SET A : PHYSICS |
|
| 1. |
The
dimension of  where
is permittivity of free space and E is electric field:- |
| (1) |
MLT-1 |
| (2) |
ML2T-2 |
| (3) |
ML-1T-2 |
| (4) |
ML2T-1 |
| Soln: |
Hence the option is (3) |
| |
|
| 2. |
A particle moves a distance x in time t according to equation x = (t + 5)-1. The acceleration of particle is proportional to: |
| (1) |
(velocity)2/3 |
| (2) |
(velocity)3/2 |
| (3) |
(distance)2 |
| (4) |
(distance)-2 |
| Soln: |
Hence the option is (2) |
| |
|
| 3. |
Six vectors , through have the magnitudes and directions indicated in the figure . Which of the following is true?
 |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Using Triangle
rule of vector addition, we have, 
Hence the option is (4) |
| |
|
| 4. |
The particle has initial velocity  and acceleration  . Its speed after 10s is |
| (1) |
10 units |
| (2) |
7 units |
| (3) |
7√2 units |
| (4) |
8.5 units |
| Soln: |
Hence the option is (3) |
| |
|
| 5. |
A block of mass m is in contact with the cart C as shown in the figure .
The coefficient of static friction between the block and
cart is  .
The acceleration 
of the cart that will prevent the block from falling satisfies
|
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
weight ≤ Frictional
force 
mg ≤ N
mg ≤ ma.
Hence the option is (4) |
| |
|
| 6. |
A man of 50 kg mass is standing
in a gravity free space at a height of 10 m above the
floor. He throws a stone of 0.5 kg mass downwards with
a speed 2 m/s. When the stone reaches the floor, the
distance of the man above the floor will be: |
| (1) |
20 m |
| (2) |
9.9 m |
| (3) |
10.1 m |
| (4) |
10 m |
| Soln: |
Let distance
of man from the floor be (10 + x) m. As centre of mass
of system remains at 10m above
the floor, so 50(x) = 0.5(10)
x = 0.1 m
distance of the man above the floor = 10 + 0.1 = 10.1
m.
Hence the option is (3) |
| |
|
| 7. |
An engine pumps water through a hose pipe. Water passes through the pip and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine? |
| (1) |
800 W |
| (2) |
400 W |
| (3) |
200 W |
| (4) |
100 W |
| Soln: |
Power = Fv =
v(m/t)v = v2(rAv) = v3 (mass per
unit length) = 800W.
(Here, r = mass/volume).
Hence the option is (1) |
| |
|
| 8. |
A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocity (in m/s) after collision will be: |
| (1) |
0, 2 |
| (2) |
0, 1 |
| (3) |
1, 1 |
| (4) |
1, 0.5 |
| Soln: |
By conservation
of linear momentum :
2m = mv1 + 2mv2 v1 + 2v2 = 2...........(1)
by definition of e : e = 0.5 =.......(2), On solving (1)
and (2) v1= 0 and v2 =1.
Hence the option is (2) |
| |
|
| 9. |
A gramophone record
is revolving with an angular velocity  .
A coin is placed at a distance rfrom the centre of the
record. The static coefficient of friction is m. The coin
will revolve with the record if |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Here frictional
force provides centripetal force so
Hence the option is (4) |
| |
|
| 10. |
A circular disk of moment of
inertia It is rotating in a horizontal plane,
about its symmetry
axis, with a constant angular speed i.
Another disk of moment of inertia Ibis dropped
coaxially onto the rotating disk. Initially the second
disk has zero angular speed. Eventually both the disks
rotate with a constant angular speed f.
The energy lost by the initially rotating disc to friction
is |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Hence the option is (1) |
| |
|
| 11. |
Two particle which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of center of mass of the system will be: |
| (1) |
v |
| (2) |
2v |
| (3) |
Zero |
| (4) |
1.5v |
| Soln: |
Net external
force on system is zero. So,  =
0
Hence the option is (3) |
| |
|
| 12. |
The radii of circule orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A and 3V, then the speed of satellite B will be: |
| (1) |
3 V/2 |
| (2) |
3 V/4 |
| (3) |
6V |
| (4) |
12 V |
| Soln: |
Speed of satellite
=
VA = 6V.
