fiziks - IIT JAM Physics

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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES JAM JOINT ADMISSION TEST FOR MSc 2015
SECTION – A: MCQ
Q1 – Q10 carry one mark each.
Q1.
A system consists of N number of particles, N >> 1 . Each particle can have only one of
the two energies E1 or E1 + ε (ε > 0 ) . If the system is in equilibrium at a temperature T ,
the average number of particles with energy E1 is
(a)
Ans:
N
2
(b)
e
+1
(c)
N
e
− ε / kT
+1
(d) Ne −ε / kT
(c)
Solution: N = Ne
Q2.
N
ε / kT
−( E2 − E1 )
kT
=e
−⎡⎣( E1 +ε ) − E1 ⎤⎦
kT
−ε
⇒ N = Ne kT
A mass m , lying on a horizontal, frictionless surface, is connected to one end of a spring.
The other end of the spring is connected to a wall, as shown in the figure. At t = 0 , the
mass is given an impulse.
m
Impulse
The time dependence of the displacement and the velocity of the mass (in terms of nonzero constants A and B ) are given by
Ans:
(a) x(t ) = A sin ωt , v(t ) = B cos ωt
(b) x(t ) = A sin ωt , v(t ) = B sin ωt
(c) x(t ) = A cos ωt , v(t ) = B sin ωt
(d) x(t ) = A cos ωt , v(t ) = B cos ωt
(a)
Solution: At time t = 0 , the mass ‘ m ’ is at rest. Thus, displacement will be zero at time t = 0 .
∴
x = A sin (ωt )
dx
= Aω cos ωt = B cos ω t
dt
Thus, x = A sin ω t and V ( t ) = B cos ωt
Velocity is v =
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
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Q3.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A particle with energy E is incident on a potential given by
x<0
⎧ 0,
V ( x) = ⎨
⎩ V0 ,
x≥0
.
The wave function of the particle for E < V0 in the region x > 0 (in terms of positive
constants A, B and k ) is
(a) Ae kx + Be − kx
Ans:
(b) Ae − kx
(d) Zero
(b)
Solution: For x > 0 ; −
d 2ψ ΙΙ
+ V0ψ ΙΙ = Eψ ΙΙ ;
2m d
2
ψ ΙΙ = Bekx + Ae − kx where k =
ψ ΙΙ → 0 as x → ∞
Q4.
(c) Ae ikx + Be − ikx
E < V0
2m (V0 − E )
2
⇒ A = 0 ⇒ ψ ΙΙ = Ae− kx
⎡
π ⎞⎤
⎛
The electric field of a light wave is given by E = E 0 ⎢iˆ sin (ωt − kz ) + ˆj sin ⎜ ωt − kz − ⎟⎥ .
4 ⎠⎦
⎝
⎣
The polarization state of the wave is
Ans:
(a) Left handed circular
(b) Right handed circular
(c) Left handed elliptical
(d) Right handed elliptical
(c)
π⎞
⎛
Solution: Ex = E0 sin (ωt − kz ) , E y = E0 sin ⎜ ωt − kz − ⎟ .
4⎠
⎝
Thus resultant is elliptically polarized wave.
π⎞
⎛
At z = 0, Ex = E0 sin (ωt ) , E y = E0 sin ⎜ ωt − ⎟
4⎠
⎝
When ωt = 0, Ex = 0, E y = −
E0
E
π
and when ωt = , Ex = 0 , E y = 0
4
2
2
Q5.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES x+ y
x− y
, y′ =
Consider the coordinate transformation x ′ =
. The relation between the
2
2
area elements dx ′dy ′ and dxdy is given by dx ′dy ′ = jdxdy . The value of j is
(a) 2
Ans:
(b) 1
(c) − 1
(d) − 2
(c)
Solution: x′ =
x+ y
x− y
, y′ =
2
2
∵ dx′dy′ = J dxdy
Q6.
⎛ ∂x′
⎜ ∂x
⇒ J =⎜
⎜ ∂y′
⎜ ∂x
⎝
The trace of
∂x′ ⎞ ⎛ 1
1 ⎞
⎟
⎜
∂y
1 1
2 ⎟⎟
⎟=⎜ 2
= − − = −1
∂y′ ⎟ ⎜ 1
1 ⎟
2 2
−
⎜
⎟
⎟
∂y ⎠ ⎝ 2
2⎠
a 2× 2 matrix is 4 and its determinant is 8 . If one of the eigenvalues is
2(1 + i ) , the other eigenvalue is
(a) 2(1 − i )
Ans:
(b) 2(1 + i )
(c) (1 + 2i )
(d) (1 − 2i )
(a)
Solution: λ1 = 2 + 2i, λ2 = 2 (1 − i ) ⇒ λ1 + λ2 = 4 and λ1 ⋅ λ 2 = 8
Q7.
Temperature dependence of resistivity of a metal can be described by
(a)
(b) R
R
T
T
(c)
R
(d)
T
R
T
Ans: (a)
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Electrical resistivity of metal varies as
ρ ∝T5
(For T << θ D )
ρ ∝T
(For T >> θ D )
where θ D is the Debye temperature. Thus, correct answer is option (a)
Q8.
A proton from outer space is moving towards earth with velocity 0.99 c as measured in
earth’s frame. A spaceship, traveling parallel to the proton, measures proton’s velocity to
be 0.97 c . The approximate velocity of the spaceship in the earth’s frame, is
(a) 0.2 c
Ans:
(b) 0.3 c
(c) 0.4 c
(d) 0.5 c
(d)
Solution: Velocity of proton w.r.t. spaceship = 0.97 c
s′
∵ u ′x = 0.99 c, v = −v, u x = 0.97 c
E
−v
u′ + v
0.99 c − v
⇒ ux = x
⇒ 0.97 c =
⇒ v = 0.5 c
0.97v
u′ v
1−
1 + x2
c
c
Q9.
s
p = 0.99 c
p = 0.99 c
A charge q is at the center of two concentric spheres. The outward electric flux through
the inner sphere is φ while that through the outer sphere is 2φ . The amount of charge
contained in the region between the two spheres is
(a) 2q
Ans:
(c) − q
(d) − 2q
(b)
Solution: φ =
Q10.
