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PHYSICS
(Solved Paper of CEE)
Assam Electronics Development Corporation Limited
(AMTRON) (A Government of Assam Undertaking)
1) Which of the following substance (A, B, C) has the highest specific heat?
A) A
B) B
C) C
D) All has equal specific heat.
Answer [C]
Hints:
Q= MsdT dQ/dt=MsdT/dt Pdt=MsdT
dT=(P/Ms)dt ------ (1)
From the graph we can observe that rise in temperature in graph-A is more
than in B & C
From -1
Rise in temperature dT ∝1/S
dT is maximum for A and minimum for C
S is maximum for C and minimum for A
2)A barometer tube of length 90 cm contains some air over mercury.The reading of the
mercury level is 74.8 cm when true atmospheric pressure is 76 cm and temperature is 30 . If
the reading is observed to be75.4 cm on some another day when temperature is 10 , then
what will be the true pressure?
A) 74.25 cm
B) 75.65 cm
C) 76.57 cm
D) 77.26 cm
Answer [C]
Hints:
2
22
1
11
/
1 4.75
2.1
768.74
T
vp
T
vp
pp
p
p
As no. of moles of gas is constant in the barometer tube
56.76166.14.75
,
166.1
283
6.14
303
2.152.1
/
1
1
p
therefore
p
p
3) For the following circuit, the potential drop across 3 Mf capacitor when switch S is open and
switch S is closed will be
A) 12 V, 10 V
B) B) 8V, 9 V
C) C) 9 V, 8 V
D) D) 12 V, 8 V
Answer: [D]
Hints:
Case(1),
When switch is opened
Potential drop across VCeqv 18
In series q is same
VV
VVV
V
V
V
V
C
C
V
V
CV
CqV
12
18
2
36
/1
/
1
21
2
1
2
1
1
2
2
1
Case (2),
If switch is closed,
Capacitors are at steady state
So no current will flow through there
Ai
i
1
1818
Potential drop across F3 is same as to resiostor8
Potential drop across resiostor8 is V818
Potential drop across F3 capacitor is 8 V
So, answer is 12V, 8V
4) What is the current passing through the 1 ohm resistor in the circuit given?
A) 1/19 Ampere
B) 2/19 Ampere
C) 3/19 Ampere
D) 4/19 Ampere
Answer: [B]
Hints:
Applying Kirchoff’s law in loop-1
1234
0234
233
1
1
1
ii
ii
ViiiV AA
Apllying Kirchoff’s lawin loop-2
Applying Kirchoff’s law in loop-3
Applying junction law
5
4
432
1 2
iii
iii
3
22
34
34
ii
ViiV cc
2232
0223
1233
14
41
41
ii
ii
ViiV BB
Solving these equations we get
Ai19
2
5) Two wires AB and AC carry 1 Ampere current each as shown in the figure. One end wires
extend to infinity and the 60BAC . The magnetic field at the point at a distance 1 m from
A and lies on the bisector of the BAC will be
A) 3244
B) 3222
C) 3228
D) 3242
Answer [A]
Hints:
Direction of current is wrong. But assuming that is correct
2]2
32[
]90sin60[sin
90&60
]sin[sin
,
1]2
32[
]90sin60[sin
90&60
]sin[sin
2
2
21
212
1
1
21
211
rkiB
rkiB
rkiB
also
rkiB
rkiB
rkiB
21 BBBnet
)324(4
)32(2
]32[2
]2
32[4
2/1
]2
32[2
]2
32[]
2
32[
iB
iB
kiB
rkiB
rkiB
rki
rkiB
net
net
net
net
net
net
6) The radiation of a wavelength 332 nm is incident on a metal of work function 1.07 eV. The
value of the stopping potential will be
A) 0.70 V
B) 1.14 V
C) 1.68 V
D) 2.06 V
. Answer [D]
Hints:
Energy incident = 7.3332
1240
hc
ev07.1
Stopping potential must be able to stop the most energetic electron
vv
evev
evev
66.2
66.2
734.3
max
max
max
7) The de-Broglie wavelength of a neutron at C27 is . What will be its de Broglie
wavelength at 927 .C
A) . /2
B) . /3
C) . /4
D) 3 . /2
. Answer [A]
Hints:
T
Tv
now
v
mv
h
1
,
1
22
2
277.92
7.9227
27
7.92
7.92
27
T
T
8) When U238
92 decays to final product Pb206
82 , the number of particle and particle
emitted are:
A) 8 particle , 6 particle
B) 6 particle , 8 particle
C) 8 particle , 8 particle
D) 6 particle , 6 particle
Answer [A]
Hints:
0
1
4
2
206
82
238
92 nnPbU
Balance mass as particle has no mass no.
238-206=32
So,
No. of particle = 32/4=8
By charge balance
92=82+28+ n (-1)
92=82+16- n
n =6 & n =8
9) A radioactive substance of half-life 6 minutes is placed near a Geiger counter which is found
to register 1024 particles per minute. How many particles per minute it will register after 42
minutes?
A) 8
B) 16
C) 24
D) 32
Answer [A]
Hints:
Half life= 6 min =
2ln
6
692.0
6
2ln
At t= 0, 1024 particles per minute
After 42 minutes, 7 half-life is complete
Therefore, no. of particles = 72
1024
No. of particle = 8
10) For the given logic circuit, the output will be 1 under which of the following conditions?
A) A=0, B=0, C= 1
B) A=1, B=0, C=1, D=0
C) A=0, B=1, C=0, D=1
D) A=0, B=1, C=1, D=1
Answer [D]
11) A particle is moving with constant acceleration and 32 ,,1
VVV are the average velocity of
the particle in three successive intervals 321 ,, ttt . Which of the following relation will be
correct?
