STWT-5
The IITians Prashikshan Kendra
A Resonance of Brilliant Minds!
JEE MAINS Full Portion Test 1In-Program Test Series for GM-14
Program
Physics
Q.31With a sawtooth alternating current of peak value i0, the
r.m.s value is ___
(a) i0 (b)
2
0
i
(c)
2
0
i
(d)
3
0
i
Q.32Percentage error in the measurement of mass and speed are 2%
and 3% respectively. The error in the estimate of kinetic energy
obtained by measuring mass and speed will be ___
(a) 12 %(b) 10 %(c) 8 %(d) 2 %
Q.33The acceleration of a particle is increasing linearly with
time t as bt. The particle starts from the origin with an initial
velocity v0. The distance travelled by the particle in time t will
be
(a)
3
0
6
1
bt
t
v
+
(b)
3
0
3
1
bt
t
v
+
(c)
2
0
3
1
bt
t
v
+
(d)
2
0
2
1
bt
t
v
+
Q.34A pole driven into the bottom of a lake extends 3 meters
above the bottom of the lake in water and 1 meter above the surface
of water
.
3
4
=
w
a
m
If the sun is 30o above the horizon, then the length of the
shadow of the pole on the bottom of the lake is approximately
(Given sin 40.6( = 0.651 and tan 40.6( = 0.8571)___
(a) 1.73 m(b) 3.00 m(c) 2.73 m(d) 3.45 m
Q.35The Sri Lanka Broadcasting Corporation broadcasts its
programmes on the 25 m band. Which one of the following is the
frequency of broadcasting ?
(a) 12 Kilo Hz(b) 1.2 Mega Hz(c) 12 Mega Hz(d) 120 Mega Hz
Q.36A point source of sound emits waves equally in all
directions. The medium surrounding the source of sound is
non-absorbing. Two points A and B are situated at a distance of 10
m and 20 m, respectively from the source. The ratio of the
amplitude of the waves at A and B is___
(a) 1 : 2(b) 4 : 1 (c) 1 : 4 (d) 2 : 1
Q.37An inverted vessel (bell) lying at the bottom of a lake;
47.6 meter deep, has 50 cm3 of air trapped in it. If the bell is
brought to the surface of the lake, the volume of the trapped air
will now be (atmospheric pressure = 70 cm of Hg, density of Hg =
13.6 gm/cm3 and g = 980 cm/sec2)
(a) 350 cm3(b) 300 cm3(c) 250 cm3(d) 200 cm3
Q.38The magnetic potential at a point distant 10 cm from the
middle point of a magnetic dipole on a line inclined at an angle of
60o with the axis is 3 C.G.S. e.m.u. Then, the magnetic moment of
the magnet is____
(a) 30 ab-amp
cm2 (b) 150 ab-amp
cm2
(c) 300 ab-amp
cm2 (d) 600 ab-amp
cm2
Q.39A silver and a copper voltmeters are connected in parallel
to a 12 V battery of negligible resistance. In 30 min, 1.0 gram of
silver and 1.8 gram of copper are deposited. The rate at which the
energy being delivered by the battery is (given that E.C.E. of Ag =
11.2
104 g/joule and E.C.E. of Cu = 6.6
104 g/joule)
(a) 42.91 J/sec.(b) 24.12 J/sec(c) 21.94 J/sec(d) 14.29
J/sec.