(b) q
q
ε0
, φ ′ = 2φ =
q + q′
ε0
⇒ q′ = q
At room temperature, the speed of sound in air is 340 m/sec. An organ pipe with both
ends open has a length L = 29 cm . An extra hole is created at the position L / 2 . The
lowest frequency of sound produced is
(a) 293 Hz
Ans:
(b) 586 Hz
(c) 1172 Hz
(d) 2344 Hz
(c)
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES v
Solution: The fundamental frequency in organ pipe with both end open is f =
2L
L/2
L
with additional rate at
f′=
L
, the fundamental frequency becomes
2
v
v
v 340 m / sec
=
= =
= 1172 Hz
2 L ′ 2 L L 29 × 10−2 m
2
Q11 – Q30 carry two marks each.
Q11.
A system comprises of three electrons. There are three single particle energy levels
accessible to each of these electrons. The number of possible configurations for this
system is
(a) 1
Ans:
(b) 3
(c) 6
(d) 7
(c)
Solution: For electron spin is
1
. So in one single state two electrons can be adjusted the number
2
of ways are
Ground
First
Second
1
2
1
0
2
2
0
1
3
1
2
0
4
1
0
2
5
0
1
2
6
0
2
1
So, number of ways are 6 .
Q12.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A rigid and thermally isolated tank is divided into two compartments of equal volume V ,
separated by a thin membrane. One compartment contains one mole of an ideal gas A
and the other compartment contains one mole of a different ideal gas B . The two gases
are in thermal equilibrium at a temperature T . If the membrane ruptures, the two gases
mix. Assume that the gases are chemically inert. The change in the total entropy of the
gases on mixing is
(a) 0
Ans:
(b) R ln 2
(c)
3
R ln 2
2
(d) 2R ln 2
(d)
Solution: For A , number of microstate after mixing is 2
For A , number of microstate before mixing is 1
A
B
⇒ ΔS A = R ln 2 − R ln1 = R ln 2
Similarly, for B ⇒ ΔS B = R ln 2
⇒ ΔS = ΔS A + ΔS B = 2 R ln 2
Q13.
A Zener regulator has an input voltage in the range 15V − 20V and a load current in the
range of 5 mA − 20 mA . If the Zener voltage is 6.8V , the value of the series resistor
RS
should be
V0
+
15 − 20 V
(a) 390 Ω
Ans:
6 .8 V
−
(b) 420 Ω
(c) 440 Ω
(d) 460 Ω
Some data is missing. (No answer is possible)
Q14.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The variation of binding energy per nucleon with respect to the mass number of nuclei is
shown in the figure.
Average binding energy per
nucleon (MeV)
9
8
7
6
5
4
3
2
1
0
20 40 60 80 100 120 140 160 180 200 220 240
Number of nucleons in nucleus, A
Consider the following reactions:
(i)
238
92
206
U →82
Pb + 10 P + 22n
(ii)
238
92
U→
Pb + 8 24He + 6e −
206
82
Which one of the following statements is true for the given decay modes of
238
92
U?
(a) Both (i) and (ii) are allowed
(b) Both (i) and (ii) are forbidden
(c) (i) is forbidden and (ii) is allowed
(d) (i) is allowed and (ii) is forbidden
Ans:
(c)
Q15.
A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods.
The constant pressure molar specific heat (C p ) of an ideal gas consisting of such
molecules is
(a) 6 R
Ans:
(b) 5 R
(c) 4 R
(d) 3R
(c)
6 RT
⎛ ∂U ⎞
⇒ CV = ⎜
⎟ = 3R ⇒ CP = CV + R = 4 R
2
⎝ ∂T ⎠V
A satellite moves around the earth in a circular orbit of radius R centered at the earth. A
Solution: D.O.F = 6 ⇒ U =
Q16.
second satellite moves in an elliptic orbit of major axis 8 R , with the earth at one of the
foci. If the former takes 1 day to complete a revolution, the latter would take
(a) 21.6 days
Ans:
(b) 8 days
(c) 3 hours
(d) 1.1 hour
(a)
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2
⎛T ⎞ ⎛ R ⎞
3/ 2
Solution: ⎜ 1 ⎟ = ⎜
⎟ ⇒ T2 = ( 8 ) T1 ≈ 22 days
⎝ T2 ⎠ ⎝ 8 R ⎠
Q17.
3
A positively charged particle, with a charge q , enters a region in which there is a uniform
electric field E and a uniform magnetic field B , both directed parallel to the positive
y -axis. At t = 0 , the particle is at the origin and has a speed v 0 directed along the
positive x - axis. The orbit of the particle, projected on the x- z plane, is a circle. Let T
be the time taken to complete one revolution of this circle. The y -coordinate of the
particle at t = T is given by
(a)
Ans:
π 2 mE
2qB
(b)
2
2π 2 mE
qB 2
(c)
π 2 mE
qB
2
+
v0π m
qB
2πmv0
qB
(d)
z
(b)
2
1 2
1 qE ⎛ 2π m ⎞
2π 2 mE
Solution: y = u y t + a y t ⇒ y =
⎜
⎟ =
2
2 m ⎝ qB ⎠
qB 2
E, B
x
Q18.
v0
y
Vibrations of diatomic molecules can be represented as those of harmonic oscillators.
Two halogen molecules X 2 and Y2 have fundamental vibrational frequencies
v X = 16.7 ×1012 Hz and vY = 26.8 × 1012 Hz , respectively. The respective force constants
are K X = 325 N / m and K Y = 446 N / m . The atomic masses of F , Cl and Br are
19.0, 35.5 and 79.9 atomic mass unit respectively. The halogen molecules X 2 and Y2 are
Ans:
(a) X 2 = F2 and Y2 = Cl2
(b) X 2 = Cl 2 and Y2 = F2
(c) X 2 = Br2 and Y2 = F2
(d) X 2 = F2 and Y2 = Br2
(b)
Solution: The oscillation frequency of diatomic molecule with reduce mass ‘ μ ’ is
f =
1
2π
k
μ
⇒μ=
1
4π
2
k
where k is force constant.
f2
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES mx mx
m
For X 2 molecule: μ =
= x
2
m x + mx
⇒ mx =
1
2π
2
×
kx
1
325 N / m
=
×
2
2
f x 2 × ( 3.14 )
16.7 × 1012 Hz
(
)
2
⇒ mx = 59.07 ×10−27 kg = 35.5 × 1.67 × 10−27 kg = 35.5 a.m.u.