A) 32
21
32
31
tt
tt
VV
VV
B)31
21
32
21
tt
tt
VV
VV
C) 32
21
32
21
tt
tt
VV
VV
D)32
21
32
21
tt
tt
VV
VV
Answer [C]
Hints:
If constant acceleration
2
fi
avg
VVV
32
21
32
21
33
22
11
11
:
322
222
,
122
2
tt
tt
VV
VV
sveequationSolvingabo
atUV
atUV
Similarly
atUV
atUUV
=
12)The maximum height attained by a projectile is increased by 10% by increasing its speed of
projection, without changing the angle of projection. The percentage of increase in the time of
flight of the projectile will be:
A) 5%
B) 10%
B) 15%
D) 20%
Answer: [A]
Hints:
Hg
U
2
sin 22
%5
20/1
sin2
20/1/
210
1
2
2
T
T
T
T
U
U
T
T
UT
g
UT
UDelU
U
delU
U
delU
H
delH
UH
13) The resultant of the three vectors shown in the figure and the angle made by the resultant
with X-axis:
A) 37&10m
B) 5.35&6.8 m
C) 37&)35( m
D) None of these
Answer [B]
Hints:
CBAR
A= 4i+3J
B= 3i
C=2j
From the above equations:
5.35&6.877
57
2334
mR
jiR
jijiR
14) A clock has its minute hand 4.0 cm long. The average velocities of the tip of the minute
hand between 6:00 a.m. and 6:30 a.m., and 6:00 a.m. and 6:30 p.m. respectively are:
A) sec/104.4 3 cm , sec/108..1 4 cm
B) sec/104.4 3 cm , sec/104.4 3 cm
C) sec/108..1 4 cm , sec/108.1 4 cm
D) sec/106.9sec,/104.5 33 cmcm
Answer [A]
Hints:
C-1
4
3
3
108.1
10177.0
60750
8
sec1044.4
6030
8
t
rV
t
rV
avg
15) A body weighs 98 Non a spring balance at the North Pole, Its weight recorded on the same
scale if it is shifted to the equator is:
A) 98 N
B) 0
C) 98.34 N
D) 97.66 N
Answer [D]
Hints:
we know:
766.9
0,
90,
cos
2
22
eff
eff
eff
g
rgg
Atequator
golesgravityatp
Atpoles
Rgg
So weight at equator:
W=mg effect=97.66
= 10*9.766
=97.66
16) A lift weighing 250 Kg is to be lifted up at constant velocity of 0.20 m/sec. What would be
the minimum horse power of the motor to be used?
A) 1.3 HP
B) 0.65 HP
C) 1.5 HP
D) 0.75 HP
Answer [B]
Hints:
P= VF.
F=mg
=250*10
=2500
V=0.2 m/sec
From (1):
P=2500*0.2
=500 watt
P=0.65 HP
17) Starting from rest, a fan takes 16 second to attain the maximum speed of 400 rpm.
Assuming constant acceleration, the time taken by the fan in attaining half the maximum speed
is:
A) 2 Seconds
B) 4Seconds
C) 8 Seconds
D) None of these
Answer: [C]
Hints:
.,
sec8
)(6/503/20
sec/3
20200,
:
6
5
48
40
1603
40
3
40
60
2*400
400
:
tan
CHence
t
t
radrpmt
Again
ngSubstituti
rpmW
Given
t
tcons
fif
f
f
if
18) The value of acceleration due to gravity in )/( 2sm at a point 5-km above the earth’s surface
and 5 km below the earth surface are respectively:
A) 9.78,9.79
B)9.78,0
C) 9.79,0
D) 9.79, 9.78
Answer [A]
Hints:
We know:
For a height h above the ground:
2sec/78.96400
5*218.9
)2
1(
mg
R
hgg
h
h
At Depth,
.,
sec/79.9
)6400/51(8.9
)/1(
2
AHence
m
rdggd
19) The three vessels shown in figure have some base area. Equal volumes of a liquid are
poured in the three vessel. The force on the base will be:
A) Maximum at Vessel A
B) Maximum at Vessel B
C) Maximum at vessel C
D) Equal in all vessels
Answer[C]
Hints:
hP
ghP
Height is maximum in “C”.
Hence pressure at base will be maximum in Vessel C.
Therefore force is maximum.
Hence, “C” is correct.
20) A water drop of radius 1 cm is broken into 1000 equal droplets. If the surface tension of
water is 0.075 N/m, then the gain in surface energy will be
A) 0
B) J4105.8
C) J4105.7
D) Infinite
Answer [B]
Hints:
Initial Volume= Final Volume (Applying conservation of Volume)
mr
nr
nr
rnR
001.0
)1000/(1/1
1
3
4
3
4
3
1
3
1
33
33
Increase in surface energy= J
ST
4
22
105.8
01.0001.010004075.0
21) A compound microscope consists of an objective of focal length 1.0 cm and an eyepiece of
focal length 5.0 cm separated by 12.2 cm. At what distance from the objective should an object
is placed to focus it properly so that the final image be formed at the least distance of distinct
vision 25 cm?
A) -1.1 cm
B) -2.1 cm
C) -1.5 cm
D) -2.5 cm
Answer [A]
Hints:
For Objective:
V=
1/
1)(
)1)((
xx
x
x
fu
uf
So, distance of image formed by objective from eyepiece is=12.2-1x
x
=1
2.122.12
x
xx
For eyepiece:
U= -1
2.122.11
x
x
V=-25 f=5
So, -25= 51
)2.112.12(/5)
1
2.112.12(
x
x
x
x
Solving and getting:
X=1.14
22) The distance between two parallel planes of a crystal is 3 A
. What should be the
wavelength of X-ray used for getting the first order Bragg reflection at a glancing angle of
7.14 ?
