K-CET PYQs
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llikhithnsKCET-2015 (Physics)
1. The ratio of the dimensions of Planck constant and that of moment of inertia has the
dimensions of
a. time b. frequency
c. angular momentum d. velocity
2. The velocity – time graph for two bodies A and B are shown. Then the acceleration of A and
B are in the ratio
Velocity
A
40°
25°
Time
a. tan 25° to tan 40° b. tan 25° to tan 50°
c. sin 25° to sin 50° d. cos 25° to cos 50°
3. A particle is projected with a velocity v so that its horizontal range is twice the greatest
height attained. The horizontal range is
v2 2v 2
a. b.
g 3g
4v 2 v2
c. d.
5g 2g
4. A stone of mass 0.05 kg is thrown vertically upwards. What is the direction and magnitude
of net force on the stone during its upward motion?
a. 0.49 N vertically upwards b. 0.49 N vertically downwards
c. 0.98 N vertically downwards d. 9.8 N vertically downwards
5. The kinetic energy of a body of mass 4 kg and momentum 6 Ns will he
a. 2.5 J b. 3.5 J
c. 4.5 J d. 5.5 J
6. The ratio of angular speed of a second-hand to the hour-hand of a watch is
a. 720 : 1 b. 60 : 1
c. 3600 : 1 d. 72 : 1
7. If the mass of a body is M on the surface of the earth, the mass of the same body on the
surface of the moon will be
a. M/6 b. M
c. 6 M d. zero
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8. Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing
through its centre is 1. If the same rod is bent into a ring and its moment of inertia about its
I
diameter is F. then the ratio will be
I
3 2 8 2
a. b.
2 3
2 2 5 2
c. d.
3 3
9. The ratio of hydraulic stress to the corresponding strain is known as
a. Compressibility b. Bulk modulus
c. Young's modulus d. Rigidity modulus
10. The efficiency of a Carnot engine which operates between the two temperatures T1 = 500 K
and T2 = 300 K is
a. 50% b. 25%
c. 75% d. 40%
11. Water is heated from 0 °C to 10 °C. then its volume
a. decreases
b. increases
c. does not change
d. first decreases and then increases
12. 1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the
mixture is
a. 0 °C b. 100 °C
c. 50 °C d. 55 °C
13. The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance
is equal to half of its amplitude, the distance being measured from its equilibrium position
will be
a. 3 : 1 b. 4 : 1
c. 2 : 1 d. 8 : 1
14. When two tuning forks A and B are sounded together. 4 beats per second are heard. The
frequency of the fork B is 384 Hz. When one of the prongs of the fork A is filed and sounded
with B. the beat frequency increases, then the frequency of the fork A is
a. 380 Hz b. 378 Hz
c. 379 Hz d. 389 Hz
15. A stretched string is vibrating in the second overtone, then the number of nodes and
antinodes between the ends of the string are respectively
a. 4 and 3 b. 3 and 2
c. 3 and 4 d. 2 and 3
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16. Two spheres carrying charges + 6 C and + 9 C separated by a distance d, experiences a
force of repulsion F. When a charge of –3 C is given to both the sphere and kept at the
same distance as before, the new force of repulsion is
a. F b. 3F
c. F/3 d. F/9
17. Pick out the statement which is incorrect.
a. The tangent drawn to a line of force represents the direction of electric field.
b. The electric field lines form closed loop.
c. A negative test charge experiences a force opposite to the direction of the field.
d. Field lines never intersect.
18. The angle between the dipole moment and electric field at any point on the equatorial
plane is
a. 0° b. 90°
c. 180° d. 45°
19. Three point charges 3nC, 6nC and 9nC are placed at the corners of an equilateral triangle of
side 0.1 in. The potential energy of the system is
a. 8910 J b. 89100 J
c. 9910 J d. 99100 J
20. A spherical shell of radius 10 cm is carrying a charge q. If the electric potential at distances
5 cm, 10 cm and 15 cm from the centre of the spherical shell is V1, V2 and V3 respectively,
then
a. V1 > V2 > V3 b. V1 < V2 < V3
c. V1 = V2 > V3 d. V1 = V2 < V3
21. A parallel plate capacitor is charged and then isolated. The effect of increasing the plate
separation on charge, potential and capacitance respectively are
a. constant, decreases, decreases
b. increases, decreases, decreases
c. constant, decreases, increases
d. constant, increases, decreases
22. Four identical cells of emf E and internal resistance r are to be connected in series. Suppose
if one of the cell is connected wrongly. the equivalent emf and effective internal resistance
of the combination is
a. 4E and 4r b. 4E and 2r
c. 2E and 4r d. 2E and 2r
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23. Three resistances 2, 3 and 4 are connected in parallel. The ratio of currents passing
through them when a potential difference is applied across its ends will be
a. 6 : 3 : 2 b. 6 : 4 : 3
c. 5 : 4 : 3 d. 4 : 3 : 2
24. Two cells of emf E1 and E2 are joined in opposition (such that E1 > E2). If r1 and r2 be the
internal resistances and R be the external resistance, then the terminal potential difference
is
r1 E1 E2 r2
+ – – +
E1 + E 2 E1 + E 2
a. R b. R
r1 + r2 r1 + r2 + R
E1 − E 2 E1 − E 2
c. R d. R
r1 + r2 r1 + r2 + R
25. In the circuit shown below, the ammeter and the voltmeter readings are 3 A and 6 V
respectively. Then the value of the resistance R is
R
A
V
a. 2 b. > 2
c. < 2 d. 2
26. In Wheatstones network P = 2 , Q = 2 , R = 2 and S = 3 . The resistance with which S
is to be shunted in order that the bridge may be balanced is
a. 1 b. 2
c. 4 d. 6
27. The resistance of the bulb filament is 100 at a temperature of 100 °C. If its temperature
co-efficient of resistance be 0.005 per °C, its resistance will become 200 at a temperature
a. 300°C b. 400°C
c. 500°C d. 200°C
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28. Two concentric coils each of radius equal to 2 cm are placed right angles to each other. If
3A and 4A are the currents flowing through the two coils respectively. The magnetic
induction (in Wb m–2) at the centre of the coils will be
a. 12 × 10–5 b. 10–5
c. 5 × 10–5 d. 7 × 10–5
29. A proton beam enters a magnetic field of 10–4 Wb m–2 normally. if the specific charge of the
proton is 1011 C kg–1 and its velocity is 109 ms–1, then the radius of the circle described will
be
a. 0.1 m b. 10 m
c. 100 m d. 1 m
30. A cyclotron is used to accelerate
a. neutron
b. only positively charged particles
c. only negatively charged particles
d. both positively and negatively charged particles
31. A galvanometer of resistance 50 gives a full-scale deflection for a current 5 × 104 A. The
resistance that should be connected in series with the galvanometer to read 3 V is
a. 595 b. 5050
c. 5059 d. 5950
32. Two parallel wires 1 m apart carry currents of 1 A and 3 A respectively in opposite
directions. The force per unit length acting between these two wires is
a. 6 × 10–7 Nm–1 repulsive b. 6 × 10–7 Nm–1 attractive
c. 6 × 10–5 Nm–1 repulsive d. 6 × 10–5 Nm–1 attractive
33. If there is no torsion in the suspension thread, then the time period of a magnet executing
SHM is
1 MB 1 1
a. T= b. T=
2 1 2 MB
1 MB
c. T = 2 d. T = 2
MB 1
34. Core of electromagnets are made of ferromagnetic material which has
a. high permeability and low retentivity
b. high permeability and high retentivity
c. low permeability and high retentivity
d. low permeability and low retentivity
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35. The magnetic susceptibility of a paramagnetic material at –73 °C is 0.0075 and its value at
–173 °C will be
a. 0.0045 b. 0.0030
c. 0.015 d. 0.0075
36. Two coils have a mutual inductance 0.005 H. The current changes in the first coil according
to the equation i = im sin t where im = 10 A and = 100 rad s–1. The maximum value of
the emf induced in the second coil is
a. 2 b. 5
c. d. 4
37. An aircraft with a wingspan of 40 m flies with a speed of 1080 km/hr in the eastward
direction at a constant altitude in the northern hemisphere, where the vertical component
of the earth's magnetic field is 1.75 × 10–5 T. Then the emf developed between the tips of
the wings is
a. 0.5V b. 0.34V
c. 0.21V d. 2.1V
38. In an LCR circuit, at resonance
a. the current and voltage are in phase
b. the impedance is maximum
c. the current is minimum
d. the current leads the voltage by /2
39. A transformer is used to light 100 W –110 V lamp from 220 V mains. If the main current is
0.5 A, the efficiency of the transformer is
a. 90% b. 95%
c. 96% d. 99%
40. The average power dissipated in a pure inductor is
1 b. VI2
a. VI
2
VI 2 d. zero
c.
4
41. If 0 and 0 are the permittivity and permeability of free space and and are the
corresponding quantities for a medium, then refractive index of the medium is
00
a. b.
00
c. 1 d. Insufficient information
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42. A person wants a real image of his own, 3 times enlarged. Where should he stand in front of
a concave mirror of radius of curvature 30 cm?
a. 10 cm b. 30 cm
c. 90 cm d. 20 cm
43. Calculate the focal length of a reading glass of a person if his distance of distinct vision is 75
cm.
a. 25.6 cm b. 37.5 cm
c. 75.2 cm d. 100.4 cm
44. In a Young's double slit experiment, the slit separation is 0.5 m from the slits. For a
monochromatic light of wavelength 500 nm, the distance of 3nd maxima from 2nd minima
on the other side is
a. 2.75 mm b. 2.5 mm
c. 22.5 mm d. 2.25 mm
45. To observe diffraction, the size of the obstacle
a. has no relation to wavelength.
b. should be /2, where is the wavelength.
c. should be much larger than the wavelength.
d. should be of the order of wavelength.
46. The polarizing angle of glass is 57°. A ray of light which is incident at this angle will have an
angle of refraction as
a. 25° b. 33°
c. 43° d. 38°
47. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV
respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio
of maximum speeds of emitted electrons will be
a. 1 : 5 b. 1 : 4
c. 1 : 2 d. 1 : 1
48. Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.
a. 95 pm b. 102 pm
c. 112 pm d. 124 pm
49. An -particle of energy 5 MeV is scattered through 180° by gold nucleus. The distance of
closest approach is of the order of
a. 10–10 cm b. 10–12 cm
c. 10 cm
–14 d. 10–16 cm
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50. If an electron in hydrogen atom jumps from an orbit of level n = 3 to an orbit of level n = 2,
the emitted radiation has a frequency (R = Rydberg constant, C = velocity of light)
3RC RC
a. b.
27 25
8RC 5RC
c. d.
9 36
51. What is the wavelength of light for the least energetic photon emitted in the Lyman series
of the hydrogen spectrum. (take he = 1240 eV nm)
a. 82 nm b. 102 nm
c. 122 nm d. 150 nm
52. A nucleus at rest splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities
are in the ratio
a. 8:1 b. 6:1
c. 4:1 d. 2:I
53. The half-life of a radioactive substance is 20 minutes. The time taken between 50 % decay
and 87.5 % decay of the substance will be
a. 30 minutes b. 40 minutes
c. 25 minutes d. 10 minutes
54. A radioactive decay can form an isotope of the original nucleus with the emission of
particles
a. one and four b. one and two
c. one and one d. four and one
55. An LED is constructed from a pn junction based on a certain semi-conducting material
whose energy gap is 1.9 eV. Then the wavelength of the emitted light is
a. 2.9 × 10–9 m b. 1.6 × 10–8 m
c. 6.5 × 10–7 m d. 9.1 × 10–5 m
56. Amplitude modulation has
a. one carrier with two side band frequencies
b. one carrier
c. one carrier with infinite frequencies
d. one carrier with high frequency
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57. The circuit has two oppositely connected ideal diodes in parallel. What is the current
flowing in the circuit?
2
D2
D1
3
+ –
4 12V
a. 1.71 A b. 2.0 A
c. 2.31 A d. 1.33 A
58. The input characteristics of a transistor in CE mode is the graph obtained by plotting
a. IB against VBE at constant VCE b. IB against VCE at constant VBE
c. IB against IC at constant VCE d. IB against IC at constant VBE
59. The given truth table is for Input Output
Input Output
A B Y
0 0 1
0 1 1
1 0 1
1 1 0
a. AND gate b. OR gate
c. NAND gate d. NOR gate
60. The waves used for line-of-sight (LOS) communication is
a. ground waves b. space waves
c. sound waves d. sky waves
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ANSWER KEYS
* G – Indicates one Grace mark awarded for the question number.
1. (b) 2. (b) 3. (c) 4. (b) 5. (c) 6. (a) 7. (b) 8. (c) 9. (b) 10. (d)
11. (d) 12. (b) 13. (a) 14. (b) 15. (a) 16. (c) 17. (b) 18. (c) 19. (G) 20. (c)
21. (d) 22. (c) 23. (b) 24. (d) 25. (c) 26. (d) 27. (b) 28. (c) 29. (c) 30. (d)
31. (d) 32. (a) 33. (c) 34. (a) 35. (c) 36. (b) 37. (c) 38. (a) 39. (a) 40. (d)
41. (b) 42. (d) 43. (b) 44. (G) 45. (d) 46. (b) 47. (c) 48. (c) 49. (b) 50. (d)
51. (c) 52. (a) 53. (b) 54. (b) 55. (c) 56. (a) 57. (a) 58. (a) 59. (c) 60. (b)
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Solution
1. (b)
E = h
E ML T
2 −2
h= = = ML2T −1 ………………… (1)
T
−1
Moment of Inertia I = mr2 = [ML2] ………………… (2)
Dividing eq. (1) and (2)
h ML2T −1
I = = T −1
ML
2
[T–1] = frequency of a particle.
2. (b)
v
Acceleration a =
t
So, acceleration of body A = aA = tan25°
Line B makes an angle of 50°. So its slope is equal to tan 50°.
aB = tan 50°
aA : aB = tan 25° : tan 50°
3. (c)
Option (3) is a correct answer not matched with answer sheet
Given R = 2H
v 2 sin 2 2v 2 sin 2
=
g 2g
sin 2 = sin 2
2 sin cos = sin2
2 sin cos = sin sin
sin sin
2=
sin cos
2 = tan
From triangle we can say that –
2 1
sin = , cos =
5 5
2v sin cos 2v 2 2
2
1
So, R = =
g g 5 5
2
4v
R=
5g
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4. (b)
F = mg
Here m = 0.05 kg
F = 0.05 × (–9.8) = – 0.49N
F = 0.49 N vertically downwards
5. (c)
Given m = 4kg, p = 6 Ns
p 2 6 6 0.36
K.E. = = =
2m 2 4 8
K.E. = 4.5 J
6. (a)
2
Angular velocity = =
60 30
For an hour hand to complete rotation i.e. 2 radians it takes 12 hrs.
2
So angular velocity of an hour hand =
12 3600
=
21600
Then the ratio is
21600
= 30 = = 720 :1
30
21600
7. (b)
As we know that the value of acceleration due to gravity is independent of mass, shape, size
etc. of the body and depends upon the mass and radius of the earth.
8. (c)
mL
mR 2
M.O.I. of ring about its diameter is given by I 2 =
2
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I2
2
mL
I1 =
12
L
L = 2R R =
2
2
mR 2
1 L mL2
I2 = = m = 2
2 2 2 8
I1 mL 82 2
8 2
22
= = =
I2 12 mL 12 3
9. (b)
The ratio of hydraulic stress to the corresponding strain is known so bulk modulus.
hydraulicstress
Bulk modulus [K] =
strain
10. (d)
T1 = 500K, T2 = 300K
T2
Efficiency of Carnot engine = 1 –
T1
300 500 − 300 200
= 1− = =
500 500 500
2
= = 0.4
5
So, the efficiency is 40%.
11. (d)
When water is heated from 0°C to 4°C, there is decreasing amount in volume and density
increases till it is a 4°C and than volume starts increasing to 10°C.
12. (b)
(1) Ice convert into water
Q1 = M1Lf = 1 × 80 = 80 cal
(2) Steam to convert into water
Q2 = M2Lv = 1 × 540 = 540 cal
(3) 1g of water at 0°C into water at 100°C
=1 × 1 × 100 = 100 cal
So clearly, whole steam is not condensed. So temperature of the mixture is 100°C.
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13. (a)
1
P.E. = Kx 2
2
Distance is half its amplitude so
1 A2
P.E. = K
2 4
K ( A2 − x 2 )
1
K.E. =
2
1 A 2 1 4A 2 − A 2
K.E. = K A 2 − = K
2 4 2 4
1 3A 2
K.E. = K
2 4
K.E. 3
=
P.E. 1
14. (b)
Beat frequency
b = A − B
4 = A − 384
A = 388Hz
15. (a)
From figure, number of nodes and antinodes will be 4 and 3.
A A A
N N N N
16. (c)
Kq1q 2 K ( 9 )( 6 )
F1 = =
r2 d2
K F
2 = 1
d 54
Now charges are 3C and 6C
K ( 6 )( 3) 18K
So, new force is F2 = = 2
d2 d
K
Put the value of 2 in F2
d
F1 F
F2 = 18 =
54 3
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17. (b)
Electric field lines do not form closed loop as line can never start and end on the same
change.
18. (c)
The angle between electric dipole moment and electric field in the equatorial line is 180° as
both of them are in opposite directions.
19. (G)
BONUS
20. (c)
Inside the spherical shell potential is same.
V1 = V2
1
V
d
V1 = V2 V3
21. (d)
0 A
From C =
d
Q
When increase the plate separation capacitance decrease and by V =
C
V is also increase and q is constant.
22. (c)
Resistance in series connection -
req = r1 + r2 + r3 + r4 = r + r + r + r = 4r
emf in series connection –
Eeq = E1 + E2 + E3 – E4 (When one is connected wrong)
Eeq = E + E + E – E = 2E
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23. (b)
2
I3
3
I2
4
I1
I
v
V
Current through 4 = I1 =
4
V
Current though 3 = I2 =
3
V
Similarly, I3 =
2
V V V
I1 : I 2 : I3 = : :
4 3 2
1 1 1
I1 : I 2 : I3 = : :
4 3 2
I1 : I 2 : I3 = 6 : 4 : 3
24. (d)
Two cells of emf E1 and E2 are joined in opposition E1 > E2
E1 − E 2
I=
r1 + r2 + R
V = IR
E − E2
V= 1 R
r1 + r2 + R
25. (c)
V 6
R= = = 2 (For ideal ammeter and voltmeter)
I 3
R < 2 when A and V are infinite
3(R + r) = 6V
R+r=2
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R=2–r
26. (d)
2 2
3
2
x
2 2
=
2 3x
3+ x
2 3− x
=
1 3x
3x = 6 + 2x
X = 6
27. (b)
R = R0 (1+T)
200 = 100(1+0.005.(T2-100))
2 = (1+0.005. T2-0.5)
2 = 0.5+0.005T2
1.5
T2 =
0.005
T2 = 300°C
28. (c)
0 2 2 2
Bnet = B12 + B22 = i1 + i 2
4 r
2
= 10−7 −2
32 + 42
210
= 5 × 10 wb/m2
–5
29. (c)
Given B = 10–4 Wb/m2
Q = 1011 C/kg, v = 109 m/sec
mV
Radius of circle described, r =
qB
1109 1013
r= −4
= = 102
10 10
11
1011
r = 100 m
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30. (d)
A cyclotron is used to accelerate both positively and negatively charged particles but a
neutral particle cannot be accelerated in cyclotron.
31. (d)
Give
Ig = 5 × 10–4
RG = 50
V = 3V
V
R = − RG
Ig
Put all these values in this formula
3
R= − 50
5 104
30000
R= − 50
5
R = 6000 – 50
R = 5950
32. (a)
I1 = 1A, I2 = 3A, d = 1m
2I I
F= 0 1 2
4x d
2 (1)( 3)
F = 10−7
1
F = 6 × 10 N/m
–7
Current flowing in opposite direction is repulsive in nature.
33. (c)
If there is no torsion in the suspension thread, then the time period of a magnet executing
I
S.H.M. is = T = 2
MB
34. (a)
Electromagnetic cores support the formation of a magnetic field because of high
permeability and low retentivity so that the magnetic field gets demagnetized easily.
35. (c)
Given x1 = 0.0075
T1 = -73ºC = 273-73 = 200k
T2 = -173ºC = 273-173 = 100k
X1 T1 = X2 T2
(0.0075) (200) = X2 (100)
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X2 = 0.015
36. (b)
Mdi d
e= =0.005 × (i0sint)
dt dt
= 0.005 × i0 cos t
e = 0.005 × 10 × 100 = 5
37. (c)
Given B=1.75 ×10-5T. = 40m
5
V = 1080 km/h = 1080 × =300 m/sec
18
E=Bv
E= (1.75×10-5)(300)(40)
E= (1.75×10-5)12000
E=2100×10-4
E=0.21 volt
38. (a)
For resonance condition
XL = XC
In an L-C-R circuit at resonance the current and voltage are in phase.
39. (a)
P=VI
Po
Efficiency of transformer = 100
Pi
Po = 100W
Pi = Vi Ii = 220 × 0.5 = 110W
100
= 100 = 90.9%
110
So, the efficiency of the transformer is 90%
40. (d)
Average power dissipated
vi
Pavg = 0 0 cos
2
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v0i0
Pavg = cos 90º
2
Pavg =0
41. (b)
C
μ=
V
1 1
Here C = and V=
μ 0ε 0 με
1 1
So, μ=
μ 0ε 0 με
με με
μ= =
μ 0ε 0 μ 0ε 0
42. (d)
R −30
f= = = –15cm
2 2
−v
m=
u
−v
–3 =
u
v = 3u
1 1 1
By mirror formula – = +
F v u
1 1 1 1+ 3 4
= + = =
−15 3u u 3u 3u
1 4
=
−15 3u
4 (−15)
So, =–20 cm
3
43. (b)
Here V= –7.5 cm
u = –D = –25 cm
It is Because of distinct vision-
Using lens formula
1 1 1
= −
f v u
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1 1 1 −1 + 3 2
= + = =
f −75 25 75 75
75
f= = 37.5 cm
2
44. (G)
Bonus
45. (d)
To observe diffraction the size of the obstacle should be of the order of wavelength because
of d.
46. (b)
Angle of incident i=57°
So, i = r = 57°
the reflected and refracted rays are mutually perpendicular to each other
So angle of refraction r' = 90°–i
r' = 90°-57°=33°
47. (c)
h = KE +
1 eV = KE1 + ()
2.5eV = K1E2 + ...(2)
From equation (1) and (2)
1eV = K.E.1 + 0.5 eV ...(3)
K.E.1 = 0.5eV
2.5 eV = K.E.2 = 0.5eV
K.E.2 = 2eV ...(4)
Dividing equation (3) and (4)
K .E.1 0.5 1
= =
K .E.2 2.0 4
1
mv12
2 v2 1
= 12 =
1
mv2 2 v2 4
2
v1 1
=
v2 2
48. (c)
h
λ=
p
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KCET-2015 (Physics)
−31
Here P= 2M.K.E = 2 9 10 120 1.6 10−19
P = 5.88 × 10-24 Kg MS-1
So, de Broglie wavelength
h 6.63 10−34
λ= = = 1.13 × 10-10 m
p 5.88 10−24
= 1.13 Å
49. (b)
Here K.E. = P.E.
1 q1q 2
5MeV=
4πε 0 r
(9×109 )(9 2e)(2e)
5×106×e =
r
(9 10 )(92)(2 e)
9
5×106 =
r
(9 10 )(92)(2 1.6 10−19 )
9
r=
5 106
r = 5.3×10 m = 5.3×10-12cm
-14
50. (d)
n1 = 2 and n2 = 3
wavelength emitted is -
1 1 1
= R 2 − 2
n1 n2
1 1 1
= R 2 − 2
2 3
9 − 4 5R
= R =
36 36
C
v =
5 RC
So, v =
36
51. (c)
Given hc=1240 ev nm, R=1.09737×107 per meter
Wavelength of light for least – energetic photon emitted in Lyman series of hydrogen atom
1 1 1
= R 2 − 2
n2 n1
here n1 = 2 and n2 = 1, R = 0.01097 nm-1
1 1 1
= 0.01097 2 − 2
1 2
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1 4 − 1
= 0.01097
4
1 3
= .01097
4
4
=
0.01097 3
= 121.5436
= 122nm
52. (a)
By conservation of linear momentum.
m1v1 = m2v2
3
v1 m 2 R 2
= =
v 2 m1 R1
3
v1 2
= = 8 :1
v2 1
53. (b)
T1/2 = 20min
N1 = 50, N2 = 100-87.5=12.5
T N 20 50
− n 1 = n
n2 N2 n2 12.5
20
n n = 40min
n2
54. (b)
2
z X ⎯⎯
A
→ z − 2 X A − 4 ⎯⎯ → z X A−4
55. (c)
Eg = 1.9 eV, hc = 1240 eV
hc
Wavelength = = nm
Eg
1240
= = 652.63 650nm
1.9
= 6.5 × 10–7m
56. (a)
In amplitude modulation modulated signal has a carrier wave with two side band
frequencies one is lower side band and other is upper side bad.