This is the atomic mass of chlorine ( Cl ) .
For Y2 molecule: μ =
⇒ my =
1
2π
2
×
ky
(f )
y
2
my my
my + my
=
=
my
2
1
2 × ( 3.14 )
2
×
446 N / m
( 26.8 ×10
12
Hz )
2
⇒ my = 31.73 ×10−27 kg = 19 × 1.67 × 10−27 kg = 19 a.m.u.
This is the atomic mass of F . Thus, correct answer is option (b)
Q19.
A hollow, conducting spherical shell of inner radius R1 and
outer radius R2 encloses a charge q inside, which is located at a
distance d (< R1 ) from the centre of the spheres. The potential at
the centre of the shell is
(a) Zero
(c)
Ans:
1 ⎛q q ⎞
⎜ − ⎟
4π ∈0 ⎝ d R1 ⎠
(b)
1 q
4π ∈0 d
(d)
1 ⎛q q
q ⎞
⎜ − + ⎟
4π ∈0 ⎝ d R2 R2 ⎠
R1
q d
R2
(d)
Solution: V =
1 ⎛q q
q ⎞
⎜ − + ⎟
4πε 0 ⎝ d R1 R2 ⎠
Q20.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Doppler effect can be used to measure the speed of blood through vessels. Sound of
frequency 1.0522 MHz is sent through the vessels along the direction of blood flow. The
reflected sound generates a beat signal of frequency 100 Hz. The speed of sound in blood
is 1545 m/sec. The speed of blood through the vessel, in m/sec, is
(a) 14.68
Ans:
(b) 1.468
(c) 0.1468
(d) 0.01468
(d)
Solution: Consider Vb , Vsound are velocities of blood cell and sound in blood. The sound of
frequency
( f0 )
is traveling towards blood cell where blood cell is moving away with
velocity Vb
f0
Vsound
Vb
Frequency of sound observed on blood cell is
⎛V
− Vb ⎞
f ′ = f 0 ⎜ sound
⎝ Vsound ⎟⎠
(i)
Sound from blood cell of frequency f ′ reflect back.
f′
observer
Vb
⎛ Vsound ⎞
The frequency observed by observer is f = f ′ ⎜
⎝ Vsound + Vb ⎟⎠
(ii)
⎛V
− Vb ⎞ ⎛ Vsound ⎞
From equation (i) and (ii), we get f = f 0 ⎜ sound
⎝ Vsound ⎟⎠ ⎜⎝ Vsound + Vb ⎟⎠
⎛V
− Vb ⎞
⇒ f = f 0 = ⎜ sound
⎟
⎝ Vsound + Vb ⎠
⎛V
− Vb ⎞
Now, Δf = f 0 − f = f 0 − f 0 ⎜ sound
⎟
⎝ Vsound + Vb ⎠
(iii)
⎛ 2Vb
⎞
= f0 ⎜
⎟
⎝ Vsound + Vb ⎠
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2Vb
V
+ Vb 2 f 0
Vsound
Δf
⇒ Vb =
⇒ sound
=
⇒
=
Vsound + Vb
f0
Δf
Vb
⎛ 2 f0
⎞
⎜⎝ Δf − 1⎟⎠
Given Vsound = 1545 m / sec, f 0 = 1.0522 × 106 Hz , Δf = 100 Hz
∴Vb =
1545
⎛ 2 × 1.0522 × 10
⎞
− 1⎟
⎜⎝
100
⎠
6
=
1545
= 0.073 ⇒ Vb = 0.073 m / sec
21043
Thus the best suitable answer is option (d).
Q21.
Which of the following circuits represent the Boolean expression
S = P + QR + Q P
(a) P
Q
(b) P
Q
S
(c) P
Q
S
(d) P
Q
S
R
S
R
Ans:
(b)
Q22.
A conducting wire is in the shape of a regular hexagon, which is
I
inscribed inside an imaginary circle of radius R , as shown. A current
I flows through the wire The magnitude of the magnetic field at the
R
C
center of the circle is
(a)
Ans:
3μ 0 I
2πR
(b)
μ0 I
2 3πR
(c)
3μ 0 I
πR
(d)
3μ 0 I
2πR
(c)
Solution: d = R cos 600 =
3
R
2
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES μI
μI
μ0 I
μ0 I
∵ B = 0 ( sin θ 2 − sin θ1 ) ⇒ B1 = 0 2sin 300 =
2sin 300 =
4π d
4π d
3
2 3π R
4π
R
2
⇒ B = 6 B1 = 6 ×
Q23.
μ0 I
3μ0 I
3μ0 I
=
=
πR
2 3π R
3π R
An observer is located on a horizontal, circular turntable which rotates about a vertical
axis passing through its center, with a uniform angular speed of 2 rad/sec . A mass of
10 grams is sliding without friction on the turntable. At an instant when the mass is at a
distance of 8 cm from the axis it is observed to move towards the center with a speed of
6 cm/sec. The net force on the mass, as seen by the observer at that instant, is
(a) 0.0024 N
Ans:
(b) 0.0032 N
(c) 0.004 N
(d) 0.006 N
(c)
Solution: Two forces will act on the particle
First coriolis force Fc = −2m(ω × v) = −240 × 10−5 N (in tangential direction)
Another force is centrifugal force Fr = mω 2 r = 320 × 10−5 N (in radial direction)
Total force F = Fc2 + Fc2 r = 0.04 N
Q24.
Miller indicates of a plane in cubic structure that contains all the directions [100], [011] and
[111] are
(a) (011)
Ans:
(b) (101)
(c) (100 )
(a)
(d) (110 )
y
Solution: The name of the plane containing all the directions
[111]
[100] , [011] & [111] is ( 0 11) or ( 01 1 )
The best suitable answer is option (a)
[ 011]
x
[100]
z
Q25.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Seven uniform disks, each of mass m and radius r , are inscribed
inside a regular hexagon as shown. The moment of inertia of this
system of seven disks, about an axis passing through the central disk
and perpendicular to the plane of the disks, is
Ans:
(a)
7 2
mr
2
(c)
13
mr2
2
(d)
55
mr2
2
(d)
Solution:
Q26.