A) 1 A
B) 1.3 A
C) 2.5 A
D) 1.5 A
Answer [D]
Hints:
A/Q:
2dsin n
."",
5.1
)7.14sin(1032 10
iscorrectDHence
A
23) The density of water and steam are 1000 3/ mkg and 0.6
3/ mkg respectively. The latent
heat of vaporisation of water= 2.25 kgj /10 6 . The increase in internal energy of 1 kg of water
at 100 degree celcius, when it is converted into steam at the same temperature and 1 atm(100
kPa) is
A) 2 J610
B) -2 J810
C) - 2 J1010
D) -2 J1210
Answer [A]
Hints:
Given:
.,
102
)mod1(
1067.1
)106/10(10
6/10
/
10
/6.0
1
/1000
6
5
35
3
33
3
3
AHence
U
ynamicsrstlawoftheWUQ
W
W
VPW
mV
mV
mV
mkg
kgm
mkg
steam
steamsteamsteam
w
steam
w
w
24) A liquid cools from 70 c to 60 c in 5 minutes. If the temperature of the surrounding is
constant at 30 c , then the time taken by the liquid to cool from 60 c to 50 c is:
A) 5 min
B) 10 min
C) 7 min
D) 8 min
Answer [C]
Hints:
A/Q, Newton’s law of cooling:
.,
min7
:&
)2/(
,
35/2
,
)2
(/
2112
2112
CHence
t
gcalculatinEquating
TTTKt
TT
Now
K
wegetngSubstituti
TTT
KtTT
25) Figure below shows a triangular loop PQR carrying a current I, The triangle is equilateral
with edge length l. A uniform magnetic field B exists in a direction parallel to PQ. The forces
acting on PQ, QR, and RP respectively are
A) 0, ilB2/3 , ilB2/3
B) 0, 1/2ilB, ilB2/3
C) 0, ilB2/3 ,1/2ilB
D) ilB2/3 , 0, 0
Answer [A]
Hints:
As we know that:
BilBliF effeqB )(
For PQ:
l
ilB
BilF
ll
ForPR
F
Bparallell
eff
pR
PQB
eq
B
2
3
2
3
:
0)(
Similarly for QR:
ilBFB2
3
26) A positively charged particle projected towards east is deflected towards north by a
magnetic field. The magnetic field may be
A) Towards west
B) Towards south
C) Upward
D) Downward
Answer [D]
Hints:
".",
ˆˆ
)ˆˆ(
ˆ
ˆˆ
DHence
KB
BiJ
jF
iV
BVQF
B
B
27) Two long straight wires each carrying a current of 5.0 A, are kept parallel to each other at a
separation of 2.5 cm. The magnitude of magnetic force experienced by 5.0 cm of wire is
A) 8 N510
B) 4.0 N510
C) 2.0 N510
D) 1.0 N510
Answer [D]
Hints:
.,
101
5.22/55104
2/
5
7
21
DHence
N
diiFB
28) A parachutist drops freely from an aeroplane for 10s before the parachute opens out. Then
he descends with a net retardation of 2.5 2/ sm . If he jumps out of the plane at a height of 2495
m and g= 10 2/ sm , then his velocity on reaching the ground will be
A) 2.5 1/ sm
B) 7.5 1/ sm
C) 5 1/ sm
D) 10 1/ sm
Answer [C]
Hints:
Using:
.,
2
2
1
22
2
CHence
asUV
atutS
29) If
kji ,, are unit vectors along X, Y an Z axes respectively, then the angle between
the vectors
KJi and
i is given by
A) = 3/1cos 1
B) = 3/1sin 1
C) = 2/3cos 1
D) =
2
31Sin
Answer [A]
Hints:
Let,
.,
)3
1(
:
.
1
AHence
Cos
Solvingfor
ABCosBA
i
A
B
kji
30) A projectile has same range R when the maximum height attained by it is either 21orhh ,
Then 211, andhhR will be related as
A) R= 21hh
B) R= 2 21hh
C) R= 3 21hh
D) R=4 21hh
Answer [D]
Hints:
Rangeg
U 2sin2
Hg
U
2
sin 22
max
Range is same for complementary angles.
Max. height/Range= tan RH /4
R=4 21hh
Hence “D”.
31) A ball of mass m is moving towards a batsman at a speed “V”. The batsman strikes the ball
and deflects it by an angle without changing its speed. The impulse imparted to the ball is
given by
A) mvcos
B) mvsin
C) 2mvcos /2
D) 2mvsin /2
Answer [D]
Hints:
Impulse= Change in momentum
R= 2ACos 2/
=2mvcos2
)180(
=2mvsin 2/ .
Hence,”D”.
32) A block of mass m is lying on a horizontal surface of coefficient of friction . A force F is
applied to the block at an angle of with the horizontal. The block will move with a minimum
force F, if
A) tan
B) cot
C) sin
D) cos
Answer [A]
Hints:
tan , by solving horizontal and vertical equation., and equating
dy/dx=0.
33) The potential energy ( in joule) of a body of mass 2 kg moving in X-Y plane is given by
U=6x+8y, where the position coordinates x and y are measured in metres. If the body is at rest
at point (6m, 4m) at time t=0, then it will cross the y-axis at time tequal to:
A) 1s
B) 2s
C) 3s
D) 4s
Answer [B]
Hints:
Potential Energy (U) =6x+8y
Mass=2 kg
Body is at rest at (6,4) at t = 0
93
86
)//(/
4
ji
j
ji
onacceleratimassi
dydudxdudxduf
Particle will cross the Y-axis, when x=0. At x=6, v=0; distance=-6, U=o, a=-3
sec2
36
3
/6
,
222
t
t
tV
atUV
smV
Weget
asUV
ngSubstituti
34) A wind-powered generator converts wind energy into electrical energy. Assume that the
generator converts a fixed fraction of the wind energy intercepted by its blades into electrical
energy. For wind speed v, the electrical power output will be proportional to:
A) v
B) v2
C) v3
D) v4
Answer [C] Hints: Bernouli’s Theorem
35) n small balls, each of mass m, impinge elastically each second on a surface with velocity u.
The force experienced by the surface will be:
A) mnu
B) 2mnu
C) 4mnu
D) mnu2
1
Answer [B]
Hints:
Let there be “n” ball striking per second.