Hence the option is (3) |
| |
|
| 13. |
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at point a/2 distance from the centre |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Potential = V
= 
Hence the option is (2) |
| |
|
| 14. |
A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed V. The two balls meet at t = 18s. What is the value of V?
(take g = 10 m/s2) |
| (1) |
60 m/s |
| (2) |
75 m/s |
| (3) |
55 m/s |
| (4) |
40 m/s |
| Soln: |
Let two balls
meet at depth h from platform
So,  ,
On solving we have, v = 75m/s.
Hence the option is (2) |
| |
|
| 15. |
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half and radius of the original rod. What is the amount of heat conducted be the new rod, when placed in thermal contact with the two reservoirs in time t? |
| (1) |
Q/2 |
| (2) |
Q/4 |
| (3) |
Q/16 |
| (4) |
2Q |
| Soln: |
Volume is same.
So, A l = A' l' we have A' = A/4, so, l' = 4l
So, 
Hence the option is (3) |
| |
|
| 16. |
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Total radiant
energy per unit area
Hence the option is (2) |
| |
|
| 17. |
If  U and W represent the increase in internal energy and work done by the syatem respectively in a thermo dynamical process, which of the following is true? |
| (1) |
U = - W, in a isothermal process |
| (2) |
U = - W, in a adiabatic process |
| (3) |
U = W, in a isothermal process |
| (4) |
U = W in a adiabatic process |
| Soln: |
First law of thermodynamics,
Q
= U
+ W.
For adiabatic process, Q = O.
So, U
= -W.
Hence the option is (2) |
| |
|
| 18. |
The displacement of particle along the x – axis is given by x = asin2t. The motion of particle corresponds to |
| (1) |
simple harmonic motion of frequency |
| (2) |
simple harmonic motion of frequency |
| (3) |
simple harmonic motion of frequency |
| (4) |
Non simple harmonic motion |
| Soln: |
x = a sin2t
Hence the option is (2) |
| |
|
| 19. |
The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be |
| (1) |
√2T |
| (2) |
T |
| (3) |
T/√2 |
| (4) |
2T |
| Soln: |
Time period T
= 
Hence the option is (1) |
| |
|
| 20. |
A transverse wave is represented by y = A sin (t – kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity? |
| (1) |
A |
| (2) |
 A/2 |
| (3) |
A |
| (4) |
2A |
| Soln: |
Wave velocity
= n= maximum particle velocity = A/n
= 
= 2A
Hence the option is (4) |
| |
|
| 21. |
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was: |
| (1) |
508 Hz |
| (2) |
510 Hz |
| (3) |
514 Hz |
| (4) |
516 Hz |
| Soln: |
Frequency of the
tuning fork is 512 Hz. Beat frequency is 4. So, the frequency
of the piano string may be 508 or 516 Hz. We know, Frequency
a
 .
On increasing tension frequency of the piano will increase
and the beat frequency is decreasing. So, frequency of
piano should be 508 Hz.
Hence the option is (1) |
| |
|
| 22. |
Which of the following statements is false for the properties of electromagnetic waves? |
| (1) |
These waves do not require any material medium for propagation |
| (2) |
Both electric and magnetic field vectors attain the maxima and minima at the same place and same time. |
| (3) |
The energy in electromagnetic wave is divided equally between electric and magnetic vectors. |
| (4) |
Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation wave |
| Soln: |
Electric and magnetic field vectors
are perpendicular to each other in electromagnetic waves.
Hence the option is (4) |
| |
|
| 23. |
A lens having focal length f
and aperture of diameter d forms an image of intensity
I. Aperture of diameter d/2 in central region of lens
is covered by a black paper. Focal length of lens and
intensity of image now will be respectively |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Focal length will be same if half of the aperture of the lens is covered by a black paper.
So, f' = f. Intensity a area. So, intensity will becomes,
I - I/4 = 3I/4.
