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C SIR NET, GATE, IIT-JAM, UGC NET , TIFR, IISc , JEST , JNU, BHU , ISM , IBPS, CSAT, SLET, NIMCET, CTET
1. Tw enty four clerk can clear 180 f iles in 15 days. Number of clerk required to clear
240 files in 12 days is
(1) 38
(2) 39
(3) 40
(4) 42
2. In the given f igure, RA = SA = 9cm and QA = 7cm. If PQ is the diameter, then radius is
AR S
Q
P
(1)65 cm7
(2) 130 cm7
(3) 8 cm
(4) None
3. If the circles are draw n with radii R1, R2, R3 w ith centre at the vertices of a triangle as show n in f igure. Side of triangle is a, b, c respectively, then R1 + R2 + R3 is equal to
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18. Find the function whose laplace transform is 2
2 2 2
s(s a )+
(1) +1[at sin at cos at]
2a
(2) 1
(a sin at cos at )2a
+ −
(3) 1
[a cosat sinat]2a
+ +
(4) 1
[at cosat sin at]2a
+
19. The Lagrange equation of motion of two rigid bodies of masses ‘m’ and ‘2m’ are connected by a light flexible spring of spring constant K. what is the Lagrange equation of motion.
(1) k
x x 0m
+ =ɺɺ
(2) k
x x 02m
+ =ɺɺ
(3) 3k
x x 02m
+ =ɺɺ
(4) 5k
x x 02m
+ =ɺɺ
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20. In the fol lowing indicator diagram, the net amount of work done will be
(1) Positive
(2) Negative
(3) Zero
(4) Infinity
21. A particle moves in a plane under the influence of a force, acting towards a centre of force whose
magnitude is 2
2 2
1 r – 2rrF 1 –
r c
=
ɺ ɺɺ where r is the distance of the particle to the centre of force, then
the Lagrangian for the motion in a plane is
(1) 2 2 2 2
2
r r 1 1 rL – –
2 2 r rcθ= +ɺɺ ɺ
(2) 2 2 2 2
2 2
r r 1 2 rL
2 2 r c rθ
= + + +ɺɺ ɺ
(3) 2 2 2
2 2
r r 2 1 rL –
2 2 r c rθ
= + +ɺɺ ɺ
(4) 2 2
2 2
r 1 1 rL –
2 r c r= +
ɺ ɺ
22. Calculate the Fermi energy in electron volt for sodium assuming that i t has one free electron per atom. Given density of sodium = 0.97 g cm–3, atomic weight of sodium is 23.
(1) 3.541 eV
(2) 3.451 eV
(3) 5.135 eV
(4) 3.145 eV
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23. The paramagnetic contribution to the magnetic susceptibili ty per m3 of potassium, for which the Fermi energy is 2.1 eV is at wt. of potassium is 39.1 gm and density of potassium is 0.86 × 103 kgm–3.
(1) 420.5 × 10–8
(2) 450.2 ×10+8
(3) 420.5× 10–6
(4) 425.2×10+6
24. The figure shows the inverse magnetic susceptibili ty (1/χ) (dimensionless) a s a function of temperature for a paramagnetic material. Calculate the concentration of magnetic ions, i f they are assumed to be Co2+ with configuration 3d7.
(1) 5·4 × 1023
(2) 5·1 × 1026
(3) 6·9 × 1023
(4) 6·9 × 1026
25. A 3D structure of current carrying wire is as shown in the figure. The magnetic force experienced by charge particle of mass m and charge q, when it is crossing origin with velocity v
along +ve Y-axis will
be
(1) 0 0q qV8R 4 R
µ Ι ν µ Ι + π i
(2) 0 q 2 ˆ ˆ1 –8R
µ Ι ν + π i k
(3) 0 q ˆ–8
µ Ι νk
(4) Zero
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26. If E f(0) and Ef are Fermi levels of a metal at 0K and 30000 K, then what is the value of f
f( 0)
EE
if
E f(0) = 7 eV.
