JEE Main 2018 Solved Paper PDF Download

 Page 1


CODE-C
1
61. The angular width of the central maximum in
a single slit diffraction pattern is 60°. The width
of the slit is 1 mm. The slit is illuminated by
monochromatic plane waves. If another slit of
same width is made near it, Young’s fringes can
be observed on a screen placed at a distance
50 cm from the slits. If the observed fringe
width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)
(1) 50 mm (2) 75 mm
(3) 100 mm (4) 25 mm
Ans. (4)
Sol. In diffraction
d sin 30º = l
60º
d
d
2
l=
Young's fringe width
[d' – separation between two slits]
l´
b=
D
d'
2
2
50 10 d
10
2 d'
-
-
´
=´
10
–2
 = 
62
10 50 10
2 d'
--
´´
´
d' 25m =m
62. An electron from various excited states of
hydrogen atom emit radiation to come to the ground
state. Let l
n
, l
g
 be the de Broglie wavelength of
the electron in the n
th
 state and the ground state
respectively. Let 
n
L be the wavelength of the
emitted photon in the transition from the n
th
 state
to the ground state. For large n, (A, B are
constants)
(1) 
nn
AB L » +l (2) 
22
nn
AB L » +l
(3) 
2
n
L »l (4) n 2
n
B
A L »+
l
PART C – PHYSICS
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
Ans. (4)
Sol.
l==
n
n
hh
mu 2mk
Þ k
n
 =  
l
2
2
n
h
2m
 ;  k
g
 = 
l
2
2
g
h
2m
Þ k
g
 – k
n
 = 
éù
-
êú
ll
êú
ëû
2
22
gn
h 11
2m
E
n
 = – k
n
for emitted photon
=-= -
L
n g gn
n
hc
EE KK
-
=
L
gn
n
KK
1
hc
L=
-
n
gn
hc
KK
 Þ 
L=
éù
-
êú
ll
êú
ëû
n
2
22
gn
hc
h 11
2m
L=
æö l -l
ç÷
ç÷
ll
èø
n
22
ng
22
gn
2mc
h
( )
ll
L=
l -l
22
gn
n
22
ng
2mc
h
as l
n
 µ n
l
n
 >> l
g
-
éù
ll æö
êú L=-
ç÷
l
êú èø
ëû
1
2
2
gg
n
n
2mc
1
h
éù
lll æö
êú L= ++
ç÷
ll
êú èø
ëû
2
2
ggg
n
nn
2mc
1 higher powersof
h
Page 2


CODE-C
1
61. The angular width of the central maximum in
a single slit diffraction pattern is 60°. The width
of the slit is 1 mm. The slit is illuminated by
monochromatic plane waves. If another slit of
same width is made near it, Young’s fringes can
be observed on a screen placed at a distance
50 cm from the slits. If the observed fringe
width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)
(1) 50 mm (2) 75 mm
(3) 100 mm (4) 25 mm
Ans. (4)
Sol. In diffraction
d sin 30º = l
60º
d
d
2
l=
Young's fringe width
[d' – separation between two slits]
l´
b=
D
d'
2
2
50 10 d
10
2 d'
-
-
´
=´
10
–2
 = 
62
10 50 10
2 d'
--
´´
´
d' 25m =m
62. An electron from various excited states of
hydrogen atom emit radiation to come to the ground
state. Let l
n
, l
g
 be the de Broglie wavelength of
the electron in the n
th
 state and the ground state
respectively. Let 
n
L be the wavelength of the
emitted photon in the transition from the n
th
 state
to the ground state. For large n, (A, B are
constants)
(1) 
nn
AB L » +l (2) 
22
nn
AB L » +l
(3) 
2
n
L »l (4) n 2
n
B
A L »+
l
PART C – PHYSICS
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
Ans. (4)
Sol.