57. (a)
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Diode D1 is forward biased but D2 is revered Biased. So no current flows through D2 but
current flows through D1.
Req = 3 + 4 = 7
V 12
I= = = 1.71A
R eq 7
58. (a)
Input characteristics circle is drawn between IB and VEB at constant VCE.
V
R = BE
I B V =constant
CE
59. (c)
Boolean expression of NAND gate = A.B
So NAND gate is right.
60. (b)
Space waves are used for the line of sight communication.
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KCET-2016 (Physics)
1. A body falls freely for 10 sec. Its average velocity during this journey (take g = 10 ms -2)
a. 100 ms-1 b. 10 ms-1
c. 50 ms -1 d. 5 ms-1
2. Three projectiles A, B and C are projected at an angle of 30°, 45°, 60° respectively. If RA, RB,
and RC are ranges of A, B and C respectively, then (velocity of projection is same for A, B &
C).
a. RA = RB = RC b. RA = RC> RB
c. RA <RB< RC d. RA = RC< RB
3. The component of a vector r along x – axis have a maximum value if
a. r is along + ve x - axis b. r is along + ve y - axis
c. r is along – ve y - axis d. r makes an angle of 45° with the x –
axis
4. Maximum acceleration of the train in which a 50 kg box lying on its floor will remain
stationary (Given : Co-efficient of static friction between the box and the trains floor is 0.3
and g = 10 ms-2)
a. 5.0 ms-2 b. 3.0 ms-2
c. 1.5 ms -2 d. 15.0 ms-2
5. A 12 kg bomb at rest explodes into two pieces of 4 kg and 8 kg piece is 20 Na, the kinetic
energy of the 8 kg piece is -
a. 25 J b. 20 J
c. 50 J d. 40 J
6. Which of the points is likely position of the centre of mass of the system shown in the
figure?
a. A b. D
c. B d. C
7. Three bodies a ring (R), a solid cylinder (C) and a solid sphere (S) having same mass and
same radius roll down the inclined plane without slipping. They start from rest if vR, vC,
and vS are velocities of respective bodies on reaching the bottom of the plane, then -
a. vR = vC = vS b. vR>vC>vS
c. vR , vC<vS d. vR = vC>vS
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8. Variation of acceleration due to gravity (g) with distance x from the centre of the earth is
best represented by (R → Radius of the earth)
a. b.
c. d.
9. A spring is stretched by applying load to its free end. The strain produced in the spring is –
a. Volumetric b. Shear
c. Longitudinal & Shear d. Longitudinal
10. An ideal fluid flow through a pipe of circular cross section with diameters 5 cm and 10 cm
as shown. The ratio of velocities of fluid at A and B is –
a. 4:1 b. 1:4
c. 2:1 d. 1:2
11. A pan filled with hot food from 94°C to 86°C in 2 minutes. When the room temperature is
20°C. How long will it cool from 74°C to 66°C?
a. 2 minutes b. 2.8 minutes
c. 2.5 minutes d. 1.8 minutes
12. Four rods with different radii r and length are used to connect to heat reservoirs at
different temperature. Which one will conduct most heat?
a. r = 1 cm, = 2 m 1
b. r = 1 cm, = m
2
c. r = 2 cm, = 2 m 1
d. r = 2 cm, = m
2
13. A Carnot engine working between 300 K and 400 K has 800 J of useful work. The amount of
heat energy supplied to the engine from the source is –
a. 2400 J b. 3200 J
c. 1200 J d. 3600 J
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KCET-2016 (Physics)
14. A particle executing SHM has a maximum speed of 0.5 ms-1 and maximum acceleration of
1.0 ms-2. The angular frequency of oscillation is –
a. 2 rad s-1 b. 0.5 rad s-1
c. 2 rad s-1 d. 0.5 rad s-1
15. A source of sound is moving with a velocity of 50 ms-1 towards stationary observer. The
observer measures the frequency of sound as 500 Hz. The apparent frequency of sound as
heard by the observer when source is moving away from him with the same speed is
(Speed of sound at room temperature 350 ms-1) –
a. 400 Hz b. 666 Hz
c. 375 Hz d. 177.5 Hz
16. If there are only one type of charge in the universe, then
( E → Electric field, ds → Area vector)
a. ∮ E.ds 0 on any surface
b. ∮ E.ds could not be defined
c. ∮ E.ds = if charge is inside
q
d. ∮ E.ds = 0 if charge is outside, inside= if charge is
0
17. An electron of mass m, charge e falls through a distance h meter in a uniform electric field
E. Then time of fall –
2hm 2hm
a. t = t=
eE b. eE
2eE 2eE
c. t= t=
hm d. hm
18. If Eax and Eeq represents electric field at a point on the axial and equatorial line of a
dipole. If points are at a distance r from the centre of the dipole, for r>>a –
Eax = Eeq Eax = − Eeq
a. b.
Eax = −2 Eeq Eeq = 2 Eax
c. d.
19. Nature of equipotential surface for a point charge is –
a. Ellipsoid with charge at foci.
b. Sphere with charge at the centre of the sphere
c. Sphere with charge on the surface of the sphere
d. Plane with charge on the surface.
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20. A particle of mass 1 gm and charge 1 C is held at rest on a frictionless horizontal surface
at distance 1 m from the fixed charge 2 mC. If the particle is released, it will be repelled. The
speed of the particle when it is at a distance of 10 m from the fixed charge –
a. 60 ms-1 b. 100 ms-1
c. 90 ms-1 d. 180ms-1
21. A capacitor of 8F is connected as shown. Charge on the plates of the capacitor –
a. 32 C b. 40 C
c. 0C d. 80 C
22. Four metal plates are arranged as shown. Capacitance between X and Y ( A → area of each
plate, d → distance between the plates)
3 0 A 2 0 A
a. b.
2 d d
2 0 A 30 A
c. d.
3 d d
23. Mobility of free electrons in a conductor is –
a. Directly proportional to electron density.
b. Directly proportional to relaxation time.
c. Inversely proportional to electron density.
d. Inversely proportional to relaxation time.
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KCET-2016 (Physics)
24. Variation of resistance of the conductor with temperature is as shown
R
Slope=m
R0
T
T0
The temperature co-efficient ( ) of the conductor is –
R0 b. mR0
a.
m
c. m2R0 R0
d. m
25. Potential difference between A and B in the following circuit –
a. 4V b. 5.6 V
c. 2.8 V d. 6V
26. In the following network potential at ‘O’
a. 4V b. 3V
c. 6V d. 4.8 V
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KCET-2016 (Physics)
27. Effective resistance between A and B in the following circuit
a. 10 b. 20
c. 5 20
d. 3
28. Two heating coils of resistances 10 and 20 are connected in parallel and connected to a
battery of emf 12V and internal resistance 1 . The power consumed by them are in the ratio –
a. 1 : 4 b. 1 : 3
c. 2 : 1 d. 4 : 1
29. A portion is projected with a uniform velocity ‘v’ along the axis of a current carrying
solenoid, then –
a. The proton will be accelerated along the axis
b. The proton path will be circular about the axis
c. The proton moves along helical path.
d. The proton will continue to move with velocity ‘v’ along the axis.
30. In the cyclotron, as radius of the circular path of the charged particle increases ( = angular
velocity, v = linear velocity)
a. both and v increases
a. only increases, v remains constant
b. v increases, remains constant
c. v increases, decreases
31. A conducting wire carrying current is arranged as shown. The magnetic field at ‘O’
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0i 1 1 0i 1 1
a. − b. +
12 R1 R2 12 R1 R2
0i 1 1 0i 1 1
c. − +
6 R1 R2 6 R1 R2
d.
32. The quantity of a charge that will be transferred by a current flow of 20 A over 1 hour 30
minutes period is –
a. 10.8 103C b. 10.8 104C
c. 5.4 103C d. 1.8 104C
33. A galvanometer coil has a resistance of 50 and the meter shows full scale deflection for a
current of 5 mA. This galvanometer is converted into voltmeter of range 0 – 20 V by
connecting.
a. 3950 in series with galvanometer
b. 4050 in series with galvanometer
c. 3950 in parallel with galvanometer
d. 4050 in parallel with galvanometer
34. x1 and x2 are susceptibility of a paramagnetic material at temperatures T1K and T2K
respectively, then
a. x1 = x2 b. x1T1 = x2T2
c. x1T2 = x2T1 x1 T1 = x2 T2
d.
35. At certain place, the horizontal component of earth’s magnetic field is 3.0 G and the angle
dip at the place is 30°. The magnetic field of earth at that location.
a. 4.5 G b. 5.1 G
c. 3.5 G d. 6.0 G
36. The process of super imposing message signal on high frequency carrier wave is called –
a. Amplification b. Demodulation
c. Transmission d. Modulation
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KCET-2016 (Physics)
37. A long solenoid with 40 turns per cm carries a current of 1 A. The magnetic energy stored
per unit volume is _____________ J/m3
a. 3.2 b. 32
c. 1.6 d. 6.4
38. A wheel with 10 spokes each of length ‘L’ m is rotated with a uniform angular velocity ‘ ’
in a plane normal to the magnetic field ‘B’. The emf induced between axle and the rim of the
wheel.
1 1
a. N BL2 BL2
2 b. 2
c. BL2
d. N BL
2
39. The rms value of current in a 50 Hz AC circuit is 6 A. The average value of AC current over a
cycle is –
a. 6 2 3
b. 2
c. Zero 6
d. 2
40. A capacitor of capacitance 10 F is connected to an AC source and an AC ammeter. If the
source voltage varies as V = 50 2 sin 100t, the reading of the ammeter is -
a. 50 mA b. 70.7 mA
c. 5.0 mA d. 7.07 mA
41. In a series L.C.R circuit, the potential drop across L, C and R respectively are 40 V, 120 V
and 60 V. Then the source voltage is –
a. 220 V b. 160 V
c. 180 V d. 100 V
42. In a series L.C.R. circuit an alternating emf (v) and current (i) are given by the equation
v = v0 sin t , i0 sin t +
3
The average power dissipated in the circuit over a cycle of AC is –
v0i0 v0i0
a. b.
2 4
3 d. Zero
v0i0
c. 2
43. Electromagnetic radiation used to sterilise milk is –
a. X-ray b. Y-ray
c. UV ray d. Radiowaves
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KCET-2016 (Physics)
44. A plane glass plate is placed over a various coloured letters (violet, green, yellow, red). The
letter which appears to raised more.
a. Red b. Yellow
c.
d. Green e. Violet
45. A ray of light passes through four transparent media with refractive index n1, n2, n3 and n4
as shown. The surfaces of all media are parallel.
If the emergent ray DE is parallel to incident ray AB, then
a. n1 = n4 b. n2 = n4
c. n3 = n4 n2 + n3 + n4
d. n1= 3
46. Focal length of a convex lens is 20 cm and its RI is 1.5. It produces an erect, enlarged image
if the distance of the object from the lens is –
a. 40 cm b. 30 cm
c. 15 cm d. 20 cm
47. A ray of light suffers a minimum deviation when incident on an equilateral prism of
refractive index 2 . The angle of incidence is –
a. 30° b. 45°
c. 60° d. 50°
48. In Young’s double slit experiment the source is white light. One slit is covered with red
filter and the other with blue filter. There shall be –
a. Alternate red & blue fringes
b. Alternate dark & pink fringes
c. Alternate dark & yellow fringes
d. No interference
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49. Light of wavelength 600 m is incident normally on a slit of width 0.2 mm. The angular
width of central maxima in the diffraction pattern is (measured from minimum to
maximum).
a. 6 10-3 rad b. 4 10-3 rad
c. 2.4 10 rad
-3 d. 4.5 10-3 rad
50. For what distance is ray optics is good approximation when the aperture is 4 mm and the
wavelength of light is 400 m?
a. 24 m b. 40 m
c. 18 m d. 30 m
51. The variation of photo – current with collector potential for different frequencies of
incident radiation v1, v2 and v3 is as shown in the graph, then-
a. v1 = v2 = v3 b. v1> v2> v3
c. v1< v2< v3 v +v
d. v3= 1 2
2
52. The de Broglie wavelength of an electron accelerated to a potential of 400 V is
approximately –
a. 0.03 nm b. 0.04 nm
c. 0.12 nm d. 0.06 nm
53. Total energy of electron in an excited state of hydrogen atom is – 3.4 eV. The kinetic and
potential energy of electron in this state –
a. K = -3.4 eV U = - 6.8 eV b. K = 3.4 eV U = - 6.8 eV
c. K = -6.8eV U = 3.4eV d. K = 10.2eV U = - 13.6eV
54. When electron jumps from n = 4 level to n = 1 level, the angular momentum of electron
changes by –
h 2h
a. b.
2 2
3h 4h
c. d.
2 2
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55. A radio – active sample of half – life 10 days contains 1000 x nuclei. Number of original
nuclei present after 5 days is –
a. 707 x b. 750 x
c. 500 x d. 250 x
56. An element X decays into element Z by two – step process.
4
X → Y + He
2
Y → Z + 2e then
a. X & Z are isobars. b. X & Y are isotopes.
c. X & Z are isotones. d. X & Z are isotopes.
57. A nucleus of mass 20 u emits a photon of energy 6 MeV. If the emission assume to occur
when nucleus is free and rest, then the nucleus will have kinetic energy nearest to
(Take 1u = 1.6 10-27 kg)
a. 10 KeV b. 1 KeV
c. 0.1 KeV d. 100 KeV
58. Constant DC voltage is required from a variable AC voltage. Which of the following is
correct order of operation?
a. Regulator, filter, rectifier b. Rectifier, regulator, filter
c. Rectifier, filter, regulator d. Filter, regulator, rectifier
59. In a transistor, the collector current varies by 0.49 mA and emitter current varies by 0.50
mA. Current gain measured is –
a. 49 b. 150
c. 99 d. 100
60. Identify the logic operation carried out by the following circuit.
a. AND b. NAND
c. NOR d. OR
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ANSWER KEYS
1. (c) 2. (d) 3. (a) 4. (b) 5. (a) 6. (b) 7. (c) 8. (d) 9. (c) 10. (a)
11. (b) 12. (d) 13. (b) 14. (a) 15. (c) 16. (G) 17. (a) 18. (c) 19. (b) 20. (d)
21. (a) 22. (c) 23. (b) 24. (d) 25. (b) 26. (d) 27. (a) 28. (c) 29. (d) 30. (c)
31. (a) 32. (b) 33. (a) 34. (b) 35. (c) 36. (d) 37. (a) 38. (b) 39. (b) 40. (a)
41. (d) 42. (b) 43. (c) 44. (d) 45. (a) 46. (c) 47. (b) 48. (d) 49. (a) 50. (abcd)
51. (c) 52. (d) 53. (b) 54. (c) 55. (a) 56. (d) 57. (b) 58. (c) 59. (a) 60. (d)
* G – Indicates One GRACE MARK awarded for the question number
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Solution
1. (c)
Given:- g =10 m/s2
t=10sec.
Initial velocity (u) = 0 m/s
1
S=ut+ gt2
2
1
S = 0+ ×10×102
2
500
Vavg. = 50 m/s
10
2. (d)
RA=RC<RB
u2 sin 2
R=
g
u2 sin 60 3 u2 u2 sin 90 u 2
RA = = ,R B = =
g 2 g g g
u2 sin120 u2 u2 3
RC = = cos 30
g g g 2
R A = RC BB
3. (a)
→
r
x
rx =| r |cos
(rx) = |r| cos max
rx = |r| cos
cos is maximum of = 0
=0
r =||–x− axis
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4. (b)
= 0.3
m=50 kg
g= 10 m/s2
f = mg
The box kept on the floor of train remains stationary it the pseudo force acting on the box is
balanced by frictional force
ma = mg
a= g = 0.3×10 3.0 m/s2
5. (a)
As initial momentum of the system is zero
According to the law of conservation of momentum
Initial momentum = final momentum
So, final momentum of the system must also be zero
Hence, Momentum of 8kg piece must be equal, opposite to the momentum of 4kg piece.
8kg. piece, P = 20 NS
P2
K.E. of 8kg piece, K =
2m
( 20)2
25 J
28
6. (b)
The centre of mass lies towards the heavier mass, from the above diagram, the mass is
more at the point D.
7. (c)
Let the body start from rest
Using work-energy equation
W = K.E
= K.Ef – K.Ei
= K.E.f [ K.Ei = 0]
1 1
mgh = mv2 + I2
2 2
But V= r
I = kmr2
1 1
mgh = mv2 + kmv2
2 2
1
mgh = mv2 [1+k]
2
2gh
v=
1+ k
As we know; kR =1, kC= 0.5, kS=0.4
KS<KC<KR
VR< VC < VS
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8. (d)
Acceleration due to gravity at a depth d
d
g = g 0 = 1 −
R
Distance from earth centre x=R-d
d = R-x
x
g = g0
R
Thus acceleration due to gravity increases linearly with the increase in distance from
centre of earth
Acceleration due to gravity at a height h
2
R
g = g0
R+h
Distance from earth centre x = R+h
h = x–R
2
R
g = g0
x
1
So acceleration due to gravity decreases as with the increase in distance from centre of
x2
earth.
R x
9. (c)
The length and shape of the spring changes and the weight of the load behaves as a
deforming force. The change in length corresponds to longitudinal strain and change in
shape corresponds to shearing strain.
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10. (a)
B
A
5cm 10cm
DA = 5cm, DB = 10cm
From continuity equation
A1V1 = A2V2
Where,
V1 = VA
V2 = VB
D2 D2 V
A VA = B B
4 4
VA DB (10)2
2
= =
VB D2A (5)2
On solving
VA 4
=
VB 1
11. (b)
Using newton’s law of cooling with approximation
dT −k
= (Tavg–Tsurrounding) … (1)
dt ms
Case I:- dt=2min, Ts = 20°C
94 + 86
Tavg = = 90°C
2
dT = 94–86 = 8°C
−k dT 1 8 1
=
ms dt (Tavg. − Ts ) 2 (90 − 70)
−k
= 0.05714
ms
Case II:-
74 + 66
Ts = 20°C, Tavg = 70C
2
dT = 74-66 = 8°C
Now value’s putting in equation (1)
8
=0.05714 × (70–20)
dt
dt = 2.8 min
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12. (d)
Heat conducted by Rod
KAtT
Q=
A= r2
So,
The rate of the flow of heat
d AT
dt
On considering above relation
1
The option r=2cm, = m
2
13. (b)
Given:-
TM = 400k
TL = 300k
W = 800J
We know
TL W
Efficiency ( ) 1− =
TH Q
300 800
1− =
400 Q
On solving
Q = 3200J
14. (a)
Maximum velocity, Vmax = A 0.5 m/s
Maximum acceleration, amax = 2A 1 m/s2
a max
Angular frequency =
v max
1
=
0.5
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= 2rad / s
15. (c)
Speed of source, Vs = 50m/s
V= 350m/s [speed of sound]
Using Doppler formula:-
v + v0
f'= f ..(1)
v − vs
Vs = 50m/s V0 =0
Source Observer
Given:- Frequency Heard by observer (f’’) = 500Hz
From equation (1)
350 + 0
500 = f
350 − 50
35
500 = f
30
15, 000
f=
35
f = 428.6
Vs = –50m/s V0 =0
Observer
Source
350 + 0
Frequency heard observer is (f'') = 428.6
350 + 50
35
428.6 = 375Hz
40
16. (G) BONUS
17. (a)
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Distance S=h
Electrostatic force F=eE
F eE
a= =
m m
1
S = ut+ at2
2
1 eE
h = 0 + t2
2 m
2hm
t=
eE
18. (c)
r
Ea
E.F. on axial line of dipole is
2p
Eax = , r>>a
40 r3
E.F. on equatorial line of dipole is
p
Eeq = , r>>a
40 r3
So, Eax = −2Eq
19. (b)
+ –
For a point charge, equipotential surface are concentric spherical shells with centre at the
point charge.
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20. (d)
Given:-
r1 = 1 m, r2 = 10m
q1 = 1
q2 = 2mc
kq1q 2 9 109 10−6 2 10−3
Ui = =
r1 1
On solving
Ui = 18J
Ki = 0
9 109 10−6 2 10−3
Uf = 1.8J
10
1
Kf = 0 + mv 2
2
1
Kf = (0.001) v 2
2
Kf = 0.0005×v2
Now Apply Energy Conservation
Kf + U f = Ki + U i
0.0005v2 + 1.8 = 0+18
0.0005v2 = 16.2
On solving
v = 180m / s
21. (a)
4V 8F 20
4
i
i 4
i
5V 1
5V 1
No current through 20
5v
v 4 = 4 ×1A i= = 1A
( 4 + 1)
= 4V
VC = 4V Q = C×VV
= 8F×4V
Q = 32C
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22. (c)
1 x
2
1 2 4 3
x y
3 y
2 3
4
A 0
Capacitance of each capacitor; c =
d
C C C 2C
x y
C
Equation capacitance between x and y
C 2C 2
Ceq = = C
C + 2C 3
2 A 0
Ceq =
3 d
23. (B)
q
Mobility of free – e–, =
m
= [q & m are constant]
Hence,
Mobility of free e– relation time
24. (d)
R
Slope=m
R0
T
T0
Resistance of conductor, R= R 0 (1+ T) … (1)
T = T − T0
From graph
R=R0 + m (T–T0)
= R0 + m T … (2)
Equation (1) & (2)
R 0 + ( T) = m( T)
So, = m/R0
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25. (b)
i 2 8
B
A
6V 4V
Using kirchhoff’s second law from A to B in direction of Anti-clock wise.
26. (d)
B
4V
A 2 O 4
8V
2
2V
C
Apply K.V.L.
8−v v −4 v −2
= +
2 4 2
8 − v v − 4 + 2v − 4
=
2 4
On solving
16-2v = 3v-8
5v=24
24
v = 4.8v
5
27. (a)
20
10 10 10
10 A B
1
The given circuit represents a balanced wheat stone bridge and each resistance is equal to
10
So, Reg = 10
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28. (c)
20
10
1 12V
Let, potential drop across 10 and 20 resistor is V
V2 V2
p1 = = watts
R 10
V2 V2
p2 = = watts
R 20
V2 V2
p1:p2 = :
R 20
p1:p2 = 2:1
29. (d)
The magnetic field due to a solenoid at the axis is along the axis.
Hence, if a proton projected with a velocity ‘v’ along the axis, its velocity will be parallel to
the magnetic field.
If there is no component of velocity of proton ⊥ to the magnetic field, net force on the
proton will be zero and proton will continue to move along the axis with velocity ‘v’
30. (c)
Bqr
v=
m
vr
Bq
Angular velocity =
m
=constant
Hence,
v increases, remains constant.
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31. (a)
R1 I
O 60° B1 B2
B0
R2
Magnetic field at point ‘O’
0
B= 0 1 ( )
4 r
= 60°= ~
3
1
B
R
I I
B1 = 0 0
4R1 3 12R1
I I
B2 = 0 0
4R 2 3 12R 2
0I 1 1
So, magnetic field at point ‘O’ B =B1 –B2 = −
12 R1 R 2
32. (b)
Given:-
Current, I= 20A
Time, T = 1hr 30min
= 1.5 ×3600
T= 5400 sec.
q=I×t
= 20×5400
On solving
q = 10.8×104 c
33. (a)
RG r
Ig = 5mA = 0.005A
V = 20 volts
RG = 50
V =Ig (RG+r)
20 = 0.005(50+r)
20
=50+r
0.005
4000 = 50+r
R = 4000–50 3950
In series with galvanometer
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KCET-2016 (Physics)
34. (b)
According to curie’s law
1
x
T
x T
1= 2
x 2 T1
x1 T1 = x2 T2
35. (c)
BH
Magnetic field of earth, B =
cos
Given:-
Magnetic field of horizontal component of earth
BH = 3.0G
= 30°
Now, value putting in equation (1)
3.0
B=
cos 30
B = 3.5G
36. (d)
The process of super imposing message signal on carrier wave is modulation.