For 1 ball-
t
pF
mnvP
mvP
2
,2
=2mnv
Hence, (B).
36) A particle falls from a height h on a fixed horizontal surface and rebounds. If e is the
coefficient of restitution, then the total distance travelled by the particle before it stops
rebounding is:
A) h 2
2
1
1
e
e
B) h 2
2
1
1
e
e
C) h/2 2
2
1
1
e
e
D) h 2
2
1
1
e
e
Answer [A]
Hints:
Total Distance travelled
.,
)1
1(
.....)1(.2
...22
2
2
22
42
AHence
e
eh
heheh
heheh
37) When W joule of work is done by a flywheel, its frequency of rotation increases from v1Hz
to v2Hz. The moment of inertia of the flywheel about its axis of rotation is given by:
A) )(2 2
1
2
2
2 vv
W
B) )(2 2
1
2
2
2 vv
W
C) )(4 2
1
2
2
2 vv
W
D) )(2 2
1
2
2
2 vv
W
Answer [A]
Hints:
Work done
.,
)(2/
2/)(2
2,
2
1
2
2
2
2
1
2
2
2
22
AHence
VVWI
VV
Now
I
if
38) A disc is rotating with angular velocity w . A force F acts at a point whose position vector
with respect to the axis of rotation is r . The power associated with the torque due to the force is
given by:
A) wFr
).*(
B) wFr
*)*(
C) )*.( wFr
D) )*(* wFr
Answer [A]
Hints:
.,
).(
.
AHence
Fr
Power
39) An extremely small and dense neutron star of mass M and radius R is rotating with angular
velocity w. If an object is placed at its equator, then it will stuck to it due to gravity, if
A)M>G
Rw
B)M>G
wR 22
C)M>G
wR 23
D)M>G
wR 32
Answer [C]
Hints:
.,
/
0/
0
,,
23
22
2
2
CHence
GRM
RRGM
Rg
ifedstayattachObject
rgg eff
40) A body of mass m is dropped from a height nR above the surface of the earth (here Ris the
radius of the earth). The speed at which the body hits the surface is:
A)1
2
n
gR
B)1
2
n
gR
C)1
2
n
gRn
D)1
2
n
gRn
Answer [D]
Hints:
By Applying conservation of energy:
.,
1/2
/2/1)/(0 2
DHence
ngnRV
RGMmmvRnRGMm
UU fi
41) The volume of a wire remains unchanged when the wire is subjected to a certain tension.
Then Poisson’s ratio of the material of the wire is:
A) 0.25
B) 0.4
C) 0.5
D) 0.75
Answer [C]
Hints:
As the length and radius is changed, Volume remains same:
.,
'5.02/1
21)1(
)1()1(
)1()1(
))((
:,'
2
222
22
CHence
sratiopoisson
l
lr
r
r
r
l
l
r
r
l
l
r
rr
l
lllr
rrlllr
rrNewR
llNewLsaysLet
42) Equal masses of two substances of density 1 and
2 are mixed together. The density of the
mixture would be:
A) 212
1
B) 21
C)21
21
D)21
212
Answer [D]
Hints:
Let us say mass of each substance is”M”.
.,
/2
/
2121
21
DHence
MMMM
etotalvolum
totalmassmix
43) A vessel contains oil (density 0.8 g cm-3) over mercury (density 13.6 g cm-3). A
homogeneous sphere floats with half volume immersed in mercury and other half immersed in
oil. The density of the material of the sphere (in g cm-3) is:
A) 3.3
B) 6.4
C) 7.2
D) 12.8
Answer [C]
Hints:
F (Body) = mg.
Let the total volume be “V”.
.,
2.7
2/8.02/6.13
CHence
gVgVgV
body
body
44) A cylindrical tank having cross-sectional area A is filled with water to a height of 2m. A
circular hole of cross sectional area a is opened at a height of 75cm from the bottom. If a/A =
2.0 then the velocity with which the water emerges from the hole is:
A) 4.9 ms-1
B) 4.95 ms-1
C) 5.0 ms-1
D) 5.5 ms-1
Answer [D]
Hints:
Velocity is given by the formula
2)(1
2
A
a
gh
Substituting h=12.5 and a/A=0.2,
We get, V=5.5m/sec
Hence, D.
45) If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are
related through the expression 4v2=25-x2, then its time period is given by:
A)
B) 2
C) 4
D) 6
Answer [C]
Hints:
Comparing the equation
.""
4/2,2/1,
)(,)25(4/1 222222
iscorrectC
THence
xAVwithxV
46) A simple pendulum attatched to the ceiling of a stationary lift has a time period T. The
distance y covered by the lift moving upwards varies with time t as y=t2, where y is in metre
and t in second. If g=10ms-2, then the time period of the pendulum will be:
A) T5
4
B) T6
5
C) T4
5
D) T5
6
Answer [B]
Hints:
T= 2/2&/2 glgl , in case the lift is stationary and accelerating
upward respectively.
Since,
.2
2/
2
y
y
a
tdtdyV
ty
.,
6/5
10/12/
BHence
TT
TT
47) The extension in a string, obeying Hooke’s law is x. The speed of the wave in the stretched
string is v. If the extension of the string is increased to 1.5x, then the speed of the wave in the
string will be:
A) 1.22 v
B) 0.61 v
C) 1.50 v
D) 0.75 v
Answer [A]
Hints:
As we know that,
.,
.22.1
5.1/
5.1
,
,
AHence
VV
x
xVV
xV
xV
extensionxV
ExtensionT
TV
48) When the temperature of a rod increases from t to t+ t , the moment of inertia of the rod
increases from I to II . If the coefficient of linear expansion of the rod is , then the ratio
of I
Iis:
A)t
t
B)t
t2
C) t
D) 2 t
Answer [D]
Hints:
.,
2/
/
/2/
2
DHence
tII
tRR
RRII
rI
49) The ratio of adiabatic bulk modulus and isothermal bulk modulus of a gas is
v
p
C
C
A) 1
B)
C)1
D)
1
Answer [B]
Hints:
We know that:
Adiabatic bulk modulus= ϒρ
Isothermal bulk modulus=ρ
Ratio=ϒ
Hence, B
50) An ideal gas is expanded isothermally from volume v1to v2 and then compressed
adiabatically to original volume v1. If the initial pressure is P1, the final pressure P3, and net
work done is W then:
A) P3 >P1, W > 0
B) P3 <P1, W < 0
C) P3 >P1, W < 0
D) P3 =P1, W = 0
Answer [C]
Hints:
Work is negative, 13 PP
Option C is correct.