(1) 0.119
(2) 0.88
(3) 1.113
(4) 1.188
27. The small (rotational) Raman displacement for HCI molecule is 41·6 cm–1. Find the internuclear distance between the atoms forming the molecule. Given : h = 6·63 × 10–34 J s, c = 3·0 × 108 m s–1 and NA = 6·023 × 1023 mol–1.
(1) 1·29 A°
(2) 2·29 A°
(3) 2·49 A°
(4) 0·64 A°
28. For the given circuit the the open loop gain is 12000 and R1 = 120 kΩ and Rf = 600 kΩ . V i = 1.2 V. Find the exact output voltage for the inverting operational amplifier.
(1) – 5 V
(2) – 6 V
(3) – 5·99 V
(4) – 4·99 V
29. Oxygen has nuclear spin of 5/2. The NMR of oxygen gives
(1) 2 lines
(2) 3 lines
(3) 4 lines
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30. A metal strain gauge factor of two. Ιts nominal resistance is 120 ohms. It undergoes strain 10–5, the value of change of resistance in response to the strain is
(1) 240 ohms
(2) 2 × 105 ohm
(3) 2.4 × 10–3 ohm
(4) 1.2 × 10–3 ohm
31. Evaluate V
Fd ,τ∫ ∫∫
where 2F xyz=
over the prism placed at origin as shown in the Figure.
(1) 13
(2) 23
(3) 19
(4) 29
32. A star initially has 1040 deuterons. It produces energy via the processes
1H2 + 1H2 → 1H3 + p and 1H2 + 1H3 → 2He4 + n
If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of
(1) 106 sec
(2) 108 sec
(3) 1012 sec
(4) 1016 sec
The masses of the nuclei are as follows: M (H2) = 2·014 amu; M (p) = 1·007 amu;
M (n) = 1·008 amu; M (He4) = 4·001 amu.
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33. A γ-ray photon produces an electron-positron pair, each moving with a K.E. of 0·01 MeV. The energy of the γ-ray photon is
(1) 1·02 MeV
(2) 1·04 MeV
(3) 2·08 MeV
(4) 1·03 MeV
34. The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T 2 and T 1 (T 2 > T 1).
2KT2 T1
x 4x
The rate of heat transferred through the slab, in a steady state is 2 1A(T – T )Kf,
x
with f equals to
(1) 1
(2) 1/2
(3) 2/3
(4) 1/3
35. In quark model what is the state of η°
(1) ud
(2) 1
(uu – dd)2
(3) 1
(uu dd)2
+
(4) 1
(us – su)2
PART C (36-55)
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38. A reversible engine works between three thermal reservoirs, A,B and C . The engine absorbs an equal amount of heat from the thermal reservoirs A and B kept at temperatures TA and TB respectively, and rejects heat to the thermal reservoir C kept at temperature TC. The efficiency of the engine is α times the efficiency of the reversible engine, which works between the two reservoirs A and C. which one of the fol lowing relation statement is correct ?
(1) ( ) ( )A A
B C
T T2 – 1 2 1 –
T T= α + α
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39. A quantum mechanical particle of mass m is confined in three-dimensional infinite square well
potential of side ‘a’. The eigen-energy of particle is given as E = π ℏ
2 2
2
9.
ma The state is
(1) 4 fold degenerate
(2) 3-fold degenerate
(3) 2-fold degenerate
(4) Non-degenerate
40. Ten grammas of water at 20° C is converted into at – 10°C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4.2 J/gK and that of ice to be half of this value, and taking the latent heat of fusion of ice at 0°C to be 335 J/g , the total entropy of the system is .
(1) Zero
(2) 16.02 JK–1
(3) – 15.63 JK–1
(4) 15.63JK–1
41. A sphere rolls down a rough included plane; i f x be the distance of the point of contact of the sphere from a fixed point on the plane, find the acceleration.