High frequency signals on the other hand can be sent over large distances with small
dissipation in power.
37. (a)
I=1A
No. of turns, n=40 turns per cm 4000 per m.
Now,
Apply formula
0 n 2 I2
Energy stored per unit volume, (u) =
2
410−7 ( 400)2 (1)2
u=
2
On solving
u=3.2 J/m3
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38. (b)
Angular velocity is constant
V=0, increase to V = L
0 + L
Vavg. =
2
L
Vavg. =
2
We know that
= BVavg L
2
L 1
= B = BL2
2 2
39. (c)
F=50Hz
Irms = 6A
Iavg. = ?
Iavg = 0
As, current is negative for same amount of time for which it is positive.
So, Average value of AC current over a cycle is zero.
40. (a)
Given:-
C= 10 10–5F
V = 50 2 sin 100t
On comparing
V=V0 sin t
V0 = 50 2
= 100
Now,
50 2
Vrms = =50volts
2
V
Irms = rms
xc
1
xc =
c
V
Irms = rms c Vrms
1 / c
Irms = 100×10–5×50
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KCET-2016 (Physics)
= 50×10–3A
= 50mA
41. (d)
Given:-
VL = 40V
VC = 120
VR = 60
Source voltage, (Vs) = (CC − VL )2 + VR2
Vs = (120 − 40)2 + (60)2
= (80)2 + (60)2
= 6400 + 3600 = 10 , 000
Vs = 100V
42. (b)
Power dissipated in A.C. circuit is
V0 i0
P= cos
2 2
= 60
V0 i0
P= cos 60
2 2
V0 i0 1
2 2
V0 i0
4
43. (c)
The electromagnetic radiation used to sterilise the milk in dairy is ultraviolet
44. (d)
Normal shift through plate of thickness t
1
d = t 1 −
Now, Refractive index ( ) is related to wavelength ( ) of light as
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B
=A+
2
45. (a)
n1 n2 n3 n4
E
D
C
B
A
Angle of incidence = Angle of refraction
i=r … (1)
Using snell’s law of refraction
n1sini = n4sini
n1 = n4
46. (c)
A convex lens forms an erect and enlarged image when the object is placed between focus
and the lens. So, object distance must be less then the focal length of the lens.
Thus, object distance must be 15cm.
47. (b)
A
i1
i2
r1 r2
Incident ray
n1 = 1, n2 = 2
A = 60° (equilateral prism)
By geometry
A = r1 + r2 … (1)
For min. deviation
A = r1 = r2 … (2)
From equation (1) & (2)
A = 2r1
60
R1 = =30°
2
Using snell’s law
n1 = sini1 = n2 sin (r1)
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KCET-2016 (Physics)
1
1.sini1 = 2
2
After solving
i1 = 45°
48. (d)
The light from two slits of young’s double-slit experiment is of different colours, and having
different wavelengths and frequencies.
Hence, there shall be no interference fringes.
49. (a)
= 600 nm 600 × 10–9m
2D
Linear width of central maxima is =
a
2
Angular width =
d
d = 0.2mm = 0.2×10–3m
So,
2(600 10−9 )
Angular width =
0.2 10−3
After solving
Angular width = 6×10-3 rad
50. (abcd)
Aperture width a=4mm 4×10-3m
Wavelength
= 400nm
= 400×10-9m
So,
Fresnel distance:-
a2 ( 4 10−3 )2
Df =
400 10−9
After solving
Df = 40m
51. (c)
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Photo-current
Saturation
Current
V03 V02 V01 Collector Potential
Retarding potential
V03>V02>V01 [stopping potential]
Minimum Energy of photon required for emission ‘ ’ stopping potential
E3>E2>E1
E=h
So,
V3>V2>V1
52. (d)
Using de-Broglie wavelength
123
= Å
Given:-
V=400v
12.3 12.3
= A
400 20
= 0.615A°
= 0.0615nm
53. (b)
–3.4ev = K.E+P.E. ….. (i) [Given]
The relation b/w P.E. & K.E
P.E. = –2 K.E. ......... (ii)
Use the relation in equation (i)
–3.4ev =K.E.–2K.E.
K.E. =3.4ev
From equation (ii)
P.E. = –2(3.4ev)
P.E. = –6.8ev
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54. (c)
From Bohr’s postulate, the angular momentum is:-
nh
mvr =
2
For n=4
4h
mvr = …(i)
2
For n=1
1h
mvr = …(ii)
2
Thus, the change in angular momentum
4h h
mvr = –
2 2
3h
mvr =
2
55. (a)
−0.693
Nuclei present after 5 days, n=No exp t … (1)
T
2 / 2
Given:-
No = 1000x
T1/2 = 10days
t = 5days
Substitute’s values in equation (1)
−0.693 5
N=1000x × exp
10
N =1000x × exp. (–0.346)
N = 1000x × 0.707
N = 707x
56. (d)
A & Z = mass number
A −4
Z x →Z − 2 y + 2 He
A 4
A −4
y →ZA − 4 Z → +2e−
Z− 2
x & z are isotopes.
57. (b)
E=6Mev 6×1.6×10-13J
m=20u 20×1.6×10-27 kg
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Momentum of –photon, P=E/C
P=P=E/C [momentum of Nucleus]
P2 E2
K=
2m 2mc2
(6 1.6 10−13 )2
K=
2 ( 20 1.6 10−27 ) (3 108 )2
After solving
K.E. = 1Kev
58. (c)
To convert AC to DC, a full wave rectifier is required the O/P of the full wave rectifier will
be positive cycles of the applied input AC voltage. To convert the positive cycles to ripples a
RC filter will be required.
59. (a)
IC
Current gain = … (1)
Ib
Given:-
Ic = 0.49mA
Ie = 0.50mA [emitter current]
Change in base current:-
Ib = Ie – Ic
= 0.50 –0.49 0.01mA
Substitute value in equation (1)
0.49
= = 49
0.01
60. (d)
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C = AB = A
D=B
Y = CD
=A+B
A B C D Y
0 0 1 1 0
0 1 1 0 1 OR Gate
1 0 0 1 1
1 1 0 0 1
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KCET-2017 (Physics)
_____________________________________________________________________________________________
1. A substance of mass 49.53 g occupies 1.5 cm3 of volume. The density of the substance
(in g cm3) with correct number of significant figures is
a. 3.302 b. 3.300
c. 3.3 d. 3.30
2. A car moving with a velocity of 20 ms–1 is stopped in a distance of 40 m. If the same car is
travelling at double the velocity, the distance travelled by it for same retardation is
a. 640 m b. 320 m
c. 1280 m d. 160 m
3. The angle between velocity and acceleration of a particle describing uniform circular
motion is
a. 45º b. 60º
c. 90º d. 180º
4. If A = 2î + 3 ĵ + 8k̂ is perpendicular to B = 4ˆj – 4iˆ + kˆ , then the value of ‘’ is
1 1
a. b. –
2 2
c. 1 d. –1
5. A body of mass 50 kg, is suspended using a spring balance inside a lift at rest. If the lift
starts falling freely, the reading of the spring balance is
a. = 50 kg b. > 50 kg
c. < 50 kg d. =0
6. A motor pump lifts 6 tones of water from a well of depth 25m to the first floor of height
35 m from the ground floor in 20 minutes. The power of the pump (in kW) is [g = 10 ms–2]
a. 3 b. 6
c. 1.5 d. 12
7. Two balls are thrown simultaneously in air. The acceleration of the centre of mass of the
two balls when in air
a. depends on the masses of the two balls
b. depends on the speeds of the two balls
c. is equal to g (Acceleration due to gravity)
d. depends on the direction of motion of the two balls
8. The value of acceleration due to gravity at a depth of 1600 km is equal to [Radius of earth
= 6400 km]
a. 9.8 ms–2 b. 19.6 ms–2
c. 4.9 ms–2 c. 7.35ms–2
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9. ‘Young’s modulus is defined as the ratio of
a. tensile stress and longitudinal strain
b. hydraulic stress and hydraulic strain
c. shearing stress and shearing strain
d. bulk stress and longitudinal strain
10. ‘Hydraulic lift’ works on the basis of
a. Stoke’s law b. Toricelli’s law
c. Pascal’s law d. Bernoulli’s law
11. The S.I. unit of specific heat capacity is
a. J mol–1 K–1 b. J kg–1 K–1
c. J K–1 d. J kg
12. For which combination of working temperatures, the efficiency of ‘Carnot’s engine’ is the
least?
a. 60K, 40 K b. 40K, 20K
c. 80K, 60K d. 100K, 80K
13. The mean energy of a molecule of an ideal gas is
a. 2 KT 3
b. KT
2
c. KT 1
d. KT
2
14. Two simple pendulums A and B are made to oscillate simultaneously and it is found that A
completes 10 oscillations in 20 sec and B completes 8 oscillations in 10 sec. The ratio of
the lengths of A and B is
8 64
a. b.
5 25
5 25
c. d.
4 64
15. The waves set up in a closed pipe are
a. Transverse and progressive b. Longitudinal and stationary
c. Transverse and stationary d. Longitudinal and progressive
16. Two spheres of electric charges +2 nC and −8 nC are placed at a distance‘d’ apart. If they
are allowed to touch each other, what is the new distance between them to get a repulsive
force of same magnitude as before?
4d 3d
a. b.
3 4
c. d d
d.
2
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17. Three point charges of +2q, + 2q and –4q are placed at the corner A, B and C of an
equilateral triangle ABC of side ‘x’. The magnitude of the electric dipole moment of this
system is
a. 2 qx b. 2 3 qx
c. 3 2 qx d. 3 qx
18. 4 × 1010 electrons are removed from a neutral metal sphere of diameter 20 cm placed in
air. The magnitude of the electric field (in NC–1) at a distance of 20 cm from its centre is
a. 5760 b. 1440
c. 640 d. zero
19. Two point charges A =+3 nC and B = 1nC are placed 5 cm apart in air. The work done to
move charge B towards A by 1 cm is
a. 1.35 × 10–7J b. 2.7 × 10–7 J
c. 2.0 × 10–7 J d. 12.1 × 10–7J
20. A system of 2 capacitor of capacitance 2F and 4F is connected in series across a
potential difference of 6V. The electric charge and energy stored in the system are
a. 10 C and 30 J b. 36 C and 108 J
c. 8 C and 24 J d. 1C and 2 J
21. The minimum value of effective capacitance that can be obtained by combining 3
capacitors of capacitances 1pF, 2pF and 4pF is
4 b. 1pF
a. pF
7
7 d. 2pF
c. pF
4
22. A cylindrical conductor of diameter 0.1 mm carries a current of 90 mA. The current
density (in Am–2) is ( 3)
a. 1.2 × 107 b. 2.4× 107
c. 3 × 106 d. 6 × 106
23. A piece of copper is to be shaped into a conducting wire of maximum resistance. The
suitable length and diameter are _________ and ____________respectively.
a. L and d b. 2 L and d
c. L/2 and 2 d d. L and d/2
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24. Of the following graphs, the one that correctly represents the I –V characteristics of a
‘Ohmic device’ is
I I
a. b.
V V
I I
c. d.
V V
25. The value of I in the figure shown below is
20A 4A
5A
3A 1
a. 8A b. 21A
c. 19A d. 4A
26. The power dissipated in 3 resistance in the following circuit is
a. – 261 kJ b. + 103 kJ
c. +261 kJ d. –103 kJ
27. In metre bridge experiment, with a standard resistance in the right gap and a resistance
coil dipped in water (in a beaker) in the left gap, the balancing length obtained is ‘I’. If the
temperature of water is increased, the new balancing length is
a. > 1 b. < 1
c. = 1 d. = 0
28. A proton, a deuteron and an α - particle are projected perpendicular to the direction of a
uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths
described by them is
a. 1 : 2 : 1 b. 1 : 2 : 2
c. 2 :1:1 d. 2 : 2 :1
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29. A galvanometer of resistance 50 Ω is connected to a battery of 3V along with a resistance
of 2950 Ω in series shows full-scale deflection of 30 divisions. The additional series
resistance required to reduce the deflection to 20 divisions is
a. 1500 b. 4440
c. 7400 d. 2950
30. The magnetic field at the center of a current carrying loop of radius 0.1 m is 5 5 times
that at a point along its axis. The distance of this point from the centre of the loop is
a. 0.2 m b. 0.1m
c. 0.05m d. 0.25m
31. A straight wire of length 50 cm carrying a current of 2.5 A is suspended in mid-air by a
uniform magnetic field of 0.5 T (as shown in figure). The mass of the wire is (g = 10 ms–2 )
I
B
a. 62.5 gm b. 250 gm
c. 125 gm d. Cu2O + FeS
32. Which of the following properties is ‘False’ for a bar magnet?
a. Its poles cannot be separated.
b. It points in North-South direction when suspended
c. It’s like poles repel and unlike poles attract.
d. It doesn’t produce magnetic field
33. A magnetic dipole of magnetic moment 6 × 10–2 Am2and moment of inertia 12 × 10–6 kgm2
performs oscillation in a magnetic field of 2 × 10–2 T. The time taken by the dipole to
complete 20 oscillations is
(~ 3)
a. 36 s b. 6s
c. 12 s d. 18 s
34. The susceptibility of a ferromagnetic substance
a. >>1 b. > 1
c. < 1 d. zero
35. A bar magnet is allowed to fall vertically through a copper coil placed in a horizontal plane.
The magnet falls with a net acceleration
S
N
a. = g b. > g
c. < g d. zero
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36. The working of magnetic braking of trains is based on
a. Alternating current b. Eddy current
c. Steady current d. Pulsating current
37. A jet plane of wing span 20 m is travelling towards west at a speed of 400 ms –1. If the
earth’s total magnetic field is 4× 10–4 T and the dip angle is 30º , at that place, the voltage
difference developed across the ends of the wing is
a. 1.6 V b. 3.2 V
c. 0.8 V d. 6.4 V
38. In the A.C. circuit shown, keeping ‘K’ pressed, if an iron rod is inserted into the coil, the
bulb in the circuit,
a. Glows more brightly
b. Glows less brightly
c. Glows with same brightness (as before the rod is inserted)
d. Gets damaged
39. The output of a step down transformer is measured to be 48 V when connected to a 12 w
bulb. The value of peak current is
a.
1
A b. 2A
2
1 1
c. A d. A
2 2 4
40. A coil of inductive reactance 1 / 3 and resistance 1 is connected to a 200 V, 50 Hz A.C.
supply. The time lag between maximum voltage and current is
1 1
a. s b. s
300 600
1 1
c. s d. s
500 200
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41. If E and B represent electric and magnetic field vectors of an electromagnetic wave, the
direction of propagation of the wave is along
a. E b. B
c. E×B d. B × E
42. According to Cartesian sign convention, in ray optics
a. all distances are taken positive
b. all distances are taken negative
c. all distances in the direction of incident ray are taken positive
d. all distances in the direction of incident ray are taken negative
43. A linear object of height 10 cm is kept in front of a concave mirror of radius of curvature 15
cm, at a distance of 10 cm. The image formed is
a. magnified b. magnified and inverted
c. diminished and erect d. diminished and inverted
44. During scattering of light, the amount of scattering is inversely proportional to _______ of
wavelength of light,
a. cube b. square
c. fourth power d. half
45. In Young’s double-slit experiment if yellow light is replaced by blue light, the interference
fringes become
a. wider b. narrower
c. brighter d. darker
46. According to Huygens’ principle, during refraction of light from air to a denser medium
a. Wavelength and speed decrease
b. Wavelength and speed increase
c. Wavelength increases but speed decreases
d. Wavelength decreases but speed increases
47. In a system of two crossed polarisers, it is found that the intensity of light from the second
polariser is half from that of first polariser. The angle between their pass axes is
a. 45º b. 60º
c. 30º d. 0º
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48. From the following graph of photo current against collector plate potential, for two
different intensities of light I1 and I2 , one can conclude
ph oto cu rrent
I2
I1
–v0 collector plate potential
a. I1 = I2 b. I1> I2
c. I1< I2 d. Comparison is not possible
49. A particle is dropped from a height ‘H’. The de’Broglie wavelength of the particle depends
on height as
a. H b. H0
c. H 1/2 d. H–1/2
50. The scientist who is credited with the discovery of ‘nucleus’ in an atom is
a. J.J. Thomson b. Rutherford
c. Niels Bohr d. Balmer
51. The energy (in eV) required to excite an electron from n = 2 to n = 4 state in hydrogen atom
is
a. + 2.55 b. – 3.4
c. – 0.85 d. + 4.25
52. In a nuclear reactor the function of the Moderator is to decrease
a. Number of neutrons b. Speed neutrons
c. Escape of neutrons d. Temperature of the reactor
53. The particles emitted in the decay of 238
92 U to 234
92 U
a. 1 and 2 b. 1 only
c. 1 and 1 d. 2 and 2
54. The mass defect of 42 He is 0.03 . The binding energy per nucleon of helium (in MeV) is
a. 27.93 b. 6.9825
c. 2.793 d. 69.825
55. The energy gap in case of which of the following is less than 3 eV?
a. copper b. Iron
c. Zener diode d. Germanium
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56. Which of the following semi-conducting devices is used as a voltage regulator?
a. Photo diode b. LASER diode
c. Zener diode d. Solar cell
57. In the three parts of a transistor, ‘Emitter is of
a. Moderate size and heavily doped b. Large size and lightly doped
c. Thin size and heavily doped d. Large size and moderately doped
58. In the figure shown, if the diode forward voltage drop is 0.2 V, the voltage difference
between A and B is
B
0.2mA
5k
5k
A
a. 1.3 V b. 2.2 V
c. 0 d. 0.5 V
59. Which of the following logic gates is considered as ‘universal’?
a. A b.
Y A Y
B
c. A d. A
Y Y
B B
60. A basic communication system consists of
(a) Transmitter
(b) Information source
(c) User of information
(d) Channel
(e) Receive
a. a, b, c, d and e b. b, a, d, e and c
c. b, d, a, c and e d. b, e, a, d and c
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KCET-2017 (Physics)
ANSWER KEYS
1. (G) 2. (d) 3. (c) 4. (b) 5. (d) 6. (a) 7. (c) 8. (d) 9. (a) 10. (c)
11. (b) 12. (d) 13. (b) 14. (b) 15. (b) 16. (b) 17. (b) 18. (b) 19. (a) 20. (c)
21. (a) 22. (a) 23. (d) 24. (c) 25. (b) 26. (a) 27. (a) 28. (a) 29. (b) 30. (a)
31. (a) 32. (d) 33. (c) 34. (a) 35. (c) 36. (b) 37. (a) 38. (b) 39. (c) 40. (b)
41. (c) 42. (c) 43. (b) 44. (c) 45. (b) 46. (a) 47. (a) 48. (c) 49. (d) 50. (b)
51. (a) 52. (b) 53. (a) 54. (b) 55. (d) 56. (c) 57. (a) 58. (b) 59. (d) 60. (b)
* G – Indicates One GRACE MARK awarded for the question number.
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KCET-2017 (Physics)
Solution
1. (G)
Bonus
2. (d)
Let; initial speed of car u = 20 m/s
Final speed of car; V = 0 m/s
Retardation = a
Case –I V2 = u2 + 2as
0 = u2 + 2as
– 400
=a
2 40
[a = –5 m/s2]
Case – II u = 40 m/s
v = 0 m/s
a = –5 m/s2
s =?
V2 = u2 +2as
0 = (40)2 + 2 (–5). S
1600
S=
10
S = 160 m
3. (c)
In circular motion, velocity is always tangential to the circular path. Also in uniform circular
motion, tangential acceleration is zero. Hence, the net acceleration will be centripetal
acceleration which always acts towards the centre of the circular path.
Hence, velocity and acceleration of a particle describing uniform circular motion are at
right angle to each other.
4. (b)
When vectors are perpendicular to each other, then their dot product is zero.
A. B = 0
( )( )
2î + 3 ĵ + 8k̂ . – 4î + 4 ĵ + k̂ = 0
–8 + 12 + 8 = 0
8 = – 4
–1
=
2
(B)
5. (d)
As the lift is falling freely, therefore its acceleration = g (downwards)
Let, the force acting on body due to spring is “F”
(50g) – F = (50) a
50g – F = 50 g
F=0
Thus, no force is acting on the body due to spring. So, the reading of spring balance is zero.
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6. (a)
work done
Power =
time taken
mgh
P=
t
m
P = gh
t
h = 25 +35 = 60m.
m
Flow rate = = 6 tones / 20 min
t
6000 kg
= = 5kg / sec
20 60 sec
P = 5 × 10 × 60
= 3000 w
= 3 kw
7. (c)
As there is no force in horizontal direction, so there will be no acceleration of COM in
horizontal direction.
As for vertical direction only gravitational force acts on both the balls, hence acceleration of
COM will be equal to acceleration due to gravity which is ‘g’.
8. (d)
Acceleration due to gravity at a depth “d”:
d
gd = g 1 –
Re
d → 1600km
Re → 6400 km
g → 9.8 m/s2
1600
gd = 9.8 1 –
6400
gd = 7.35 m/s2
9. (a)
Young’s modulus is defined as the ratio of tensile stress to tensile strain or longitudinal
strain
10. (c)
Hydraulic lift works on the principle of equal transmission throughout a fluid i.e. pascal’s
law.
11. (b)
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Q
Specific heat capacity: S =
mT
Q → Amount of heat absorbed
M → Mass of the substance
T → Change in temperature of the substance
SI unit of S → J Kg–1 K–1
12. (d)
TLower
Efficiency of carnot’s engine; = 1 –
THigher
Now, let’s check options.
40
(A) → A = 1 – = 0.33
60
20
(B) → B = 1 – = 0.5
40
60
(C) → c = 1 – = 0.25
80
80
(D) → d 1 – =0.2
100
13. (b)
Mean energy of a molecule of an ideal gas:
1
E = mVrms2
2
3KT
Rms velocity of molecules: Vrms =
m
1 3KT
E= ×m×
2 m
3
E = KT
2
14. (b)
Time period of simple pendulum: T = 2
g
2
A TA
=
B TB
20
Time period of A: TA = = 2 sec
10
10 5
Time period of B: TB = = sec
8 4
A 64
=
B 25
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15. (b)
The waves set up in a closed pipe are longitudinal and stationary because of superposition
of incident wave and reflected wave.
16. (b)
Case – I:
A B
+2nc d – 8nc
Kq A q B
F=
r2
K.2.8
F=
d2
16K
F=
d2
Case – II: A B
2–8 –6
qA = qB = = = –3nc
2 2
A B
–3nc r – 3nc
K (3)(3)
F =
r2
16k
But F=
d2
16k 9k
= 2
d2 r
9 2
r2 = d
16
3
r= d
4
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17. (b)
C (–4q) C
(– 2q) (– 2q)
60º 60º
x x x x
60º 60º
60º 60º
A B A B
x x
(+ 2q) (+ 2q) (+ 2q) (+ 2q)
As evident from the diagram above, we can consider two dipoles one CA and CB.
M1: Dipole moment of CA = 2q x (from C to A)
M2: Dipole moment of CB = 2qx (from C to B)
C
60º
m1 m2
A B
Net dipole moment = m12 + m 22 + 2m1m 2 cos 60 º
1
= m12 + m 22 + 2m1m 2
2
= m12 + m 22 + m1m 2
= 4q 2 x 2 + 4q 2 x 2 + 4q 2 x 2
= 2 3qx
18. (b)
Total charge removed from sphere = ne
= (4 × 1010) × (1.6 × 10–19)
= 6.4 × 10–9 C
Distance of point from centre of sphere = 20 cm =0.2 m
kq
Electric field: E = 2
r
9 10 (6.4 10 –9 )
9
E=
(0.2) 2
= 1440 N/C
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19. (a)
KQ A Q B
Potential energy of system: u =
d
Initial distance between charges → 5 cm = 0.05m
(9 109 ) (3 10 –9 ) (110 –9 )
Uinitial =
0.05
= 5.4 × 10 J –7
Final distance between charged → 5 – 1 = 4 cm =0.04m
(9 109 ) (3 10 –9 ) (110 –9 )
Ufinal =
0.04
= 6.75 ×10–7 J
Work done: W = uf – ui
W = (6.75 – 5.4) × 107
= 1.35 × 107 J
20. (c)
C1C 2 2 4
Equivalent capacitance: Ceq = =
C1 + C 2 2+4
2F 4F
6v
4F
Ceq =
3
Charge: q = cv
4
Q = × 6 = 8 C
3
1 1 4
Energy stored: E = CV2 = × (62) = 24 J
2 2 3
21. (a)
Minimum capacitance can be obtained by connecting these three capacitors in series
1 1 1 1
= + +
Ceq C1 C 2 C3
1 1 1
= + +
1 2 4
7
=
4
4
Ceq= pF
7
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22. (a)
I
Current density: J =
A
Current: I ⎯→ 90 mA = 90 × 10–3A
Diameter of wire: d = 0.1mm = 10–4 m2
d 2
Cross section area of wire; A =
4
3 (10 )
–4 2
A= = 0.75 × 10–8 m2
4
90 10 –3
J= = 1.2 × 107 A/m2
0.75 10 –8
23. (d)
L
Resistance of copper wire: R =
A
→ Resistivity of material and is constant
L → Length of wire
A → Cross sectional area of conductor
d 2
A=
4
D → diameter of wire
L
R 2
d
Now, let’s check the options
L
(A) RA 2
d
L
(B) RB 2 2
d
1 L
(C) RB
8 d2
L
(D) RD 8 2
d
24. (c)
Ohmic device, obey ohm’s law, i.e.