(b) 7mr 2
⎛ mr 2
⎞ mr 2 54mr 2 55mr 2
mr 2
+6×⎜
+ 4mr 2 ⎟ =
+
=
2
2
2
2
⎝ 2
⎠
A nucleus has a size of 10 −15 m . Consider an electron bound within a nucleus. The
estimated energy of this electron is of the order of
(b) 10 2 MeV
(a) 1 MeV
Ans:
(c) 10 4 MeV
(d) 10 6 MeV
(d)
6.6 ×10−34
= 6.6 × 10−19 kgm / sec
Solution: p = =
−15
10
λ
h
p2
44 ×10−38
=
= 2.4 × 10−7 Joule
∵E =
−31
2me 2 × 9.1× 10
⇒E=
Q27.
2.4 × 10−7
eV = 1.5 × 1012 eV = 1.5 ×106 MeV
−19
1.6 ×10
Consider a vector field F = yiˆ + xz 3 ˆj − zykˆ . Let C be the circle x 2 + y 2 = 4 on the
plane z = 2 , oriented counter-clockwise. The value of the contour integral
(a) 28 π
Ans:
(b) 4 π
(c) − 4 π
∫
F ⋅ d r is
C
(d) − 28 π
(a)
Solution:
(
)
∵ ∫ F .d r = ∫ ∇ × F .d a
C
S
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES xˆ
yˆ
zˆ
∇ × F = ∂ / ∂x ∂ / ∂y ∂ / ∂z
y
− zy
xz 3
3
⎛ ∂ ( − yz ) ∂ ( xz 3 ) ⎞
⎛ ∂y ∂ ( − zy ) ⎞ ⎛ ∂ ( xz ) ∂y ⎞
⎟ + yˆ ⎜ −
⇒ ∇ × F = xˆ ⎜
−
− ⎟
⎟ + zˆ ⎜
⎜ ∂y
∂z ⎟
∂z
∂x ⎠ ⎜ ∂x
∂y ⎟
⎝
⎝
⎠
⎝
⎠
2
3
⇒ ∇ × F = xˆ ( − z − 3xz ) + yˆ ( 0 − 0 ) + zˆ ( z − 1)
∵ z = 2 ⇒ ∇ × F = − ( 2 + 12 x ) xˆ + 7 zˆ
(
)
∵ d a = rdrdφ zˆ ⇒ ∇ × F .d a = ⎡⎣ − ( 2 + 12 x ) xˆ + 7 zˆ ⎤⎦ .rdrdφ zˆ = 7 rdrdφ
(
)
2
2π
0
0
⇒ ∫ ∇ × F .d a = 7 ∫ rdr ∫ dφ = 28π
S
Q28.
Consider the equation
dy y 2
=
with the boundary condition y (1) = 1 . Out of the
dx
x
following the range of x in which y is real and finite is
(a) − ∞ ≤ x ≤ −3
(b) − 3 ≤ x ≤ 0
(c) 0 ≤ x ≤ 3
(d) 3 ≤ x ≤ ∞
Ans:
Solution of the differential equation is satisfied by options (c) and (d).
Q29.
The Fourier series for an arbitrary periodic function with period 2 L , is given by
f (x ) =
a0
nπ x
nπ x
∞
∞
+ ∑n =1 a n cos
+ ∑n =1 bn sin
. For the particular periodic function
L
2
L
shown in the figure the value of a0 is
f (x )
1
1/ 2
−2
(a) 0
Ans:
(b) 0.5
−1
0
1
(c) 1
2
x
(d) 2
(c)
Solution: The wavefunction of the given function can be written as
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 0 < x <1
⎧x
f ( x) = ⎨
−1 < x < 0
⎩− x
Coefficient a0 is defined as
a0 = 1∫ − x dx + 1∫ x dx
0
1
−1
0
ω
1
2
2
⎡
⎤
−1) ⎤ ⎡ (1)
⎡ x2 ⎤
⎡ x2 ⎤
(
1 1
= − ⎢ ⎥ + ⎢ ⎥ = − ⎢0 −
− 0⎥ = + + − 1
⎥+⎢
2 ⎦⎥ ⎣⎢ 2
2 2
⎢⎣
⎥⎦
⎣ 2 ⎦ −1 ⎣ 2 ⎦ 0
∴ a0 = 1
Q30.
The phase of the complex number (1 + i ) i in the polar representation is
(a)
Ans:
π
4
(b)
π
2
(c)
3π
4
(d)
5π
4
(c)
Solution: z = (1 + i ) i ⇒ z = ( −1 + i ) ⇒ z = x + iy
tan θ =
y
3π
= −1 ⇒ θ = tan −1 ( −1) ⇒ θ =
x
4
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES SECTION – B: MSQ
Q1 – Q10 carry two marks each.
Q1.
For an electromagnetic wave traveling in free space, the electric field is given by
V
E = 100 cos 10 8 t + kx ˆj . Which of the following statements are true?
m
(a) The wavelength of the wave in meter is 6π
(
)
(b) The corresponding magnetic field is directed along the positive z direction
(c) The Poynting vector is directed along the positive z direction
(d) The wave is linearly polarized
Ans:
(a) and (d)
Solution: E = 100 cos (108 t + kx ) ˆj V / m
Option (a) is true
ω = 108 ⇒
2π c
λ
= 108 ⇒ λ =
2π × 3 × 108
= 6π
108
Option (b) is wrong
(
)
B ∝ kˆ × E ∝ ( − xˆ × yˆ ) ∝ − zˆ
Option (c) is wrong
S ∝ kˆ ∝ − xˆ
Option (d) is true
Q2.