Q.40Two metal spheres (radii a and b) are very far apart but are
connected by a thin wire. Their combined charge is q. Their
absolute potential is proportional to____
(a)
2
)
(
1
b
a
+
(b)
2
)
(
1
b
a
-
(c)
)
(
1
b
a
-
(d)
)
(
1
b
a
+
Q.41A glass tube of 1.0 m length is filled with water. The water
can be drained out slowly at the bottom of the tube. If a vibrating
tuning fork of frequency 500 c/sec is brought at the upper end of
the tube and the velocity of sound is 330 m/sec. Then, the total
number of resonances obtained will be___
(a) 4(b) 3(c) 2(d) 1
Q.42The Btu is defined as the heat needed to raise 1 lb of water
through 1F. Then the conversion factor between Btu and cal will
be___
(a) 1 Btu = 252 Cal(b) 1 Cal = 252 Btu(c) 1 Btu = 522 Cal(d)1
Cal = 522 Btu
Q.43A slightly conical wire of length L and end radii r1 and r2
is stretched by two forces F and F applied parallel to the length
in opposite directions and normal to the end faces. If Y denotes
the Youngs modulus then the extension produced is____
(a)
2
1
r
r
FLY
p
(b)
Y
r
FL
1
p
(c)
Y
r
r
FL
2
1
p
(d)
Y
r
FL
2
1
p
Q.44A circular plate of uniform thickness has a diameter of 56
cm. A circular portion of diameter 42 cm is removed from one edge
of the plate as shown in fig. The position of centre of mass of the
remaining portion will be at____
(a) 9 cm from the centre of the plate(b) 6 cm from the centre of
plate
(c) 11 cm from the centre of plate(d) None of these
Q.45In the situation shown in the figure, the system is in
equilibrium. All the strings, spring and pulley are light. Just
after cutting the string AB,
(a) the tension in the string GH is 6 g
(b) the tension in the string GH is zero
(c) the tension in string GH is
4
21
g
(d) the 2 kg block remains in equilibrium
Q.46A ball is projected upwards from the top of the tower with a
velocity 50 m/sec, making angle 30 with the horizontal. The height
of the tower is 70 m. After how many seconds from the instant of
throwing will the ball reach the ground ?
(a) 2 sec(b) 5 sec(c) 7 sec(d) 9 sec
Q.47A pair of electron and positron is produced by a
g
-ray photon. If the rest mass energy of an electron is 0.5 MeV
and the total kinetic energy of electron-position pair is 0.78 MeV,
then the energy of
g
-ray photon will be____
(a) 1.28 MeV(b) 0.78 MeV(c) 0.28 MeV(d) 1.78 MeV
Q.48An
a
-particle is fired towards a nucleus with a momentum P and the
distance of closest approach is r. If its momentum is increased to
4P, the distance of closest approach will be____
(a) 4 r(b)
2
r
(c) 16 r(d)
16
r
Q.49The electric field associated with a light wave is
)]
(
10
57
.
1
sin[
7
0
ct
x
E
E
-
=
where, x is in meter and t is in second. If this light is used
to produce photo-electric emission from the surface of a metal of
work function 1.9 eV, then the stopping potential will be_____
(a) 1.2 eV(b) 1.5 eV(c) 1.75 eV(d) 1.9 eV
Q.50In a hydrogen atom, the electron jumps from the state n to
(n1), where n >>> 1. The wavelength of the emitted
radiation is proportional to_____
(a) n0(b) n3(c) n2(d) None of these
Q.51Two circular coils can be arranged in any of three
situations shown in figure. Their mutual inductance will be___
(a) maximum in situation (i)(b) maximum in situation (ii)
(c) maximum in situation (iii)(d) the same in all situations
Q.52If two bulbs of wattage 25 and 100 respectively each rated
at 220 volt are connected in series with the supply of 440 volt,
which bulb will fuse?
(a) 25 watt bulb(b) 100 watt bulb
(c) both of them(d) none of these
Q.53A body A experiences perfectly elastic collision with a
stationary body B. If after collision, the bodies fly apart in
opposite direction with equal magnitude of velocities, then the
mass ratio of mA and mB is__
(a)
2
1
(b)
3
1
(c)
4
1
(d)
5
1
Q.54Three equal weights of mass 2 kg each are hanging on a
string passing over a fixed pulley as shown below. What is the
tension in the string connecting the weights B and C ?
