2.Classical Mechanics GATE

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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

( )

2 3

3 2 1 x kx x V = −λ , where k, λ > 0. Then, (A) x = 0 and λ k

x= are points of stable equilibrium

(B) x = 0 is a point of stable equilibrium and λ k

x= is a point of unstable equilibrium

(C) x = 0 and λ k

x= are points of unstable equilibrium (D) There are no points of stable or unstable equilibrium Ans: (b) Solution: 3 2 1 3 2 x kx V = −λ ₌ ₋ 2 ₌0 ∂ ∂ ⇒ kx x x V _λ λ k x x= = ⇒ 0, . x k x V _λ 2 2 2 − = ∂ ∂ ⇒ ve x V x At =+ ∂ ∂ = ⇒ 0, 2₂ (Stable) and ve x V k x At =− ∂ ∂ = ⇒ , 2₂ λ (unstable)

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Q3. A π0 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 (B) 0 s (C) 10 s (D) 20 s Ans: (a) Solution: c vc v t t − + = 1 1 0 1 1 0.6 10 1 0.6 + = − = × =10 2 20sec, c v c v t t + − = 1 1 0 2 1 0.6 10 1 0.6 − = + 1 10 5sec 2 = × = 1 2 t t 15sec ⇒ − =

Statement for Linked Answer Questions 4 and 5:

The Lagrangian for a simple pendulum is given by θ

(

1 cosθ

)

2

1 2 2 ₋ ₋

= ml mgl

L Q4. Hamilton’s equations are then given by

(A) sin ; ₂ ml p mgl p_θ =− θ θ = θ (B) sin ; ₂ ml p mgl p_θ = θ θ = θ (C) m p m p θ θ =− θ; θ = (D) _ml p l g p θ θ ⎟θ θ = ⎠ ⎞ ⎜ ⎝ ⎛ − = ; Ans: (b) Solution: θ

(

1 cosθ

)

2 2 2 − + = mgl ml P H ⇒∂H= − ⇒P_θ P_θ =mglsin ;θ ∂θ _ml2 P P H θ θ θ θ ⇒ = = ∂ ∂ .Q5. The Poisson bracket between θ and θis

(A)

{ }

θ,θ =1 (B)

{ }

, 1₂ ml = θ θ (C)

{ }

m 1 ,θ = θ (D)

{ }

l g = θ θ,

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{ }

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ = , ₂ , ml P_θ θ θ θ where ₂ ml Pθ θ = . _⎟⎟= ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ ⇒ θ θ θ θ θ θ θ θ P P P ml2 1 0 1 1⋅ ₂ − ml 2 1 ml = . GATE- 2011

Q6. A particle is moving under the action of a generalized potential

( )

, 1 ₂

q q q

q

V = + . The

magnitude of the generalized force is (A) 2

(

1₃

)

q q + _(B)

(

)

3 1 2 q q − _(C) 3 2 q (D) _q3 q Ans: (c) Solution: q d V V F dt q q ⎛∂ ⎞ ∂ − = ⎜_∂ ⎟ _∂ ⎝ ⎠ q 3 2 F q ⇒ = .

Q7. Two bodies of mass m and 2m are connected by a spring constant k. The frequency of the normal mode is (A) 3k 2/ m (B) k /m (C) 2k 3/ m (D) k 2/ m Ans: (a) Solution: μ ω= k m k m k 2 3 3 2 =

= where reduce mass

m m mm + = 2 2 μ 3 2m = .

Q8. Let (p, q) and (P, Q) be two pairs of canonical variables. The transformation Q=qα cos

( )

βp _,P=qαsin

( )

βp is canonical for (A) α = 2, β = 1/2 (B) α = 2, β =2 (C) α = 1, β = 1 (D) α = 1/2, β = 2 Ans: (d) Solution: =1 ∂ ∂ ⋅ ∂ ∂ − ∂ ∂ ⋅ ∂ ∂ q P p Q p P q Q

( )

cos

( )

(

sin

( )

)

sin

( )

1

cos 1 1 _× ₋ ₋ _× ₌ ⇒_α_qα− _β_p _qα_β _β_p _qα_β _β_p _α_qα− _β_p

(

cos2 sin2

)

1 2 1 1 1 2α− _β _β ₊ _β ₌ _⇒_αβ α− ₌ αq p p q , 2 2 1 = = ⇒α β . m ₂_m k

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Q9. 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 (B) 2m (C) 5m / 2 (D) 25 m / 8 Ans: (c)

Solution: From conservation of energy

2 1 2 2 2 2 2 2 1 1 c m c v mc c v mc ₌ − + − 2 1 2 2 2 1 2 c m c v mc ₌ − ⇒ 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 εcosθ

)

1 2 + = L m r

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 (b) ε = 1 (c) 0 < ε < 1 (d) ε > 1 Ans: (b) Solution: = ₂

(

1+εcosθ

)

l m r

l _{for parabolic trajectory}_ε ₌₁

.