l==
n
n
hh
mu 2mk
Þ k
n
 =  
l
2
2
n
h
2m
 ;  k
g
 = 
l
2
2
g
h
2m
Þ k
g
 – k
n
 = 
éù
-
êú
ll
êú
ëû
2
22
gn
h 11
2m
E
n
 = – k
n
for emitted photon
=-= -
L
n g gn
n
hc
EE KK
-
=
L
gn
n
KK
1
hc
L=
-
n
gn
hc
KK
 Þ 
L=
éù
-
êú
ll
êú
ëû
n
2
22
gn
hc
h 11
2m
L=
æö l -l
ç÷
ç÷
ll
èø
n
22
ng
22
gn
2mc
h
( )
ll
L=
l -l
22
gn
n
22
ng
2mc
h
as l
n
 µ n
l
n
 >> l
g
-
éù
ll æö
êú L=-
ç÷
l
êú èø
ëû
1
2
2
gg
n
n
2mc
1
h
éù
lll æö
êú L= ++
ç÷
ll
êú èø
ëû
2
2
ggg
n
nn
2mc
1 higher powersof
h
JEE(MAIN)-2018
2
L »+
l
n 2
n
B
A
where A = 
l
2
g
2mc
h
& B = 
l
4
g
2mc
h
63. The reading of the ammeter for a silicon diode in
the given circuit is :-
3V
200W
(1) 15 mA (2) 11.5 mA
(3) 13.5 mA (4) 0
Ans. (2)
Sol.
3V
200W
Silicon diode is in forward bias.
Hence across diode potential barrier
DV = 0.7 volts
I = 
-D-
=
V V 3 0.7
R 200
  = 
2.3
200
 = 11.5 mA
64. The density of a material in the shape of a cube
is determined by measuring three sides of the cube
and its mass. If the relative errors in measuring
the mass and length are respectively 1.5% and
1%, the maximum error in determining the density
is :-
(1) 3.5 % (2) 4.5 %
(3) 6 % (4) 2.5 %
Ans. (2)
Sol. Density = 
Mass
Volume
1d 1M 3 L
d ML
D DD
=+
= 1.5 + 3(1)
= 4.5 %
65. An electron, a proton and an alpha particle
having the same kinetic energy are moving in
circular orbits of radii r
e
, r
p
, r
a
 respectively in
a uniform magnetic field B. The relation
between r
e
, r
p
, r
a
 is:-
(1) r
e
 < r
p
 = r
a
(2) r
e
 < r
p
 < r
a
(3) r
e
 < r
a
 < r
p
(4) r
e
 > r
p
 = r
a
Ans. (1)
Sol. Radius of circular path in magnetic field is given
by  =
2Km
R
qB
where K = kinetic energy of particle
m = mass of particle
q = charge on particle
B = magnetic field intensity
R = radius of path
For electron
e
e
2Km
r
eB
= ...(i)
For proton
p
p
2Km
r
eB
= ...(ii)
For a particle
pp
2K 4m 2K m
2Km
r
q B 2eB eB
a
a
a
= ==
...(iii)
as m
e
 < m
p
   so    r
e
 < r
p
 = r
a
66. Three concentric metal shells A, B and C of
respective radii a, b and c (a < b < c) have surface
charge densities +s, –s and +s respectively. The
potential of shell B is :-
Page 3


CODE-C
1
61. The angular width of the central maximum in
a single slit diffraction pattern is 60°. The width
of the slit is 1 mm. The slit is illuminated by
monochromatic plane waves. If another slit of
same width is made near it, Young’s fringes can
be observed on a screen placed at a distance
50 cm from the slits. If the observed fringe
width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)
(1) 50 mm (2) 75 mm
(3) 100 mm (4) 25 mm
Ans. (4)
Sol. In diffraction
d sin 30º = l
60º
d
d
2
l=
Young's fringe width
[d' – separation between two slits]
l´
b=
D
d'
2
2
50 10 d
10
2 d'
-
-
´
=´
10
–2
 = 
62
10 50 10
2 d'
--
´´
´
d' 25m =m
62. An electron from various excited states of
hydrogen atom emit radiation to come to the ground
state. Let l
n
, l
g
 be the de Broglie wavelength of
the electron in the n
th
 state and the ground state
respectively. Let 
n
L be the wavelength of the
emitted photon in the transition from the n
th
 state
to the ground state. For large n, (A, B are
constants)
(1) 
nn
AB L » +l (2) 
22
nn
AB L » +l
(3) 
2
n
L »l (4) n 2
n
B
A L »+
l
PART C – PHYSICS
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
Ans. (4)
Sol.