(a) 3.3 N(b) 13 N(c) 19.6 N(d) zero
Q.55The ratio of the pressure exerted by electromagnetic waves
on reflecting surface and the absorbing surface will be
(a) 2 : 1(b) 1 : 2(c) 3 : 1(d) 1 : 3
Q.56Sun appears reddish at sunrise or sunset whereas it appears
white at noon. The reason is_____
(a) reflection
(b) diffraction
(c) that the sun is less hot at sunrise or sunset than at
noon
(d) scattering due to dust particles and air molecules
Q.57A convex lens (R1 = R2 = 30 cm) is made of a material having
refractive index 1.5. It is cut into two parts along AB. Now, one
of its parts is immersed in water of refractive index
,
3
4
then its focal lengths will be___
(a) 120 m(b) 60 m(c) 30 m(d) 240 m
Q.58The impedance of the circuit will be____
(a)
2
2
1
-
+
L
C
R
w
w
(b)
(
)
2
2
2
1
-
+
L
C
R
w
w
(c)
2
2
1
1
1
-
+
L
C
R
w
w
(d)
2
2
1
1
1
-
+
L
C
R
w
w
Q.59When a gas in a closed vessel is heated through 1oC, its
pressure increases by 0.4%. The initial temperature of the gas
was____
(a) 250 K(b) 2500 K(c) 250oC(d) 25oC
Q.60Two parallel plate air filled capacitors each of capacitance
C are joined in series to a source of constant e.m.f. V. The space
between the plates of one of the capacitors is then completely
filled up with a uniform dielectric having dielectric constant K.
The amount of charge which flows through the battery is_____
(a)
+
-
1
1
2
K
K
CV
(b)
-
+
1
1
2
K
K
CV
(c)
+
-
1
1
2
K
K
CV
(d)
-
+
1
1
2
K
K
CV
Answerkey and Detail Solution for
JEE MAINS Full Portion Test - 1
In-Program Test Series for GM-2014 Program
Physics
31
D
32
C
33
A
34
D
35
C
36
D
37
B
38
D
39
B
40
D
41
B
42
A
43
C
44
A
45
C
46
C
47
D
48
D
49
A
50
D
51
A
52
A
53
B
54
B
55
A
56
D
57
D
58
C
59
A
60
A
Physics
Sol.31(d) We know that,
=
T
T
dt
dt
i
i
0
0
2
2
But, for a saw tooth curve,
=
T
t
i
i
0
\
[
]
T
T
T
T
t
t
T
i
dt
dt
T
t
i
i
0
0
3
2
2
0
0
0
2
2
2
0
2
]
[
3
=
=
3
2
0
2
i
i
=
\
3
0
2
i
i
i
rms
=
=
Sol.32(c) We know that, Kinetic energy
2
2
1
mv
E
K
=
\
Fractional error in kinetic energy
v
v
m
m
E
E
K
K
D
+
D
=
D
2
\
% error in kinetic energy =
%
100
2
%
100
D
+
D
v
v
m
m
=
%
8
6
2
3
2
2
=
+
=
+
Sol.33(a) Given that, the acceleration of a particle is
increasing linearly with time t as bt
i.e.,a = bt
but
dt
dv
a
=
= bt
On integrating, we get
C
bt
v
+
=
2
2
(1)
where C is a constant of integration
initially when t = 0, v = v0
\
from (1), we get
0
0
0
v
C
C
v
=
+
=
\
(1) becomes
0
2
2
v
bt
v
+
=
But
dt
dx
v
=
\
0
2
2
v
bt
dt
dx
+
=
Integrating again,
D
t
v
t
b
x
+
+
=
0
3
3
2
(2)
Where D is a constant of integration
Given that at t = 0, x = 0
\
(2)
0 = 0 + 0 + D
D = 0
\
(2) becomes
t
v
bt
x
0
3
6
+
=
or
6
3
0
bt
t
v
x
+
=
Sol.34(d) The figure of the problem is shown. EC will be the
shadow of the pole AC.
EC = EF + FC =
1
2
x
x
+
Now in
D
ADB, we have
3
60
tan
1
1
=
=
x
\
732
.
1
3
1
=
=
x
But
33
.
1
sin
60
sin
sin
sin
=
=
=
r
r
i
m
or sin r = 0.651 r = 40.6o
Also from
D
DEF, we have
8571
.
0
6
.
40
tan
tan
2
2
=
=
=
r
x
(given)
\
714
.
1
7142
.
1
8571
.
0
2
2
=
=
=
x
m
\
Shadow EC =
45
.
