) 4π
mr03 a
(a)
a ar02 Solution: V ( r ) = − + 3 for equilibrium r 3r
∂ 2V 2a 4ar02 =− 3 + 5 ∂r 2 r r
ω= Q26.
(c) 2π
∂ 2V ∂r 2 m
r0
r0
∂V a 3ar02 = − =0 ∂r r 2 3r 4
r = ± r0
2a 4ar02 2a =− 3 + 5 = 3 r0 r0 r0
⇒ T = 2π
mr03 2a
A planet of mass m moves in a circular orbit of radius r0 in the gravitational potential V (r ) = −
k , where k is a positive constant. The orbit angular momentum of the planet is r
(a) 2r0 km
(b) 2r0 km
(c) r0 km
(d)
r0 km
Ans: (d) Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 2 dVeffect k J J k Solution: Veffctive = = − 3 + 2 =0 at r = r0 − ⇒ 2 r dr mr r 2mr
so J = r0 km Q27.
Given that the linear transformation of a generalized coordinate q and the corresponding momentum p ,
Q = q + 4ap P = q + 2p is canonical, the value of the constant a is _________________ Ans:
0.5
Solution: Q28.
∂Q ∂P ∂Q ∂P . − . = 0 ⇒ 1.2 − 4a.1 = 0 ⇒ a = 0.5 ∂q ∂p ∂p ∂q
The Hamiltonian of particle of mass m is given by H =
p2 α q2 .which one of the − 2m 2
following figure describes the motion of the particle in phase space? (a)
(b)
p
p
q
(c)
(d)
p
q
Ans:
q
p
q
(d)
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES GATE- 2015
Q29.
A satellite is moving in a circular orbit around the Earth. If T ,V and E are its average kinetic, average potential and total energies, respectively, then which one of the following options is correct? (a) V = −2T ; E = −T (c) V = −
(b) V = −T ; E = 0
T T ;E = 2 2
(d) V =
− 3T −T ;E = 2 2
Ans.: (a) Solution: From Virial theorem T =
∵V = Q30.
n +1 V where V ∝ r n +1 2
−k 1 ⇒ V ∝ ⇒ n = −2 ⇒ V = −2 T r r
In an inertial frame S , two events A and B take place at
(ct B
(ct A = 0, rA = 0) and
= 0, rB = 2 yˆ ) , respectively. The times at which these events take place in a frame
S ′ moving with a velocity 0.6cyˆ with respect to S are given by
(a) ct ′A = 0; ct B′ = − (c) ct ′A = 0; ct B′ =
3 2
3 2
(b) ct ′A = 0; ct ′B = 0 (d) ct ′A = 0; ct B′ =
1 2
Ans.: (a) Solution: Velocity of S ' with respect to S is v = .6c t A' =
t B' =
v y c2 v2 1− 2 c
for event A t A = 0, y = 0 so ct A' = 0
v y c2 v2 1− 2 c
for event B t B = 0, y = 2 so ct B' = −
tA −
tB −
3 2
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fiziks Q31.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The Lagrangian for a particle of mass m at a position r moving with a velocity v is given m by L = v 2 + Cr .v − V (r ) , where V (r ) is a potential and C is a constant. If pc is the 2 canonical momentum, then its Hamiltonian is given by 1 1 ( pc + Cr )2 + V (r ) ( pc − Cr )2 + V (r ) (a) (b) 2m 2m
(c)
pc2 + V (r ) 2m
(d)
1 2 pc + C 2 r 2 + V (r ) 2m
Ans.: (b) Solution: L =
m 2 v + Cr.v − V ( r ) 2
H = ∑ r pc − L = rpc − L
⇒
where v = r where L =
m 2 r + Cr.r − V ( r ) 2
p − Cr ∂L = pc = ( mr + Cr ) ⇒ r = c ∂r m 2
m ⎛ pc − Cr ⎞ ⎛ p − Cr ⎞ ⎛ pc − Cr ⎞ ⇒ H =⎜ c ⎟ pc − ⎜ ⎟ − cr ⎜ ⎟ +V (r ) 2⎝ m ⎠ ⎝ m ⎠ ⎝ m ⎠ 2
m ⎛ pc − Cr ⎞ ⎛ p − Cr ⎞ ⇒ H =⎜ c ⎟ ( pc − Cr ) − ⎜ ⎟ +V (r ) 2⎝ m ⎠ ⎝ m ⎠
( p − Cr ) ⇒H = c m
Q32.