Q11. A particle of unit mass moves along the x-axis under the influence of a potential,

( ) (

_x _{= x}_x ₋₂

)

2

V . The particle is found to be in stable equilibrium at the point x = 2. The time period of oscillation of the particle is

(a) 2 π _(b)_π _(c) 2 3π _(d)₂_π Ans: (b)

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( ) (

_x _{= x}_x ₋₂

)

2 V =

(

−2

)

2 +2

(

−2

)

=0 ∂ ∂ ⇒ x x x x V 2 x 2, x 3 ⇒ = =

(

x

)

(

x

)

x x V 2 2 2 2 2 2 2 2 + − + − = ∂ ∂ 4 2 2 2 2 = × = ∂ ∂ ⇒ x V 2 2 2 = ∂ ∂ = ⇒ x x V ω ⇒ = 2 =2 T π ω ⇒ T =π _.

Q12. 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

(a) 0.35l0 (b) 0.48l0 (c) 0.87l0 (d) 0.97l0 Ans: (d) Solution: 2 0 1 2 0.97 0 x u l l l c = − =

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

Lagrangian of the particle is

( )

θ φ

(

θ sin θφ

)

cosθ

2 1

, _ma2 2 2 2 _mga

L = + − where θ andφ are the

polar angles. The Hamiltonian of the particles is

(a) θ θ φ θ cos sin 2 1 2 2 2 2 mga p p ma H _⎟⎟− ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = (b) θ θ φ θ cos sin 2 1 2 2 2 2 mga p p ma H _⎟⎟+ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = (c)

(

_θ _φ

)

cosθ 2 1 2 2 2 p p mga ma H = + − (d)

(

_θ _φ

)

cosθ 2 1 2 2 2 p p mga ma H = + + Ans: (b) z θ θ a m g

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Solution: H =Pθθ +Pφφ−L θθ φφ

(

θ sin θφ

)

cosθ

2 1_ma2 2 2 2 _mga P P + − + + = θ θ P L = ∂ ∂ 2 2 P ma P ma θ θ ⇒ θ = ⇒ θ = and 2 2 2 2 P L P ma sin ma sin φ φ ∂ = = θφ ⇒ φ = ∂φ θ

Put the value of θ and φ

θ θ θ θ φ θ φ φ θ θ cos sin sin 2 1 sin 2 2 2 2 2 2 2 2 2 2 _ma mga P ma P ma ma P P ma P P H _⎟⎟+ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − × + × = θ θ θ φ φ θ θ _cos sin 2 sin 2 2 2 2 2 2 2 2 2 2 2 mga ma P ma P ma P ma P H = − + − + θ θ θ θ cos sin 2 1 2 2 2 2 mga P P ma H _⎟⎟+ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + =

Statement for Linked Answer Questions 14 and 15:

Q14. A particle of mass m slides under the gravity without friction along the parabolic path

2

ax

y = axis shown in the figure. Here a is a constant.

The Lagrangian for this particle is given by

(a) 2 2 2 1 mgax x m L= − (b)

(

₁ ₄ 2 2

)

2 2 2 1 mgax x x a m L= + − (c) 2 2 2 1 mgax x m L= + (d)

(

₁ ₄ 2 2

)

2 2 2 1 mgax x x a m L= + + Ans: (d) m x y

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Solution: Equation of constrain is given by _{y ax , K.E T}2 1_{m x}

(

2 _y2

)

2 = = +

(

2 2 2

)

1 y 2axx T m x 4ax x 2 = ⇒ = + 2

(

₁ ₄ 2

)

2 1 ax x m + = 2 mgax mgy

V =− =− . Since particle is moving downward direction so potential V is negative.