V = IR
V I (R → constant)
I – V characteristic will be straight line
I
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25. (b)
We will use concepts of KCL i.e. Iin = Iout at a junction
A I1 D
20A 4A
5A I3
B C
3A I2 I
Using KCL at A
20 = 5 + I1
I1 = 15A
Using KCL at B 5 = 3 + I2
I2 = 2A
Using KCL at D I1 + 4 = I3
I3 = 15 +4
I3 = 19A
Using KCL at C I = I2 + I3
I = 2 + 19
I = 21A
26. (a)
The given circuit can be redrawn as
3
2
6
4
1 + –
6 3 6 3
4.5V
2 2 4
4 4
1 6 3 1
4.5V 4.5V
2
V 4.5
I= =
R 2 +1
1
4.5V
4.5
I= =1.5A
3
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I1 3
0.75A
6
0.75A 2
0.75A I2
4
Again 1.5A
1 + –
1.5A
4.5V
6 6
I1 = I × = 0.75 ×
3+ 6 9
I1 = 0.5A
Power in 3 = I12 R = (0.5)2 × 3 = 0.75W
27. (a)
R unknown
=
R s tan dard (1 – )
As temperature increases Þ Resistance increases
R unknown
Increases Increases
R s tan dard (1 – )
(1 – ) should decrease ( > 1)
28. (a)
qBr mV
v= r=
m qB
P = mV = 2mE , E is same (kinetic energy)
2mE
r=
qB
m
r
q
m
Or r =
q2
m 2m 4m
rp : rd : r = 2
: 2
:
e e 4e2
= 1: 2 :1
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29. (b)
Total internal resistance = G + R
= 50 + 2950 = 3000
3
Current; I = = 10–3A = 1 mA.
3000
If the deflection is to be reduced to 20 divisions
1mA 2
I = 20 = mA
30 3
Let x be the effective resistance of the circuit.
2
3V = 3000 × 1mA = x × mA
3
x = 4500
Resistance to be added = (4500 – 50)
= 4450
4440
30. (a)
For a circular current carrying loop of radius R
i
Bcentre = 0
2R
0iR 2
Baxis = [at a distance x from the centre]
2(R 2 + x 2 )3/2
Given Bcentre = 5 5 Baxis
0i 0iR 2
=5 5
2R 2(R 2 + x 2 ) 3 / 2
(R 2 + x 2 )3
= 125
R6
R2 + x2
=5
R2
x2
=4
R2
x = 2R
x = 2 × 10 = 20 cm = 0.2m
31. (a)
For equilibrium
FB = mg
Bil = mg
Bil
m=
g
(0.5) (2.5) (50 10 –2 )
m= = 62.5g
10
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32. (d)
A bar magnet can produce magnetic field. Option (d) is False
33. (c)
Time period of oscillation
I
T = 2
MB
12 10 –6
= 2
(6 10 – 2 )(2 10 – 2 )
12 10 –6
= 2
12 10 – 4
= 2 × 10–1
For 20 oscillations
T ` = 20 × 2 × 10–1
T ` = 12 sec
34. (a)
Susceptibility of Ferromagnetic substance is >>>1, because it is strongly attracted in
an external magnetic field
35. (c)
If the magnet is falling vertically down with its north pole towards the coil, then
upper face of coil will become north pole, So that it can oppose the approaching north
pole of the magnet. Here, the acceleration will be less than “g”
36. (b)
Working of magnetic breaking in train is based on eddy current
37. (a)
Voltage difference = BV sin
= (4 × 10–4) × (20) × (400) sin30º
= 1.6 V
38. (b)
When iron rod is inserted in the coil, then the inductive increases, hence the current
flowing through the bulb decrease, therefore brightness decreases.
39. (c)
P = V.I.
P
I=
V
12
I= = 0.25 A
48
Peak current = 2 I
1
= 2
4
1
= A
2 2
40. (b)
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= 2f
= 2 × 3.14 × 50
= 314
xL L 1/ 3 1
tan = = = =
R R 1 3
= 30º
t =
6
t=
6
3.14
t=
6 314
1
t= sec
600
41. (c)
As we know, by property of electro-magnetic wave, the electric field, magnetic field &
the direction of propagation, all the three are mutually perpendicular.
There by right hand thumb rule using the figure. E B
E
Direction of
propagation
B
42. (c)
In ray optics, all the directions in the direction of incident ray are taken as positive
43. (b)
Using mirror formula
1 1 1
= +
f u v
R 15
f= =
2 2
1 1 1
= +
– 15 / 2 – 10 v
V = –30cm
hi = – 3 × 10
hi = – 30
Image will be magnified & inverted
44. (c)
During scattering of light
1
Intensity: I 4
45. (b)
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D
Fringe width: w =
d
w
But yellow light > Blue light
Interference pattern will become narrower
46. (a)
Refractive index of medium
c
=
v
C → speed of light in vacuum or air
V → speed of light in medium
Above equation implies that speed of light decrease when it goes from rarer medium
(air) to denser medium
v
Also; =
f
= wave length, f → frequency
Frequency remains the same for all medium unless source of light is disturbed
V
Wave length decreases.
47. (a)
From, Malus law
I = I0 cos2
I
There; I = 0
2
I0
= I0 cos2
2
1
Cos =
2
= 45º
48. (c)
The value of photo current depends on intensity of incident light. So, when the
intensity of incident light increases, then the value of photo current is increases.
Intensity of light photo current
Photo
current
I2
I1
–V0 Collector plate potential
I1 < I2
49. (d)
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From de-broglie expression
= h/mv
h
=
m 2gH
H–1/2
50. (b)
Rutherford is credited with the discovery of nucleus in an atom.
51. (a)
The expression of energy is
13.6
En = 2
n
required energy
E = E2 – E4
13.6 13.6
= –
(2) 2 (4) 2
= +2.55
52. (b)
In a nuclear reactor, the function of moderator is to decrease the speed of neutrons.
53. (a)
92 U ⎯
⎯→ 90 U ⎯
⎯→ 91 U ⎯⎯→
238 234 234 234
92 U
1 and 2 particles
54. (b)
Mass defect of 42 He ; M = 0.03u
Binding energy of helium: E = M × 931 MeV
E = 0.03 × 931.5 = 27.93 MeV
Number of nucleons in helium; n = 4
E 27.93
Binding energy per nucleon = = = 6.9825
n 4
55. (d)
In Germanium energy gap is less than 3eV
56. (c)
Zener diode is used as a voltage regulated as in reverse biased condition it exhibits
constant voltage
57. (a)
In case of transistor, emitter is of moderate size and heavily doped so as to supply
large number of majority charge carriers for current flow.
58. (b)
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The voltage drop across AB by using KVL is as follows:
VAB = (0.2 10–3 5 103 ) + 0.2 + (0.2 10–3 5 103 )
= 1 + 0.2 + 1
= 2.2V
59. (d)
NAND and NOR gate are considered as universal gates. Out of the given options,
an option (D) is NAND gate.
60. (b)
The block diagram of communication system is as follows:
Information
Transmitter Channel
source
User of
Receiver
information
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_____________________________________________________________________
1. A particle show distance-time curve as shown in the figure. The maximum
instantaneous velocity of the particle is around the point.
S
Distance
Q
P
Time
a. P b. S
c. R d. Q
2. Which of the following graphs correctly represents the variation of ‘g’ on the Earth?
g g
a. b.
r r
R R
g g
c. d.
r r
R R
3. A cup of tea cools from 65.5°C to 62.5°C in 1 minute in a room at 22.5°C. How long will it
take to cool from 46.5°C to 40.5°C in the same room?
a. 4 minutes b. 2 minutes
c. 1 minute d. 3 minutes
4. The dimensions of the ratio of magnetic flux () and permeability () are
a. [M0L1T0A1] b. [M0L–3T0A1]
c. [M0L1T1A–1] d. [M0L2T0A1]
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5. A mass ‘m’ on the surface of the Earth is shifted to a target equal to the radius of the
Earth. If ‘R’ is the radius and ‘M’ is the mass of the Earth, then work done in this process
is
mgR b. mgR
a.
2
c. 2 mgR mgR
d.
4
6. First overtone frequency of a closed pipe of length ‘l1’ is equal to the 2nd harmonic
l
frequency of an open pipe of length ‘l2’. The ratio 1 =
l2
3 4
a. b.
4 3
3 2
c. d.
2 3
V
7. The resistance R = where V = (100 5) V and I = (10 0.2) A.
I
The percentage error in R is
a. 5.2% b. 4.8%
c. 7% d. 3%
8. A block rests on a rough inclined plane making an angle of 30° with the horizontal. The
coefficient of static friction between the block and the plane is 0.8. If the frictional force
on the block is 10 N, the mass of the block is (g = 10 ms–2)
a. 1 kg b. 2 kg
c. 3 kg d. 4 kg
9. Two particles of masses m1 and m2 have equal kinetic energies. The ratio of their
moments is
a. m1 : m2 b. m2 : m1
c. m1 : m2 d. m12 : m22
10. The pressure at the bottom of a liquid tank is not proportional to the
a. Acceleration due to gravity b. Density of the liquid
c. Height of the liquid d. Area of the liquid surface
11. A Carnot engine takes 300 calories of heat from a source at 500 K and rejects 150
calories of heat to the sink. The temperature of the sink is
a. 125 K b. 250 K
c. 750 K d. 1000 K
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12. Pressure of an ideal gas is increased by keeping temperature constant. The kinetic
energy of molecules
a. Decreases
b. Increases
c. Remains same
d. Increases or decreases depending on the nature of gas
13. A man weighing 60 kg is in a lift moving down with an acceleration of 1.8 ms –2. The
force exerted by the floor on him is
a. 588 N b. 480 N
c. Zero d. 696 N
14. Moment of inertia of a body about two perpendicular axes X and Y in the plane of
lamina are 20 kg m2 respectively. Its moment of inertia about an axis perpendicular to
the plane of the lamina and passing through the point of intersection of X and Y axes is
a. 5 kg m2 b. 45 kg m2
c. 12.5 kg m2 d. 500 kg m2
15. Two wires A and B are stretched by the same load. If the area of cross-section of wire ‘A’
is double that of ‘B’ , then the stress on ‘B’ is
a. Equal to that on A b. Twice that on A
c. Half that on A d. Four times that on A
16. The magnitude of point charge due to which the electric field 30 cm away has the
magnitude 2 NC–1 will be
a. 2 × 10–11C b. 3 × 10–11 C
c. 5 × 10–11C d. 9 × 10–11C
17. A mass of 1 kg carrying a charge of 2 C is accelerated through a potential of 1 V. The
velocity acquired by it is
a. 2 ms–1 b. 2 ms–1
1 1
c. ms–1 d. ms–1
2 2
18. The force of repulsion between two identical positive charges when kept, with a
separation ‘r’ in air is ‘F’. Half the gap between the two charges is filled by a dielectric
slab of dielectric constant = 4. Then, the new force of repulsion between those two
charges becomes
F F
a. b.
3 2
F 4F
c. d.
4 9
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19. For the arrangement of capacitors as shown in the circuit, the effective capacitance
between the point A and B is (capacitance of each capacitor is 4 F)
4F 4F
A B 4F
4F 4F
a. 4 F b. 2 F
c. 1F d. 8F
20. The work done to move a charge on an equipotential surface is
a. Infinity b. Less than 1
c. Greater than 1 d. Zero
21. Two capacitors of 3 F and 6F are connected in series and a potential difference of 900
V is applied across the combination. They are then disconnected and reconnected in
parallel. The potential difference across the combination is
a. Zero b. 100 V
c. 200 V d. 400 V
22. Ohm’s Law is applicable to
a. Diode b. Transistor
c. Electrolyte d. Conductor
23. If the last band on the carbon resistor is absent, then the tolerance is
a. 5% b. 20 %
c. 10 % d. 15 %
24. The effective resistance between P and Q for the following network is
R
3 3
6
4 5
P Q
1 b. 21Ω
a. Ω
12
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c. 12 Ω 1
d. Ω
21
25. Five identical resistors each of resistance R = 1500 are connected to a 300 V battery
as shown in the circuit. The reading of the ideal ammeter A is
+
300 V R R R R R
–
A
1 3
a. A b. A
5 5
2 4
c. A d. A
5 5
26. Two cells of internal resistances r1 and r2 and of same emf are connected in series,
across a resistor of resistance R. If the terminal potential difference across the cells of
internal resistance r1 is zero, then the value of R is
a. R = 2(r1 + r2) b. R = r2 – r1
c. R = r1 – r2 d. R = 2 (r1 – r2)
27. The I – V graphs for two different electrical appliances P and Q are shown in the
diagram. If RP and RQ be the resistances of the devices, then
P
I
O V
a. RP = RQ b. RP> RQ
c. RP< RQ RQ
d. RP =
2
28. The correct Biot-Savart law in vector form is
0 I(d l r) 0 I(d l r)
a. dB = b. dB =
4 r2 4 r3
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0 Id l 0 Id l
c. dB = d. dB = ·
4 r 2 4 r 3
29. An electron is moving in a circle of radius r in a uniform magnetic field B. Suddenly, the
B
field is reduced to . The radius of the circular path now becomes
2
r b. 2r
a.
2
r d. 4r
c.
4
30. A charge q is accelerated through a potential difference V. It is then passed normally
through a uniform magnetic field, where it moves in a circle of radius r. The potential
difference required to move it in a circle of radius 2r is
a. 2V b. 4V
c. 1V d. 3 V
31. A cyclotron’s oscillator frequency is 10 MHz and the operating magnetic field is 0.66 T.
If the radius of its dees is 60 cm, then the kinetic energy of the proton beam produced
by the accelerator is
a. 9 MeV b. 10 MeV
c. 7 MeV d. 11 MeV
32. Needles N1 , N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagnetic
substance respectively. A magnet when brought close to them will
a. Attract all three of them
b. Attract N1 strongly, N2 weakly and repel N3 weakly
c. Attract N1 strongly but repel N2 and N3 weakly
d. Attract N1 and N2 strongly but repel N3
33. The strength of the Earth’s magnetic field is
a. Constant everywhere
b. Zero everywhere
c. Having very high value
d. Varying from place to place on the Earth’s surface
34. A jet plane having a wing-span of 25 m is travelling horizontally towards east with a
speed of 3600 km/hour. If the Earth’s magnetic field at the location is 4 × 10–4 T and the
angle of dip is 30° , then, the potential difference between the ends of the wing is
a. 4V b. 5V
c. 2V d. 2.5 V
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35. Which of the following represents the variation of inductive reactance (XL) with the
frequency of voltage source (v)?
XL XL
a. b.
v v
XL XL
c. d.
v v
36. The magnetic flux linked with a coil varies as = 3t2 + 4t + 9. The magnitude of the emf
induced at t = 2 seconds is
a. 8V b. 16 V
c. 32 V d. 64 V
37. A 100 W bulb is connected to an AC source of 220 V, 50 Hz. Then the current flowing
through the bulb is
5 1
a. A b. A
11 2
c. 2A 3
d. A
4
38. In the series LCR circuit, the power dissipation is through
a. R b. L
c. C d. Both L and C
39. In Karnataka, the normal domestic power supply AC is 220 V, 50 Hz. Here 220 V and 50
Hz refer to
a. Peak value of voltage and frequency
b. Rms value of voltage and frequency
c. Mean value of voltage and frequency
d. Peak value of voltage and angular frequency
40. A step-up transformer operates on a 230 V line and l load current of 2 A. The ratio of
primary and secondary windings is 1:25. Then the current in the primary is
a. 25 A b. 50 A
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c. 15 A d. 12.5 A
41. The number of photons falling per second on a completely darkened plate to produce a
force of 6.62 × 10–5 N is ‘n’. If the wavelength of the light falling is 5 × 10–7 m, then n =
_____× 1022. (h = 6.62 × 10–34 J–s)
a. 1 b. 5
c. 0.2 d. 3.3
42. An object is placed at the principal focus of a convex mirror. The image will be at
a. Centre of curvature b. Principal focus
c. Infinity d. No image will be formed
43. An object is placed at a distance of 20 cm from the pole of a concave mirror of focal
length 10 cm. The distance of the image formed is
a. + 20 cm b. + 10 cm
c. –20 cm d. – 10 cm
44. A candle placed 25 cm from a lens forms an image on screen placed 75 cm on the other
side of the lens. The focal length and type of the lens should be
a. + 18.75 cm and convex lens b. – 18.75 cm and concave lens
c. + 20.25 cm and convex lens d. –20.25 cm and concave lens
45. A plane wavefront of wavelength is incident on a single slit of width a. The angular
width of principal maximum is
2
a. b.
a a
a a
c. d.
2
46. In a Fraunhofer diffraction at a single slit, if yellow light illuminating the slit is replaced
by blue light, then diffraction bands
a. Remain unchanged b. Become wider
c. Disappear d. Become narrower
47. In Young’s double slit experiment, two wavelengths 1 = 780 nm and 2 = 520 nm are
used to obtain interference fringes. If the nth bright band due to 1 coincides with (n +
1)th bright band due to 2, then the value of n is
a. 4 b. 3
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KCET-2018 (Physics)
c. 2 d. 6
48. In Young’s double slit experiment, slits are separated by 2 mm and the screen is placed
at a distance of 1.2 m from the slits. Light consisting of two wavelengths 6500 Å and
5200 Å are used to obtain interference fringes. Then the separation between the fourth
bright fringes of two different patterns produced by the two wavelengths is
a. 0.312 mm b. 0.123 mm
c. 0.213 mm d. 0.412 mm
49. The maximum kinetic energy of emitted photoelectrons depends on
a. Intensity of incident radiation
b. Frequency of incident radiation
c. Speed of incident radiation
d. Number of photons in the incident radiation
50. A proton and an particle are accelerated through the same potential difference V. The
ratio of their de Broglie wavelengths is
a. 2 b. 2 2
c. 3 d. 2 3
51. The total energy of an electron revolving in the second orbit of hydrogen atom is
a. – 13.6 eV b. –1.51 eV
c. –3.4 eV d. Zero
52. The period of revolution of an electron in the ground state of hydrogen atom is T. The
period of revolution of the electron in the first excited state is
a. 2T b. 4T
c. 6T d. 8 T
53. The energy equivalent to a substance of mass 1 g is
a. 18 × 1013 J b. 9 × 1013 J
c. 18 × 106 J d. 9 × 106 J
54. The half-life of tritium is 12.5 years. What mass of tritium of initial mass 64 mg will
remain undecayed after 50 years?
a. 32 mg b. 8 mg
c. 16 mg d. 4 mg
55. In a CE amplifier, the input ac signal to be amplified is applied across
a. Forward biased emitter-base junction
b. Reverse biased collector-base junction
c. Reverse biased emitter-base junction
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KCET-2018 (Physics)
d. Forward biased collector-base junction
56. If A = 1 and B = 0, then in terms of Boolean algebra, A + B =
a. B b. B
c. A d. A
57. The density of an electron-hole pair in a pure germanium is 3 × 1016 m–3 at room
temperature. On doping with aluminium, the hole density increases to 4.5 × 10 22 m–3.
Now the electron density (in m–3) in doped germanium will be
a. 1 × 1010 b. 2 × 1010
c. 0.5 × 1010 d. 4 × 1010
58. The dc common emitter current gain of a n-p-n transistor is 50. The potential difference
applied across the collector and emitter of a transistor used in CE configuration is, VCE =
2 V. If the collector resistance, RC = 4 k, the base current (IB) and the collector current
(IC) are
a. IB = 10 A, IC = 0.5 mA b. IB = 0.5 A , IC = 10 mA
c. IB = 5 A, IC = 1 mA d. IB = 1 A, IC = 0.5 mA
59. The radius of the Earth is 6400 km. If the height of an antenna is 500 m, then its range is
a. 800 km b. 100 km
c. 80 km d. 10 km
60. A space station is at a height equal to the radius of the Earth. If ‘VE’ is the escape velocity
on the surface of the Earth, the same on the space station is ______ times VE.
1 1
a. b.
2 4
1 1
c. d.
2 3
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KCET-2018 (Physics)
ANSWER KEYS
1. (C) 2. (B) 3. (A) 4. (A) 5. (A) 6. (A) 7. (C) 8. (B) 9. (C) 10. (D)
11. (B) 12. (C) 13. (B) 14. (B) 15. (B) 16. (A) 17. (B) 18. (D) 19. (A) 20. (D)
21. (C) 22. (D) 23. (B) 24. (C) 25. (B) 26. (C) 27. (B) 28. (B) 29. (B) 30. (B)
31. (C) 32. (B) 33. (D) 34. (B) 35. (A) 36. (B) 37. (A) 38. (A) 39. (B) 40. (B)
41. (B) 42. (C) 43. (A) 44. (A) 45. (A) 46. (D) 47. (C) 48. (A) 49. (B) 50. (B)
51. (C) 52. (D) 53. (B) 54. (D) 55. (A) 56. (A) 57. (B) 58. (B) 59. (C) 60. (C)
KCET-2018 (Physics) Page | 11
KCET-2018 (Physics)
Solution
1. [C]
S
Distance
Q
P
Time
dx
Vm ax = = maximum slope
dt
From the figure, at point R the slope is maximum, hence at this point velocity is
maximum.
∴ Maximum instantaneous velocity of the particle is around the point ‘R’.
2. [B]
g
r
R
For an isothermal process,
1
B= c
P
Graph will be rectangular hyperbola
3. [A]
T1 = 65.5°C, T2 = 62.5°C
Room temperature (T0) = 22.5°C
Ist Case: -
Using Newton’s law cooling
T1 − T2 T − T
= −k 1 2 − T0
t t
65.5 − 62.5 65.5 − 62.5
= −k − 22.5
1 2
3 = - k [64 – 22.5]
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KCET-2018 (Physics)
3 = -k [41.5] ------- (1)
IInd-Case:-
T1 = 46.5°C, T2 = 40.5°C
46.5 − 40.5 46.5 + 40.5
= −k − 22.5
t 2
6
= −k 43.5 − 22.5
t
6
= -k [21] --------- (2)
t
Eq. (1) dividing by Eq. (2)
3 −k[41.5]
=
6 −k[21]
t
t 41.5
=
2 21
After solving
t = 4 min.
4. [A]
Magnetic flux (φ) = BA
Permeability (μ) = B/H
1 0 IA
Φ = B.A × μ0 A =
d d
IL [M0L1A1T0]
5. [A]
−GMm
(2) P. E2 = 2R
h= R
(1)
P. E1 = −GMm
R
GMm
P.E. of the earth surface (P.E1) = −
R
P.E. of h height on the surface of earth
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KCET-2018 (Physics)
−GNM GMm
P.E,2 = −
R+R 2R
GMm GMm
∴ W = Uf – Vi P.E.2 – P.E.1 − −−
2R R
After solving
MgR
W= −
2
6. [A]
First overtone frequency for closed
Organ pipe f = 3V/4L1
2u
Frequency for open pipe f =
2 L2
L1 3
=
L2 4
7. [C]
V
R=
I
V = (100 ±5) V, I = (10± 0.2) A.
100
R= = 10
10
∆𝑅 ∆𝑉 ∆𝐼
= ±( 𝑉 + )
𝑅 𝐼
∆𝑅 ∆𝑉 ∆𝐼
( 𝑅 × 100)= ±( 𝑉 × 100 )+ ( 𝐼 × 100 )
5 0.2
=± 100 + 100
100 10
R=±7%
8. [B]
N = mg cosθ
Mg sin θ = fs
M × 10 × sin 30 = 10
∴ m = 2 kg
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KCET-2018 (Physics)
N
f5
mg sin mg cos
mg
9. [C]
P2
Kinetic energy, k =
2M
∴ P2 = 2 mK
P= 2mK
P1 2m1 K
=
P2 2m2 K
m1
=
m2
∴ P1:P2 = m1 : m2
10. [D]
The pressure depends on acceleration, height of the liquid and the density of liquid.