Hence the option is (4) |
| |
|
| 24. |
A ray of light traveling in a transparent medium of refractive index m, falls on a surface separating the medium from air at an angle of incidence of 450. for which of the following value of m the ray can undergo total internal reflection? |
| (1) |
m = 1.25 |
| (2) |
m = 1.33 |
| (3) |
m = 1.40 |
| (4) |
m = 1.50 |
| Soln: |
For total internal reflection, incidence angle,
45° > qc Þ sin 45° > sinΘc
Hence the option is (4) |
| |
|
| 25. |
Two positive ions, each carrying
a charge q, are separated by a distance d. If F is the
force of repulsion between the ions, the number of electrons
missing from each ion will be (e being the charge on
an electron) |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
According to coulomb
law, Force of repulsion 
( no. of electrons missing in each of the ions)
Hence the option is (4) |
| |
|
| 26. |
A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle q with the horizontal side of the square as shown in figure. The electric flux linked to the surface, in units of volt-m, is
 |
| (1) |
zero |
| (2) |
EL2 |
| (3) |
EL2CosΘ |
| (4) |
EL2CosΘ |
| Soln: |
Total electric flux =  .
Here E is perpendicular to area vector. So, total flux
will be zero.
Hence the option is (1) |
| |
|
| 27. |
A series combination
of n capacitors, each of value C1 is charged
by a source of potential
difference 4V. When another parallel combination of n2
capacitors, each of value C2 is
charged by a source of potential difference V, it has
the same (total) energy stored in it, as the first combination
has. The value of C2 in terms of C1
is then: |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Equivalent capacitance for n1 series combination = Cs = 
Equivalent capacitance for n2 parallel combination = Cp = n2C2
so, 
Hence the option is (1) |
| |
|
| 28. |
A potentiometer
circuit is setup as shown. The potential gradient, across
the potentiometer wire, is k volt/cm and the ammeter,
present in the circuit, reads 1.0 A when two way key is
switched off. The balance points, when the key between
the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in,
are found to be at lengths l1 cm and l2
cm
respectively. The magnitudes, of the resistors R and X,
in ohms, are then, equal, respectively, to:
|
| (1) |
kl1 and kl2 |
| (2) |
k(l2 - l1) and kl2 |
| (3) |
kl1 and k(l2 -l1) |
| (4) |
k(l2 -l1) and kl1 |
| Soln: |
R = kl1
and R + X = kl2
R = k(l2- l1).
Hence the option is (3) |
| |
|
| 29. |
A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. if it is to work as a voltmeter of 30 volt range, the resistance required to be added will be: |
| (1) |
1000 ohm |
| (2) |
900 ohm |
| (3) |
1800 ohm |
| (4) |
500 ohm |
| Soln: |
Let required resistance be R then, (R + Rg) Ig = V = 30.
 .
Putting the values, R = 900 Ohm
Hence the option is (2) |
| |
|
| 30. |
Consider the following two statements:
(A) Kirchhoff’s junction law follows from the conservation of charge.
(B) Kirchhoff’s loop law follows from the conservation of energy
Which of the following is correct? |
| (1) |
Both (A) and (B) are correct |
| (2) |
Both (A) and (B) are wrong |
| (3) |
(A) is correct and (B) is wrong |
| (4) |
(A) is wrong and (B) is correct |
| Soln: |
Kirchchoff's junction law or current
law deals with conservation of charge. Kirchchoff's loop
law or voltage law deals with conservation of energy.
Hence the option is (1) |
| |
|
| 31. |
In producing chlorine by electrolysis
100 kW power at 125 V is being consumed. How much. chlorine
per minute is liberated (E.C.E. of chlorine is of chlorine
is 0.367 × 10-6 kg/C) |
| (1) |
3.67 x 10-3
kg |
| (2) |
1.76 x 10-3
kg |
| (3) |
9.67 x 10-3
kg |
| (4) |
17.61 x 10-3
kg |
| Soln: |
According to Faraday's first law
of electrolysis, m = zIt = z (P/V)t.
