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AIEEE/2011/PHYSICS
S.
No
Q.1
Questions
The transverse displacement y (x,t) of a wave on a string is given by y(x,t)
2
2
=e−(ax +bt +2 ab xt ) This represents a
(a) wave moving in – x direction with speed
(b) standing wave of frequency
(c) standing wave of frequency
1
Q.3
Q.4
Q.5
Q.6
Q.7
b
a
b
Sol: 1 (a)
y(x,t) = e− ( ax +
bt )2 V =
b
a
Wave is moving in negative x - dir n
b
(d) wave moving in + x direction with
Q.2
Solutions
a
b
A screw gauge gives the following reading when used to measure the
diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions of the
circular scale. The diameter of wire from the above date is :
(1) 0.052 cm
(2) 0.026 cm
(3) 0.005 cm
(4) 0.52 cm
A mass m hangs with the help of a string wrapped around a pulley on a
frictionless bearing. The pulley has mass m and radius R. Assuming pulley
to be a perfect uniform circular disc, the acceleration of the mass m, if the
string does not slip on the pulley, is
2
g
3
(a) g
(b) 3g
(c) 3
(d) 2g
Sol: 2 (a)
Work done in increasing the size of a soap bubble from a radius of 3 cm to
5 cm is nearly (Surface tension of soap solution = 0.03 Nm−1):
(a) 0.2π mJ
(b) 2π mJ
(c) 0.4 π mJ
(d) 4π mJ
A thin horizontal circular disc is rotating about a vertical axis passing
through its centre. An insect is at rest at a point near the rim of the disc.
The insect now moves along a diameter of the disc to reach its other end.
During the journey of the insect, the angular speed of the disc:
(a) continuously decreases
(b) continuously increases
(c) first increases and then decreases (d) remains unchanged
Two particles are executing simple harmonic motion of the same
amplitude A and frequency ω along the x-axis. Their mean position is
separated by distance X0 (X0 > A) . If the maximum separation between
them is (X0 + A) , the phase difference between their motion is :
π
π
π
π
(a) 3
(b) 4
(c) 6
(d) 2
Two bodies of masses m and 4 m are placed at a distance r. The
gravitational potential at a point on the line joining them where the
gravitational field is zero is:
4Gm
6Gm
9Gm
(a) - r
(b) - r
(c) - r
(d) zero
Sol: 4 (c)
Diameter of wire =
1
100
×
52
10
= 0.052 cm
Sol: 3 (b)
1
1
T × R = I ∝ = 2 MR2 ∝ (I = 2 MR2 )
T = ½ Ma (a = ∝ R) ……………….. (i)
Mg – T = Ma …………………..(ii)
2
Form (i) & (ii) a = 3g
W = T × ΔA = T × 8π (r2 2 – r1 2 ) = 0.4πmJ
Sol: 5 (c)
τ=0
Angular momentum is conserve
I ω1
I1ω1 = I2ω2 ⇒ ω2 = 1I
2
I2 is first decreasing and then increasing ∴ω first increases and then decreases.
Sol:6 (d)
ψ1 = 0
π
ψ2= 2
Sol: 7 (c)
Distance between the null point and mass m, x =
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r
4m
1+
m
=
r
3
V = - Gm
12
2r
+
3
2
=-
9Gm
r
1
AIEEE/2011/PHYSICS
Q.8
Q.9
Q.10
Q.11
Q.12
Q.13
Q.14
Two identical charged spheres suspended from a common point by two
massless strings of length l are initially a distance d (d << 1) apart because
of their mutual repulsion. The charge begins to leak from both the spheres
at a constant rate. As a result the charges approach each other with a
velocity v. Then as a function of distance x between them,
(a) v ∝ x−1
(b) v ∝ x1/ 2
(c) v ∝ x
(d) v ∝ x−1/ 2
Sol: 8 (d)
Let the separation b/w the charges at any instant is x
A boat is moving due east in a region where the earth’s magnetic field is
5.0 ×10−5NA−1m−1 due north and horizontal. The boat carries a vertical
aerial 2m long. If the speed of the boat is 1.50 ms−1, the magnitude of the
induced emf in the wire of aerial is :
(a) 0.75 mV
(b) 0.50 mV
(c) 0.15 mV
(d) 1 mV
An object, moving with a speed of 6.25 m/s, decelerates at a rate given by
dv
: = - 2.5 v Where v is the instantaneous speed. The time taken by the
dt
object, to come to rest, would be :
(a) 2 s
(b) 4 s
(c) 8 s
(d) 1 s
Sol: 9 (c)
A fully charged capacitor C with initial charge q0 is connected to a coil of
self inductance L at t = 0. The time at which the energy is stored equally
between the electric and the magnetic field is :
π
(a) 4 LC
(b) 2π LC
(c) LC
(d) π LC
Let the x – z plane be the boundary between two transparent media.