In an ideal Op-Amp circuit shown below R1 = 3k Ω, R2 = 1k Ω and Vi = 0.5sin ω t
(in Volt). Which of the following statements are true? V
i
(a) The current through R1 = The current through R2
R
(b) The potential at P is V0 2
P
R1
(c) The amplitude of V0 is 2V
R2
+
−
V0
R1
(d) The output voltage V0 is in phase with Vi
Ans:
(b), (c) and (d)
Option (a) is wrong
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES V −V
V
Current through R1 is I1 = o i and Current through R2 is I 2 = i
R1
R2
Option (b) is true
The potential at P is V0
R2
(voltage divider rule)
R1
Option (c) is true
⎛ R ⎞
⎛ 3⎞
V0 = ⎜ 1 + 2 ⎟ Vi = ⎜1 + ⎟ 0.5sin ωt = 2sin ωt ⇒ Vm = 2 V
R1 ⎠
⎝ 1⎠
⎝
Option (d) is true
Q3.
A particle of mass m is moving in x − y plane. At any given time t , its position vector is
given by r ( t ) = A cos ωt i + B sin ωt ˆj where A, B and ω are constants with A ≠ B .
Which of the following statements are true?
(a) Orbit of the particle is an ellipse
(b) Speed of the particle is constant
(c) At any given time t the particle experiences a force towards origin
(d) The angular momentum of the particle is mω ABkˆ
Ans:
(a), (c) and (d)
x
y
Solution: (a) r ( t ) = A cos ω t iˆ + B sin ωt ˆj ⇒ x = A cos ωt , y = B sin ωt ⇒ = cos ωt , = sin ωt
A
B
x2 y2
⇒ 2 + 2 = 1 (Ellipse)
A B
(b)
dr
= − Aω sin ωt iˆ + Bω cos ωt ˆj
dt
Speed =
dr
= A2ω 2 sin 2 ωt + B 2ω 2 cos 2 ωt . Speed is function of time, so not constant.
dt
2
dr
(c) 2 = − Aω 2 cos ωt iˆ − Bω 2 sin ωt ˆj = −ω 2 r . Force act towards origin.
dt
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ˆj
⎛
i
kˆ⎞
⎜
⎟
(d) L = ( r × p ) = m ⎜ A cos ω t
B sin ω t 0⎟ ⇒ L = mω ABk
⎜⎝ − Aω sin ω t Bω cos ω t 0⎟⎠
Q4.
A rod is hanging vertically from a pivot. A partic1e traveling in horizontal
direction, collides with the rod as shown in the figure. For the rod-particle
system, consider the linear momentum and the angular momentum about
the pivot .Which of the following statements are NOT true?
(a) Both linear momentum and angular momentum are conserved
(b) Linear momentum is conserved but angular momentum is not
(c) Linear momentum is not conserved but angular momentum is conserved
(d) Neither linear momentum nor annular momentum are conserved
Ans:
(b), (c) and (d)
Q5.
A particle is moving in a two dimensional potential well
V ( x, y ) = 0,
0 ≤ x ≤ L, 0 ≤ y ≤ 2 L
= ∞, elsewhere
which of the following statements about the ground state energy E1 and ground state
eigenfunction ϕ 0 are true?
(a) E1 =
(c) ϕ0 =
Ans:
2
π2
(b) E1 =
mL2
2
πx πy
sin
sin
L
L
2L
5 2π 2
8mL2
(d) ϕ 0 =
2
πx
πy
cos cos
L
L
2L
(b) and (c)
⎛ nx2 n y2 ⎞
Solution: En =
⎜ +
⎟
2m ⎜⎝ L2 4 L2 ⎟⎠
π2
2
Ground state nx = 1, n y = 1 ⇒ Ex =
Wave function ψ =
π2
1 ⎞ 5π 2 2
⎛ 1
+
⎜
⎟=
2m ⎝ L2 4 L2 ⎠ 8mL2
2
2
2 sin π x sin π y
⋅
⋅
L 2L
L
2L
Q6.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Consider the circuit, consisting of an AC function generator V (t ) = V0 sin 2πvt with
V0 = 5V an inductor L = 8.0mH , resistor R = 5Ω and a capacitor C = 100 μF . Which of
the following statements are true if we vary the frequency?
L
C
R
(a) The current in the circuit would be maximum at ν = 178Hz
(b) The capacitive reactance increases with frequency
(c) At resonance, the impedance of the circuit is equal to the resistance in the circuit
(d) At resonance, the current in the circuit is out of phase with the source voltage
Ans:
(a) and (c)
Solution: Option (a) is true
ν=
1
2π LC
=
1
2 × 3.14
(8 ×10−3 )(100 ×10−6 )
= 178 Hz
Option (b) is wrong
XC =
1
⇒ X C ↓ as ω ↑
ωC
Option (c) is true
Option (d) is wrong
Q7.
Muons are elementary particles produced in the upper atmosphere. They have a life time
of 2.2μs . Consider muons which are traveling vertically towards the earth’s surface at a
speed of 0.998c . For an observer on earth, the height of the atmosphere above the
surface of the earth is 10.4 km . Which of the following statements are true?
(a) The muons can never reach earth’s surface
(b) The apparent thickness of earth’s atmosphere in muon’s frame of reference is 0.96 km
(c) The lifetime of muons in earth’s frame of reference is 34.8μs
(d) Muons traveling at a speed greater than 0.998 c reach the earth’s surface
Ans: (c) and (d)
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES −6
Δt0
2.2 ×10
Solution: Δt =
⇒ Δt =
= 34.8 × 10−6 sec
2
2
v
1 − ( 0.998 )
1− 2
c
Now distance will be = Δt × v = 34.8 × 10−6 × 0.998 × 3 × 108 = 10.4192 km
Apparent thickness ΔX = Δt × v = 2.2 ×10−6 × 0.998 × 3 × 108 = 0.658 km
Q8.
As shown in the P − V diagram AB and CD are two isotherms at temperatures T1 and
T2 , respectively (T1 > T2 ) . AC and BD are two reversible adiabats. In this Carnot cycle,
which of the following statements are true?
P
Q Q
(a) 1 = 2
T1 T2
Q1
A
B
T1
(b) The entropy of the source decreases
(c) The entropy of the system increases
(d) Work done by the system W = Q1 − Q2
C
Q2 D
T2
V
Ans:
(a), (b) and (d)
Q9.
The following figure shows a double slit Fraunhofer diffraction pattern produced by two
slits, each of width a separated by a distance b, a < b .
Secondary maxima
Primary maxima
Which of the following statements are correct?