51) A vessel of volume V contains an ideal gas at temperature T and pressure P. The gas is
allowed to leak till its pressure falls to P’. Assuming that the temperature remains constant
during leakage, the number of moles of the gas that have leaked is:
A) )( PPRT
V
B) )(2
PPRT
V
C) )( PPRT
V
D) )(2
PPRT
V
Answer [C]
Hints:
From Ideal gas equation:
PV= nRT
Initially:
.,
/
:,.
),(
21
2
1
CHence
PPRTVnn
leakedofmolesNO
leakageAfterRTnVP
RTnPV
52) Two spheres A and B having radii 3 cm and 5 cm respectively are coated with carbon black
on their outer surfaces. The wavelengths of maximum intensity of emitted radiations are 300
nm and 500 nm respectively. If the powers radiated are QA and QB respectively then QA/QB is:
A)3
5
B)3
5
C)
2
3
5
D)
4
3
5
Answer: [C]
Hints:
Wein’s displacement law
Now,As we know that:
Power Ratio,
".",3/5/
:&,4,
///
2
2
424244
CHenceQQ
RTgSustitutinrAWhere
TrTreTAeTAQQ
BA
BBAAAAba BB
53) A metallic sphere of radius R is charged to a potential V. The magnitude of electric field at
a distance r (r>R) from the centre of the sphere is:
A)r
V
B)2R
Vr
C)2r
VR
D) 0
Answer [C]
Hints:
At a distance “r” from the Centre, where r>R:
3/5300/500//, maxmax
max
ABBA TTHence
bT
.,
/
:""
/
/
/&/
2
2
CHence
rVRE
inEQngSubstituti
KVRQ
RKQV
rKQERKQV
54) A particle of mass m and charge +q is midway between two charged particles each having
a charge +q, and at a distance 2L apart. The middle charge is displaced slightly along the line
joining the fixed charges and released. The time period of oscillation is proportional to:
A) L1/2
B) L
C) L3/2
D) L2
Answer [C]
Hints:
As the particle is placed mid-way between two charge +q. Force on the
central charge by both the charge particle will be equal and away from it.
If the central particle is displaced by “x”,then the net force acting on it will
be.
Fnet= F2-F1
.,
:,
/2
)..(,
/4
,
/4
:,
)/()/(
2/3
2
32
32
2222
CHence
LT
wegetByequating
T
MHforSxaNow
Lxkqa
maFnetNow
LxKqFnet
wegetSolving
xLKqxLKq
55) If n drops each of capacitance C and charged to a potential V, coalesce to form a big drop,
the ratio of the energy stored in the big drop to that in each small drop will be:
A) n:1
B) n4/3 : 1
C) n5/3 : 1
D) n2 : 1
Answer [C]
Hints:
Let the Charge on each drop is q.
C=q/V
Charge on final drop=Q= nq
Let radius of each small drop is “r” and big drop is “R”.
V=nv (Volume)
.,
1/2/1/'2/1/'
:
//'
)(
3/4.3/4
3/5
3/2
3/1
33
CHence
nqvQVuU
rgyRatioofene
rKqnRKQV
rnR
rnR
56) Two identical capacitors have the same capacitance C. One of them is charged to potential
v1and the other to v2. The negative ends of the capacitors are connected together. When the
positive ends are also connected the decrease in energy of the combined system is:
A) 2
2
2
14
1vvC
B) 2
2
2
14
1vvC
C) 22
2
2
14
1vvC
D) 22
2
2
14
1vvC
Answer [C]
Hints:
From charge conservation:
.,
)(4/1,
)(
2/
2
2
21
21
2121
1
CHence
VVCWeget
UUSystemU
VVV
CVCVCVCV
QQqq
fi
57) What is the total current supplied by the battery to the circuit shown in the figure below?
A) 1 A
B) 4 A
C) 6 A
D) 8 A
Answer [B]
Hints:
We know that:
:
6
eq
eqb
eqbb
gforRCalculatin
Ri
RiV
2 ohm is parallel to 6 ohm and in series with 1.5 ohm. The combination is
parallel to 3 ohm resistance.
R (Effective)= 3/2 Ω
I (Current in the circuit)=6/1.5=4 Ampere
Hence “B”.
58) An electric bulb has a rating of 500 W, 100 V. It is used in a circuit having a 200 V supply.
What resistance must be connected in series with the bulb so that it delivers 500 W?
A) 10 ohm
B) 20 ohm
C) 30 ohm
D) 40 ohm
Answer [B]
Hints:
Let us say resistance of bulb is “R”,
.,
20
200)(5
200)(
)(5
20500
:
20
/)100(500
/
,
2
2
2
2
BHence
R
RR
VVoltageSupply
currentcircuitAi
i
RiP
Now
R
R
RVP
ext
bulbext
59) A charged particle is released from rest in a region of steady and uniform electric and
magnetic fields which are parallel to each other. The particle will move in a
A) Straight line
B) Circle
C) Helix
D) Cycloid
Answer [A]
Hints:
As the charge particle is released from rest:
enet
b
b
e
benet
FF
FV
BVqF
EqF
FFF
BE
0,0
)(
Since Electric field is acting along the straight line, and force on particle
will be always in the direction of electric field. Hence the particle will show
straight-line motion.
Hence, A.