2
( p − Cr ) − c 2m
2
+V (r )
⇒H =
1 2 ( pc − Cr ) + V ( r ) 2m
The Hamiltonian for a system of two particles of masses m1 and m2 at r1 and r2 having velocities v1 and v2 is given by H =
1 1 C m1v12 + m2v22 + zˆ ⋅ ( r1 × r2 ) , wrong where 2 2 2 ( r1 − r2 )
C is constant. Which one of the following statements is correct?
(a) The total energy and total momentum are conserved (b) Only the total energy is conserved (c) The total energy and the z - component of the total angular momentum are conserved (d) The total energy and total angular momentum are conserved Ans.: (c) Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Lagrangian is not function of time so energy is conserve and component of ( r1 × r2 ) are
Only in zˆ direction means potential is symmetric under φ so Lz is conserve. Q33.
A particle of mass 0.01 kg falls freely in the earth’s gravitational field with an initial velocity (0) = 10ms −1 . If the air exerts a frictional force of the form, f = −kv , then for k = 0.05 Nm −1 s , the velocity (in ms −1 ) at time t = 0.2 s is _________ (upto two decimal places). (use g = 10 ms −2 and e = 2.72 )
Ans.: Data given is incorrect u
dv dv k dv dv = dt ⇒ ∫ = Solution: m = mg − kv ⇒ =g− v ⇒ k k dt dt m 10 g− v g− v m m
0.2
∫ dt 0
u
m⎡ ⎡ k ⎤⎤ m ⎧⎡ ⎛ k ⎞⎤ 10k ⎞ ⎫ 0.2 ⎛ ⇒ − ⎢ ln ⎢ g − v ⎥ ⎥ = [t ]0 ⇒ − ⎨ ⎢ln ⎜ g − u ⎟ ⎥ − ln ⎜ g − ⎟ ⎬ = 0.2 k ⎩⎣ ⎝ m ⎠⎦ m ⎠⎭ k ⎣ ⎣ m ⎦ ⎦10 ⎝ ⇒− ⇒−
m⎧ ⎛ 0.05 ⎞ .05 ⎞ ⎫ ⎛ u ⎟ − ln ⎜10 − 10 × ⎨ln ⎜10 − ⎟ ⎬ = 0.2 k⎩ ⎝ 0.01 ⎠ .01 ⎠ ⎭ ⎝
m {ln (10 − 5u ) − ln ( −40 )} = 0.2 k
∵ ln ( −40 ) can not be defined. So given data are not correct.
Q34.
Consider the motion of the Sun with respect to the rotation of the Earth about its axis. If Fc and FCo denote the centrifugal and the Coriolis forces, respectively, acting on the
Sun, then (a) Fc is radially outward and FCo = Fc (b) Fc is radially inward and FCo = −2 Fc (c) Fc is radially outward and FCo = −2 Fc (d) Fc is radially outward and FCo = 2 Fc Ans.: (b)
Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
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fiziks Q35.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A particle with rest mass M is at rest and decays into two particles of equal rest masses
3 M which move along the z axis. Their velocities are given by 10 (a) v1 = v 2 = (0.8c )zˆ (c) v1 = −v 2 = (0.6c )zˆ
(b) v1 = −v 2 = (0.8c )zˆ (d) v1 = (0.6c )zˆ; v 2 = (− 0.8c )zˆ
Ans.: (b) M→
Solution:
3 3 M+ M 10 10
From momentum conservation 0 = P1 + P 2 ⇒ P1 = − P 2 ⇒ P1 = P2 From energy conservation E = E1 + E2 ⇒ Mc 2 =
3 Mc 2 3 Mc 2 + 10 v 2 10 v2 1− 2 1− 2 c c
⇒ Mc 2 =
3 Mc 2 5 v2 1− 2 c
⎛ v2 ⎞ 9 v 2 16 ⇒ 2 = ⇒ v = 0.8c ⎜1 − 2 ⎟ = 25 v ⎝ v ⎠ 25
GATE-2016
Q36.