Medium 1 in z ≥ 0 has a refractive index of 2 and medium 2 with z < 0
has a refractive index of 3 . A ray of light in medium 1 given by the
vector A = 6 3i +8 3 j −10k is incident on the plane of separation. The
angle of refraction in medium 2 is
(a) 450
(b) 600
(c) 750
(d) 300
A current I flows in an infinitely long wire with cross section in the form
of a semicircular ring of radius R. The magnitude of the magnetic
induction along its axis is
μ l
μ0l
μ l
μ l
(a) 2π02 R
(b) 2πR
(c) 4π02 R
(d) π 20R
Sol:11 (a)
Charge oscillates with simple harmonic motion q = q0 sin ωt,
A thermally insulated vessel contains an ideal gas of molecular mass M
and ratio of specific heats γ. It is moving with speed υ and is suddenly
brought to rest. Assuming no heat is lost to the surroundings, its
temperature increases by :
Sol: 14 (c)
W=∆ ∪
(a)
Q.15
(γ−l)
2γR
Mυ2 K
(b)
γMυ 2
2R
K
(c)
(γ−1)
2R
Mυ2 K
2
(a)
M+m
M
(b)
M
(c)
M
M+m
1/2
X
dQ
⇒ 2Q
dt
= C3X2
dx
dx
⇒
dt
dt
∝
x 3/2
x2
∝ x −1/2
E = BHℓV = 5.0 × 10−5 × 2 × 1.50
= 0.15 mV
Sol: 10 (a)
dv
= - 2.5 v ⇒
dt
1
dv
2.5
v
=
⇒2 v = - 2.5t+c
1 q2
∪=2
5
at v = 0 ⇒ t 2.5 = 2s
at t = 0 v = 6.25 C = 5
q=
C
dt
q0
2
π
T
⇒ ω= 4
⇒t = 8 =
2π
LC
8
Sol: 12 (a)
Normal to the plane is z –axis
A
10
1
Cosθ1 = Az = 20 = 2, θ1 =600
μ1 sinθ1 = μ2 sinθ2 ⇒ 2 ×
Sol: 13 (d)
μ 0 di
dB = 2πR
−cosθi − sinθj
T
di = πR Rdθ
I
= π dθ
1
mv2 = nCv dT
2
3
2
= 3 sin θ2 ⇒ θ2 = 450
μ I
μ I
dB = 2π02 R −cosθi − sinθj
=
m
M
R
dT
y−1
B = - π 20R j
dT =
M y−1 v 2
2R
K
(γ−1)
(d) 2(γ−1)R Mυ2 K
A mass M, attached to a horizontal spring, executes S.H.M. with
amplitude A1. When the mass M passes through its mean position then a
smaller mass m is placed over it and both of them move together with
A
amplitude A2. The ratio o A 1 is:
M+m 1/2
Q2
equilibrium condition = K 2 = ω
X
2l
⇒Q2 = CX3
Sol: 15 (c)
Simple harmonic oscillator has constant Energy
2
A
1
M +m
2
2
⇒2Mω2 A = (m+M) ω2 A
⇒ 12 = M
1
2
A
2
M
(d) M+m
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A
∴ A1 =
2
M+m
m
2
AIEEE/2011/PHYSICS
Q.16
Q.17
Q.18
Water is flowing continuously from a tap having an internal diameter 8
×10−3m. The water velocity as it leaves the tap is 0.4 ms−1 . The diameter
of the water stream at a distance 2×10−1m below the lap is close to :
(a) 7.5 ×10−3m
(b) 9.6 ×10−3m
−3
(c) 3.6 ×10 m
(d) 5.0 ×10−3m
Sol: 16 (c)
(Vb )2 − (0.4)2 = 2 × 9.8 × 0.2
From continuity Equation
a1 v1 = a 2 v2
π [8 × 10−3 ] × 0.4 = πd2 × 4
On The basis of given statement, choose the correct answer
Statement-1 : Sky wave signals are used for long distance radio
communication. These signals are in general, less stable than ground wave
signals.
Statement-2 : The state of ionosphere varies from hour to hour, day to
day and season to season.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is the
correct explanation of Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not the
correct explanation of Statement-1.
(c) Statement-1 is false, Statement-2 is true.(d) Statement-1 is true,
Statement-2 is false.