(1) 5
g sin7
α
(2) 5
g sin14
α
(3) mg sin α
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42. A symmetrical top with moments of inertia Ιn = Ιy and Ιz in the body axes frame is described by the Hamiltonian
( )2 2 2x y z
x
1 1H L L L
2I 2Ix= + +
Here moments of inertia are parameters and not operators. Lx Ly and Lz are the angular momentum operator in the body axes frame.
(i) The eigenvalues of the Hamiltonian
(1) 2
2 2
x z x
1 1( + 1) + – m
2 2 2
Ι Ι Ι
ℏℓ ℓ ℏ
(2) 2
2 2
x z x
1 1( + 1) + m
2 2 2
+ Ι Ι Ι
ℏℓ ℓ ℏ
(3) 2 2
2
x z x
1 1 m– ( 1)
2 2 2
+ + Ι Ι Ι
ℏℏ ℓ ℓ
(4) 2 2
2
x z x
1 1 m( 1)
2 2 2
+ + + Ι Ι Ι
ℏℏ ℓ ℓ
(ii) Expected value for a measurement of Lx + Ly + Lz for any state is
(1) Zero
(2) –ℏm
(3) ℏm
(4) m2ℏ
43. If we take in the semi-empirical mass formula ac = 0.58 MeV and aa = 19.3 MeV. Then possible atomic number of most stable nuclei of mass number 64.
(1) 26
(2) 29
(3) 32
(4) 33
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44. 36 g of water at 30°C are converted into steam at 250°C at constant atmospheric pressure. The specific heat of water is assumed constant at 4.2 J/g K and the latent heat of vaporization at 100°C is
2260 J/g. For water vapour, assume pV = mRT where R = 0.4619 kJ/ kg K, and pC
R = a + bT + cT°,
where a = 3.634,
b = 1.195 × 10–3 K–1 and c = 0.135 × 10–6 K–2
Calculate the entropy change of the system.
(1) 0.2181 kJ/K
(2) 0.0235 kJ/K
(3) 273.1 J/K
(4) 314.3 J/K
45. A perpendicularly polarized wave propagates from region 1(εr1 = 8.5, µr1=1, σ1 = 0) to region 2, free space, with an angle of incidence of 15°. Given i
0E 1.0 V /m= µ , then r0E , is–
(1) 1.62 µV/m
(2) 0.623 µV/m
(3) 4.23 µV/m
(4) 7.75 mV/m
46. A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on coll ision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?
(1) 9 m
(2) 6 m
(3) 3 m
(4) 2 m
47. Calculate the amount of energy released if all the deuterium atoms in the water in the lake of area
about 105 sq. miles and of depth 1
20the mile area used up in fusion.
(1) 2.18 × 1038 MeV
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48. The maximum wave length of photons that can be detected by a photo diode made of a semiconductor of band gap 2 eV is about
(1) 620 nm
(2) 700 nm
(3) 740 nm
(4) 860 nm
49. The three electronic circuits marked (i), (i i ) and (ii i ) in the figure below can al l work as logic gates, where the input signals are either 0V or 5V and the output is V0.
Identify the correct combination of logic gates (i), (i i ), (ii i ) in the options given below.
(1) NOR, XOR, AND
(2) OR, NAND, NOR
(3) NAND, AND, XOR
(4) AND, OR, NOR
Statement for Linked Answer Question 50(i) and 50(ii)
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This particle has a charge q and is placed in a uniform electric field E parallel to the x – axis, E = ˆEx .