(

2 2

)

2 2 4 1 2 1 mgax x x a m L V T L= − ⇒ = + + ∵

Q15. The Lagrange’s equation of motion of the particle for above question is given by (a) x=2gax (b) _m

(

_{1 4}₊ _{a x x}2 2

)

_{= −}₂_mgax₋₄_{ma xx}2 2

(c) _m

(

₁₊₄_a2_x2

)

_x₌₂_mgax₊₄_ma2_x_x2_(d)_x₌₋₂_gax Ans: (c) Solution: d dL dL _m_{(1 4}_{a x x}2 2₎ ₄_{ma xx}2 2 ₂_mgax 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

(a) c (b) c− (c) 0.15c (d) −0.15c Ans: (b)

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Q18. The Lagrangian of a system with one degree of freedom q is given by_L₌_α_q2 ₊_β_q2_,

where αand β are non-zero constants. If pq denotes the canonical momentum conjugate to q then which one of the following statements is CORRECT?

(a) pq =2βq and it is a conserved quantity.

(b) p_q =2βq and it is not a conserved quantity. (c) p_q =2αq and it is a conserved quantity. (d) p_q =2αq and it is not a conserved quantity. Ans: (d) Solution: L p_q q ∂ ₌ ∂ but 0 L q ∂ _≠ ∂

Q19. The relativistic form of Newton’s second law of motion is (a) dt dv v c mc F 2 2 ₋ = (b) dt dv c v c m F 2 2 ₋ = (c) dt dv v c mc F ₂ ₂ 2 − = (d) dt dv c v c m F ₂ 2 2 ₋ = Ans: Solution: 2 2 1 c v mv P − = _3/2 ₂ 2 2 2 2 1 1 1 2 2 1 1 dP dv v dv F m mv dt dt _v _v c dt c c − ⎛ ⎞ ⇒ = = ⋅ + _⎜− _⎟⋅ ⋅ ⎝ ⎠ ⎛ ⎞ − _⎜ − _⎟ ⎝ ⎠ 2 2 2 2 3 2 2 ₂ ₂ 2 2 ₂ 1 1 1 ₂ 1 2 1 1 ₁ v v dv _c dv _c F m m dt v v dt _v c c _c ⎛ ⎞ ⎛ ⎞ _⎜ _⎟ ⎜ ⎟ _⎜ ₋ _⎟ ⎜ ⎟ ⇒ = _⎜ + _⎟= _⎜ _⎟ ⎛ ⎞ _⎜_⎛ _⎞ _⎟ − ⎜ ⎟ − ⎜_⎝ ⎜_⎝ ⎟_⎠⎟_⎠ _⎜_⎜ − _⎟ _⎟ ⎝ ⎠ ⎝ ⎠

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Q20. 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,

(a) M k 2 2 3+ (b) M k 2 3 3+ (c) M k 2 5 3+ (d) M k 2 6 3+ Ans: (c) Solution: 2 2 2 1 2 1 2 1 x m x m T = + ,

(

)

2 1 2 2 1 2 1 2 1 x x k kx V = + −

(

2 ₂ ₁

)

1 2 2 2 1 2 2 1 2 1 x x x x k kx + + − =

(

₂ 2 2 ₂ ₂ ₁

)

2 1 x x x x k + − = 0 2 ; 0 m k k T V m k k − ⎛ ⎞ ⎛ ⎞ =_⎜ _⎟ = _⎜ _⎟ − ⎝ ⎠ ⎝ ⎠ 2 2 2 0 k m k k k m ω ω − − = − − ⇒

(

)(

)

2 2 2 2k−ω m k−ω m −k =0⇒ω 3 5 2 k m + = GATE- 2014

Q21. If the half-life of an elementary particle moving with speed 0.9c in the laboratory frame is

, 10

5_× −8_s _{then the proper half-life is _______________}_×₁₀−8 _s_.