Putting the values, m = 17.61 10-3kg.
Hence the option is (4) |
| |
|
| 32. |
A square current carrying loop
is suspended in a uniform magnetic field acting in the
plane of the loop. If the force on one arm of the loop
is  ,
the net force on the remaining three arms of the loop
is: |
| (1) |
|
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Net force on loop
is zero. So net force on the remaining arm is
Hence the option is (3) |
| |
|
| 33. |
A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency off revolutions/s. The value of magnetic induction in Wb/m at the centre of the ring is: |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Magnetic induction
= B = 
Hence the option is (1) |
| |
|
| 34. |
Electromagnets are made of soft iron because soft iron has: |
| (1) |
high retentivity and low coercive force |
| (2) |
low retentivity and high coercive force |
| (3) |
high retentivity and high coercive force |
| (4) |
low retentivity and low coercive force |
| Soln: |
Electromagnets
are made of soft iron because soft iron has low retentivity
and low coercive force.
Hence the option is (4) |
| |
|
| 35. |
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth’s horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be:
|
| (1) |
4s |
| (2) |
1s |
| (3) |
2s |
| (4) |
3s |
| Soln: |
Time period of
a bar magnet in magnetic induction B is 
So, 
= 4s
Hence the option is (1) |
| |
|
| 36. |
A conducting circular loop is
placed in a uniform magnetic field, B = .025 T with
its plane perpendicular to the loop. The radius of the
loop is made toshrinkata constant rateofi mms The radius
of the loop is made to shrink at a constant rate of
1 mms–1. The induced emf when the radius is 2
cm is
|
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Here, B = 0.025T , r = 2cm and dr/dt = 1mm. Putting the
values, e = 
Hence the option is (3) |
| |
|
| 37. |
In the given circuit the reading
of voltmeter V1 and
V2 are 300 volts each.
The reading of the voltmeter V3
and ammeter A are respectively
|
| (1) |
100 V , 2.0 A |
| (2) |
150 V , 2.2 A |
| (3) |
220 V, 2.2 A |
| (4) |
220 V, 2.0 A |
| Soln: |
Potential across
R is 220V. So current through R is (220/100) = 2.2A.
Hence the option is (3) |
| |
|
| 38. |
A 220-volt input is supplied
to a transformer. The output circuit draws a current
of 2.0 ampere at 440 volts. If the efficiency of the
transformer is 80%, the current drawn by the primary
windings of the transformer is |
| (1) |
5.0 ampere |
| (2) |
3.6 ampere |
| (3) |
2.8 ampere |
| (4) |
2.5 ampere |
| Soln: |
efficiency
= 0.8
Ip=
Putting the values, Ip = 5A.
Hence the option is (1) |
| |
|
| 39. |
A source S1 is producing
1015 photons per second of wavelength 5000
A0. Another source S2 producing
1.02 x 1015 photons per second of wavelength
5100 A0 then (power of S2) /(
power of S1) |
| (1) |
0.98 |
| (2) |
1.00 |
| (3) |
1.02 |
| (4) |
1.04 |
| Soln: |
On putting the values, the ratio will be 1.
Hence the option is (2) |
| |
|
| 40. |
A beam of cathode rays is subjected
to crossed Electric (E) and Magnetic field (B). The fields
are adjusted such that the beam is not deflected. The
specific charge of the cathode rays is given by (where
V is the potential difference between cathode and anode) |
| (1) |
 |
| (2) |
 |
| (3) |
 |
| (4) |
 |
| Soln: |
Force due to magnetic
field = Force due to electric field
qvB = qE
v = E/B
K.E = P.E due to electric potential 
= qv (V= electric potential).
Specific charge = q/m = 
Hence the option is (1) |
| |
|
| 41. |
The potential difference that
must be applied to stop the fastest photo electrons
emitted by a nickel surface, having work function 5.01
eV, when ultraviolet light of 200 nm falls on it, must
be |
| (1) |
1.2 V |
| (2) |
2.4 V |
| (3) |
-1.2 V |
| (4) |
-2.4 V |
| Soln: |
According to,
Einstein's photo?electric equation
= W + eV0 (W = work function = 5.01eV and V0
= stopping potential)
= (hc/e )
eV = 6.2 eV. So, V0= 6.2 - 5.01 i.e. equal
to 1.2V. So, stopping potential is -1.2V.