Pressure does not depend on the area of the liquid surface.
Force ( F ) = ( h. A ) .g
F = hA g
F
Pressure (P) =
A
P = h g
11. [B]
Given: -
T1 = 500 k
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KCET-2018 (Physics)
Q1 = 300 calories
Q3 = 150 calories
T2=?
Q2 T2
=
Q1 T1
Q2 T1
T2 =
Q1
150 500
T2 =
300
T2 = 250 K
12. [C]
PV = nRT
Hence PV
Temperature is constant there is no change in Kinetic energy of molecules.
13. [B]
Lift –
Now – N + F = mg
[Where F is a Pseudo force]
N = mg – F
1.8 m/s2
= mg – ma
Lift N F= ma (pseudo force)
= m (g-a)
= 60 (9.8 – 1.8)
= 60 × 8 mg
N = 480 N
14. [B]
Using perpendicular axis theorem: - Y
Iz = Ix + Iy … (1)
Given: -
Ix = 20 kg m2 X
Iy = 25 kg m2
Substitute values in Eqn (1)
I2 = 20 + 25
I2 = 45 kg m2
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KCET-2018 (Physics)
15. [B]
F
StressA AA AB
Stress = F/A = =
StressB F AA
AB
Given: -
Area of cross section of wire ‘A’ is double that of ‘B’
∴ AA = 2AB
StressA AB 1
= =
StressB 2 AB 2
∴ Stress B = twice on stress ‘A’
16. [A]
Magnitude of electric field is 2 N/C.
Distance r = 30 cm 30×10-2 m
Apply formula of E.F.
Kq
E= 2
r
9 109 q
2=
( )
2
30 10−2
( )
2
2 30 10−2
q=
9 109
q=
(
2 900 10−4 )
9 10 9
After solving
Q = 2 × 10-11 C
17. [B]
Given: -
m = 1 kg
q = 2C
Potential (V) = 1 V
Electrostatic Potential Energy = K.E.
q × V = ½ MV2
2 × 1 = ½×1 ×V2
V = 2 m/s
18. [D]
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KCET-2018 (Physics)
Case Ist: -
kq1q2
Force (f) = …. (1)
r2
1 q1q2
F=
4 0 r 2
Case 2nd: - New distance after dielectric slab is inserted between 2 changes,
dnew = +
r r
4
2 2
r 3r
+r
2 2
Thus,
1 q1q2
New force [F1] = ….(2)
4 0 3r 2
2
Equation (ii) dividing by equation (i)
1 q1q2
4 0 9r 2
F1 4
=
F 1 q1q2
4 0 r 2
After solving,
4F
F =
9
19. [A]
4F 4F
A B 4F
4F 4F
From balanced wheat stone bridge,
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KCET-2018 (Physics)
4F 4F
A 4F B
4F 4F
4F 4F
A B
4F 4F
2F
A B
2F
A B
4F
20. [D]
An equipotential surface is one in which all the points are at the same electric
potential so, the work done to move a charge on an equipotential surface is zero.
According to the formula
dw = q.dv
21. [C]
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KCET-2018 (Physics)
3f 6f
900V
C1 = 3μF, C2 = 6μF When there capacitor disconnected
The series combination charge on & reconnected in parallel the
Both capacitors is same charge & P.D. is same
Q1 Q2
∴ C1V1 = C2V2 Q …(1) ∴ = =V …(4)
C1 C2
V1 C2 6 Q1 C1 1
= = 2 ..(2) = = … (5)
V2 C1 3 Q2 C2 2
V1 + V2 = 900 V … (3) Q1 + Q2 = Q 1800 μC … (6)
From Eqn (2) & (3) Hence Eqn (5) & (6)
V2= 300 V, V1 = 600 V Q1 = 600 μC & Q2 = 1200 μC
Charge on each capacitor P.D.
600
Q = CV 300 6 V1 = = 200 V
3
1200
600 3 V2 = = 200 V
6
= 1800 μC
22. [D]
Ohm’s law is applicable only to conductors.
23. [B]
If the last bond on the carbon resistor is absent, there is no tolerance band, it is 20%.
24. [C]
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KCET-2018 (Physics)
R
3 3 Series Req =6
6
4 5
P Q
6
6
6
4 5 3
4 5
Req. = 4 + 3 + 5 = 12
25. [B]
+
300 V R R R R R
–
A
Using formula
V = IR
I = V/R
All resistances are in parallel.
V 300 5
∴ I= =
R /5 R
Given:- R 1500
3
Current through (A) = A.
5
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KCET-2018 (Physics)
26. [C] 2–V=0
r1 r2
E1 E2
I
R
According to Ohm’s law Terminal potential difference across the cells of
V = IR Internal resistance r1 = 0
E + E = 2I (r1 + r2 + R) V = E - Ir1
2E
I= 0 = E1 – Ir
R + r1 + r2
E1
I=
r1
2E E
=
R + r1 + r2 r1
2r1 = R + r1 + r2
R = r1 - r2
27. [B]
P
I
O V
Slope of this graph is reciprocal of resistance
Q (slope) > P (slope)
1 1
RP RQ
RQ RP
28. [B]
Biot–Savart law: -
Idl sin
dB = 0
4 r2
Vector form: -
Now we know,
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KCET-2018 (Physics)
r
rˆ = , Using this
r
Idl sin
dB = 0 rˆ
4 r2
dB =
(
0 I dl r )
4 r3
29. [B]
Radius of circular path,
mV
R=
qB
Thus,
r1 B1 = r2 B2
B
r. B = r1.
2
r = 2r
1
30. [B]
Radius of circular path
mV
r=
qB
For acceleration energy
ev = ½ mv2
2eV
v2 =
m
2eV
v=
m
r V
V r2
So,
2
V2 r2
=
V1 r1
After solving
V2 = 4V1
31. [C]
Frequency, = 10 MHz 107Hz
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KCET-2018 (Physics)
Magnetic field: -
2 mv
B=
q
22
2 1.6 10−27 107
7 10.4876
∴ B= −19
1.6 10 1.6
After solving
B = 0.656 T
Now energy: -
B2q2 R2
Emax =
2M
( 0.656 ) ( ) ( )
2 2
1.67 10−27 60 10−2
2
==
2 1.67 10−27
After solving,
= 7 MeV
32. [B]
Very strong Pulled in Pushed up
pull
N S N S N S
[A] [B] [C]
a) These are strongly attracted in an external magnetic field [In ferromagnetic
substance]
b) These are feebly attracted in an external magnetic field (in Paramagnetic
substance).
c) These are repelled in an external magnetic field. (In diamagnetic Substance)
So,
A magnet will attract N1 strongly, N2 weakly and repel N3 weakly.
33. [D]
A magnetic field extends infinitely. The strength of the earth’s magnetic field is not
constant. It varies from one place to other place on the surface of earth.
34. [B]
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KCET-2018 (Physics)
Length (l) = 25 m
Speed (V) = 3600 KM/hour
5
= 3600
18
= 200 × 5
V = 1000 m/s
Magnetic field B = 4 × 10–4 T
θ = 30°
Vertical component of earth (BV)
BV = B. sin θ
Ε= (BV) × l × V B. Sin θ× l × V
= 4 × 10-4 × ½ × 25 × 1000
After solving,
E = 5 Volt
35. [A]
We know that
Inductive reactance XL
XL = L
Frequency =
XL = 2πL v
XL = 2πL ×
This equation can be compared to the equation of straight line
Y = m ×C
36. [B]
d
e=
dt
φ = 3t2 + 4t + 9
So,
e = 6t + 4
t = 2 sec.
e=6×2+4
e = 16 v
37. [A]
Power (P) = 100 w
V = 220
We know that
P = VI
KCET-2018 (Physics) Page | 25
KCET-2018 (Physics)
100
I = P/V
220
5
I= Amp
11
38. [A]
Formula used power dissipated in an LCR circuit
P = Vrms. Irmscos
39. [B]
V0
irms =
2
rms value of voltage & frequency
40. [B]
No. of primary winding = 1
No. of secondary winding = 25
Used formula
NP IS
=
NS IP
So,
NS
IP = IS
NP
Substitute values
25
IP = 2
1
IP = 50 A
41. [B]
Wavelength ( ) = 5 × 10-7
Force (F) = 6.62 × 10 -5 N
h = 6.62 × 10-34 J.S
Used formula
n hc
P= .
t
n hc
Fc = .
t
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KCET-2018 (Physics)
n F
=
t h
6.62 10−5 5 10−7
=
6.62 10−34
= 5 1022 V
42. [C]
Convex mirror
Light
P
When the object is placed at the focus the image is formed at infinity & highly
emerged.
43. [A]
Apply mirror formula u = - 20 cm
F = 10 cm
1 1 1
= +
f u
44. [A]
Using mirror formula
1 1 1
= −
f u
Given: -
Object initial distance (u) = – 25 cm
Image distance ( ) = +75 cm
Substitute values
1 1 1
= −
f 75 −25
1 1 1 1 1+ 3 1 4
= + = =
f 75 25 f 75 f 75
After solving
f = + 18.75 cm and convex lens
45. [A]
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KCET-2018 (Physics)
D
Fringe width =
d
Angular width
D
D
∴ θ=
dD
=
d
d = a (given)
Angular width =
a
46. [D]
When a blue light is used instead of yellow, decreases & hence, diffraction bands
become narrower.
47. [C]
1 = 780 nm
2 = 520 nm
We know that,
The distance from the central maxima to nth bright band
n D n1 D D
yn = = n +1 2
d d d
n1 1 = (n + 1) 2
n + 1 1 780
=
n 2 520
n + 1 78
=
n 52
52 n + 52 = 78 n
26 n = 52
n=2
48. [A]
1 = 6500A
41 D
1 =
d
2 = 5200A
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KCET-2018 (Physics)
42 D
2 =
d
4D
1 − 2 = ( 1 − 2 )
d
4 1.2
= ( 6500 − 5200 )
2 16−3
4 1.2 1300 10−10
=
2 10−3
After solving
= 0.312 mm.
49. [B]
By Einstein’s photoelectric equation
K.E. of photoelectron is
½ mv2 = h
K.E. of emitted photoelectrons depends on frequency of incident radiations.
50. [B]
De-Broglie wavelength
[∴P=m ]
h
=
m
De-Broglie wavelength of a proton: -
Mass = m1
h P = 2mk
1 =
2m1k
[k = qv]
h
∴ 1 = … (i)
2m1k
For an α-particle
Mass =m2
Charge =q0
h
∴ 2 =
2m2 q0
For α – particle
2 He q0 = 2 q
4
M2 =4m1
h
2 = … (ii)
2 4m1 2q V
Eqn (i) & (ii)
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KCET-2018 (Physics)
1 h 2 m1 4 2q
= =
2 2m1q h
After solving,
1
=2 2
2
51. [C]
Energy: -
−13.62
En =
n2
For hydrogen atom
Z = 1; n = 2
−13.6 12
En =
22
−13.6
=
4
=-3.4eV
52. [D]
2 r
Time period (T) =
v
n2
= r
Z
So,
n3
T
z2
First excited state, n = 2
3
T2 n2 2
3
= =
T1 n1 1
T2 = 8T
53. [B]
M = 1g 10-3 kg
Using Einstein’s equation: -
E = mc2
=10-3 × (3×108)2
=10-3 × 9×1016
E = 9×1013
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KCET-2018 (Physics)
54. [D]
Un-decayed
N 1
=
N0 2 x
50
X= =4
12.5
N 0 64
N= = = 4mg
16 16
55. [A]
Ie
VCC RL
output
IB signal
Input signal VBB E
IE
.
The input ac signal to be amplified is applied across forward biased emitter – base
junction.
56. [A]
A = 1, B=0
A+ B = I +0 = 0+0 = 0
A+ B =B
57. [B]
n1 = 3× 1016 m-3
nn= 4.5 × 1022m-3
n12 =nnne
n12
ne =
nn
( 3 10 )
2
16
=
4.5 1022
9 1032
ne =
4.5 1022
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KCET-2018 (Physics)
ne= 2× 1010
58. [B]
VCE
IC =
RC
Given; -
VCE = 2V
RC = 4 K
Substituting values
2
IC = = 0.5 10−3 A
4 10 3
IC = 0.5 mA
I I
β= C I B = C
IB
0.5 10−3
= = 10−5 A
50
IB = 10 μA
59. [C]
Range = 2Rh
= 2 6400 103 500
= 80 × 103
= 80 cm
60. [C]
−GMm 1 2
+ mv = 0
( R + R) 2
1 2 −GMm
mv =
2 2R
GM
V = gR
R
VE = 2 gR
VE 2 gR
=
V gR
1
V = VE
2
KCET-2018 (Physics) Page | 32
KCET-2019 (Physics)
1. Which one of the following nuclei has shorter mean life?
a. B b. Same for all
c. A d. C
2. The conductivity of semiconductor increases with increase in temperature because
a. Relaxation time increases
b. Number density of current carriers increases, relaxation time decreases but effect of
decrease in relaxation time is much less than increase in number density
c. Number density of charge carriers increases
d. Both number density of charge carriers and relaxation time increase
3. For a transistor amplifier, the voltage gain
a. Is high at high and low frequencies and constant in the middle frequency range
b. Constant at high frequencies and low at low frequencies
c. Remains constant for all frequencies
d. Is low at high and low frequencies and constant at mid frequencies
4. In the following circuit, what are P and Q?
a. P = 1, Q = 0 b. P = 1, Q =1
c. P = 0, Q = 0 d. P = 0, Q =1
5. An antenna uses electromagnetic waves of frequency 5 MHz. For proper working, the size
of the antenna should be
a. 300 m b. 3 km
c. 15 m d. 15 km
KCET-2019 (Physics) Page | 1
KCET-2019 (Physics)
6. A magnetic needle has a magnetic moment of 5 x 10-2 Am2 and moment of inertia 8 x 10-6
⃗ . The magnitude of magnetic
kgm2. It has a period of oscillation of 2s in a magnetic field 𝐵
field is approximately.
a. 0.4 × 10-4T b. 0.8 × 10-4T
c. 1.6 × 10 T
-4 d. 3.2 × 10-4T
7. A toroid has 500 turns per metre length. If it carries a current of 2A, the magnetic energy
density inside the toroid is
a. 0.314 J/m3 b. 3.14 J/m3
c. 0.628 J/m 3 d. 6.28 J/m3
8. Consider the situation given in figure. The wire AB is slid on the fixed rails with a constant
velocity. If the wire AB is replaced by a semicircular wire, the magnitude of the induced
current will
V
B
a. Remain same
b. Increase or decrease depending on whether the semicircle bulges towards the
resistance or away from it
c. Increase
d. Decrease
9. The frequency of an alternating current is 50 Hz. What is the minimum time taken by
current to reach its peak value from rms value?
a. 2.5 × 10-3s b. 10 × 10-3s
c. 5 × 10 s
-3 d. 0.02s
10. The readings of ammeter and voltmeter in the following circuit are respectively
a. 1.5 A, 100 V b. 2.2 A, 220 V
c. 1.2 A, 120 V d. 2.7 A, 220 V
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11. Two metal plates are separated by 2 cm. The potentials of the plates are -10 V and + 30 V.
The electric field between the two plates is
a. 1000 V/m b. 3000 V/m
c. 500 V/m d. 2000 V/m
12. The equivalent capacitance between A and B is,
a.
100
pF b. 300 pF
3
c. 50 pF d. 150 pF
13. A capacitor of capacitance C charged by an amount Q is connected in parallel with an
uncharged capacitor of capacitance 2C. The final charges on the capacitors are
Q 3Q Q 4Q
a. , b. ,
4 4 5 5
Q Q Q 2Q
c. , d. ,
2 2 3 3
14. Though the electron drift velocity is small and electron charge is very small, a conductor
can carry an appreciably large current because
a. Drift velocity of electron is very large
b. Relaxation time is small
c. Electron number density is very large
d. Electron number density depends on temperature
15. Masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio
5:3:1. The ratio of their electrical resistance are
a. 5:3:1 b. 125:15:1
c. 1:3:5 d. 1:15:125
16. If P, Q and R are physical quantities having different dimensions, which of the following
combinations can never be a meaningful quantity?
a. PQ - R PR − Q
2
b.
R
P−Q PQ
c. d.
R R
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17. The given graph shows the variation of velocity (v) with position (x) for a particle moving
along a straight line
Which of the following graph shows the variation of acceleration (a) with position (x)?
a. b.
c. d.
18. The trajectory of a projectile projected from origin is given by the equation y = x − 2x . The
2
5
initial velocity of the projectile is?
a. 5 ms-1 b. 5 ms−1
2
2 d. 25 ms−1
c. ms−1
5
19. An object with mass 5 kg is acted upon by a force, 𝐹 = (−3𝑖̂ + 4𝑗̂) N. If it’s initial velocity at
𝑡 = 0 𝑖𝑠 𝑣 = (6𝑖̂ − 12𝑗̂) ms-1, the time at which it will just have a velocity along y-axis is
a. 10 𝑠 b. 15 s
c. 5 𝑠 d. 2 s
20. During inelastic collision between two objects, which of the following quantity always
remains conserved?
a. Total mechanical energy b. Speed of each body
c. Total kinetic energy d. Total linear momentum
21. In Rutherford experiment, for head-on collision of α-particles with a gold nucleus, the
impact parameter is
a. of the order of 10-14 m
b. of the order of 10-6 m
c. zero
d. of the order of 10-10 m
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22. Frequency of revolution of an electron revolving in nth orbit of H-atom is proportional to
a. n b.
1
n3
c.
1 d. n independent of n
n2
23. A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular
momentum of the electron is increased by
a. 2.11 × 10-34 Js b. 4.22 × 10-34 Js
c. 1.05 × 10-34 Js d. 3.16 × 10-34 Js
24. The end product of decay of 90 Th232 is 82 Pb208 . The number of and particles emitted are
respectively.
a. 6, 4 b. 4, 6
c. 3, 3 d. 6, 0
25. Two protons are kept at a separation of 10 nm. Let Fn and Fe be the nuclear force and the
electromagnetic force between them
a. Fe >> Fn b. Fe and Fn differ only slightly
c. Fe = Fn d. Fe << Fn
26. Two particles which are initially at rest move towards each other under the action of their
mutual attraction. If their speeds are v and 2v at any instant, then the speed of center of
mass of the system is,
a. Zero b. v
c. 2v d. 1.5 v
27. A particle is moving uniformly along a straight line as shown in the figure. During the
motion of the particle from A to B, the angular momentum of the particle about 'O'
a. decreases b. first increases then decreases
c. increases d. Remains constant
28. A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra
kinetic energy required by it to just overcome the gravitation pull of the earth is
a. 2K b. 2 2K
c. K d. 3K
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29. A wire is stretched such that it’s volume remains constant. The Poisson's ratio of the
material of the wire is
a. –0.50 b. –0.25
c. 0.50 d. 0.25
30. A cylindrical container containing water has a small hole at height of H = 8 cm from the
bottom and at a depth of 2 cm from the top surface of the liquid. The maximum horizontal
distance travelled by the water before it hits the ground (x) is
a. 4 2 cm b. 6 cm
c. 8 cm d. 4 cm
31. A transparent medium shows relation between i and r as shown. If the speed of light in
vaccum is c, the Brewster angle for the medium is
sin r
30°
sin i
a. 45° b. 90°
c. 30° d. 60°
32. In Young's double slit experiment, using monochromatic light of wavelength , the intensity
of light at a point on the screen where path difference is is K units. The intensity of light
at a point where path difference is is
3
a. K b. 2K
4
c. K d. 4K
33. Due to Doppler's effect, the shift in wavelength observed is 0.1 Å for a star producing
wavelength 6000 Å. Velocity of recession of the star will be
a. 10 km/s b. 20 km/s
c. 25 km/s d. 5 km/s
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34. An electron is moving with an initial velocity V = V0 î and is in a uniform magnetic field
B = B0 ĵ . Then, it’s de Broglie wavelength
a. Increases with time
b. Increases and decreases periodically
c. Remains constant
d. Decreases with time
35. Light of certain frequency and intensity incident on a photosensitive material causes
photoelectric effect. If both the frequency and intensity are doubled. The photoelectric
saturation current becomes.
a. doubled b. unchanged
c. quadrupled d. halved
36. A certain charge 2Q is divided at first into two parts q1 and q2. Later, the charges are placed
at a certain distance. If the force of interaction between two charges is maximum then
Q
= ______ .
q1
a. 2 b. 0.5
c. 4 d. 1
37. A particle of mass m and charge q is placed at rest in uniform electric field E and then
released. The kinetic energy attained by the particle after moving a distance y is
a. qE2y b. q2Ey
c. qEy 2 d. qEy
38. An electric dipole is kept in non-uniform electric field. It generally experiences
a. A force but not a torque b. Neither a force nor a torque
c. A force and torque d. A torque but not a force
39. The figure gives the electric potential V as a function of distance through four regions on x-
axis. Which of the following is true for the magnitude of the electric field E in these regions?
a. EA = EC and EB < ED b. EA < EB < EC < ED
c. EA > EB > EC > ED d. EB = ED and EA < EC
40. A system of two charges separated by a certain distance apart stores electrical potential
energy. If the distance between them is increased, the potential energy of the system,
a. Decreases in any case b. Remains the same
c. Increases in any case d. May increase or decrease
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41. In a cyclotron, a charged particle
a. Speeds up between the dees because of the magnetic field.
b. Slows down with in a dee and speeds up between dees
c. Undergoes acceleration all the time
d. Speeds up in dee
42. The number of turns in a coil of Galvanometer is tripled, then
a. Voltage sensitivity remains constant and current sensitivity increases 3 times
b. Both voltage and current sensitivity decreases by 33%
c. Voltage sensitivity increases 3 times and current sensitivity remains constant
d. Both voltage and current sensitivity remains constant
43. A circular current loop of magnetic moment M is in an arbitrary orientation in an external
uniform magnetic field B . The work done to rotate the loop by 30° about an axis
perpendicular to its plane is
a. 3 MB b. Zero
2
c. MB d. MB
2
44. In a permanent magnet at room temperature
a. The individual molecules have nonzero magnetic moment which are all perfectly
aligned.
b. Domains are all perfectly aligned.
c. Magnetic moment of each molecule is zero.
d. Domains are partially aligned.
45. Coersivity of a magnet where the ferromagnet gets completely demagnetized is 3 × 103 Am-1.
The minimum current required to be passed in a solenoid having 1000 turns per metre, so
that the magnet gets completely demagnetized when placed inside the solenoid is
a. 60 mA b. 6A
c. 30 mA d. 3A
46. An inductor of inductance L and resistor R are joined together in series and connected by a
source of frequency. The power dissipated in the circuit is
V2R V2R
a. b.
2
R + 2L2 R2 + 2L2
R 2 + 2L2 V
c. d.
V R 2 + 2L2
47. An electromagnetic wave is travelling in x-direction with electric field vector given by.
Ey = E0 sin (kx − t ) j. The correct expression for magnetic field vector is
a. Bz = E0C sin (kx − t ) k b. Bz =
E0
sin (kx − t ) k
C
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c. By = E0C sin (kx − t ) j d. By =
E0
sin (kx − t ) j
C
48. The phenomenon involved in the reflection of radio-waves by ionosphere is similar to
a. Total internal reflection of light in air during a mirage
b. Scattering of light by air particles
c. Reflection of light by plane mirror
d. Dispersion of light by water molecules during the formation of a rainbow
49. A point object is moving uniformly towards the pole of a concave mirror of focal length 25
cm along it’s axis as shown below. The speed of the object is 1ms-1. At t = 0, the distance of
the object from the mirror is 50 cm. The average velocity of the image formed by the mirror
between time t = 0 and t = 0.25 s is :
a. 20 cm s-1 b. Infinity
c. 40 cm s-1 d. Zero
50. A certain prism is found to produce a minimum deviation of 38°. It produces a deviation of
44° when the angle of incidence is either 42° or 62°. What is the angle of incidence when it
is undergoing minimum deviation?
a. 40° b. 60°
c. 30° d. 49°
51. In the given circuit, the current through 2 resistor is
a. 0.3A b. 0.1A
c. 0.2𝐴 d. 0.4𝐴
52. Kirchhoff's junction rule is a reflection of
a. Conservation of energy
b. Conservation of charges
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c. Conservation of current density vector
d. Conservation of momentum
53. The variation of terminal potential difference (V) with current flowing through a cell is as
shown
3V
V
(in Volt)
6A
I in A
The emf and internal resistance of the cell are
a. 3V, 0.5 b. 6V, 0.5
c. 3V, 2 d. 6V, 2
54. In a potentiometer experiment, the balancing point with a cell is at a length 240 cm. On
shunting the cell with a resistance of 2 , the balancing length becomes 120 cm. The
internal resistance of the cell is
a. 2 b. 0.5
c. 4 d. 1
55. The magnetic field at the centre 'O' in the given figure is
5 0 I 0I
a. b.