52.(i). The Hamiltonian of the particle
(1) 2
2 2P 1m x x
2m 2+ ω −ε
(2) 2
2 2P 1m x x
2m 2+ ω +ε
(3) 2
2 2P 1m x x
2m 2ω − ε
(4) 2
2 2P 1m x x
2m 2ω + ε
52(ii). Perform a coordinate transformation y = ax+b (where a and b are constant / such that in the y coordinate the Hamiltonian is similar to that of a one – dimensional harmonic oscil lator (with no charge) What are a and b
(1) a = 1 , b = ε / mω2
(2) a = 1 , b = – ε / mω2
(3) a = ε / mω2 , b = 1
(4) a =– ε / mω 2 , b = 1
52(iii). The energy eigenvalues of the system is
(1) 2
2
1 1n
2 2 mε ω + − ω
ℏ
(2) 2
2
1 3n
2 2 mε ω + ω
ℏ
(3) 2
2
1 1 ew n
2 2 2mwℏ
+ −
(4) 2
2
1 3n
2 2 2mε ω + − ω
ℏ
53. The equation x3 – x2 + 4x – 4 = 0 is to be solved using the Newton-Rephson method. If x = 2 is taken as the initial approximation using this method will be
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20.(2) Cycle ‘1’ is clockwise so work done during cycle ‘1’ is positive . Similarly cycle ‘2’ is anticlockwise and work done during cycle ‘2’ becomes negative.
But area of cycle ‘2’ is greater than area of cycle ‘1’. So resultant work is negative.
21.(1) Here the expression for F represents the force between two charges in Weber’s electrodynamics.
We have 2
r 2 2
1 r – 2rrF 1 –
r c
=
ɺ ɺɺ
Taking U = qφ – q(A ⋅⋅⋅⋅ v ) and Fr = 2
U d U–
r dt r∂ ∂ + ∂ ∂ ɺ
in usual notation,
2
r 2 2
1 r – 2r rF 1 –
r c
=
ɺ ɺɺ=
2 2 2 2
2 2 2 2
1 c – r 2r r 1 c – r 2rrr c c r
+ +=
ɺ ɺɺ ɺɺ
2p
V
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46.(4) Since the kinetic energy of A after coll ision is one-ninths of i ts initial kinetic energy, the momentum of A after collision is one-third of i ts initial momentum.
Since the momentum is to be conserved, we have
p = p’ – p/3 where p is initial momentum of A and p’ is the momentum of B after the coll ision.
[The final momentum of particle A is negative since its direction is reversed].
Therefore, p’ = 4p/3
The kinetic energy gained by particle B due to the coll ision is p ’2/2M where M is the mass of particle B.
The kinetic energy lost by particle A due to the collision is (8/9)×p2/2m.
[Note that the initial kinetic energy of particle A is p2/2m and its final kinetic energy is (1/9) p2/2m].
Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have
p ’2/2M = (8/9) p2/2m
Substituting for p’ = 4p/3, we have
(16/9) (p2/2M) = (8/9) p2/2m from which M = 2 m.
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∴ Mass of water = 5000 × (1.609 × 103)3 × 103 kg = 2.08 × 1016 kg.
or No. of molecules of water = 2.08 × 1016 × 6.02214 × 1026/18 = 6.97 × 1041 molecules.
As the abundance of deuterium is 0.0156% so that the total number of deuterium atoms
= 6.97 × 1041 × 2 × 0.0156 × 10–2 = 2.18 × 1038.
As the fusion of 6 deuterium atoms gives an energy release of 43 MeV, hence the total energy released = 2.18 × 1038 × (43/6).
= 1.56 × 1039 MeV.
48.(1) The wave length λ (in Angstrom unit) of a photon of energy E (in electron volt) is given by
λE = 12400, very nearly.
Therefore, λ = 12400/E
[The above expression can be easily obtained by remembering that a photon of energy 1 eV has wave length 12400 Ǻ and the energy is inversely proportional to the wave length].
Since E = 2 eV we have λ = 12400/2 = 6200 Ǻ = 620 nanometre.
Photons with wave length greater than 640 nm wil l have energy less than 2 eV so that they will be unable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is (1).