Three perfect gases at absolute temperatures T1, T2 and T3 are mixed. The
masses of molecules are m1, m2 and m3 and the number of molecules are
n1 , n2 and n3 respectively. Assuming no loss of energy, the final
temperature of the mixture is :
Sol: 17 (a)
(a)
(c)
Q.19
n 1 T 1 +n 2 T 2 +n 3 T 3
n 1 +n 2 +n 3
2 2 2 2 2 2
n T +n T +n T
1 1 2 2 3 3
n 1 T 1 +n 2 T 2 +n 3 T 3
2
2
n 1 T 1 +n 2 T +n 3 T
2
3
1 T 1 +n 2 T 2 +n 3 T 3
d ≈ 3.6 × 10−3 m
Ionospheric property does not remain same with time. So generally sky wave signals are less stable than
ground wave signals
Sol: 18 (a)
Data ⇒n1k t1+ n2k T2+ n3k T3 = (n1+n2+n3)kT
n T +n T 2 +n 3 T 3
∴ T = 1 1n +n2 +n
1
2
3
(T 1 +T 2 +T 3 )
3
A pulley of radius 2 m is rotated about its axis by a force F = (20t − 5t2)
Newton (where t is in seconds) applied tangentially. If the moment of
inertia of the pulley about its axis of rotation made by the pulley before its
direction of motion if reversed, is :
(a) more than 3 but less than 6
(b) more than 6 but less than 9
(c) more than 9
(d) less than 3
Q.20
⇒ Vb = 2 m/s,
(b) n
(d)
A resistor ‘R’ and 2μF capacitor in series is connected through a switch to
200 V direct supply. Across the capacitor is a neon bulb that lights up at
120 V. Calculate the value of R to make the bulb light up 5 s after the
switch has been closed. (log10 2.5 = 0.4)
(a) 1.7 ×105Ω (b) 2.7 ×106Ω (c) 3.3 ×107Ω (d) 1.3 ×104Ω
3
Sol: 19 (a)
r×F=I×α
2(20t-5t2) = 10 α ⇒ α = 4t – t2
dω
=4t −t2
dt
dω= (4t −t 2 )dt
t3
ω =2t2− 3 (on integration)
ω = 0 ⇒ t = 6s
2t 3
t4
⇒ θ = 3 − 12 (on integration)
θ (t = 6sec) = 36 rad
⇒ 2πn = 36
Sol: 20 (b)
VC = E(1 - e−t/Rc )
120
3
1 - e−t/Rc = 200 = 5
5
⇒ R = 1.84 ×10 −6 =2.7 × 106 Ω
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ω=
dθ
dt
= 2t2 -
36
t3
n = 2π =<6
3
dθ = 2t 2 −
t3
3
dt
AIEEE/2011/PHYSICS
Q.21
Q.22
Q.23
Q.24
Q.25
Q.26
A Carnot engine operating between temperatures T1 and T2 has efficiency
1
1
When T2 is lowered by 62 K, its efficiency increases to 3 . Then T1 and
6
T2 are, respectively :
(a) 372 K and 330 K
(b) 330 K and 268 K
(c) 310 K and 248 K
(d) 372 K and 310 K
Sol: 21 (d)
T (T − 62) 1
η2 = 1− 2
=
If a wire is stretched to make it 0.1% longer, its resistance will :
(1) increase by 0.2%
(2) decrease by 0.2%
(3) decrease by 0.05%
(4) increases by 0.05%
Sol: 22 (a)
R∝ ℓ2 (for a given volume)
∆R
2∆ℓ
⇒ R%= ℓ %
When wire is stretched by 0.1% then resistance will increases by 0.2%
Direction:
The question has a paragraph followed by two statements, Statement – 1
and statement – 2. Of the given four alternatives after the statements,
choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a plane – convex
lens over a plane glass plate. With monochromatic light, this film gives an
interference pattern due to light reflected from the top (convex) surface
and the bottom (glass plate) surface of the film.
Statement-1: When light reflects from the air-glass plate interface, the
reflected wave suffers a phase change of π.
Statement-2: The centre of the interference pattern is dark.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is the correct
explanation
of Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct
explanation of Statement-1.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is false.