12 R 12R
7 0 I 3 0 I
c. d.
14 R 10 R
56. An aluminium sphere is dipped into water. Which of the following is true?
a. Buoyancy will be more in water at 0 °C than that in water at 4 °C
b. Buoyancy may be more or less in water at 4 °C depending on the radius of the sphere
c. Buoyancy will be less in water at 0 °C than that in water at 4 °C
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d. Buoyancy in water at 0°C will be same as that in water at 4 °C
57. A thermodynamic system undergoes a cyclic process ABC as shown in the diagram. The
work done by the system per cycle is
a. – 1250 J b. 1250 J
c. 750 J d. –750 J
58. One mole of O2 gas is heated at constant pressure starting at 27 °C. How much energy must
be added to the gas as heat to double it’s volume?
a. 450 R b. 1050 R
c. Zero d. 750 R
59. A piston is performing S.H.M. in the vertical direction with a frequency of 0.5 Hz. A block of
10 kg is placed on the piston. The maximum amplitude of the system such that the block
remains in contact with the piston is
a. 0.5 m b. 0.1 m
c. 1 m d. 1.5 m
x
60. The equation of a stationary wave is y = 2 sin cos (48πt). The distance between a node
15
and it’s next antinode is
a. 1.5 units b. 30 units
c. 7.5 units d. 22.5 units
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ANSWER KEYS
1. (c) 2. (b) 3. (d) 4. (d) 5. (c) 6. (c) 7. (c) 8. (a) 9. (c) 10. (b)
11. (d) 12. (a) 13. (d) 14. (c) 15. (b) 16. (c) 17. (d) 18. (a) 19. (a) 20. (d)
21. (c) 22. (b) 23. (c) 24. (a) 25. (a) 26. (a) 27. (d) 28. (c) 29. (c) 30. (c)
31. (d) 32. (a) 33. (d) 34. (c) 35. (a) 36. (d) 37. (d) 38. (c) 39. (a) 40. (d)
41. (c) 42. (a) 43. (b) 44. (d) 45. (d) 46. (a) 47. (b) 48. (a) 49. (b) 50. (d)
51. (d) 52. (b) 53. (a) 54. (a) 55. (a) 56. (c) 57. (d) 58. (b) 59. (c) 60. (c)
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Solution
1. (c)
Activity is given by slope of the graph, higher activity implies shorter mean life.
Slope of A is greatest higher activity shorter mean life
2. (b)
As temperature increases, the number density of charge carriers increases according
to the equation
E
− KT
3/2 2
n = CT e
Thus as temperature increases, the number density of charge carriers increases
resulting in increased conductivity of semiconductors.
3. (d)
The frequency response curve of the transistor amplifier shows that the voltage gain
of a transistor amplifier is constant at mid frequency range only. It is low at high and
low frequencies.
V
Frequency
4. (d)
For both NOR gates
Output: Y = A + B
For gate 1; A = 1 P=0
For gate 2; B = 0 Q =1
5. (c)
C = f
3 108 3
= c/f = = 10 = 60m
2
5 10 6
5
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60
Antenna length = = = 15 m
4 4
6. (c)
I
T = 2
MB
8 10 –6
2 = 2
5 10 – 2 B
B = 1.6 × 10–4 T
7. (c)
Magnetic field inside a toroid
B = 0 NI
Magnetic energy density inside the toroid
B2
=
2 0
02 N 2 I 2
=
2 0
0N2I2
=
2
4 10–7 (500)2 22
=
2
= 0.628 J/m 3
8. (a)
Induced emf.: = . (V B)
= (BV sin)
= BV
Induced current: I =
R
If the straight wire is replaced by a semicircular wire resistance remain the same,
therefore current remains same
9. (c)
1 1
Time period: T = = sec
f 50
T
Time taken by current to reach peak value in Ac =
4
T 1 1
= =
4 50 4 200
= 5 × 10–3 S
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10. (b)
As evident from question
VL = VC
Circuit is purely resistive
V = 220 V
V 220
Also; I = =
R 100
I = 2.2 A
11. (d)
V
Electric field: E =
d
V → Potential difference
d → separation between plates
30 – (–10)
E=
2 10 – 2
E = 2000 V/m
12. (a)
50pF 50pF
C1
100pF 100pF A A
B B
50pF 50pF
100 100
C1 = = 50 pF
100 + 100
C2
A
B
50pF
C2 = 50 + C1
= 50 + 50
= 100pF
Now; C2 & 50pF are in series across AB
50 100 5000
C AB = =
50 + 100 150
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100
CAB = pF
3
13. (d)
Let the common potential attained by the two capacitor be “Vc”
Totalcharge Q+0
Vc = =
Totalcapacitance C + 2C
Q
VC =
3C
Final charge on capacitors
CQ Q
Q1 = CVc = =
3C 3
2Q
Q2 = 2CVc =
3
14. (c)
Current: I = neAVd
The large value of I is due to number density of free electron in a conductor (n),
Which is of the order of n= 1028 per m3
15. (b)
L
Resistance of wire: R =
A
→ resistivity of wire
Volume V Mass m
Area: A = = , Volume = =
Length L Density d
L2
R = d
m
L2
R
m
L21 L22 L23 52 32 12
R1 : R2 : R3 = : : = : :
m1 m 2 m 3 1 3 5
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R1 : R2 : R3 = 125 : 15 : 1
16. (c)
By principle of Homogeneity; Addition & subtraction of physical quantities having
different dimensions is not possible
I2
17. (d) current Collectol plate
Photo potential
The equation for the given graph can be written as.
V
V0
V = – mx + v0 ………… (1)
m → slope of graph which is –ve
V0 → Intercept of line which is +V and a constant
dv dx d
a= =–m + V0
dt dt dt
a = – mV + O
But fromI2eq (1)
a = – m (–mx +VCollectol
Photo current 0) plate
a = + m2x – mV0 potential …. (2)
Slope for given eq. (2): +Ve
Intercept for eq. (2):–Ve
a
18. (a)
2x 2
Given; y = x –
5
Comparing with standard equation of trajectory
gx 2
i.e. y = x tan – 2
2u cos2
tan = 1
= 45º
g 2
Also =
2u cos 5
2 2
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g 2
=
1 5
2u2
2
u = 5 m/s
19. (a)
F = –3î + 4 ĵ
F –3 4ˆ
a= == î + j
m 5 5
–3
ax =
5
At t = 0; V0 = 6î – 12 ĵ
Vox at t = 0 =6
For body to have velocity only along y-axis, velocity along x-axis should be zero
Vx = Vox + axt
3
O = 6 – t
5
t = 10 sec
20. (d)
Linear momentum always remains conserved for any type of collision
21. (c)
For head on collision impact parameter is zero as it retraces its path
22. (b)
2r
Time period of revolution; T =
V
v
frequency; f =
2 r
1
Also; V
n
And r n 2
(By Bohr’s model)
1
f 3
n
23. (c)
By absorbing 10.2 eV electron goes to 2nd orbit, ie n = 2
Angular momentum increased by
2h – h
L2 – L1 =
2
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h 6.6 10 –34
= = = 1.05 × 10–34 JS
2 2 3.14
24. (a)
90 Th
232
⎯⎯
→82 Pb208
Hence ; Mass number changes by 24 this implies 6 particles are emitted as
1 particle emitted decreases mass number by 4, then proton number should
decrease by 12, but change in proton number is by 8 only, this implies 4 particles
have been emitted.
25. (a)
As separation is less than 10nm, therefore Fe >> Fn as nuclear force is short range
force.
26. (a)
Initially both A & B are at rest, hence initial momentum is zero, Also no external force
is acting on the two particles as they are moving under mutual attraction so final
momentum should also be zero, by conservation of linear momentum.
Vcom = 0, as mass cannot be zero
27. (d)
Angular momentum = mvr
Here r is perpendicular distance of particle from reference point i.e. O, so as m, v, r
are constant for position A & B.
Angular momentum remains constant.
28. (c)
We known that escape velocity
Ve = 2 V0
V0 → orbital velocity
1
When satellite is in orbit; K = mv 02
2
1
Kinetic energy required to escape; K = mv e2 = mv 02 = 2 K
2
Extra Kinetic energy = 2K – K = K
29. (c)
Let L be the length and r be the radius of wire.
Volume: V = r2L
Differentiating both sides
V = (2rr)L + r2L
As volume remains unchanged L = 0
O = 2rLr + r2L
r / r –1
=
L / L 2
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Lateralstrain
Poisson’s ratio =
Longitudinal strain
r / r 1
=– = 0.5.
L / L 2
30. (c)
Vx = 2gh (h = 2 × 10–2)
= 2 10 2 10−2
4 2
= =
10 5
−2
H = 8 10 m
g = 10m / s2
2H
t =
g
2 8 10−2
t=
10
2
t= sec
5 10
2 2 2 2 2
Maximum horizontal distance; x = v × t = = = m = 8 cm
5 5 10 5 50 25
31. (d)
By Brewster’s law; n = tanp
sin i
Also; n =
sin r
sin i 1
But =
sin r Slope
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sin i 1
n= = = 3
sin r tan 30º
n = tanp = 3
p = tan–1 3
p = 60º
32. (a)
We know that
x =
2
=
3 2
2
Phase difference; =
3
Intensity; I = I0 cos2
2
I = K cos2
3
2
–1
I=K
2
K
I=
4
33. (d)
V
Doppler shift: =
C
V= .C
0 .1
V= × 3 × 108
6000
V = 5km/s
34. (c)
Magnetic force on moving electron:
F = –e (V0 î B0 ĵ)
F = –eV0 B0 k̂
So, Force is perpendicular to both V & B so the magnitude of V will not change it
means momentum will not change.
Now; by De – Broglie
h h
= =
mv p
As momentum remains constant
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De Broglie wavelength remains constant
35. (a)
We know that photo electric current is directly proportional to the intensity of light,
given that incident frequency is greater than threshold frequency.
When intensity is doubled.
Photo electric saturation current doubles
36. (d)
Force of interaction between two charges q1 & q2 will be maximum when
2Q
q1 = q2 = =Q
2
Q Q
= =1
q1 Q
37. (d)
Velocity gained after moving a distance y:
V2 = u2 + 2ay
Here; u = 0,
F = qE
ma = qE
qE
a=
m
qE
V2 = O + 2 y
m
Kinetic energy
1
E= mv2
2
1 qE
E= × m × 2 y
2 m
E = qEy
38. (c)
An electric dipole in non-uniform electric field generally experiences a force and a
torque.
39. (a)
dV
In region A, V → constant EA = 0 as E =
dx
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4–2
In region B, EB = = 1 v/m EB = 1V / m
2
In region C, V → constant EC = 0
4–2
In region D, ED = = 2v/m ED = 2 V/M
1
EA = EC & ED > EB
40. (d)
Potential energy of system of 2 charges
1 q1q 2
U=
40 r
r → separation between 2 charges q1 & q2
If both charges are of same type, then on increasing the separation, potential energy
decreases.
If the charges are of opposite type, then on increasing separation, potential energy
increases.
41. (c)
In a cyclotron a charged particle experiences electric force between the dees and a
magnetic force inside the dees while circulating, hence it always experiences a force
and hence acceleration.
42. (a)
NBA
Current sensitivity: SI = =
I K
NIBA NBA
Voltage sensitivity: SV = = =
V KV KR
When we triple “N” then R is also tripled
Hence; → becomes 3 times
I
But → Remain unchanged
V
43. (b)
Even when the coil is rotated, about an axis perpendicular to its plane, the potential
energy does not change.
Hence, work done is zero
44. (d)
Even in case of permanent magnets all the domain are not perfectly aligned due to
thermal agitation
45. (d)
Coercivity refers to the intensity of magnetic field (B) where the magnets get
demagnetized
For solenoid; B = 0 n I
n → No of turns per unit length
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I → current through each turn
0 → Absolute permeability
B → magnetic field in solenoid
Also 0 H = 0nI
H → intensity of magnetic field
I = H/n
3 103
I= = 3A
100
0.1
6 3
46. (A) L
R
~
V
Z= R 2 + X 2L
Z= R 2 + (L)2
P = I2R
V
I=
R + 2L22
V
P= 2
(R + 2L2 )
47. (B)
Ey = E0 sin (Kx –t) ˆj
E0
C=
B0
E0
B0 =
C
E0
BZ = sin (Kx – t) k̂
C
48. (A)
The phenomenon involved in the reflection of radio-waves by ionosphere is similar
to total internal reflection of light in air during a mirage
49. (B)
Focal length = 2.5cm
At t = 0, object is at a distance of xi =50cm
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At t = 0.25 object is at a distance of xf = speed × time
=0.1 × 0.25
= 0.25 m
= 25 cm
At t = 0.25 image will beat infinity
x
Vavg = =
t
50. (d)
min = 38º
A+
i=
2
→ deviation
i → angle of incidence
A → Angle of prism
= (r1 + r2) –A
–A = 44 – 104
A = 60º
60 + 38
i= = 49
2
3
51. (d) 4.5V
The circuit can be redrawn as.
1 1
1.2V 3 1
1
2
1 1
Req = 1 +2 = 3
V 1 .2
I= = = 0.41A
R eq 3
52. (b)
Kirchhoff's junction rule is based on conservation of charges.
53. (a)
From the given graph, we can form the following equation
V = E – Ir
When I = 0, E = V = 3V
When V = 0 I = 6A
E 3
r= = = 0.5
I 6
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54. (a)
Internal resistance of a cell is given by:
r = R 1 – 1
2
240
r = 2 – 1
120
r=2×1
r = 2
55. (a)
Here = 300° as can be seen from given diagram
300 5
Number of turns; n = =
360 6
nI 5 I
B = 0 = 0
2R 6 2R
5 0 I
B=
12R
56. (c)
Buoyancy at 0°C; FB = V0 g
Buoyancy at 4°C; FB = V4 g
FB
= 0
FB 4
As we know at 4°C is more than at 0°C
4 > 0
FB < FB
57. (d)
Work done = PV
Also; work done = area of ABC
1
W= (10–5) × (100 – 400)
2
W = – 750 J
58. (b)
If at constant pressure, volume is doubled then temperature also gets doubled.
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T1 = 300K, T2 = 600 K
T = 600 – 300 = 300 K
7
Q = nCpT = 1 × R×300
2
= 1050 R
59. (c)
f = 0.5 Hz
= 2f
1
=2××
2
=
For block to remain in contact with the piston, at amplitude position of SHM
Weight of block = force due to oscillation
mg = ma
mg = m (2A)
g 10
A= 2 = 2
A 1m
60. (c)
x
y = 2 sin cos (48t)
15
2
=
15
= 30
Distance between node & next antinodes
30
= = 7.5
4 4
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1. In a permanent magnet at room temperature
a. Domains are all perfectly aligned.
b. Magnetic moment of each molecule is zero.
c. The individual molecules have non-zero magnetic moment which are all perfectly
aligned.
d. Domains are partially aligned.
2. A rod of length 2 m slides with a speed of 5 𝑚𝑠 −1 on a rectangular conducting frame as shown
in figure. There exists a uniform magnetic field of 0.04 T perpendicular to the plane of the
figure. If the resistance of the rod is 3 Ω. The current through the rod is
a. 1.33 A b. 75 mA
c. 133 mA d. 0.75 A
3. The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic
moment is ‘𝑥’. If the current and the radius both are doubled. The new ratio will become
𝑥
a. 8 b. 2𝑥
𝑥
c. 4𝑥 d. 4
4. In the given circuit the peak voltages across C, L and R are 30 V, 110 V and 60 V respectively.
The rms value of the applied voltage is
a. 141 V b. 100 V
c. 200 V d. 70.7 V
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1
5. The power factor of R-L circuit is . If the inductive reactance is 2 Ω. The value of resistance
√3
is
a.
1
Ω b. 2 Ω
√2
c. √2 Ω d. 0.5 Ω
6. In the given circuit, the resonant frequency is
a. 15910 Hz b. 15.92 Hz
c. 159.2 Hz d. 1592 Hz
7. The current in a coil of inductance 0.2 H changes from 5 A to 2 A in 0.5 sec. The magnitude of
the average induced emf in the coil is
a. 0.3 V b. 0.6 V
c. 1.2 V d. 30 V
𝑚
8. An object approaches a convergent lens from the left of the lens with a uniform speed 5 𝑠
and stops at the focus, the image
a. Moves towards the lens with a non-uniform acceleration.
𝑚
b. Moves away from the lens with uniform speed 5 𝑠 .
c. Moves away from the lens with uniform acceleration.
d. Moves away from the lens with a non-uniform acceleration.
𝐴
9. The refracting angle of a prism is A and refractive index of material of prism is cot The
2.
angle of minimum deviation is
a. 180∘ − 2𝐴 b. 180∘ − 3𝐴
c. 180∘ + 2𝐴 d. 90∘ − 𝐴
𝑊
10. A light beam of intensity 20 𝑐𝑚2 is incident normally on a perfectly reflecting surface of sides
25 𝑐𝑚 × 15 𝑐𝑚. The momentum imparted to the surface by the light per second is
a. 1.2 × 10−5 𝑘𝑔 𝑚𝑠 −1 b. 2 × 10−5 𝑘𝑔 𝑚𝑠 −1
c. 1 × 10−5 𝑘𝑔 𝑚𝑠 −1 d. 5 × 10−5 𝑘𝑔 𝑚𝑠 −1
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11. Three polaroid sheets 𝑃1 , 𝑃2 𝑎𝑛𝑑 𝑃3 are kept parallel to each other such that the angle
between pass axes of 𝑃1 and 𝑃2 is 45𝑜 and that between 𝑃2 and 𝑃3 is 45𝑜 . If unpolarised beam
of light of intensity 128 𝑊𝑚−2 is incident on 𝑃1 . What is the intensity of light coming out of
𝑃3 ?
a. 64 𝑊𝑚−2 b. 128 𝑊𝑚−2
c. 0 d. 16 𝑊𝑚−2
12. Two poles are by a distance of 3.14 𝑚. The resolving power of human eye is 1 𝑚𝑖𝑛𝑢𝑡𝑒 of an
arc. The maximum distance from which he can identify the two poles distinctly is
a. 376 𝑚 b. 10.8 𝑘𝑚
c. 5.4 𝑘𝑚 d. 188 𝑚
13. The following figure shows a beam of light converging at point P. When a concave lens of
focal length 16 cm is introduced in the path of the beam at a place shown by dotted line such
that OP becomes the axis of the lens, the beam converges at a distance 𝑥 from the lens. The
value of x will be equal to
a. 48 cm b. 12 cm
c. 24 cm d. 36 cm
14. The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is
a. 10 𝐴𝑜 b. 0.3 𝐴𝑜
c. 3.3 𝐴𝑜 d. 6.26 𝐴𝑜
15. The following graph represents the variation of photo current with anode potential for a
metal surface. Here 𝐼1 , 𝐼2 and 𝐼3 represents intensities and 𝛾1 , 𝛾2 , 𝛾3 represent frequency for
curves 1, 2 and 3 respectively, then
a. 𝛾2 = 𝛾3 𝑎𝑛𝑑 𝐼1 = 𝐼3 b. 𝛾1 = 𝛾2 𝑎𝑛𝑑 𝐼1 ≠ 𝐼2
c. 𝛾1 = 𝛾3 𝑎𝑛𝑑 𝐼1 = 𝐼3 d. 𝛾1 = 𝛾2 , 𝑎𝑛𝑑 𝐼1 = 𝐼2
16. In Young’s Double Slit Experiment, the distance between the slits and the screen is 1.2 m and
the distance between the two slits is 2.4 mm. If a thin transparent mica sheet of thickness
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1 𝜇𝑚 𝑎𝑛𝑑 𝑅. 𝐼. 1.5 is introduced between one of the interfering beams, the shift in the
position of central bright fringe is
a. 0.25 mm b. 2 mm
c. 0.5 mm d. 0.125 mm
3ℎ
17. Angular momentum of an electron in hydrogen atom is 2𝜋 (h is the Planck’s constant). The
K.E. of the electron is
a. 6.8 eV b. 4.35 eV
c. 1.51 eV d. 3.4 eV
18. A beam of fast-moving alpha particles were directed towards a thin film of gold. The parts A,
B and C of the transmitted and reflected beams corresponding to the incident parts A, B and
C of the beam are shown in the adjoining diagram. The number of alpha particles in
a. C’ will be minimum and in B’ maximum
b. B’ will be minimum and in C’ maximum
c. A’ will be maximum and in B’ minimum
d. A’ will be minimum and in B’ maximum
19. The period of revolution of an electron revolving in 𝑛𝑡ℎ orbit of H-atom is proportional to
a. Independent of n b. 𝑛2
c.
1
d. 𝑛3
𝑛
20. During a 𝛽 − decay
a. A proton in the nucleus decays emitting an electron.
b. An atomic electron is electron is ejected.
c. An electron which is already present within the nucleus is ejected.
d. A neutron in the nucleus decays emitting an electron.
21. A radio-active element has half-life of 15 years. What is the fraction that will decay in 30
years?
a. 0.85 b. 0.25
c. 0.5 d. 0.75
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22. Two protons are kept at a separation of 10 nm. Let 𝐹𝑛 and 𝐹𝑒 be the nuclear force and the
electromagnetic force between them
a. 𝐹𝑒 𝑎𝑛𝑑 𝐹𝑛 differ only slightly b. 𝐹𝑒 = 𝐹𝑛
c. 𝐹𝑒 ≫ 𝐹𝑛 d. 𝐹𝑒 ≪ 𝐹𝑛
23. In the following circuit what are P and Q:
a. P = 1, Q = 1 b. P = 1, Q = 0
c. P = 0, Q = 1 d. P = 0, Q = 0
24. A positive hole in a semiconductor is
a. an artificially created particle.
b. an anti-particle of electron.
c. a vacancy created when an electron leaves a covalent bond.
d. absence of free electrons.
25. A 220 V A.C. supply is connected between points A and B as shown in figure what will be the
potential difference V across the capacitor?
a. 220√2 V b. 220 V
c. 110 V d. 0
26. Two long straight parallel wires are a distance 2d apart. They carry steady equal currents
flowing out of the plane of the paper. The variation of magnetic field B along the line xx’ is
given by
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a. b.
c. d.
27. At a metro station, a girl walks up a stationary escalator in 20 sec. If she remains stationary
on the escalator, then the escalator take her up in 30 sec. The time taken by her to walk up
on the moving escalator will be
a. 10 sec b. 25 sec
c. 60 sec d. 12 sec
28. Rain is falling vertically with a speed of 12 𝑚𝑠 −1 . A woman rides a bicycles with a speed of
12 𝑚𝑠 −1 in east to west direction. What is the direction in which she should hold her
umbrella?
a. 45𝑜 towards West b. 30𝑜 towards East
c. 45 towards East
𝑜
d. 30𝑜 towards West
29. A cylindrical wire has a mass (0.3 ± 0.003) 𝑔, radius (0.5 ± 0.005) mm and length (6 ±
0.06) cm. The maximum percentage error in the measurement of its density is
a. 4 b. 1
c. 2 d. 3
30. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration.