60) A particle of charge q moves with a velocity v =ai in a magnetic field B=bj +ck where a, b
and c are constants. The magnitude of the force experienced by the particle is:
A) 0
B) qa(b+c)
C) qa(b2-c2)1/2
D) qa(b2+c2)1/2
Answer [D]
Hints:
.,
)ˆˆ(
)ˆˆ(ˆ(
)(
22
DHence
cbqaF
jackabq
kcjbiaq
BVqF
B
B
61) An L-C-R series circuit with R =100 is connected to a 200 V, 50Hz a.c. source When
only the capacitance is removed, the voltage leads the current by 60 C and when only the
inductance is removed, the current leads the voltage by 60 C . The current in the circuit is:
A) 3
2A
B) 2
3A
C) 1 A
D) 2 A
Answer [D]
Hints:
Consider cases like, When capacitance is removed the circuit is L-R Circuit,
and when Inductor is removed the circuit is R-C Circuit, and calculate :
3100,3100.& clcl XXXX
Current in L-C-R circuit is
:
.,
2
2
DHence
Ai
XXRZ
XXRZ
ziV
cl
cl
62) When a glass prism of refracting angle 60 C is immersed in a liquid, its angle of minimum
deviation is 30 C . The critical angle of glass prism with respect to the liquid medium (in
degree)
A) 42
B) 45
C) 50
D) 52
Answer [B]
Hints:
For minimum deviation:
Angle of incidence(i)= angle of emergence(e)
Angle of refraction(r)= prism angle(A)/2
Now as we know that:
"."
45
)2/1(sin
2/1/
:,
)2/sin()2/sin(
)sin(sin
1
BHence
gleCriticalan
nn
valuesabovengSubstituti
AnAn
rnin
prismmedium
prismmmedium
prismmedium
63) A convex lens and a concave lens are placed in contact. The ratio of magnitude of the
power of the convex lens to that of the concave lens is 4 : 3. If the focal length of the convex
lens is 12 cm, then the focal length of the combination will be:
A) 16 cm
B) 24 cm
C) 32 cm
D) 48 cm
Answer [D]
Hints:
As we know that:
Power (P)=1/focal length(f)
Here 21 & ff are focal lengths of convex and concave lens respectively.
cmf
ff
pP
16123/4
3/4/
3/4/
2
12
21
Hence focal length of concave lens=16 cm
Focal length of Combination can be calculated by the formula:
21 /1/1/1 fff ncombinatio
Keeping f2 negative, as for concave lens focal length is negative.
We get focal length of combination= 48 cm.
Hence,D.
64) Monochromatic light is refracted from air into glass of refractive index . The ratio of the
wavelength of the incident and refracted waves is:
A) 1 : 1
B) 1 : µ
C) µ : 1
D) µ2 : 1
Answer [C]
Hints:
Speed of light in any medium is given by:
/c Where “c” is the speed of light in vacuum, and is the refractive
index of the medium.
Velocity of light in air = c
Velocity of light in glass =c/
We know that:
".",
/
)(,
/
)(/)()(
CHence
c
cVV
sameremainsfrequecy
f
V
f
V
frequencyfspeedvwavelength
ri
r
r
i
i
refractedincident
65) Two coherent sources of light of intensity ratio n are employed in an interference
experiment. The ratio of the intensities of the maxima and minima in the interference pattern is:
A)
1
1
n
n
B)
2
1
1
n
n
C)
1
1
n
n
D)
2
1
1
n
n
Answer [D]
Hints:
Given:
221min
2
21max )(
/
III
III
nII BA
After calculating their ratio, and substituting “n”, we get
2)1
1(
n
n
Hence,D.
66) The wavelength of K line from an element of atomic number 41 is . Then the
wavelength of K line of an element of atomic number 21 is:
A) 4 λ
B) λ/4
C) 3.08 λ
D) 0.26 λ
Answer [A]
Hints:
Apply Moseley’s law.
67) If a hydron atom at rest emits a photon of wavelength , then the recoil speed of the atom
of mass m is given by:
A) h/mλ
B) mh/λ
C) mhλ
D) none of these
Answer: (A), self explanatory.
68) A nucleus ruptures into two parts which have their velocities in the ratio of 2: 1. What will
be the ratio of their nuclear sizes (radii)?
A) 21/3 : 1
B) 1 : 21/3
C) 31/2 : 1
D) 1 : 31/2
Answer [B]
Hints:
CBA
momentum of B & C are equal.
)(2// givenVVmm
VmVm
bccb
ccbb
Hence there area will also be in the same ratio.
Calculating the ratio, we get:
BHence
RR bc
,
)2
1(/ 3/1
69) The half-life of a radioactive sample is 6.93 days. After how many days will only 20
1 of
the sample be left over? [Take loge (20) =3]
A) 20 days
B) 27 days
C) 30 days
D) 35 days
Answer [C]
Hints:
Use teNN
, Hence(c) is the correct option.
70) Three forces start acting simultaneously on a particle moving with velocity v. These forces
are represented in magnitude and direction by three sides of a triangle taken in the same order.
The particle will now move with a velocity:
A) less than ν
B) more than ν
C) ν only
D) cannot say
Answer [C]
Hints:
As the three forces are acting along the three sides of the triangle in the
same order, hence their net forces will cancel out each other and will be
zero, Hence acceleration=0, thus velocity will remain constant.
71) Two small satellites move in circular orbits around the earth respectively, at distances
r and (r + r) from the centre of the earth. If their time periods are T and T+ T
( r >>r, T>>T), then
A) r
rTT
2
3
B) r
rTT
2
3
C) r
rTT
3
2
D) r
rTT
Answer [A]
Hints:
According to kepler’s law:
32 RT
Then, take log on both sides and differentiate with respect to “x”.
r
rTT
2/3
Hence,(A).
72) If n1, n2 and n3 are the fundamental frequencies of three segments into which a string
is divided, then the original fundamental frequency n of the string is given by:
A) 321
1111
nnnn
B)
321
1111
nnnn
C) 321 nnnn
D) 321 nnnn
Answer [A]
Hints:
Let the length of three segments be
3,21 &, xxx Respectively.