3
446
.
3
714
.
1
732
.
1
2
1
=
=
+
=
+
x
x
m
Sol.35 (c) We know that,
l
n
v
=
Given that,
8
10
3
=
v
m/sec.,
25
=
l
m
\
6
8
10
12
25
10
3
=
=
=
l
v
n
Hz
orn = 12 MHz
Sol.36(d) We know that,
2
1
d
I
But
I
(amplitude)2
\
amplitude
d
1
or
2
1
20
10
=
=
=
B
A
A
B
d
d
a
a
\
1
:
2
=
B
A
a
a
Sol.37(b) At the bottom of the lake the total pressure
P1 = Pressure due to water 47.6 deep + Pressure due to
atmosphere
Also the height of water equivalent to atmospheric pressure is
given by
g
h
g
h
m
m
w
w
.
.
.
.
r
r
=
1000
1000
6
.
13
70
.
0
.
=
=
w
m
m
w
h
h
r
r
= 9.52 m
\
12
.
57
52
.
9
6
.
47
1
=
+
=
P
m and
50
1
=
V
cm3
At the surface of the lake
=
2
P
atmospheric pressure = 9.52 m and V2 = ?
\
From Boyles law, we have
2
2
1
1
V
P
V
P
=
or
300
50
6
52
.
9
50
12
.
57
2
1
1
2
=
=
=
=
P
V
P
V
cm3
Sol.38(d) We know that, magnetic potential at a point due to a
dipole is
2
0
cos
4
r
M
V
q
p
m
=
But in C.G.S. system,
1
4
0
=
p
m
\
q
q
cos
cos
2
2
Vr
M
r
M
V
=
=
Given that,V = 3, r = 10 and
=
60
q
\
600
2
1
100
3
=
=
M
ab-amp
cm2
Sol.39(b) We know that, m = Zit or
Zt
m
i
=
\
For silver voltameter
A
i
49
.
0
60
30
10
2
.
11
1
4
1
=
=
-
and for copper voltameter
A
i
52
.
1
60
30
10
6
.
6
8
.
1
4
2
=
=
-
\
the total current
A
i
i
i
01
.
2
52
.
1
49
.
0
2
1
=
+
=
+
=
( Rate of energy delivered
12
.
24
01
.
2
12
=
=
=
VI
J/sec
Sol.40(d) Since both the spheres are connected by a thin wire,
so the potential of each sphere will be equal. Let V1 be the
potential of first sphere and V2 be of the second.
\
a
q
V
1
0
1
.
4
1
=
p
and
b
q
V
2
0
2
.
4
1
=
p
where q1 and q2 are the charges on the sphere.
\
Total charge will be
2
1
q
q
q
+
=
If
,
2
1
V
V
V
=
=
then
a
q
1
0
.
4
1
p
=
b
q
2
0
.
4
1
p
b
q
a
q
2
1
=
or
b
q
q
a
q
1
1
-
=
or
+
=
+
=
ab
b
a
q
b
q
a
q
b
q
1
1
1
or
)
(
.
1
b
a
a
q
q
+
=
\
)
(
1
)
(
.
4
1
0
b
a
V
b
a
q
V
+
+
=
p
Sol.41(b) We know that,
500
330
=
=
=
n
v
n
v
l
l
m = 66 cm
\
First resonance will occur at
5
.
16
4
1
=
=
l
l
cm
\
Second resonance will occur at
5
.
49
4
3
2
=
=
l
l
cm
\
Third resonance will occur at
5
.
82
4
5
3
=
=
l
l
cm
\
Fourth resonance will occur at
5
.
115
4
7
4
=
=
l
l
cm
Clearly, fourth resonance will not occur because the lengths of
the tube is only 1.0 m
or 100 cm.
Sol.42(a) We have,
=
=
C
9
5
gm)
(453.6
Btu
1
1F
lb
1
Btu
1
C
1
gm
1
cal
1
1 Btu = 453.6
cal
9
5
= 252 cal.