Hence the option is (3) |
| |
|
| 42. |
The activity of
a radioactive sample is measured as N0 counts
per minute at t = 0 and N0/e counts per minute
at t = 5 minutes. The time (in minutes) at which the activity
reduces to half its value is |
| (1) |
5 loge 2 |
| (2) |
 |
| (3) |
 |
| (4) |
5 log10
5 |
| Soln: |
Hence the option is (1) |
| |
|
| 43. |
The energy of a hydrogen atom
in the ground state is –13.6 eV. The energy of
a He+ ion in the first excited state will be |
| (1) |
- 6.8eV |
| (2) |
—13.6eV |
| (3) |
—27.2eV |
| (4) |
—54.4eV |
| Soln: |
En
= -13.6
= -13.6eV. ( Z= 2, first excited state = n = 2).
Hence the option is (2) |
| |
|
| 44. |
The mass of 
nucleus is 0.042 u less than the sum of the masses of
all its nucleons.
The binding energy per nucleon of
nucleus is nearly |
| (1) |
23 MeV |
| (2) |
46 MeV |
| (3) |
5.6 MeV |
| (4) |
3.9 MeV |
| Soln: |
B.E = 0.042 931 MeV
B.E/nucleon = (0.042 931)/7 = 5.6 MeV
Hence the option is (3) |
| |
|
| 45. |
An alpha nucleus of energy 1/2mc2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to: |
| (1) |
1 / v2 |
| (2) |
1 / Ze |
| (3) |
v2 |
| (4) |
1 / m |
| Soln: |
At closest distance,
d the K.E of the particle = electrostatic potential energy
So, d 1/m.
Hence the option is (4) |
| |
|
| 46. |
A common emitter amplifier has a voltage gain of 50, an input impendance of 100 Ω and an output impendance of 200 Ω. The power gain of the amplifier is : |
| (1) |
50 |
| (2) |
500 |
| (3) |
1000 |
| (4) |
1250 |
| Soln: |
Voltage gain =
current gain 
Power gain = 
x Voltage gain = 25 x 50 = 1250.
Hence the option is (4) |
| |
|
| 47. |
Which one of the following bonds
produces a solid that reflects light in the visible
region and whose electrical conductivity decreases with
temperature and has high melting point? |
| (1) |
covalent bonding |
| (2) |
Metallic bonding |
| (3) |
van der Waal’s bonding |
| (4) |
ionic bonding |
| Soln: |
Metallic bond
produces a solid that reflects light in the visible region
and whose electrical conductivity decreases with temperature
and has high melting point.
Hence the option is (2) |
| |
|
| 48. |
The device that can act as a complete electronic circuit is: |
| (1) |
Zener diode |
| (2) |
Junctin diode |
| (3) |
Integrated circuit |
| (4) |
Junction transistor |
| Soln: |
Integrated circuit
can act as a complete electronic circuit.
Hence the option is (3) |
| |
|
| 49. |
Which one of the following statements is FALSE? |
| (1) |
The resistance of intrinsic semiconductor decreases with increase of temperature. |
| (2) |
Pure Si doped with trivalent impurities gives a p-type semiconductor |
| (3) |
Majority carriers in a n-type semiconductor are holes |
| (4) |
Minority carriers in a p-type semiconductor are electrons |
| Soln: |
Majority carriers
in a n-type semiconductor are electrons.
Hence the option is (3) |
| |
|
| 50. |
To get an output Y = 1 from the
circuit shown below, the input must be:
A B
C
|
| (1) |
1
0 0 |
| (2) |
0
1
0 |
| (3) |
0 0
1 |
| (4) |
1 0
1 |
| Soln: |
Y = (A + B).C
Hence the option is (4) |
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