).(,
/1/1/1/1,
2/2/2/2/
2/
2/
2/
:
2/
:
,
321
321
33
22
11
321
AThus
nnnnHence
nvnvnvnv
xvn
xvn
xvn
Similarly
xvn
hatNowweknowt
xxxx
73) If the electric flux entering and leaving an enclosed surface are 1 and 2 respectively,
then the electric charge inside the surface will be:
A) (ɸ2-ɸ1)/ε0
B) (ɸ1+ɸ2)/ε0
C) (ɸ1-ɸ2)/ε0
D) ε0(ɸ1+ɸ2)
Answer: (D)
74) What is the reading of ammeter shown in the figure below?
A) 3 A
B) 4 A
C) 1.5A
D) 6 A
Answer [A]
Hints:
Current through capacitor is zero in steady state. So current will not flow
through capacitor. 2Ω &2Ω Re in parallel, and their equivalent will be 1Ω.
Now, I=V/R=3/1=3 Ampere’
Hence(A).
75) An astronaut is looking down on earth’s surface from a space shuttle at an altitude of
500 km. Assuming that the astronaut’s pupil’s diameter is 5 mm and the wavelength of visible
light is 500 nm, the astronaut will be able to resolve linear objects of the size of about:
A) 0.5 m
B) 5 m
C) 50 m
D) 500 m
Answer [C]
Hints:
50 m is the correct answer.
76) The velocity of the most energetic electrons emitted from metallic surface is double when
the frequency v of the incident radiation is doubled. The work function of the metal is
A) 0
B) hν/3
C) hν/2
D) 2hν/3
Answer [D]
Hints:
We know that,
Kinetic energy= hmv2
2/1
Using this equation, we get:
.,
3/2
dHence
hd
77) Two identical photocathodes receive light of frequencies f1 and f2. If the velocities of the
photoelectrons (of mass m) coming out are v1 and v2 respectively, then
A) )(2
21
2
2
2
1 ffm
hvv
B)
2/1
2121 )(2
ff
m
hvv
C) )(2
21
2
2
2
1 ffm
hvv
D)
2/1
2121 )(2
ff
m
hvv
Answer [A]
Hints:
Same as above question, take reference.
78) If the time period of a simple pendulum is glT /2 , then the fractional error in
acceleration due to gravity is:
A) 2
24
T
l
B) T
T
l
l
2
C) T
T
l
l
2
D) none of these.
Answer [C]
Hints:
Given:
"."
/2//
)/(4
)/(4
:),(
/2
22
22
cHence
TTllgg
Tlg
glT
sidesbothSquaring
glT
79) A ball moving with a momentum 5 kg ms-1 strikes a wall. If the initial and final
momenta make equal angles of 45o, then the magnitude in change in momentum is
A) 5.05 kg ms-1
B) 7.07 kg ms-1
C) 10 kg ms-1
D) 0 kg ms-1
Answer [B]
Hints:
Change in momentum= if pp
As both makes 45 degree each, hence both will be perpendicular to each
other.
22 PPP
Hence “B” is the correct option.
80) A force Fy (3x + 2)N is acting on a body. The work done by this force if it tends to
displace the body from x= 0 m to x= 4 m will be
A) 32 J
B) 16 J
C) 0 J
D) (12x + 8) J
Answer [C]
Hints:
As here yF is the acting force .Therefore the force is acting along y-
direction. Thus no work will be done as displacement is along x-direction.
It’s a tricky Question.
“C” is the correct option.
81) A motorcyclist is moving along a circular path of radius 500 m. If his speed is 30 ms-1
then the type(s) of acceleration and the resultant acceleration will be
A) one and 18 ms-2
B) two and 2 ms-2
C) two and 2.7 ms-1
D) none of these
Answer [A,D]
Hints:
In circular motion, there are two types of acceleration.
Tangential and centripetal.
Tangential acceleration is responsible for change in speed, as here the
speed is constant. Hence tangential acceleration =0.
Centripetal acceleration is given by formula= Rv /2
By calculating we get “a”= 1.82sec
m
Here in option “A” there is a printing mistake, as its 1.8 not 18. So “A” and
“D” are correct.
82) Two heavy spheres of mass m are kept separated by a distance 2r. The gravitational
field and potential at the midpoint of the line joining the centres of the spheres are:
A) 2
2
r
Gm and 0
B) 0 and r
Gm2
C) 0 and r
Gm2
D) none of these
Answer [B]
Hints:
As we know that, gravitational forces are always attractive, and hence will
act in opposite direction, thus will be zero.
Hence gravitational field will not be present at the midpoint of the line
joining.
Vnet= -Gm/R + (-Gm/R)
=-2Gm/R.
Hence “B” is the correct answer.
83) The transverse displacement of a vibrating string is
)120cos(3
2sin06.0 txy
If mass per unit length of the string is 2x10-2 kg m-1, the tension in the string will be
A) 648 N
B) 864 N
C) 684 N
D) 468 N
Answer [A]
Hints:
Use kVTV /&/
84) A and B are two charged bodies with charges q1 and q2, and with a total charge q. If the
force between A and B has to be maximum, then the ratio of charge on A to the total charge
will be
A) 1
B) 0.25
C) 2
D) 0.5
Answer [B]
85) From the diagram given below, it can be calculated that the resistivity of the material of the
rod is
A) 3 x 10-6 Ω cm
B) 2.8 x 10-5 Ω m
C) 1.5 x 10-3 Ω cm
D) 12 x 10-4 Ω m-1
Answer [B]
Hints:
We know that:
V=iR
R=V/i
=40/3Ω
lRA
A
lR
/
By calculating, we get:
“B” as the correct option.
86) If an electron has to be accelerated by a cyclotron to a kinetic energy of 5 MeV with a
magnetic field of strength 2T, which one of the following will be correct?