Sol.43(c) Given that, the wire is conical
\
The average area of cross-section of the wire may be taken
as
2
1
2
2
2
1
2
1
r
r
r
r
A
A
A
p
p
p
=
=
=
Also
L
L
r
r
F
L
L
A
F
Y
/
/
/
/
strain
stress
2
1
D
=
D
=
=
p
or
Y
r
r
FL
L
2
1
p
=
D
Sol.44(a) Let C1 be the centre of mass of the whole plate, C2 be
the centre of mass of the removed plate and C be the centre of mass
of the remaining portion.
Let weight of unit area is = W
\
Area of the whole plate =
p
p
784
2
56
2
=
cm2
\
Weight of the whole plate = 784
W
p
Area of the removed plate =
p
p
441
2
42
2
=
cm2
\
Weight of the removed plate = 441
W
p
\
Weight of the remaining portion =
W
W
W
p
p
p
343
441
784
=
-
Now, suppose C1 is a fixed point.
\
C.M. of the remaining portion from fixed point will be
C1C =
plate
remaining
of
wt.
point)
fixed
from
distance
plate
removed
of
weight
-
point
fixed
from
C.M.
of
distance
plate
whole
of
Wt..
W
W
W
p
p
p
343
7
441
0
784
-
=
00
.
9
-
=
cm i.e., 9 cm left from the fixed point.
Sol.45(c) In equilibrium, just before the string AB in cut the
tension in spring is 5g.
Just after the string AB is cut the spring force wont change but
all blocks except of
5 kg start accelerating.
Rough free body diagram is shown in figure.
mass
Total
force
pulling
Net
=
a
=
4
3
4
2
5
g
g
g
g
g
=
-
-
+
For 1 kg block (connected to spring)
Free body diagram would be as shown
4
3
1
5
g
T
g
g
=
-
+
4
21
4
3
6
g
g
g
T
=
-
=
Sol.46(c)
2
10
2
1
30
sin
50
t
t
h
-
=
2
2
10
2
1
50
70
t
t
-
=
-
2
5
14
t
t
-
=
-
0
14
5
2
=
-
-
t
t
(
2
56
25
5
+
=
t
7
2
9
5
=
=
sec
Sol.47(d) Given that,
MeV
E
MeV
c
m
K
78
.
0
,
5
.
0
2
0
=
=
We know that,
2
0
2
c
m
E
E
E
+
+
=
+
-
b
b
g
78
.
1
5
.
0
2
78
.
0
=
+
=
MeV
Sol.48(d) We have,
m
p
mv
r
q
q
K
2
2
1
2
2
2
1
=
=
Clearly,
2
1
p
r
\
16
'
16
1
)
4
(
'
'
2
2
2
r
r
p
p
p
p
r
r
=
=
=
=
Sol.49(a) Given that,
)]
(
10
57
.
1
sin[
7
0
ct
x
E
E
-
=
We know that,
-
=
)
(
2
sin
0
ct
x
E
E
l
p
Comparing, we get
7
10
57
.
1
2
=
l
p
7
10
57
.
1
2
=
p
l
m
\
Energy of photon =
19
7
8
34
10
6
.
1
10
57
.
1
2
)
10
3
)(
10
64
.
6
(
-
-
=
p
l
hc
eV = 3.1 eV
So, maximum K.E. of photo-electron will be
= 3.1 1.9 = 1.2 eV
Sol.50(d) We know that hydrogen atoms
,
1
1
1
2
2
2
1
-
=
n
n
R
l
where
2
1
n
n
>> 1
\
3
n
l
Sol.51(a) We know that, mutual induction depends on flux linked.
In situation (i), the flux linked will be maximum.
Sol.52(a) Resistance of 25 watt bulb R1 =
W
=
=
1936
25
)
220
(
2
2
P
V
Resistance of 100 watt bulb R2 =
W
=
=
484
100
)
220
(
2
2
P
V
When connected in series, their series resistance will be
W
=
+
=
+
=
2420
484
1936
2
1
R
R
R
\
Current through the series combination
11
2
2420
440
resistance
total
e.m.f.
total
=
=
=
i
\
Potential difference across 25 watt bulb
352
1936
11
2
1
1
=
=
=
iR
V
volt
Potential difference across 100 watt bulb
88
484
11
2
2
2
=
=
=
iR
V
volt
Clearly, the bulb of 25 watt will be fused because it can
tolerate only 220V while the voltage across it is 352 volt.