The power delivered to it at time ‘t’ is proportional to
a. 𝑡 2 1
b. 𝑡 2
c. 𝑡 d. 𝑡 3/2
31. A thin uniform rectangular plate of mass 2 kg is placed in 𝑥 − 𝑦 plane as shown in figure. The
moment of inertia about x-axis is 𝐼𝑥 = 0.2 𝑘𝑔𝑚2 and the moment of inertia about y-axis is
𝐼𝑦 = 0.3 𝑘𝑔 𝑚2 . The radius of gyration of the plate about the axis passing through O and
perpendicular to the plane of the plate is
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a. 31.6 cm b. 50 cm
c. 5 cm d. 38.7 cm
32. One end of a string of length ′𝐼′ is connected to a particle of mass ‘m’ and the other to a small
peg on a smooth horizontal table. If the particle moves in a circle with speed ′𝑣′, the net force
on the particle (directed towards the centre) is: (T is the tension in the string)
a. 0 b. T
𝑚𝑣 2 𝑚𝑣 2
c. 𝑇 − d. 𝑇 +
𝐼 𝐼
33. Young’s modulus of a perfect rigid body is
a. Between zero and unity b. Zero
c. Unity d. Infinity
34. A wheel starting from rest gains an angular velocity of 10 rad/s after uniformly accelerated
for 5 sec. The total angle through which it has turned is
a. 50 π rad about a vertical axis b. 25 rad
c. 100 rad d. 25 π rad
35. Iceberg floats in water with part of it submerged. What is the fraction of the volume of
iceberg submerged if the density of ice is 𝜌𝑖 = 0.917 𝑔 𝑐𝑚−3 ?
a. 0 b. 0.917
c. 1 d. 0.458
36. The value of acceleration due to gravity at a height of 10 km from the surface of earth is x. At
what depth inside the earth is the value of the acceleration due to gravity has the same value
x?
a. 15 km b. 5 km
c. 20 km d. 10 km
37. In an adiabatic expansion of an ideal gas the product of pressure and volume.
a. At first increases and then decreases
b. Decreases
c. Increases
d. Remains constant
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38. A certain amount of heat energy is supplied to a monoatomic ideal gas which expands at
constant pressure. What fraction of the heat energy is converted into work?
a.
5 b. 1
7
2 2
c. d.
3 5
39. A sphere, a cube and a thin circular plate all of same material and same mass initially heated
to same high temperature are allowed to cool down under similar conditions. Then the
a. Cube will cool the fastest and plate the slowest.
b. Plate will cool the fastest and cube the slowest.
c. Sphere will cool the fastest and cube the slowest.
d. Plate will cool the fastest and sphere the slowest.
40. A train whistling at constant frequency ‘n’ is moving towards a station at a constant speed V.
The train goes past a stationary observer on the station. The frequency ‘n’ of the sound as
heard by the observer is plotted as a function of time ‘t’. Identify the correct curve
a. b.
c. d.
41. A tray of mass 12 kg is supported by two identical springs as shown in figure. When the tray
is pressed down slightly and then released, it executes SHM with a time period of 1.5 s. The
spring constant of each spring is
a. ∞ b. 50 𝑁𝑚−1
c. 0 d. 105 𝑁𝑚−1
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42. The electric field lines on the left have twice the separation on those on the right as shown
in figure. If the magnitude of the field at A is 40 𝑉𝑚−1 , what is the force on 20 𝜇𝐶 charge kept
at B?
a. 1 × 10−4 𝑉𝑚−1 b. 4 × 10−4 𝑉𝑚−1
c. 8 × 10−4 𝑉𝑚−1 d. 16 × 10−4 𝑉𝑚−1
1
43. An infinitely long thin straight wire has uniform charge density of 4 × 10−2 𝑐𝑚−1 . What is
the magnitude of electric field at a distance 20 cm from the axis of the wire?
a. 9 × 108 𝑁𝐶 −1 b. 1.12 × 108 𝑁𝐶 −1
c. 4.5 × 108 𝑁𝐶 −1 d. 2.25 × 108 𝑁𝐶 −1
44. A point charge ‘q’ is placed at the corner of a cube side ‘a’ as shown in the figure. What is the
electric flux through the face ABCD?
𝑞
a. b. 0
72 𝜖0
𝑞 𝑞
c. d.
24 𝜖0 6𝜖0
45. The difference between equivalent capacitances of two identical capacitors connected in
parallel to that in series is 6 𝜇𝐹. The value of capacitance of each capacitor is
a. 6 𝜇𝐹 b. 2 𝜇𝐹
c. 3 𝜇𝐹 d. 4 𝜇𝐹
46. Figure shows three points A, B and C in a region of uniform electric field 𝐸⃗ . The line AB is
perpendicular and BC is parallel to the field lines. Then which of the following holds good?
(𝑉𝐴 , 𝑉𝐵 𝑎𝑛𝑑 𝑉𝐶 ) represent the electric potential at points A, B and C respectively)
a. 𝑉𝐴 > 𝑉𝐵 = 𝑉𝐶 b. 𝑉𝐴 = 𝑉𝐵 = 𝑉𝐶
c. 𝑉𝐴 = 𝑉𝐵 > 𝑉𝐶 d. 𝑉𝐴 = 𝑉𝐵 < 𝑉𝐶
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47. A dipole of dipole moment ‘p’ and moment of inertia I is placed in a uniform electric field 𝐸⃗ .
If it is displaced slightly from its stable equilibrium position, the period of oscillation of
dipole is
𝐼 𝑃𝐸
a. 𝜋√𝑃𝐸 b. √ 𝐼
𝐼 1 𝑃𝐸
c. 2𝜋√𝑃𝐸 d. √
2𝜋 𝐼
48. A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 V.
The speed of the electron when it strikes the anode is
a. 2.5 × 108 𝑚𝑠 −1 b. 1.5 × 105 𝑚𝑠 −1
c. 2.5 × 10 𝑚𝑠
6 −1
d. 2.1 × 107 𝑚𝑠 −1
1
49. A metal rod of length 10 cm and a rectangular cross – section of 1 cm × 2 cm is connected to
a battery across opposite faces. The resistance will be
a. Same irrespective of the three faces.
1
b. Maximum when the battery is connected across 1 cm × 2 cm faces.
1
c. Maximum when the battery is connected across 10 cm × 2 cm faces.
d. Maximum when the battery is connected across 10 cm × 1 cm faces.
50. A car has a fresh storage battery of e.m.f. 12 𝑉 and internal resistance 2 × 10−2 Ω. If the
starter motor draws a current of 80 A. Then the terminal voltage when the starter is on is
a. 9.3 V b. 12 V
c. 8.4 V d. 10.4 V
51. When a soap bubble is charged?
a. Its radius may increase or decrease. b. Its radius increases.
c. Its radius decreases. d. The radius remains the same.
52. The colour code for a carbon resistor of resistance 0.25 𝑘Ω ± 10% is
a. Red, Green, Silver b. Red, Grey, Brown, Silver
c. Red, Green, Brown, Silver d. Red, Grey, Silver, Silver
53. Each resistance in the given cubical network has resistance of 1Ω and equivalent resistance
between 𝐴 and 𝐵 is
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12 5
a. Ω b. Ω
5 6
6 5
c. Ω d. Ω
5 12
54. A potentiometer has a uniform wire of length 5 𝑚. A battery of emf 10 𝑉 and negligible
internal resistance is connected between its ends. A secondary cell connected to the circuit
gives balancing length at 200 𝑐𝑚. The emf of the secondary cell is
a. 8 𝑉 b. 4 𝑉
c. 6 𝑉 d. 2 𝑉
55. In the given figure, the magnetic field at ‘𝑂’
3 𝜇0 𝐼 𝜇 𝐼 3 𝜇0 𝐼 𝜇 𝐼
a. 0
− 4𝜋𝑟 b. 0
+ 4𝜋𝑟
8 𝑟 4 𝑟
3 𝜇0 𝐼 𝜇_𝐼 3 𝜇0 𝐼 𝜇 𝐼
c. − 4𝜋𝑟 d. 0
+ 4𝜋𝑟
10 𝑟 8 𝑟
⃗⃗⃗ placed at a point with vector
56. The magnetic field at the origin due to a current element 𝑖𝑑𝑙
position 𝑟 is
⃗⃗⃗⃗
𝜇0 𝑖 𝑟 ×𝑑𝑙 ⃗⃗⃗⃗ ×𝑟
𝜇0 𝑖 𝑑𝑙
a. b.
4𝜋 𝑟2 4𝜋 𝑟 3
⃗⃗⃗⃗
𝜇0 𝑖 𝑟 ×𝑑𝑙 ⃗⃗⃗⃗ ×𝑟
𝜇0 𝑖 𝑑𝑙
c. d.
4𝜋 𝑟3 4𝜋 𝑟2
57. 𝐼 − 𝑉 characteristic of a copper wire of length 𝐿 and area of cross-section 𝐴 is shown in figure.
The slope of the curve becomes
a. Less if the length of the wire is increased.
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b. More if experiment is performed at higher temperature.
c. More if a wire of steel of same dimension is used.
d. Less if the area of the wire is increased.
58. A cyclotron is used to accelerate protons( 11 𝐻), Deuterons ( 12 𝐻) and 𝛼 − 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 ( 42 𝐻𝑒).
While exiting under similar conditions, the minimum K.E. is gained by
a. Same for all b. 𝛼 − 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒
c. Proton d. Deuteron
59. A paramagnetic sample shows a net magnetization of 8 𝐴𝑚−1 when placed in an external
magnetic field of 0.6 𝑇 at a temperature of 4 𝐾. When the same sample is placed in an external
magnetic field of 0.2 𝑇 at a temperature of 16 𝐾. The magnetization will be
a. 2.4 𝐴𝑚−1 32
b. 3 𝐴𝑚−1
c.
2
𝐴𝑚−1 d. 6 𝐴𝑚−1
3
60. A long cylindrical wire of radius 𝑅 carries a uniform current 𝐼 flowing through it. The
variation of magnetic field with distance ‘𝑟’ from the axis of the wire is shown by
a. b.
c. d.
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ANSWER KEYS
1. (a) 2. (c) 3. (a) 4. (d) 5. (c) 6. (d) 7. (c) 8. (d) 9. (a) 10. (d)
11. (d) 12. (b) 13. (a) 14. (c) 15. (b) 16. (a) 17. (c) 18. (c) 19. (d) 20. (d)
21. (d) 22. (c) 23. (c) 24. (c) 25. (a) 26. (c) 27. (d) 28. (a) 29. (a) 30. (c)
31. (b) 32. (b) 33. (b) 34. (b) 35. (b) 36. (c) 37. (b) 38. (d) 39. (d) 40. (a)
41. (d) 42. (b) 43. (d) 44. (b) 45. (d) 46. (c) 47. (c) 48. (d) 49. (b) 50. (d)
51. (b) 52. (b) 53. (b) 54. (b) 55. (d) 56. (b) 57. (a) 58. (d) 59. (c) 60. (a)
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Solution
1. (a)
Due to thermal agitation, with increase in temperature randomness of the molecules
increases and magnetic field moments get partially aligned. Hence the Domains are partially
aligned.
2. (c)
𝐵𝑙𝑣 0.04×2×5
We know that the current 𝑖 = = = 133 𝑚𝐴
𝑅 3
3. (a)
𝜇 2𝜋𝑖 𝜇0 𝑖
Magnetic field at the center of the current carrying loop is given by 𝐵 = 4𝜋0 × = .
𝑎 2𝑎
Magnetic moment at the center of the current carrying loop is given by 𝑀 = 𝑖𝜋𝑎 2
𝐵 𝜇 𝑥
thus 𝑀 = 2𝜋𝑎0 3 = 𝑥 (given) when both current and radius are doubled ratio become 8 times.
4. (d)
𝑉0 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2 = √602 + (1102 − 302 ) = 100,
𝑉0 100 100 √2 √2
𝑉𝑟𝑚𝑠 = = = × = 100 × = 50 × 1.414 = 70.7 𝑉
√2 √2 √2 √2 2
5. (c)
1 √2 𝑋𝐿 2
cos 𝜙 = , tan 𝜙 = tan 𝜙 = √2 = 𝑅 ⇒ 𝑅 = √2 Ω
√3 1 𝑅
6. (d)
Resonant frequency of an 𝐿𝐶 circuit is given by
1
𝑓 = 2𝜋√𝐿𝐶
1
=
2𝜋√0.5×10−3 𝐻×20×10−6 𝐹
= 1592 𝐻𝑧
7. (c)
𝑑𝐼
Average induced emf 𝜖 = 𝐿 𝑑𝑡
5−2 2
= 0.2 ( 0.5 ) = 5 × 3 = 1.2 𝑉
8. (d)
𝑓 2
Velocity of image in convex lens 𝑉𝑖 = (𝑓+𝑢) 𝑉𝑜 and image will move from focus to infinity (i.e.
away from the lens).
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From the formula, velocity of image
𝑑𝑉𝑖 𝑑𝑉𝑖 𝑑𝑢 𝑓 𝑑𝑢
= × 𝑑𝑡 = 2 (𝑓+𝑢) 𝑓 ln (𝑓 + 𝑢) × 5 (because 𝑑𝑡 = 5 𝑚/𝑠)
𝑑𝑡 𝑑𝑢
𝑑𝑉𝑖 ln(𝑓+𝑢)
⇒ = 10𝑓 2
𝑑𝑡 𝑓+𝑢
We can see that acceleration of image ∝ ln(𝑓 + 𝑢), i.e varies with distance of object
So the image moves away from the lens with a non-uniform acceleration.
9. (a)
Refractive index of a prism is given by
𝐴+𝑑𝑚
sin( )
2
𝑛= 𝐴
sin( )
2
𝐴+𝑑𝑚
𝐴 sin( )
2
⇒ cot (2 ) = 𝐴
sin( )
2
𝐴 𝐴+𝑑𝑚
cos( ) sin( )
2 2
⇒ 𝐴 = 𝐴
sin( ) sin( )
2 2
𝐴 𝐴+𝑑𝑚
⇒ sin (90 − 2 ) = sin ( )
2
𝐴 𝐴+𝑑𝑚
⇒ 90 − 2 = 2
⇒ 180 − 2𝐴 = 𝑑𝑚
10. (d)
𝐸
Intensity 𝐼 = 𝐴 where 𝐸 is energy of radiation and 𝐴 is incident area
⇒ 𝐸 = 𝐼𝐴
2𝐸 2𝐼𝐴
Momentum of radiation is given by 𝑃 = 𝑐 = 𝑐
Where 𝑐 is speed of light.
𝑊
2×20×104 2 ×375×10−4 𝑚2
∴𝑃= 𝑚
= 5 × 10−5 𝑘𝑔𝑚𝑠 −1
3×108 𝑚/𝑠
11. (d)
We have unpolarised beam’s intensity 𝐼0 = 128𝑤/𝑚2
Using Malu’s law we have 𝐼 = 𝐼0 𝑐𝑜𝑠 2 θ
𝐼
When beam passed from the first polaroid, 𝐼 = 20
Again, as the angle between 𝑝1 𝑎𝑛𝑑 𝑝2 is 45∘ , beam intensity when it will pass 𝑝2 would be
𝐼 𝐼
𝐼1 = 20 𝑐𝑜𝑠 2 45∘ = 40
And, also the angle between 𝑝2 𝑎𝑛𝑑 𝑝3 is 45∘ , beam intensity when it will come out of 𝑝3 will
𝐼 𝐼 128
be 𝐼2 = 40 𝑐𝑜𝑠 2 45∘ = 80 = 8 = 16𝑤/𝑚2
12. (b)
Given the lateral separation of the poles 𝑑 = 3.14 𝑚
π
The resolving power of the eye is θ = 1𝑚𝑖𝑛 = 60×180 𝑟𝑎𝑑
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𝑑
The maximum distance from which poles are distinctly visible is 𝐷 = θ
3.14 × 60 × 180
⇒𝐷= = 10800𝑚 = 10.8 𝑘𝑚
π
13. (a)
So, here when we put the concave lens, let the beam will converge at a distance 𝑥 = 𝑣
1 1 1
Using lens formulae, we have, 𝑓 = 𝑣 − 𝑢
Where 𝑢 = 12 𝑐𝑚 and 𝑓 = −16 𝑐𝑚 is given
1 1 1 −1 1 1
∴ 𝑣 = 𝑓 + 𝑢 = 16 + 12 = 48 𝑐𝑚 ⇒ 𝑣 = 48 𝑐𝑚
Hence, 𝑥 = 48 𝑐𝑚
14. (c)
12.27
De-Broglie wavelength is given by λ = , where 𝐸0 is the ground state energy of the
√𝐸0
hydrogen atom whose value is 13.6 V
12.27
∴λ= = 3.33 Ǻ
√13.6
15. (b)
Here in the graph we can see that, the Stopping potential is same for 1 and 2. So,
frequencies will be same i.e. γ1 = γ2 .And, currents are different So, intensity are different
i.e. 𝐼1 ≠ 𝐼2 .
16. (a)
Path difference due to insertion of mica sheet Δ𝑥 = (𝜇 − 1)𝑡
Let the shift in the fringe pattern be ′𝑦′
𝑑 𝑑
Also, path difference Δ𝑥 = 𝑦 × 𝐷 , so comparing both (𝜇 − 1)𝑡 = 𝑦 × 𝐷
𝐷
𝑦 = (𝜇 − 1)𝑡 × 𝑑 , where 𝜇 = 1.5, 𝐷 = 2.4 and 𝑑 = 1.2 putting the values, we get 𝑦 =
0.25 𝑚𝑚.
17. (c)
Angular momentum,
𝑛ℎ 3ℎ
𝐿 = 2𝜋 = 2𝜋 , which is given in the problem, also 𝑛 = 3
13.6 13.6
∴𝐸= 𝑒𝑉 ⇒ = 1.51 𝑒𝑉.
𝑛2 32
18. (c)
𝐴′ will be maximum and 𝐵 ′ will be minimum, because atom is hollow and whole mass of the
atom is concentrated in a small centre called nucleus.
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19. (d)
2𝜋𝑟
As we know that time period of revolution of an electron is 𝑇 = 𝑣 ,
𝑛2 𝑍 𝑛3
And 𝑟 ∝ 𝑍 2 and 𝑣 ∝ 𝑛 ∴ 𝑇 ∝ 𝑍 2 and for 𝐻 −atom 𝑍 = 1, 𝑇 ∝ 𝑛3
20. (d)
During 𝛽 − decay, a neutron in the nucleus decays emitting an electron.
21. (d)
Fraction un decayed is given as,
𝑁 1 𝑡/𝑇
=( )
𝑁0 2
Here, 𝑡 = 30 𝑦𝑒𝑎𝑟𝑠 and𝑇 = 15 𝑦𝑒𝑎𝑟𝑠
So,
30 𝑦𝑒𝑎𝑟𝑠
𝑁 1 15 𝑦𝑒𝑎𝑟𝑠
=( )
𝑁0 2
𝑁 1 2
or, 𝑁 = (2)
0
𝑁 1
or, 𝑁 = 4
0
Therefore,
𝑁
Fraction un decayed = 1 − 𝑁
0
1
or, Fraction un decayed = 1 − 4
3
or, Fraction un decayed = 4 = 𝟎. 𝟕𝟓
22. (c)
Nuclear forces are short range force existing in the range of a 10−15 𝑚. If the separation
between the particles is greater than 10−15 𝑚, nuclear force is negligible.
Therefore, at a separation of 10 𝑛𝑚, the electromagnetic force is greater than the nuclear
force.
Hence, 𝑭𝒆 ≫ 𝑭𝒏
23. (c)
A NAND gate gives an output 1 if at least one of the inputs is zero.
Hence Q is 1.
Therefore, the inputs for the upper NAND gate are 1, 1.
Hence P=0.
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24. (c)
A positive hole in a semiconductor is a vacancy which is created at the site of a covalent bond
when an electron leaves a covalent bond.
25. (a)
During the positive half cycles of input 𝑎𝑐, a 𝑝 − 𝑛 junction diode is forward biased and hence
it conducts.}
While during the negative half cycles, the diode will be reverse biased and hence does not
conduct.
The capacitor is charged to maximum potential difference.
Therefore, the potential difference across capacitor 𝐶 is equal to the peak value of the applied
ac voltage, i.e.,
𝑉 = 𝑉𝑚𝑎𝑥
or, 𝑉 = √2𝑉𝑟𝑚𝑠
or, 𝑉 = √2 × 220 𝑉 = 200 √2 𝑉
26. (c)
The magnetic field due to a long straight current carrying wire is given by,
𝜇0 𝐼
𝐵=
2𝜋𝑟
1
or, 𝐵 ∝ 𝑟
At the point exactly mid-way between the conductors, the net magnetic field is zero. Using
right hand thumb rule, we find that the magnetic field due to left wire will be in 𝑗̂ direction
while due to the right wire is in (−𝑗̂) direction.
Magnetic field at a distance x from the left wire, lying between the wires.
𝜇0 𝐼 𝜇0 𝐼
𝐵= 𝑗̂ + (−𝑗̂)
2𝜋𝑥 2𝜋(2𝑑 − 𝑥)
𝜇0 𝐼 1 1
or, 𝐵 = (𝑥 − 2𝑑−𝑥)
2𝜋
At 𝑥 = 𝑑, 𝐵 = 0
For 𝑥 < 𝑑, 𝐵 is along 𝑗̂
For 𝑥 > 𝑑, 𝐵 is along (−𝑗̂)
On the left side of the left conductor, magnetic fields due to the currents will add up and the
net magnetic field will be along (−𝑗̂) direction.
To the right side of second conductor, the total magnetic field will be along 𝑗̂ direction.
27. (d)
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Let, the length of the escalator be 𝐿.
𝐿 𝐿
Velocity of the girl, 𝑣𝑔 = 𝑡 = 20
1
𝐿 𝐿
Velocity of the escalator, 𝑣𝑒 = 𝑡 = 30
2
On moving escalator,
𝑣𝑛𝑒𝑡 = 𝑣𝑔 + 𝑣𝑒
𝐿 𝐿
or, 𝑣𝑛𝑒𝑡 = 20 + 30
𝐿
or, 𝑣𝑛𝑒𝑡 = 12
𝐿
Since, 𝑣𝑛𝑒𝑡 = 𝑡
Therefore,
𝐿 𝐿
=
𝑡 12
𝑡 = 12 𝑠
28. (a)
To protect herself from the rain, the woman must hold her umbrella in the direction of the
relative velocity of the rain with respect to the woman.
𝑣𝑤 12 𝑚/𝑠
tan 𝜃 = =
𝑣𝑓 12 𝑚/𝑠
or, 𝑡𝑎𝑛𝜃 = 1
or, 𝜽 = 𝟒𝟓∘
Therefore, the direction in which she should hold her umbrella is 𝟒𝟓∘ toward west.
29. (a)
𝑚 𝑚
𝜌= = 2
𝑉 𝜋𝑟 𝑙
Δ𝜌 Δ𝑚 Δr Δl
or, 𝜌 = 𝑚 + 2 𝑟 + 𝑙
Δ𝜌 0.003 0.005 0.06
or, × 100 = ( +2 + ) × 100
𝜌 0.3 0.5 6
Δ𝜌
or, × 100 = (0.01 + 0.02 + 0.01) × 100
𝜌
Δ𝜌
or, × 100 = 0.04 × 100
𝜌
Δ𝜌
× 100 = 4%
𝜌
Therefore, the percentage error in the measurement of density is 4.
30. (c)
As we know,
𝑃 = 𝐹𝑣
where, 𝑃 is the power, 𝐹 is the force and 𝑣 is the velocity.
or, 𝑃 = 𝑚𝑎𝑣
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or, 𝑃 = 𝑚𝑎2 𝑡
or, 𝑷 ∝ 𝒕
31. (b)
By perpendicular axis theorem we know that,
𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦 = 0.3 + 0.2 = 0.5 𝑘𝑔𝑚2
as we know that MOI 𝐼 = 𝑚𝐾 2 ⇒ 0.5 = 2 × 𝐾 2
1
𝐾 = 2 = 0.5 𝑚 = 50 𝑐𝑚.
32. (b)
When a particle connected to a string revolves in a circular path around a centre, the
centripetal force is provided by the tension produced in the string. Hence, in the given case,
𝑚𝑣 2
the net force on the particle is the tension 𝑇. (𝑇 = )
𝑙
33. (b)
Young’s modulus = stress/strain, and for a perfect rigid body we know that strain is equal to
0 and hence Young’s modulus of a perfect rigid body becomes Infinity.
34. (b)
𝜔1 = 0
𝑟𝑎𝑑
𝜔2 = 10 𝑡 = 5 𝑠𝑒𝑐
𝑠𝑒𝑐
𝜔1 + 𝜔2 10
𝜃= ×𝑡 = × 5 = 25 𝑟𝑎𝑑
2 2
35. (b)
Let 𝑉 be the total volume of the iceberg and 𝑉′ of its volume be submerged into water
Floatation condition weight of iceberg = Weight of water displaced by submerged part by
ice
𝑉′ 𝜌𝑖 0.917
𝑉𝜌𝑖 𝑔 = 𝑉 ′ 𝜌𝑤 𝑔 ⇒ = = = 0.917
𝑉 𝜌𝑤 1
36. (c)
2ℎ
𝑔ℎ = 𝑔 (1 − )
𝑅
𝑑
𝑔𝑑 = 𝑔 (1 − 𝑅) ⇒ 𝑔ℎ = 𝑔𝑑
2ℎ 𝑑
𝑔 (1 − )= 𝑔 (1 − 𝑅) ⇒ 𝑑 = 2ℎ = 2 × 10 = 20 𝑘𝑚
𝑅
37. (b)
In an adiabatic expansion as temperature decreases from an ideal gas equation 𝑃𝑉 = 𝑛𝑅𝑇.