A) Cyclotron frequency, f=5.6 x 1014 Hz
B) Speed, v= 1.32 x 109 ms-1
C) r can be calculated using r=(mv/qB)
D) None of the above
Answer [C]
Hints:
“C” is the correct option as we know that, r=mV/qB.
87) The inductance of two coils are 10 H and 20 H respectively. If the mutual inductance
between the coils is 5 H and the mutual fluxes aid, each other, then the total inductance of the
circuit will be:
A) 15 H
B) 25 H
C) 9.37 H
D) none of these
Answer [D]
Hints:
MLLLeq 221
Substituting the values, we get:
L (equivelant) = 20H, 40H
Hence ”D”.
88) In Young’s double-slit experiment, if the superimposing waves have amplitude a0 and
intensity I0, then the average intensity of the light in the fringe pattern formed on a screen will
be
A) 6 I0
B) 4 I0
C) 2 I0
D) I0 /2
Answer [C]
Hints:
As I is intensity of superimposing waves.
."",
2
2/
0
4
minmax
min
max
iscorrectCHence
I
IIIavg
I
II
89) Sodium light has two wavelengths and they are 5890 A and 5896 A . This light falls on a
slit of width 2 m. The diffraction pattern formed is observed on a screen. The spacing
between the first maxima for the two wavelengths is:
A) 9 x 10-4 m
B) 3 x 10-8 m
C) 6 x 10-4 m
D) 8 x 10-6 m
90) A transistor is connected in CE configuration. In the collector circuit VCC = 8 V, RC= 800
. If the voltage drop across RC is 0.5 V and =0.96, then IC, VCE and IB will be:
A) 0.625 mA, 7.5 V and 0.026 mA
B) 0.625 mA, 0.026 V and 7.5 mA
C) 1 mA, 8.5 V and 0.028 mA
D) 6.91 mA, 9.6 V and 0.29 mA
Answer [A]
Hints:
.,
026.0
651.0
96.0/
625.0
800/5.0
5.0
AHence
mA
III
mAI
II
III
mA
I
RI
CEB
E
Ec
EBc
c
c
91) The output (Y) of the combination of gates is:
A) Y= A . B
B) Y = A + B
C) Y= BA.
D) Y= BA
Answer [A]
92) From the diagram given below, it can be calculated that the modulation factor is:
A) 4
B) 2
C) 0.6
D) 8
Answer [C]
Hints:
As we know:
Modulation factor= Amplitude change of carrier wave/Amplitude of normal
carrier wave
Amplitude change of carrier wave= 8-5= 3
Amplitude of Normal Carrier wave=5
m-factor = 3/5=0.6
Hence, “C”.
93) If the velocity of surface wave (v) depends upon surface tension (T), coefficient of
Viscosity ( ) and density ( ), then the expression for v will be:
A) n
T
2
B) n
T
C) 2T
n
D) n
Answer [B]
Hints:
By Dimensional Analysis,
94) Two balls are thrown horizontally in opposite directions from the same point from a height
h with velocities 4 m/s and 3 m/s. The separation between the two balls when their velocities
are perpendicular will be
A) 6.5 m
B) 5.25 m
C) 2.45 m
D) none of these
Answer [C]
Hints:
".",
422.2
10/327
10/32
:,
,0.
ˆˆ3
ˆˆ4
2/1
CHence
m
tUS
t
WegetSolving
VV
jgtiV
jgtiV
BA
B
A
95) Three blocks of mass 3 kg, 2kg and 1 kg are placed side-by-side on a smooth surface as
shown in figure. If a horizontal force of 24 N is applied on 3 kg block, then the netforce on 2 kg
block will be
A) 2 N
B) 4 N
C) 8 N
D) 12 N
Answer [C]
Hints:
Net Acceleration= Net Force/Net Mass=24/6=4
Fnet on 2 kg block i= 2* 4 N
=8 N .
96) A block of mass 10 kg is moving in x-axis direction with constant speed of 10 m/sis
subjected to a retarding force F = -0.1 x J/m during its travel from x=20 m to x=30m. What will
be the final KE of the block?
A) 500 J
B) 475 J
C) 450 J
D) 400 J
Answer [B]
Hints:
As we know that:
W
joulek
k
kk
KWnet
N
xdx
dxFW
f
f
if
xnet
475
50025
25
25
1.0
30
20
97) A satellite is orbiting close to the earth above the equator with a period of rotation of 1.5
hours. 1f it is above a point x on the equator at some time, then after how much time will it be
above x again?
A) 1.5 hours if it is rotating from east to west
B) 1.6 hours if it is rotating from west to east
C) 1.6 hours if it is rotating from east to west
D) 1.8 hours if it is rotating from west to east
(B)
98) The speed of the wind passing over the wings of a small aeroplane is 70 m/s and below the
wing is 60 m/s. If the mass of the plane is 1000 kg and the area of the wing is 14 m2, then what
will be the net vertical force on the aeroplane?
A) 620 N upward
B) 920 N upward
C) 620 N downward
D) 920 N downward
Answer [B]
Hints:
Here pressure below aeroplane and above aeroplane will be termed as
ab PP & respectively.
Bernoulli’s equation states that:
"",
920
)()(
780
)(2/
2/12/1
22
22
BHence
Newton
weightAreaPPupwardFnet
UUPP
zUPZUP
ab
baab
aaabbb
99) A liquid is contained in a vertical tube of semi-circular cross-section as shown in figure. If
the angle of contact is zero, then the ratio of the force of surface tension on flat side to curved
side of the tube will be
A) 1/π
B) 3/2π
C) 1/4π
D) 2/π
Answer [D]
Hints:
Force on flat side/Force on curved side= rTrT /2 = /2
Hence,D.
100) A railway engine whistling at a constant frequency moves with a constant speed and it
goes past a stationary observer standing beside the railway track. Then the frequency of (n’) of
the sound heard by the observer with respect to time (t) can be best represented by which of the
following curve?
A)