Sol.53(b) We know that, when a body A collides elastically with
a stationary body B, then the final velocities of the bodies
are
A
B
A
B
A
u
m
m
m
m
v
+
-
=
1
and
A
B
A
A
u
m
m
m
v
+
=
2
2
where
=
A
u
velocity of body A.
Given that
2
1
v
v
=
-
\
A
B
A
A
A
B
A
B
A
u
m
m
m
u
m
m
m
m
+
=
+
-
-
2
or
(
)
A
B
A
m
m
m
2
=
-
-
3
:
1
:
=
B
A
m
m
Sol.54(b) Acceleration of
2
/
3
10
10
2
2
2
2
2
2
s
m
c
=
+
+
-
+
=
For body c,
=
-
3
10
2
10
2
T
(
13
3
20
2
3
20
20
@
=
-
=
T
N
Sol.55(a) We know that, average radiation pressure exerted by
electromagnetic waves on reflecting surface is
2
0
3
2
E
P
r
=
... (i)
and average radiation pressure in absorbing surface is
3
2
0
E
P
a
=
... (ii)
From (i) and (ii), we get
1
:
2
=
a
r
P
P
Sol.56(d) We know that, sun rays are obliquely incident on earth
at rise or set and have to travel greater distance. All shorter
wavelengths are scattered according
4
1
l
I
. Only red-orange rays reach us. During noon sun rays normally
incident on earth. This time(both longer and shorter) all
wavelengths reach us and hence it appears white.
Sol.57(d) We know that, for a lens
-
-
=
2
1
1
1
)
1
(
1
R
R
f
m
= (1.5 1)
R
R
R
1
1
1
=
-
-
\
f = R = 30 cm
When the lens cut across AB, new focal length will be
\
R
R
f
2
1
1
1
)
1
5
.
1
(
'
1
=
-
-
=
60
2
'
=
=
f
f
cm
Now
240
60
1
3
/
4
5
.
1
1
5
.
1
.
1
)
1
(
=
-
-
-
-
=
air
w
a
g
a
g
a
med
f
f
m
m
m
cm
Sol.58(c) We have,
C
j
L
j
R
Z
w
w
1
1
1
1
1
+
+
=
L
j
C
j
R
w
w
-
+
=
1
)
1
(
-
=
j
Q
-
+
=
L
C
j
R
w
w
1
1
or
-
+
=
L
C
j
R
Z
w
w
1
1
1
\
2
2
1
1
1
|
|
-
+
=
L
C
R
Z
w
w
Sol.59(a) We know that,
T
P
\
T
T
P
P
D
=
D
or
K
T
T
250
100
4
.
0
1
=
=
Sol.60(a) Before, filling the dielectric. The equivalent
capacitance is
2
1
C
C
C
C
C
C
=
+
=
and
V
C
q
2
1
=
When dielectric is completely filled, the equivalent capacitance
is
K
KC
KC
C
KC
C
C
+
=
+
=
1
)
(
)
(
2
and
V
K
KC
q
+
=
1
2
\
Charge flown =
2
1
1
2
CV
V
K
KC
q
q
-
+
=
-
=
+
-
1
1
2
K
K
CV
A
~
C
L
Water
2 m
k
5 kg
x1
R
C2
C1
60o
C
D
A
C
B
A
( iii )
1kg
x1
x2
D
56cm cm
42cm cm
C
C
F
G
A
B
2kg
C
E
D
B
A
B
1kg
BV
E
F
H
u1
m2
m1
Q
1kg
E = V
q2
q2
C
C
BV
A
T
5g 1g
1g
E = V
5g 1g
2g
a
a
q1
q1
F
G
C
C
( ii )
( i )
60o
r
1 m
Air
Page 1 of 30
The IITians Prashikshan Kendra
A Resonance of Brilliant Minds! Page 15 of 15
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