Product of pressure and volume also decreases.
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38. (d)
The heat 𝑄 is converted into the internal energy and work. According to the first law of
thermodynamics, 𝑄 = 𝑈 + 𝑊 ⇒ 𝑛𝐶𝑝 Δ𝑡 = 𝑛𝐶𝑣 Δ𝑡 + 𝑊
𝑊 𝑛𝐶𝑝 Δ𝑡 − 𝑛𝐶𝑣 Δ𝑡
=
𝑄 𝑛𝐶𝑝 Δ𝑡
𝐶𝑝 −𝐶𝑣 1 1 2
= 1−𝛾 =1− 5 =5
𝐶𝑝
3
39. (d)
Surface area is more for plate and less for sphere. Hence plate will cool the fastest and sphere
the slowest.
40. (a)
As the train approaches the observer the apparent frequency is higher than the actual
frequency. As the train moves away from the observer, the apparent frequency is lower than
the actual frequency. Hence the correct option is (a).
41. (d)
We know that
𝑚
𝑇 = 2𝜋√
𝑘𝑒𝑓𝑓
3 12 9 12
= 2𝜋√2𝑘 ⇒ 4 = 4𝜋 2 × 2𝑘 𝑘 ≈ 105 𝑁𝑚−1
2
42. (b)
We know that 𝐹 = 𝑞𝐸 = 20 × 20 × 10−6 = 4 × 10−4 𝑉/𝑚.
43. (d)
We know that
𝜆 1 10−2
𝐸= = × −2 × 18 × 109 × 5
2𝜋ε0 r 4 10
2.25 × 108 𝑁/𝐶
44. (b)
Since the flux passing through face ABCD is parallel to that surface and area vector is
perpendicular. Hence flux passing through ABCD is zero.
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45. (d)
𝐶𝑃 − 𝐶𝑆 = 6 𝜇𝐹
𝐶
2𝐶 − = 6
2
𝐶 = 4 𝜇𝐹
46. (c)
Electric lines of force in an electric field always flow from a higher potential to a lower
potential. Hence, 𝑉𝐴 = 𝑉𝐵 > 𝑉𝐶
47. (c)
𝐼
𝑇 = 2𝜋√
𝑃. 𝐸
48. (d)
1
𝑚𝑣 2 = 𝑉𝑞
2
2𝑉𝑞
𝑣=√ = 2.1 × 107 𝑚𝑠 −1
𝑚
49. (b)
Resistance is inversely proportional to area.
1
Hence resistance will be maximum when the battery is connected across 1 cm × 2 cm faces.
50. (d)
𝑉 = 𝐸 − 𝑖𝑟 = 10.4 𝑉
51. (b)
When the radius of a soap bubble increases, the soap bubble gets charged.
52. (b)
Red, Grey, Brown, Silver
53. (b)
5
Ω
6
54. (b)
Emf of cell in the secondary circuit =potential gradient x balancing length
10
⇒5 x2=4V
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55. (d)
𝐵𝑛𝑒𝑡 = 𝐵1 + 𝐵2 + 𝐵3
3 𝜇0 𝐼 𝜇0 𝐼
= + +0
8 𝑟 4𝜋𝑟
56. (b)
𝜇0 𝑖 ⃗⃗⃗
𝑑𝑙 × 𝑟
4𝜋 𝑟 3
57. (a)
1 𝐴
Slope = 𝑅 = 𝜌𝑙
Hence the slope becomes less if the length of the wire is increased.
58. (d)
𝑞2
K.E. = 𝑚
Hence, a deuteron gains the least kinetic energy.
59. (c)
𝐵
𝐼∝
𝑇
𝐼2 𝐵2 𝑇1
= ×
𝐼1 𝐵1 𝑇2
2
𝐼2 = 𝐴/𝑚
3
60. (a)
The magnetic field increases as the point moves closer to the boundary of the wire and
decreases as it moves away from the boundary of the wire.
Hence the correct option is (a).
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KCET EXAMINATION – 2022
SUBJECT : PHYSICS (VERSION – B3)
DATE :- 17-06-2022 TIME : 10.30 AM TO 11.50 AM
1. The centre of mass of an extended body on the and the density of oil is 900 kg m-3. The
surface of the earth and its centre of gravity magnitude of ‘q’ is
1) Can never be at the same point 1) 1.6 10 19 C 2) 3.2 10 19 C
2) Centre of mass coincides with the centre of 3) 0.8 10 19 C 4) 8 10 19 C
gravity of a body if the size of the body is Ans. 4
negligible as compared to the size (or radius) of Sol. F=mg, Eq=mg
the earth F 6nr v t
3) Are always at the same point for any size of
the body 2r2 .g
vt
4) Are always at the same point only for 9
spherical bodies
Ans. 2 4. “Heat cannot be itself flow from a body at lower
Sol. Conceptual temperature to a body at higher temperature”.
This statement corresponds to
2. A metallic rod breaks when strain produced is 1) Conservation of mass
0.2%. The Young’s modulus of the material of 2) First law of thermodynamics
the rod is 7 109 N / m2 . The area of cross 3) Second law of Thermodynamics
4) Conservation of momentum
section of support a load of 104 N is
Ans. 3
1) 7.1 104 m2 2) 7.1 10 2 m2
Sol. Conceptual
3) 7.1 108 m2 4) 7.1 10 6 m2
Ans. 1 5. A smooth chain of length 2 m is kept on a table
F F such that its length of 60 cm hangs freely from
Sol. Y A
A. Y the edge of the table. The total mass of the
chain is 4 kg. The work done in pulling the
3. A tiny spherical oil drop carrying a net charge entire chain on the table is, (Take g=10 m/s2)
q is balanced in still air, with a vertical uniform 1) 3.6J 2) 2.0 J 3) 12.9 J 4)6.3 J
81 Ans. 1
electric field of strength 105 V / m . When mg
7 Sol. W
the field is switched off, the drops is observed 2n2
to fall with terminal velocity 2 10 3 ms 1 . Here
g=9.8 m/s2, Viscosity of air is 1.8 10 5 Ns / m2
1 - DR ACADEMY
6. The angular speed of a motor wheel is 10. The electric field and the potential of an electric
increased from 1200 rpm to 3120 rpm in 16 dipole vary with distance r as
seconds. The angular acceleration of the moto 1 1 1 1
1) 2 and 3 2) 3 and 2
wheel is r r r r
1) 6 rad /s2 2) 8 rad /s2 1 1 1 1
3) and 2 4) 2 and
3) 2 rad /s2 4) 4 rad /s2 r r r r
Ans. 2
Ans. 4
2p 1
W2 W 1 2n2 2n1 Sol. E K 3 3
Sol. r r
t t
p cos 1
V K 2
r2 r
7. Four charges +a, +2q, +q and -2q are placed at
the corners of a square ABCD respectively. The
11. The displacement of a particle executing SHM
force on a unit positive charge kept at the
centre ‘O’ is is given by X 3 sin 2t where ‘x’ is in
4
1) Along the diagonal AC
2) Perpendicular to AD meters and ‘t’ is in seconds. The amplitude and
3) Zero maximum speed of the particle is
4) Along the diagonal BD 1) 3m,6 ms 1 2) 3m,8 ms 1
Ans. 4 3) 3m,2 ms 1 4) 3m, 4 ms 1
Sol. Conceptual
Ans. 1
Sol. A 3m
8. An electric dipole with dipole moment
Vmax A 3 2 6
4 10 9 C m is aligned at 300 with the direction
of a uniform electric field of magnitude
12. Electric as well as gravitational affects can be
5 10 4 NC 1 , the magnitude of the torque
thought to be caused by fields. Which of the
acting on the dipole is following is true for an electrical or
1) 10 5 Nm 2) 10 10 3 Nm gravitational field?
3) 10 4 Nm 4) 3 104 Nm 1) Fields are useful for understanding forces
Ans. 3 acting through a distance
Sol. P.E.sin 2) There is no way to verify the existence of a
force field since it is just a concept
9. A charged particle of mass ‘m’ and charge ‘q’ is 3) The field concept is often used to describe
released from rest in an uniform electric field contact forces
E . Neglecting the effect of gravity, the kinetic 4) Gravitational or Electric fields does not exist
energy of the charged particle after ‘t’ seconds in the space around an object
is Ans. 1
Sol. Conceptual
Eqm E2q 2 t2 2E2 t 2 E2q 2 m
1) 2) 3) 4)
t 2m mq 2t 2
13. A charged particle is moving in an electric field
Ans. 2
of 3 10 10 Vm1 with mobility
1
Sol. K.E mv 2 6 2
2.5 10 m / v / s , its drift velocity is
2
2 1) 2.5 104 m /s 2) 1.2 104 m / s
1 Eq
m0 t
2 m 3) 7.5 104 m /s 4) 8.33 104 m /s
Ans. 3
Vd
Sol. Vd E
E
2 - DR ACADEMY
14. Wire bound resistors are made by 19. A parallel plate capacitor is charged by
1) Winding the wires of an alloy of Ge, Au, GA connecting a 2V battery across it. It is then
2) Winding the wires of an alloy of Manganin, disconnected form the battery and a glass slab
constantan, Nichrome is introduced between plates. Which of the
3) Winding the wires of an alloy of Cu, Al, Ag following pairs of quantities decrease?
4) Winding the wires of an alloy of Si, Tu, Fe 1) Energy stored and capacitance
Ans. 2 2) Capacitance and charge
Sol. Conceptual 3) Charge and potential difference
4) Potential difference and energy stored.
15. Ten identical cells each of potential ‘E’ and Ans. 4
internal resistance ‘r’, are connected in series Sol. Conceptual
to form a closed circuit. An ideal voltmeter
connected across three cells, will read 20. A proton moves with a velocity of 5 10 6 ˆj ms 1
1) 13E 2) 7E 3) 10E 4) 3E through the uniform electric field,
Ans. 4
E 4 10 2iˆ 0.2ˆj 0.1k
6 ˆ Vm
1
and the
Sol. V=NE
uniform magnetic field B 0.2 ˆi 0.2ˆj kˆ T.
16. In an atom electron revolve around the
0 The approximate net force acting on the proton
nucleus along a path of radius 0.72 A making is
9.4 1018 revolutions per second. The 1) 2.2 10 13 N 2) 20 10 13 N
equivalent current is
3) 5 10 13 N 4) 25 10 13 N
[Given e=1.6x10-19C]
Ans. 14.4 10 13 N
1) 1.4A 2) 1.8A 3) 1.2A 4) 1.5A
Ans. 4
Sol. F q E V B
e = 14.4 10 13 N
Sol. i ef
T
21. A solenoid of length 50cm having 100 turns
17. When a metal conductor connected to left gap
carries a current of 2.5 A. The magnetic field
of a meter bridge is heated, the balancing point
at one end of the solenoid is
1) Remains unchanged
1) 1.57 10 4 T 2) 9.42 10 4 T
2) Shifts to the center
3) Shifts towards right 3) 3.14 10 4 T 4) 6.28 10 4 T
4) Shifts towards left Ans. 3
Ans. 3 ni
Sol. B 0
R S 2
Sol.
100
If temperature increases, resistance increases. 22. A galvanometer of resistance 50 is
As R increases, balancing length also connected to a battery of 3 V along with a
increases. It will shift towards Right resistance 2950 in series. A full scale
deflection of 30 divisions is obtained in the
18. Two tiny spheres carrying charges 1.8C and galvanometer. In order to reduce this
deflection to 20 divisions, the resistance in
2.8C are located at 40 cm apart. The
series should be
potential at the mid-point of the line joining the 1) 5050 2) 4450
two charges is 3) 6050 4) 5550
1) 4.3 104 V 2) 3.6 105 V Ans. 2
4
3) 3.8 10 V 4) 2.1 105 V Sol. R ' n 1 G R
Ans. 4
30
kq1 kq 2 1 3000 1500
Sol. V 20
r1 r2
Total resistance 2950 1500 4450
3 - DR ACADEMY
23. A circular coil of wire of radius ‘r’ has ‘n’ turns 1 2 1 1 2
LI LImax
and carries a current ‘I’. The magnetic 2 2 2
induction ‘B’ at a point on the axis of the coil Imax
I
at a distance 3r from its centre is 2
0 nI 0 nI 0 nI 0 nI I
1) 2) 3) 4) Imax sin t max
16r 4r 32r 8r 2
Ans. 1
t
0 nir2 4
Sol. B 3/2
2 x 2 r2 t LC
4
24. If voltage across a bulb rated 220V, 100 W 27. A magnetic field of flux density 1.0 Wb m2 acts
drops by 2.5 % of its rated value, the normal to a 80 turn coil of 0.01 m2 area. If this
percentage of the rated value by which the coil is removed from the field in 0.2 second, the
power would decrease is emf induced in it is
1) 5% 2) 10% 3) 20% 4) 2.5% 1) 0.8V 2) 5V 3) 4V 4) 8v
Ans. 1 Ans. 3
V2 Sol. 1 BAN 1 0.01 80
Sol. P
R 1 0.8 wb
P V2
2 0
P V
100 2 100 2 2.5 5% (2 1 ) 0 0.8
P V e 4V
25. A wore of certain material is stretched slowly t 2
by 10%. Its new resistance and specific
resistance becomes respectively 28. An alternating current is given by
1) 1.21 times, same i i1 sin t i2 cos t. The r.m.s current is given
2) both remains the same by
3) 1.1 times, 1.1 times
i12 i22 i12 i22
4) 1.2 times, 1.1 times 1) 2)
2 2
Ans. 1
Sol. Let l1 100 , l2 110 i1 i2 i1 i2
3) 4)
R l2 2 2
2 2 Ans. 1
R 2 l2 110
1.21 i12 i22
R1 l1 100 Sol. irms
2
R 2 1.21R1
Specific resistance remains same
29. Which of the following statements proves that
26. A fully charged capacitor ‘C’ with initial charge
Earth has a magnetic field?
‘ q 0 ’ is connected to a coil of self inductance ‘L’ 1) Earth is surrounded by ionosphere
at t=0. The time at which the energy is stored 2) A large quantity of iron-ore is found in the
equally between the electric and the magnetic Earth
field is 3) The intensity of cosmic rays stream of
charged particles is more at the poles than at
1) LC 2) LC 3) 2 LC 4) LC
4 the equator.
Ans. 2 4) Earth is a planet rotating about the North
1 2 q2 south axis
Sol. LImax
2 2C Ans. 3
Sol. Conceptual
4 - DR ACADEMY
30. A long solenoid has 500 turns, When a current 34. A convex lens of focal length ‘f’ is placed
of 2 A is passed through it, the resulting somewhere in between an object and a screen,
magnetic flux linked with each turn of the the distance between the object and the screen
solenoid is 4 10 3 Wb, then self induction of is ‘x’. If the numerical value of the
the solenoid is magnification produced by the lens is ‘m’, then
1) 2.0 henry 2) 1.0 henry the focal length of the lens is
2 2
3) 4.0 henry 4) 2.5 henry m 1 x m 1 x
1) 2)
Ans. 2 m m
Sol. 500 4 10 3 2Wb mx mx
3) 2
4) 2
Li N m 1 m 1
2 Ans. 3
L 1H
2 Sol. u v x
v
31. Which of the following radiations of m
u
electromagnetic waves has the highest
1 1 1
wavelength?
v u f
1) IR-rays 2) Microwaves
3) X-rays 4) UV-rays
35. A series resonant ac circuit contains a
Ans. 2
Sol. Micro waves capacitance 10 6 F and an inductor of 10 4 H.
The frequency of electrical oscillations will be
32. The power of a equi-concave lens is -4.5 and is 105 10
1) Hz 2) Hz 3) 105 Hz 4) 10Hz
made of an material of R.I. 1.6, the radii of 2 2
curvature of the lens is Ans. 1
1) -2.66 cm 2) 115.44cm 1
Sol. f
3) -26.6 cm 4) +36.6 cm 2 LC
Ans. 3
1 1 1 36. In a series LCR circuit R 300 , L 0.9H,
Sol. p 1
f R R C 2.0F and 1000 rad/sec., then
impedance of the circuit is
33. A ray of light passes through an equilateral
1) 500 2) 400 3) 1300 4) 900
glass prism in such a manner that the angle of
Ans. 1
incidence is equal to the angle of emergence
2
3 Sol. Z R 2 XC XL
and each of these angles is equal to of the
4
angle of the prism. The angle of deviation is 37. For light diverging form a finite point source
1) 20 0 2) 30 0 3) 45 0 4) 390 1) The wave front is parabolic
Ans. 2 2) The intensity at the wave front does not
Sol. i1 i2 A D depend on the distance
3) the wave front is cylindrical
4) the intensity decreases in proportion to the
distance squared.
Ans. 4
1
Sol. I 2
d
5 - DR ACADEMY
38. The fringe width for red colour as compared to 42. Focal length of a convex lens will be maximum
that for violet colour is approximately for
1) 4 times 2) 8 times 1) Green light 2) Red light
3) 3 times 4) Double 3) Blue light 4) Yellow light
Ans. 4 Ans. 2
D 1 1 1
Sol. , Sol. f ,
d 2 2 1 v r
39. In case of Fraunhoffer diffraction at a single
slit the diffraction pattern on the screen is 43. The de-Broglie wavelength of a particle of
correct for which of the following statements? kinetic energy ‘K’ is ; the wavelength of the
1) Central dark band having uniform K
particle, if its kinetic energy is is
brightness on either side. 4
2) Central bright band having dark bands on
either side. 1) 2) 4 3) 4) 2
2
3) Central dark band having alternate dark Ans. 4
and bright bands of decreasing intensity on 1
either side. Sol.
k
4) Central bright band having alternate dark
and bright bands of decreasing intensity on 1 k2 k 1
,
either side. 2 k1 4k 2
Ans. 4 2 2
Sol. Conceptual
44. The radius of hydrogen atom in the ground
40. When a Compact Disc (CD) is illuminated by 0
state is 0.53 A . After collision with an electron,
small source of white light coloured bands 0
observed. This due to it is found to have a radius of 2.12 A , the
1) Interference 2) Reflection principle quantum number ‘n’ of the final state
3) Scattering 4) Diffraction of the atom is
Ans. 4 1) n 3 2) n 4 3) n 1 4) n 2
Sol. Conceptual Ans. 4
Sol. r n2
41. Consider a glass slab which is silvered at one r1 n1
2
side and the other side is transparent. Given
r2 n2
the refractive index of the glass slab to be1.5.
If a ray of light is incident at an angle of 450 on 1
0.25 2
the transparent side, the deviation of the ray n2
of light from its initial path, when it comes out 1 100
n22 4
of the slab is 0.25 25
1) 1200 2) 450 3) 900 4) 1800 n2 2
Ans. 3 45. In accordance with the Bohr’s model, the
Sol. quantum number that characterises the
Earth’s revolution around the sun in an orbit
of radius 1.5 1011 m with orbital speed
3 10 4 ms 1 is
[given mass of Earth 6 1024 kg ]
1) 8.57 1064 2) 2.57 1074
3) 5.98 1086 4) 2.57 1038
Ans. 2
h
Sol. mv0 r n
2
6 - DR ACADEMY
46. If an electron is revolving in its Bohr orbit 51. The forbidden energy gap for ‘Ge’ crystal at ‘0’
0 K is
having Bohr radius of 0.529 A , then the radius
1) 2.57 eV 2) 6.57 eV
of third orbit is
0
3) 0.071 eV 4) 0.71 eV
1) 4.761A 2) 5125 nm Ans. 4
0 Sol. Conceptual
3) 4234 nm 4) 4496 A
Ans. 1
52. Which logic gate is represented by the
n2 following combination of logic gates?
Sol. rn 0.529 n3
z
0
0.529 9 4.761A
47. Binding energy of a Nitrogen nucleus 147 N , 1) AND 2) NOR 3) OR 4) NAND
14 Ans. 1
given m N 14.00307u
7
Sol. Conceptual
1) 206.5 MeV 2) 78 MeV
3) 104.7 MeV 4) 85 MeV
53. A metallic rod of mass unit length 0.5 kg m 1
Ans. 3
is lying horizontally on a smooth inclined plane
Sol. BE Zmp A Z mn mX 931.5 which makes an angle of 300 with the
Be 7.05481 7.06069 14.0030 931.5 horizontal. A magnetic field of strength 0.25 T
is acting on it in the vertical direction. When a
104.7MeV
current ‘T’ is flowing through it, the rod is not
allowed to slide down. The quantity of current
48. In a photo electric experiment, if both the required to keep the rod stationary is
intensity and frequency of the incident light 1) 14.76 A 2) 11.32 A
are doubled, then the saturation photo electric 3) 7.14 A 4) 5.98 A
current Ans. 2
1) Is doubled 2) Becomes four times Sol. F Bil
3) Remains constant 4) Is halved Bil cos mg sin
Ans. 1
Sol. i Intensity 3 1
0.25 I 0.5 10
2 2
49. The kinetic energy of the photoelectrons 5 100 20
I A
increases by 0.52 eV when the wavelength of 25 3 3
incident light is changed from 500 nm to I 11.32 A
another wavelength which is approximately
1) 1250 nm 2) 1000 nm 54. A nuclear reactor delivers a power of 10 9 W ,
3) 700 nm 4) 400 nm
the amount of fuel consumed by the reactor in
Ans. 3
one hour is
hc hc
Sol. KE1 KE2 1) 0.72 g 2) 0.96 g 3) 0.04 g 4) 0.08 g
1 2 Ans. 3
1 1 E mv2 m 9 1016
KE hc Sol. P 109
1 2 t t 3600
m 4 10 5 kg
50. The resistivity of a semiconductor at room m 4 10 5 103 g
temperature is in between
m 4 10 2 g
1) 106 to 108 cm 2) 1010 to 1012 cm
m 0.04 g
3) 102 to 10 5 cm 4) 103 to 106 cm
Ans. 4
Sol. Conceptual
7 - DR ACADEMY
55. Which of the following radiations is deflected 59. The Vernier scale of a travelling microscope
by electric field? has 50 divisions which coincides with 49 main
1) rays 2) particles scale divisions. If each main scale division is
3) X rays 4) Neutrons 0.5 mm, then the lease count of the
Ans. 2 microscope is
Sol. Conceptual 1) 0.01 mm 2) 0.5 cm
3) 0.01 cm 4) 0.5 mm
Ans. 1
56. Two objects are projected at an angle 00 and
Sol. L.C 1MSD 1VSD
(90 )0 , to the horizontal with the same
or
speed. The ratio of their maximum vertical 1MSD 0.05 mm
L.C
heights is No.of.vernier scale division 50
1) 1 : tan 2) tan2 :1 L.C 0.01mm
3) 1:1 4) tan :1
Ans. 2 60. The displacement ‘x’ (in meter) of a particle of
H1 tan 2 mass ‘m’ (in kg) moving in one dimension
Sol. under the action of a force, is related to time ‘t’
H2 1
(in sec) by, t x 3 . The displacement of the
particle when its velocity is zero, will be
57. A car is moving in a circular horizontal track
1) 6 m 2) 2 m 3) 4 m 4) 0 m
of radius 10 m with a constant speed of
Ans. 4
10 ms 1 . A bob is suspended from the roof of
Sol. t x 3
the car by a light wire of length 1.0m. The angle
made by the wire with the vertical is (in radian) x t3
x t 2 6t 9
1) 0 2) 3) 4)
3 6 4 dx
V 2t 6
Ans. 4 dt
V 2 10 10 As V 0
Sol. tan 1
2g 10 10 2t 6 0
t 3 sec
x (3)2 6(3) 9 0 m
4
58. Two masses of 5 kg and 3 kg are suspended
with the help of massless inextensible strings
as shown in figure, when whole system is going
upwards with acceleration 2 m / s 2 , the value
of T1 is (use g 9.8 m / s2 )
1) 23.6 N 2) 59 N 3) 94.4 N 4) 35.4 N
Ans. 3
Sol. T1 m1 m2 g a
8 - DR ACADEMY
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