(a) Reducing a increases the separation between consecutive primary maxima
(b) Reducing a increases the separation between consecutive secondary maxima
(c) Reducing b increases the separation between consecutive primary maxima
(d) Reducing b increases the separation between consecutive secondary maxima
Ans:
(a) and (d)
Solution: The minima condition for double slit Fraunhofer diffraction is
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES nλ
a sin θ = nλ ⇒ sin θ =
where a is the width of slit.
a
Reducing ‘ a ’ increases the separation between diffraction minima i.e. increases the
separation between consecutive primary maxima.
The condition of interference maxima is
mλ
b sin θ = mλ ⇒ sin θ =
where b is the separation between slits.
b
The position of interference maxima gives the separation between secondary maxima.
Reducing ‘ b ’ increases the separation between consecutive secondary maxima.
The correct answer is option (a) and (d).
Q10.
A unit cube made of a dielectric material has a polarization P = 3iˆ + 4 ˆj units. The edges
of the cube are parallel to the Cartesian axes. Which of the following statements are true?
(a) The cube carries a volume bound charge of magnitude 5 units
(b) There is a charge of magnitude 3 units on both the surfaces parallel to the y − z plane
(c) There is a charge of magnitude 4 units on both the surfaces parallel to the x − z plane
(d) There is a net non-zero induced charge on the cube
Ans:
(b) and (c)
Solution:
∵ P = 3iˆ + 4 ˆj
Option (a) is wrong
ρb = −∇.P = 0
Option (b) is true
(
)( )
(
)( )
At x = 0 , σ b = P.nˆ = 3iˆ + 4 ˆj . −iˆ = −3
At x = 1 , σ b = P.nˆ = 3iˆ + 4 ˆj . iˆ = 3
Option (c) is true
(
)( )
At y = 0 , σ b = P.nˆ = 3iˆ + 4 ˆj . − ˆj = −4
(
)( )
At y = 1 , σ b = P.nˆ = 3iˆ + 4 ˆj . ˆj = 4
Option (d) is wrong
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES SECTION – C: NAT
Q1 – Q10 carry one mark each.
Q1.
The power radiated by sun is 3.8 × 10 26 W and its radius is 7 × 10 5 km . The magnitude of
the Poynting vector (in
Ans:
W
) at the surface of the sun is _________________
cm 2
6174
P
3.8 × 1026
Solution: I = =
W / cm 2 = 6174 W / cm 2
10
A 4π × ( 7 ×10 )
Q2.
A particle is in a state which is a superposition of the ground state ϕ 0 and the first
excited state ϕ1 of a one-dimensional quantum harmonic oscillator. The state is given by
Φ=
1
5
ϕ0 +
2
5
ϕ1 . The expectation value of the energy of the particle in this state (in
units of ω , ω being the frequency of the oscillator) is ________
Ans:
1.3
1⎞
⎛ 3 ω⎞ 4
⎛
⎛ ω⎞ 1
Solution: ∵ En = ⎜ n + ⎟ ω and P ⎜
⎟ = , P ⎜⎝ 2 ⎟⎠ = 5
2⎠
⎝ 2 ⎠ 5
⎝
⇒ E =
Q3.
ω 1 3 ω 4 13 ω
2
× +
× =
= 1.3 ω
5
2
5
10
In an experiment on charging of an initially uncharged capacitor, an RC circuit is made
with the resistance R = 10kΩ and the capacitor C = 1000 μF along with a voltage source
of 6V . The magnitude of the displacement current through the capacitor (in μA ),
5 seconds after the charging has started, is _______________
Ans:
364
Solution: I =
V − t / RC
6
6
6
6
−5/10×103 ×1000×10−6
e
=
e
= 4 e−5/10 =
=
= 364 μ A
3
4
4
10 ×10
10
R
e × 10 1.65 × 10
Q4.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES In the given circuit VCC = 10V and β = 100 for n − p − n transistor. The collector voltage
VCC
VC (in volts) is __________.
1K
100 K
VC
+
5V
−
Ans:
5.7
Solution: I B =
5 − 0.7
= 4.3 × 10−5 A ⇒ I C = β I B = 4.3 mA
3
100 × 10
⇒ VC = VCC − I C RC = 10 − 4.3 = 5.7 V
Q5.
Unpolarized light is incident on a calcite plate at an angle of incidence 50 o as shown in
the figure. Take n0 = 1.6584 and ne = 1.4864 for calcite. The angular separation
( in degrees) between the two emerging rays within the plate is
Air
Optic axis
Ans:
50 0
Calcite
3.51
Solution: Inside the crystal incident light split into two components, ordinary ray and extraordinary ray
According to Snell’s law
i = 500
sin i
=n
sin r
For ordinary ray i = 500 , no = 1.6584
∴ sin ro =
⎛ sin i ⎞
sin i
⇒ ro = sin −1 ⎜
no
⎝ no ⎟⎠
ro
re
e- ray
o- ray
⎡ sin 50 ⎤
−1 ⎡ 0.766 ⎤
⇒ ro = sin −1 ⎢
= sin −1 [ 0.462] ⇒ r0 = 27.510
⎥ = sin ⎢
⎥
⎣1.6584 ⎦
⎣ no ⎦
For extra-ordinary ray i = 500 , ne = 1.4864
∴ sin re =
⎛ sin i ⎞
sin i
⇒ re = sin −1 ⎜
ne
⎝ ne ⎟⎠
⎡ sin 500 ⎤
−1 ⎡ 0.766 ⎤
⇒ re = sin −1 ⎢
= sin −1 [ 0.515] ⇒ re = 31.020
⎥ = sin ⎢
⎥
⎣1.4864 ⎦
⎣ ne ⎦
Thus, the angular separation between the o - ray and e - ray is θ = re − ro = 3.510
Q6.
In the hydrogen atom spectrum. the ratio of the longest wavelength in the Lyman series
(final state n = 1 ) to that in the Balmer series (final State n = 2 ) is ____________
Ans:
0.185
Solution: According to Bohr Theory
⎛ 1
1⎞
= R⎜ 2 − 2 ⎟
λL
⎝ n f ni ⎠
1
The longest wavelength in the Lyman series is
⇒
4
⎛1 1 ⎞
⎛3⎞
= R ⎜ − 2 ⎟ = R ⎜ ⎟ ⇒ λL =
3R
λL
⎝1 2 ⎠
⎝4⎠
n=3
n=2
Hα
1
n =1
Lα
The longest wavelength in the Balmer series is
Q7.