The kinetic energy of a particle of rest mass m0 is equal to its rest mass energy. Its momentum in units of m0 c , where c is the speed of light in vacuum, is _______. (Give your answer upto two decimal places
Ans. :
1.73
Solution:
m0c 2 1−
2
v c2
= 2m0 c 2 ⇒ E
E 2 = p 2 c 2 + m02 c 4 ⇒ 4m02c 4 − m02c 4 = p 2 c 2 ⇒ p = 3m0 c = 1.732m0 c
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fiziks Q37.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES In an inertial frame of reference S , an observer finds two events occurring at the same
time at coordinates x1 = 0 and x 2 = d A different inertial frame S ′ moves with velocity v with respect to S along the positive x -axis. An observer in S ′ also notices these two
events and finds them to occur at times t1′ and t 2′ and at positions x1′ and x2′ respectively. 1
If Δt ′ = t 2′ − t1′ , Δx ′ = x 2′ − x1′ and γ =
v2 1− 2 c
(a) Δt ′ = 0, Δx ′ = γd (c) Δt ′ =
, which of the following statements is true?
(b) Δt ′ = 0, Δx ′ =
− γvd , Δx ′ = γd c2
(d) Δt ′ =
d
γ
− γvd d , Δx ′ = 2 γ c
Ans.: (c) ⎛ vx ⎜ t2 − 22 c Solution: t2' − t1' = ⎜ ⎜ v2 ⎜⎜ 1 − 2 c ⎝ ⇒ Δt ' = −
⎞ ⎛ vx ⎟ ⎜ t1 − 21 c ⎟−⎜ ⎟ ⎜ v2 ⎟⎟ ⎜⎜ 1 − 2 c ⎠ ⎝
⎞ ⎟ ⎟ ⇒ Δt ' = γΔt − γ vΔx ⎟ c2 ⎟⎟ ⎠
it is given
Δt = 0, Δx = d
⇒ Δx ' = γ ( Δx − vΔt ) it is given
Δt = 0, Δx = d
γ vΔx c2
⎛ ⎜ x − vt2 ' ' x2 − x1 = ⎜ 2 ⎜ v2 ⎜⎜ 1 − 2 c ⎝
⎞ ⎛ ⎟ ⎜ ⎟ − ⎜ x1 − vt1 ⎟ ⎜ v2 ⎟⎟ ⎜⎜ 1 − 2 c ⎠ ⎝
⎞ ⎟ ⎟ ⎟ ⎟⎟ ⎠
⇒ Δx ' = γ d Q38.
The Lagrangian of a system is given by L=
[
]
1 2 2 ml θ + sin 2 θϕ 2 − mgl cos θ , where m, l and g are constants. 2
Which of the following is conserved? (a) ϕ sin 2 θ
(b) ϕ sin θ
(c)
ϕ sin θ
(d)
ϕ sin 2 θ
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fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Ans.: (a) Solution: ϕ is cyclic coordinate so
∂L = pϕ ⇒ ml 2 sin 2 ϕ is constant hence m, l and g are ∂ϕ
constants. Then ϕ sin 2 θ Q39.
A particle of rest mass M is moving along the positive x -direction. It decays into two photons γ 1 and γ 2 as shown in the figure. The energy of γ 1 is 1 GeV and the energy of
γ 2 is 0.82 GeV . The value of M (in units of upto two decimal places) M
GeV ) is ________. (Give your answer c2 γ1 450 600
Ans.:
1.40
Solution: p=
γ2
p 2 c 2 + M 2 c 4 = E1 + E2 = 1.82GeV
E1 E 1GeV 1 .82GeV 1 1.11GeV + = cos θ1 + 2 cos θ 2 = c 2 c c c c 2
⇒ p 2 c 2 + m 2c 4 = 3.312 ⇒ m 2 c 4 = 3.312 − 1.21 = 2.077
⇒ m = 2.076 = 1.40
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