(

_c₌₃_×₁₀8_m_/_s

)

Ans: 2.18

M M

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Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: 2 2 0 1 c v t t − = , ₀ 1 ₂2 c v t t = × − = 5 10 8 .19 0 = × − × t _⇒_{2.18 10 s}_× −8

Q22. Two masses mand 3mare attached to the two ends of a massless spring with force constant K . If m 100= gand K =0.3N/m, then the natural angular frequency of oscillation is ________ Hz . Ans: 0.318 Solution: 1 2 k f π μ = 1 2 1 2 . 3 . 3 4 4 m m m m m m m m μ = = = + 4 2 3 k m ω= = =0.318 Hz

Q23. 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 Ans: (a) Solution: df f . q f . p f dt q t p t t ∂ ∂ ∂ ∂ ∂ = + + ∂ ∂ ∂ ∂ ∂ . . df f H f H f dt q p p q t ∂ ∂ ∂ ∂ ∂ = − + ∂ ∂ ∂ ∂ ∂

{

,

}

df f f H dt t ∂ ⇒ = + ∂

{

,

}

dq q H dt = and

{

,

}

dp p H dt =

Q24. A bead of mass mcan slide without friction along a mass less rod kept at _{45 with the}o

vertical as shown in the figure. The rod is rotating about the vertical axis with a constant angular speedω . At any instant ris the distance of the bead from the origin. The momentum conjugate to ris (a) mr (b) mr 2 1 (c) mr 2 1 zˆ ω m o 45

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Ans: (a)

Solution: L=1₂m(r2 +r2θ2 +r2sin2θφ2)−mgrcosθ equation of constrain is 4 π θ = and it is given φ =ω L m r r mgr 2 1 ) 2 1 ( 2 1 2 ₊ 2 2 ₋ = ω

the momentum conjugate to r is

r L p_r ∂ ∂ = = p_r =mr Q25. A particle of mass mis in a potential given by

( )

02 3 3 ar a V r r r = − +

when aand r₀are positive constants. When disturbed slightly from its stable equilibrium position it undergoes a simple harmonic oscillation. The time period of oscillation is (a) a mr 2 2 3 0 π (b) a r m 3 0 2π (c) a r m 3 0 2 2π (d) 3 0 4 mr a π Ans: (a) Solution:

( )

2 0 3 3 ar a V r r r = − + for equilibrium 2 0 2 4 3 0 3 ar V a r r r ∂ ₌ ₋ ₌ ∂ r= ± r0 0 2 2 2 0 0 2 3 5 3 5 3 0 0 0 4 4 2 2 2 r ar ar V a a a r r r r r r ∂ _{= −} ₊ _{= −} ₊ ₌ ∂ 0 2 2 ₃ 0 2 2 r V r _mr T m a ω π ∂ ∂ = ⇒ =

Q26. A planet of mass mmoves in a circular orbit of radius r₀ in the gravitational potential

( )

r k r

V =− , where kis a positive constant. The orbit angular momentum of the planet is (a) 2r₀km (b) 2r0km (c)r0km (d) r km 0

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Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: r k mr J Veffctive = ₂ − 2 2 2 3 2 effect dV _J _k dr mr r ⇒ = − + =0 at r=r₀ so J = r₀km

Q27. Given that the linear transformation of a generalized coordinate q and the corresponding momentum p , ap q Q= +4 p q P= +2

is canonical, the value of the constant ais _________________ Ans: 0.5 Solution: . . =0 ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ q P p Q p P q Q 1.2 4 .1 0a ⇒ − = ⇒ =a 0.5

Q28. The Hamiltonian of particle of mass mis given by

2 2 2 2 _q m p

H = −α .which one of the following figure describes the motion of the particle in phase space?

(a) (b) (c) (d) Ans: (d) p q p q p q p q

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GATE- 2015

Q29. A satellite is moving in a circular orbit around the Earth. If VT, 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 (b) V =−T;E =0 (c) 2 ; 2 T E T V =− = (d) 2 ; 2 3 T E T V = − = − Ans.: (a)

Solution: From Virial theorem 1 2 n T = + V where _V _∝_rn+1 1 2 k V V n r r − = ⇒ ∝ ⇒ = − ∵ ⇒ V = −2 T

Q30. In an inertial frame S, two events A and B take place at

(

ct_A =0,r_A =0

)

and

(

ct_B =0,r_B =2yˆ

)

, respectively. The times at which these events take place in a frame

S′moving with a velocity0.6cˆywith respect to Sare given by (a) 2 3 ; 0 ′ =− = ′_A ct_B t c (b)ct′_A =0; ct′_B =0 (c) 2 3 ; 0 ′ = = ′_A ct_B t c (d) 2 1 ; 0 ′ = = ′_A ct_B t c Ans.: (a)