⇒
1
36
⎛ 1 1⎞
⎛1 1⎞
⎛9−4⎞
⎛ 5 ⎞
= R⎜ 2 − 2 ⎟ = R⎜ − ⎟ = R⎜
= R ⎜ ⎟ ⇒ λB =
⎟⇒
λB
5R
λB
⎝2 3 ⎠
⎝4 9⎠
⎝ 36 ⎠
⎝ 36 ⎠
⇒
λL
4 5R 5
=
×
=
= 0.185
λB 3R 36 27
1
A rod is moving with a speed of 0.8c in a direction at 60 o to its own length. The
percentage contraction in the length of the rod is __________
Ans:
9
Solution: lx = l0 x 1 −
l 3
v2
1
2
= l0 cos θ 1 − ( 0.8) ⇒ lx = l0 × × 0.6 = 0.3l0 and l y = l0 sin θ = 0
2
2
c
2
⎛ 3l0 ⎞
3
New length l = ( 0.3l0 ) + ⎜⎜
⎟⎟ = l0 0.09 + = 0.916 l0
4
⎝ 2 ⎠
(1 − 0.91) l0 × 100 = 0.09 × 100 = 9%
% change in length
l0
2
Q8.
X − rays of wavelength 0.24 nm are Compton scattered and the scattered beam is
observed at an angle of 60 o relative to the incident beam. The Compton wavelength of
the electron is 0.00243 nm . The kinetic energy of scattered elections in eV
is___________
Ans:
25
Solution: λ = 0.24 nm, λC = 0.00243 and θ = 600
∵ λ ′ − λ = λC (1 − cos θ ) ⇒ λ ′ = λ + λC (1 − cos θ )
1
⎛ 1⎞
⇒ λ ′ = 0.24 + 0.00243 ⎜1 − ⎟ = 0.24 + 0.00243 × = 0.24 + 0.00121 = 0.2412nm
2
⎝ 2⎠
Kinetic Energy of scattered electron
K .E. =
Q9.
hc
λ
−
hc
1 ⎞ 1
⎛ 1
= 6.6 ×10−34 × 3 × 108 ⎜
−
⎟ × −9 Joules
λ′
⎝ 0.24 0.2412 ⎠ 10
⇒ K .E. =
19.8 × 10−26
19.8 × 10−26
4.17
−
4.15
=
× 0.02 = 396 × 10−20 Joules
(
)
−9
−9
10
10
⇒ K .E. =
396 ×10−20
eV = 24.75 eV
1.6 ×10−19
A diode at room temperature (kT = 0.025 eV ) with a current of 1μA has a forward bias
voltage VF = 0.4V . For VF = 0.5V , the value of the diode current (in μA ) is _________
Ans:
54.5
Solution: I = I 0 ( e
V / VT
Q10.
V2 / VT
− 1) ( e0.5/ 0.025 − 1) ( e 20 − 1)
I2 (e
− 1) ⇒ = V / V
=
=
= 54.5 ⇒ I 2 = 54.5 μ A
I1 ( e 1 T − 1) ( e0.4 / 0.025 − 1) ( e16 − 1)
GaAs has a diamond structure. The number of Ga-As bonds per atom which have to be
broken to fracture the crystal in the (001) plane is _______
Ans:
4
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Diamond structure has tetrahedral bond. To fracture the diamond structure along
( 0 0 1)
plane, four bonds need to be broken.
Q.11 – Q.20 Carry two marks each.
Q11.
In the thermodynamic cycle shown in the figure, one mole of a monatomic ideal gas is
taken through a cycle. AB is a reversible isothermal expansion at a temperature of 800 K
reduced to 300 K . CA is a constant volume process in which
Pressure
in which the volume of the gas is doubled. BC is an isobaric
the pressure and temperature return to their initial values. The
P1
net amount of heat (in Joules) absorbed by the gas in one
P2
contraction to the original volume in which the temperature is
complete cycle is _____________
Ans:
A
C
V
B
2V Volume
452
Solution: Process A → B is isothermal expansion
TA = 800 K , VA , PA and TB = 800 K , VB = 2VA , PB =
PA
2
Process B → C is isobaric
PC = PB =
PA
, VC = VA , TC = 300 K
2
C → A is Isochoric
⎛V ⎞
ΔQ1 = nRTA ln ⎜ B ⎟ = 4602 J
⎝ VA ⎠
ΔQ2 = nCP ΔT =
ΔQ3 =
nγ RΔT ⎛ γ ⎞
=
R ( 300 − 800) = −10344 J
(γ − 1) ⎜⎝ γ − 1⎟⎠
R
R
× 500 = 6194 J
(800 − 300) =
(γ − 1)
(γ − 1)
Total heat exchange is Q1 + Q2 + Q3 = 452
Q12.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES In a region of space, a time dependent magnetic field B(t ) = 0.4t Tesla points vertically
upwards. Consider a horizontal, circular loop of radius 2 cm in this region. The
magnitude of the electric field (in mV / m ) induced in the loop is ___________.
Ans:
4
∂B
r ∂B 2 ×10−2
2
=
0.4 = 4 mV / m
Solution: E × 2π r = − × π r ⇒ E =
∂t
2 ∂t
2
Q13. A plane electromagnetic wave of frequency 5 × 1014 Hz and amplitude 103 V / m traveling
in a homogeneous dielectric medium of dielectric constant 1.69, is incident normally at
the interface with a second dielectric medium of dielectric constant 2.25. The ratio of the
amplitude of the transmitted wave to that of the incident wave is __________.
Ans:
0.93
⎞
⎛ 2n1 ⎞
E0T ⎛ 2 ε r1 ⎞ ⎛
2 1.69
=⎜
Solution: E0T = ⎜
⎟ = ⎜⎜
⎟ = 0.93
⎟ E0 I ⇒
E0 I ⎜⎝ ε r1 + ε r 2 ⎟⎠ ⎝ 1.69 + 2.25 ⎟⎠
⎝ n1 + n2 ⎠
Q14. For the arrangement given in the following figure, the coherent light sources A, B and C
have individual intensities of 2 mW / m 2, 2 mW / m 2 and 5 mW / m 2 respectively at point P .