49.(4) Circuit (i) is shown the logic circuit of AND GATE and here output
Y = A ⋅⋅⋅⋅ B
Circuit (ii ) is shown the logic circuit of OR GATE and output
Y = A + B
Similarly circuit (iii ) is shown the logic circuit of NOR GATE and output is
Y = A B+
50(i).(3)Since f(z) = cosz sinz
–z
to find zeros of f(z) put f(z) = 0
⇒ cos zsin z
–z
= 0
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The Lagrangian equation for conservative system in the variable φ is given by
d L L
– 0dt ∂ ∂
= ∂φ∂φ ɺ
Again equation (1) gives
2 2L V L– and mr sin
∂ ∂ ∂= = θφ∂φ ∂φ ∂φ
ɺɺ
…(4)
With these substitutions equation (4) becomes
2 2d V(mr sin ) 0
dt∂
θφ + =∂φ
ɺ
52(i).(1) We have E = ˆEx and we seek ( )x,tφ such that
E = – ∇φ …(1)
Since B = 0, we seek a gauge in which A = 0. Intergrating (1) we obtain φ(x) = - εx + c, where c is a constant of integration. Let us choose c = 0; then
( )x xφ = −ε …(2)
The total Hamiltonian is
2
2 2p 1H m x x
2m 2= + ω − ε …(3)
The first term on the right-hand side of (3) is the standard kinetic term, the second term is the harmonic osci llator potential energy, and the third term is the electrical potential energy.
52(ii).(2) We will now write part (i) eq.(3) in the following form:
Hy = 2y 2 2
y 0
p 1m y H
2m 2+ ω + …(4)
Where H0 is a constant and y = ax + b. Consider the kinetic term. We see that py = px, so a = 1. Now we can substitute y = x + b into (4) and obtain
Hy = ( )2 2
22 2 2 2x x0 0
p p1 1 1m x b H m bx m b H
2m 2 2m 2 2+ ω + + = + ω + ω + …(5)
From Eq. (3) of part (i) we see that Hx = Hy only i f b = –ε/mω2 and H0 = – ε2 / 2mω 2 .
52(iii).(3) To conclude, i f we perform the coordinate transformation y = x –ε/mω2 , we get a one-dimensional harmonic osci llator with no charge, and the energy eigenvalues of a one-dimensional harmonic osci llator are
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Corresponding to the eigenstate n| ψ . We have a shifted harmonic oscil lator; thus, the energy eigenvalues are now.
E n = 2
2
1 1 1n
2 2 2 mε ω + − ω
ℏ
53.(2) f(x) = x3 – x2 + 4x – 4
f’(x) = 3x2 – 2x + 4
f(2) = 8 – 4 + 8 – 4 = 8
f’(2) = 12 – 4 + 4 = 12
nn 1 n
n
f(x )x x –
f '(x )+ =
xn + 1 = ( )82 – 2 – 2/3 4 /3
12= =
54.(1) Taking earth as reference level for zero potential energy, we have
V(q) = mgz ..(1)
The Hamilton-Jacobi equation for Hamilton’s principal function is
2
1 S SV(q) 0
2m q t
∂ ∂ + + = ∂ ∂
..(2)
The first term in bracket is function of q only, while the second term is function of t only, therefore each term must be equal to the same constant with opposite signs.
21 S
i.e. V(q)2m q
Sand –
t
∂+ = α ∂
∂ = α ∂
...(3)
Then we have S = W(q, α)–αt, ...(4)
W being a constant of integration.
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55.(3) From Maxwell-Boltzmann distribution law, the number of molecules in the νth state relative to that in the ν = 0 (lowest) state at T Kelvin temperature is given by
0–G ( )h c/ kT
0
Ne
Nνν =
= –G( ) – G(0 )hc /kTe ,ν
where G(ν) = 2
e e e
1 1– x
2 2 ω ν + ω ν +
∴ 2
e e e e e–( n – x n – x n)hc /kTn
0
Ne
Nω ω ω=
For the n = 1 level, we have
e e e–( – 2 x )hc / kT1
0
Ne
Nω ω=
Here –1 –11e e e
0
N 1, 214.6cm and x 0.6 cm (given)
N 10= ω = ω =
∴ –1 – 1–( 214.6 cm – 1 .2 cm )hc /kT1
e10
=
or –1 –1(213 .4 cm )hc /kT (2134 0m )hc / kT10 e e= =
Taking natural logarithm:
Loge 10 = (21340 m–1) hc/kT.
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