Solution: Velocity of_{S with respect to}' _S_is_v₌_.6_c 2 ' 2 2 1 A A v t y c t v c − = − for event A t_A =0,y= so 0 ' ₀ A ct = 2 ' 2 2 1 B B v t y c t v c − = − for event B t_B =0,y= so 2 ' 3 2 B ct = −

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Q31. The Lagrangian for a particle of mass mat a position r moving with a velocity vis given by L= mv +Cr.v−V

( )

r

2

2 _{, where}_V

( )

_r _{is a potential and}_C_{is a constant. If}

c

p is the canonical momentum, then its Hamiltonian is given by

(a)

(

p Cr

)

V

( )

r m c + + 2 2 1 (b)

(

p Cr

)

V

( )

r m c− + 2 2 1 (c) V

( )

r m p_c + 2 2 (d) p C r V

( )

r m c + + 2 2 2 2 1 Ans.: (b) Solution: 2 .

( )

2 m L= v +Cr v V r− where v r= c c H =

∑

r p − =L rp −L where 2 _.

( )

2 m L= r +Cr r V r−

(

)

c c p Cr L p mr Cr r r m − ∂ ⇒ = = + ⇒ = ∂

( )

2 2 c c c c p Cr m p Cr p Cr H p cr V r m m m − − − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⇒ =_⎜ _⎟ − _⎜ _⎟ − _⎜ _⎟+ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(

)

2

( )

2 c c c p Cr m p Cr H p Cr V r m m − − ⎛ ⎞ ⎛ ⎞ ⇒ =_⎜ _⎟ − − _⎜ _⎟ + ⎝ ⎠ ⎝ ⎠

(

) (

2

)

2

_{( )}

2 c c p Cr p Cr H V r m m − − ⇒ = − + 1

(

)

2

( )

2 c H p Cr V r m ⇒ = − +

Q32. 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

(

)

(

)

2 2 1 1 2 2 2 1 2 1 2 1 1 _ˆ 2 2 C H m v m v z r r r r = + + ⋅ × − , wrong where

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

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Solution: Lagrangian is not function of time so energy is conserve and component of

(

r r₁× ₂

)

are Only in ˆz direction means potential is symmetric under φ so L is conserve. _z

Q33. A particle of mass 0.01 kg falls freely in the earth’s gravitational field with an initial velocity

( )

₀ _{= ms . If the air exerts a frictional force of the form,}₁₀ −1 _f ₌₋_kv_{, then for}

s Nm

k ₌₀_.₀₅ −1 _{, the velocity (in}_{ms ) at time}−1 _t ₌₀_.₂_s_{is _________ (upto two decimal}

places). (use _g ₌₁₀_ms−2_and_e₌₂_.₇₂₎

Ans.: Data given is incorrect Solution: mdv mg kv dt = − dv k g v dt m ⇒ = − dv dt k g v m ⇒ = − 0.2 10 0 u _dv dt k g v m ⇒ = −

∫

[ ]

0.2 0 10 ln u m k g v t k m ⎡ ⎡ ⎤⎤ ⇒ − _⎢ _⎢ − _⎥_⎥ = ⎣ ⎦ ⎣ ⎦ 10 ln ln 0.2 m k k g u g k m m ⎧⎡ ⎛ ⎞⎤ ⎛ ⎞⎫ ⇒ − _⎨_⎢ _⎜ − _⎟_⎥− _⎜ − _⎟_⎬= ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎩ ⎭ 0.05 .05 ln 10 ln 10 10 0.2 0.01 .01 m u k ⎧ ⎛ ⎞ ⎛ ⎞⎫ ⇒ − _⎨ _⎜ − _⎟− _⎜ − × _⎟_⎬= ⎝ ⎠ ⎝ ⎠ ⎩ ⎭

(

)

(

)

{

ln 10 5 ln 40

}

0.2 m u 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 c

F and F_Co denote the centrifugal and the Coriolis forces, respectively, acting on the Sun, then

(a) F_c is radially outward and F_Co =F_c

(b) F_c is radially inward and F_Co =−2F_c

(c) F_c is radially outward and F_Co =−2F_c

(d) F_c is radially outward and F_Co =2F_c

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Q35. A particle with rest mass M is at rest and decays into two particles of equal rest masses M