The wavelength of each of the sources is 600 nm . The resultant intensity at point P
P
(in mW / m 2 ) is ___________.
15 mm
A
3.22 mm
B
2.04 mm
1m
C
Ans:
9.23 mw / m
2
p
Solution: The electric field on the screen is the sum of the fields
produced by the slits individually.
E = E1 + E2 + E3
iδ
= A + Ae + Be
where δ =
y
A
d
iaδ
2πd
sin θ
λ
B
D
O
ad
C
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The total intensity at θ is
I = EE * = 2 A2 + B 2 + 2 A2 cos δ + 2 AB ⎡⎣ cos ( aδ ) + cos (1 − a ) δ ⎤⎦
2π d
2π d
2π d
y
3.22 × 10−3 15 × 10−3
sin θ ≅
×
= 505.7
θ=
× = 2π ×
where δ =
λ
λ
λ
D
6 ×10−7
1
δ = 145.80
given, A2 = 2 mw / m , B 2 = 5 mw / m 2 , d = 3.22 mm, ad = 2.04 mm, a = 0.6335 mm
∴ I = 2 × 2 × 10−3 + 5 × 10−3 + 2 × 2 × 10−3 cos (δ ) + 2 2 5 × 10−3 ⎡⎣cos aδ + cos (1 − a ) δ ⎤⎦
= 9.23 × 10−3 w / m 2
I = 9.23 mw / m 2
Q15.
One gram of ice at 0 o C is melted and heated to water at 39 o C . Assume that the specific
heat remains constant over the entire process. The latent heat of fusion of ice is
80 Calories/gm. The entropy change in the process (in Calories per degree) is _________.
Ans:
0.39
Solution: ΔS1 =
302 dT
302
ML 1× 80
=
, ΔS2 = MC ∫
⇒ ΔS2 = 1.1ln
273 T
273
T
273
⇒ ΔS = ΔS1 + ΔS 2 ⇒ ΔS =
Q16.
80
302
+ 1.1ln
= 0.29 + 0.1 = 0.39
273
273
A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane
inclined at an angle 30 o to the horizontal. The linear acceleration of the disk (in m / sec 2 )
is _____________.
Ans:
3.266
Solution: Equation of Motion mg sin θ − f = ma
Torque = fR = I α
mg sin θ −
a=
Iα
= ma ,
R
⎛
a
mR 2 ⎞
=
=
I
,
α
⎜
⎟
R
2 ⎠
⎝
g
= 3.266
3
Q17.
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A nozzle is in the shape of a truncated cone, as shown in the figure.
The area at the wide end is 25cm 2 and the narrow end has an area
of 1 cm 2 . Water enters the wider end at a rate of 500 gm / sec . The
50 cm
height of the nozzle is 50 cm and it is kept vertical with the wider
end at the bottom. The magnitude of the pressure difference in kPa
( 1 kPa = 10 3 N / m 2 ) between the two ends of the nozzle is __________
Ans:
17.5
Solution: According to Bernoulli’s equation
Pt , Vt , At
1
1
Pb + ρ ghb + ρVb2 = Pt + ρ ght + ρVt 2
2
2
(
1
⇒ Pb − Pt = ρ g ( ht − hb ) + ρ Vt 2 − Vb2
2
)
ht − hb = 50 cm
Now given ρ AV
t t = 500 gm / sec
500 × 10−3 kg / sec
500 × 10−3 kg / sec
⇒ Vt =
=
ρ × At
1000 kg / m3 × 10−4 m 2
Pb , Vb , Ab
⇒ Vt = 5 m / sec
According to equation of continuity
At
1 cm 2
AV
Vt ⇒ Vb =
× 5 m / sec = 0.2 m / sec
t t = AbVb ⇒ Vb =
25 cm 2
Ab
∴ ΔP = Pb − Pt = 1000 × 10 × 50 × 10 −2 +
1
2
× 1000 × ⎡52 − ( 0.2) ⎤
⎣
⎦
2
⇒ ΔP = 5000 + 500 ( 25 − 0.04 ) = 5000 + 12480 = 17480 N / m 2 ⇒ ΔP = 17.5 kPa
Q18.
A block of mass 2 kg is at rest on a horizontal table The coefficient of friction between
the block and the table is 0.1. A horizontal force 3 N is applied to the block The speed of
the block (in m/s) after it has moved a distance 10 m is ________________.
Ans:
3.225
Solution: f r = μ N = 0.1 × 2 × 10 = 2 N
∵ m = 2 kg
Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Applied force is more than friction
ma = F − μ N = 3 − 2 = 1 ⇒ a =
1 1
= = 0.5 m / s 2
m 2
∵ v = u 2 + 2as ⇒ v = 2as = 2 × 0.5 × 10 = 10 = 3.225 m / s
∵ u = 0, s = 10m
Q19.
A homogeneous semi-circular plate of radius R = 3m is shown in the figure. The
distance of the center of mass of the p1ate (in meter) from the point O is _______.
0
Ans:
3m
1.3
Solution: In problem R = 3m
The area of the shaded part is π rdr . The area of the
r + dr
plate is π R 2 / 2 . As the plate is uniform, the mass per
M
unit area is
. Hence the mass of the
π R2 / 2
r
R
semicircular element
M
2Mrdr
(π rdr ) = 2
2
πR /2
R
The y - coordinate of the centre of mass of this wire is 2r / π . The y - coordinate of the
centre of the plate is, therefore,
1 ⎛ 2r ⎞ ⎛ 2 Mr ⎞ 1 4 M R 3 4 R 4
dr ⎟ =
⋅
=
= = 1.3
⎜ ⎟⎜
2
M ∫0 ⎝ π ⎠ ⎝ R 2
⎠ M πR 3 3 π π
R
Y=
The x - coordinate of the centre of mass is zero by symmetry.
Q20.
Consider a 20μm diameter p − n junction fabricated in silicon. The donor density is
1016 per cm3 . The charge developed on the n − side is 1.6 × 10 −13 C . Then the width
(in μm ) of the depletion region on the n − side of the p − n junction is _________.
Ans:
Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 30

fiziks - IIT JAM Physics

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