A car is fitted with a convex side-view mirror of focal length 20cm. A
second car 2.8 m behind the first car is overtaking the first car at relative
speed of 15 m/s. The speed of the image of the second car as seen in the
mirror of the first one is :
1
1
(a) 15 m/s
(b) 10m/ s
(c) 15m/ s (d) 10 m/s
Sol: 23 (a)
Energy required for the electron excitation in Li++ from the first to the
third Bohr orbit is :
(a) 36.3 eV
(b) 108.8 eV
(c) 122.4 eV
(d) 12.1 eV
Sol: 25 (b)
The electrostatic potential inside a charged spherical ball is given by ϕ =
ar 2 + b where r is the distance from the centre; a, b are constants. Then the
charge density inside ball is
Sol: 26 (c)
Potential inside (ϕ) = ar 2 + b
δv
∴ Er = - = - 2ar
δr
Electric field inside uniformly charged solid volume varies with ‘r’. So charge density is constant
(a) −6aε0r
(b) −24πaε0 r
(c) −6aε0
(d) −24πaε0 r
η1 =
T1
T 1− T 2
1
T 1− T 2
T2
T1
⇒ 1-T =
=6
T1
⇒
3
1
+
1
62
=
T1
6
T2
1
=
3
372
5
=6
62
T1
=
1
6
⇒ T1 = 372 K
T2 = 310K
When light enter changes its medium (air to glass) a phase change π occurs and hence destructive
interference will occur at centre.
Sol: 24 (a)
1
v
1
1
1 dv
+u = f
-v 2
1
f = 20 cm
vI = -
280
15×280
2
u
dt
1 du
- u2
1
dt
du
=0
dt
1
⇒v=
+ −280 = 20
× 15
=
1
15
v2
= − u2
du
dt
280
15
cm
m/s
Z2
En = - 13.6 n 2
9
ELI ++ = - 13.6 × 1= - 122.4ev
∆E= - 13.6 – (- 122.4)
ϕ net = −(2ar) 4πr 2 = −8πar 3 − 8πar3 =
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9
ELI +++ = - 13.6 × 9= - 13.6ev
= 108.8 eV
4
3
σ× πr 3
ε0
∴ σ = - 6aε0
4
AIEEE/2011/PHYSICS
Q.27
A water fountain on the ground sprinkles water all around it. If the speed
of water coming out of the fountain is v, the total area around the fountain
that gets wet is :
Sol: 27 (a)
(a) π
Area covered = π
v4
(b)
g2
π v4
2 g2
v2
(c) π
g2
v4
(d) π
g
Max. range =
u2
g
v2
i.e., g (radius of circle)
v2
g
2
v4
= π g2
Q.28
100g of water is heated from 300C to 500C. Ignoring the slight expansion
of the water, the change in its internal energy is (specific heat of water is
4148 J/kg/K):
(a) 8.4 kJ
(b) 84 kJ
(c) 2.1 kJ
(d) 4.2 kJ
Sol: 28 (a)
ΔQ = ΔU+ ΔW (ignoring expansion)
ΔU = mCv ΔT = 0.1× 4.184× (50 – 30) = 8.368 kJ
Q.29
The half life of a radioactive substance is 20 minutes. The approximate
2
time interval (t 2 − t1 )between the time t2 when of it has decayed and
Sol: 29 (b)
t 1 = 20 minutes
3
1
2
time t1 and of it had decayed is :
3
(a) 14 min
Q.30
(b) 20 min
3
(c) 28 min
(d) 7 min
This question has Statement – 1 and Statement – 2. Of the four choices
given after the statements, choose the one that best describes the two
statements.
Statement-1 : A metallic surface is irradiated by a monochromatic light of
frequency v > v0 (the threshold frequency). The maximum kinetic energy
and the stopping potential are Kmax and V0 respectively. If the frequency
incident on the surface doubled, both the Kmax and V0 are also doubled.
Statement-2 : The maximum kinetic energy and the stopping potential of
photoelectrons emitted from a surface are linearly dependent on the
frequency of incident light.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is the
correct explanation of Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not the
correct explanation of Statement-1.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is false.
2
−λt 2
N0 = N0 e
1
1
λ
2
= In
=
KE max
KE max
λ
2
1
N0 = N0 e−λt2 t1 = In3
λ
t 2 - t1 =
hυ − hυ0 = e × ∆v
may not be equal to 2
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3
3
1
λ
= (∆V)′ =
⇒
hυ
e
V0′
V0
-
⇒ ∆V =
hυ
e
–
hυ 0
e
hυ 0
e
may not be equal to 2
3
In − In3
= 20 min
λ
KE max = hυ − hυ0
hυ
hυ
V = e − e0
1
t 2 = In
0.693
Sol: 30 (c)
KEmax = hυ − hυ0
ʻυʼ is doubled
KEmax = 2hυ – hυ0
2
N = N0 e−λt2 λt1 = In3
……………(i)
2
5