10 3

which move along the z axis. Their velocities are given by (a) v₁ =v₂ =

(

0.8c

)

zˆ (b) v₁ =−v₂ =

(

0.8c

)

zˆ (c) v₁ =−v₂ =

(

0.6c

)

zˆ (d) v₁ =

(

0.6c

)

zˆ; v₂ =

(

−0.8c

)

zˆ Ans.: (b) Solution: 3 3 10 10 M → M+ M

From momentum conservation

0 P= 1+P2⇒P1= −P2 ⇒ P₁ = P₂

From energy conservation E E= ₁+E₂

2 2 2 2 2 2 2 3 3 10 10 1 1 Mc Mc Mc v v c c ⇒ = + − − 2 2 2 2 3 5 1 Mc Mc v c ⇒ = − 2 2 2 2 9 16 1 0.8 25 25 v v v c v v ⎛ ⎞ − = ⇒ = ⇒ = ⎜ ⎟ ⎝ ⎠ GATE-2016

Q36. The kinetic energy of a particle of rest mass m ₀ is equal to its rest mass energy. Its momentum in units of m₀c, where c is the speed of light in vacuum, is _______.

(Give your answer upto two decimal places Ans. : 1.73 Solution: 2 2 0 0 2 2 2 1 m c m c E v c = ⇒ − 2 2 2 2 4 2 4 2 4 2 2 0 4 0 0 3 0 1.732 0 E = p c +m c ⇒ m c −m c = p c ⇒ =p m c= m c

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Q37. In an inertial frame of reference S, an observer finds two events occurring at the same time at coordinates x₁=0 and x₂ =dA 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 t′₁ and t′₂ and at positions x′₁ and x′₂ respectively. If Δt′=t′₂ −t₁′,Δx′=x₂′ −x₁′and 2 2 1 1 c v − =

γ , which of the following statements is true?

(a) Δt′=0,Δx′=γd (b) γ d x t′= Δ ′= Δ 0, (c) x d c vd t′= −γ Δ ′=γ Δ ₂ , (d) γ γ d x c vd t′= − Δ ′= Δ ₂ , Ans.: (c) Solution: 2 1 2 2 1 2 ' ' 2 1 ₂ ₂ 2 2 1 1 vx vx t t c c t t v v c c ⎛ ⎞ ⎛ ⎞ ⎜ − ⎟ ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ − =_⎜ _{⎟ ⎜}− _⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ' 2 v x t t c γ γ Δ ⇒ Δ = Δ − it is given Δ = Δ = t 0, x d ' 2 v x t c γ Δ ⇒ Δ = − ' ' 2 2 1 1 2 1 ₂ ₂ 2 2 1 1 x vt x vt x x v v c c ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ − − ⎜ ⎟ ⎜ ⎟ − =_⎜ _{⎟ ⎜}− _⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ _{⇒ Δ =}_x' _γ

(

_{Δ − Δ it is given}_{x v t}

)

_{Δ = Δ =}_t _0, _{x d} ' x γd ⇒ Δ =

Q38. The Lagrangian of a system is given by

[

θ sin θϕ

]

cosθ

2

1_ml2 2 2 2 _mgl

L= + − , where m, and g are constants. l Which of the following is conserved?

(a) ϕ_sin2θ _(b)_ϕ_sin_θ _(c)

θ ϕ sin (d) θ ϕ 2 sin

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Ans.: (a)

Solution: ϕis cyclic coordinate so L p_ϕ ml2sin2ϕ ϕ

∂ ₌ _⇒

∂ is constant hence m, and g are l constants. Then ϕ_sin2θ

Q39. A particle of rest mass M is moving along the positive x-direction. It decays into two photons γ₁ and γ₂ as shown in the figure. The energy of γ₁ is GeV1 and the energy of

2

γ is 0.82GeV. The value of M (in units of ₂ c GeV

) is ________. (Give your answer upto two decimal places)

Ans.: 1.40 Solution: p c2 2+M c2 4 =E₁+E₂ =1.82GeV 1 2 1 2 1 1 .82 1 cos cos 2 2 E E GeV GeV p c θ c θ c c = + = + =1.11GeV c 2 2 2 4 _3.312 p c m c ⇒ + = 2 4 _{3.312 1.21 2.077} m c ⇒ = − = 2.076 1.40 m ⇒ = = M ₄₅0 0 60 2 γ 1 γ