l==
n
n
hh
mu 2mk
Þ k
n
 =  
l
2
2
n
h
2m
 ;  k
g
 = 
l
2
2
g
h
2m
Þ k
g
 – k
n
 = 
éù
-
êú
ll
êú
ëû
2
22
gn
h 11
2m
E
n
 = – k
n
for emitted photon
=-= -
L
n g gn
n
hc
EE KK
-
=
L
gn
n
KK
1
hc
L=
-
n
gn
hc
KK
 Þ 
L=
éù
-
êú
ll
êú
ëû
n
2
22
gn
hc
h 11
2m
L=
æö l -l
ç÷
ç÷
ll
èø
n
22
ng
22
gn
2mc
h
( )
ll
L=
l -l
22
gn
n
22
ng
2mc
h
as l
n
 µ n
l
n
 >> l
g
-
éù
ll æö
êú L=-
ç÷
l
êú èø
ëû
1
2
2
gg
n
n
2mc
1
h
éù
lll æö
êú L= ++
ç÷
ll
êú èø
ëû
2
2
ggg
n
nn
2mc
1 higher powersof
h
JEE(MAIN)-2018
2
L »+
l
n 2
n
B
A
where A = 
l
2
g
2mc
h
& B = 
l
4
g
2mc
h
63. The reading of the ammeter for a silicon diode in
the given circuit is :-
3V
200W
(1) 15 mA (2) 11.5 mA
(3) 13.5 mA (4) 0
Ans. (2)
Sol.
3V
200W
Silicon diode is in forward bias.
Hence across diode potential barrier
DV = 0.7 volts
I = 
-D-
=
V V 3 0.7
R 200
  = 
2.3
200
 = 11.5 mA
64. The density of a material in the shape of a cube
is determined by measuring three sides of the cube
and its mass. If the relative errors in measuring
the mass and length are respectively 1.5% and
1%, the maximum error in determining the density
is :-
(1) 3.5 % (2) 4.5 %
(3) 6 % (4) 2.5 %
Ans. (2)
Sol. Density = 
Mass
Volume
1d 1M 3 L
d ML
D DD
=+
= 1.5 + 3(1)
= 4.5 %
65. An electron, a proton and an alpha particle
having the same kinetic energy are moving in
circular orbits of radii r
e
, r
p
, r
a
 respectively in
a uniform magnetic field B. The relation
between r
e
, r
p
, r
a
 is:-
(1) r
e
 < r
p
 = r
a
(2) r
e
 < r
p
 < r
a
(3) r
e
 < r
a
 < r
p
(4) r
e
 > r
p
 = r
a
Ans. (1)
Sol. Radius of circular path in magnetic field is given
by  =
2Km
R
qB
where K = kinetic energy of particle
m = mass of particle
q = charge on particle
B = magnetic field intensity
R = radius of path
For electron
e
e
2Km
r
eB
= ...(i)
For proton
p
p
2Km
r
eB
= ...(ii)
For a particle
pp
2K 4m 2K m
2Km
r
q B 2eB eB
a
a
a
= ==
...(iii)
as m
e
 < m
p
   so    r
e
 < r
p
 = r
a
66. Three concentric metal shells A, B and C of
respective radii a, b and c (a < b < c) have surface
charge densities +s, –s and +s respectively. The
potential of shell B is :-
CODE-C
3
(1) 
22
0
ab
c
b
éù s-
+
êú
e
ëû
(2) 
22
0
bc
a
b
éù s-
+
êú
e
ëû
(3) 
22
0
bc
a
c
éù s-
+
êú
e
ëû
(4) 
22
0
ab
c
a
éù s-
+
êú
e
ëû
Ans. (1)
Sol.
s
–s
s
A
B
C
V
outside
 = 
Q K
r
where r is distance of point from the centre of
shell V
inside 
 = 
Q K
R
where 'R' is radius of the shell
C AB
B
b bc
Kq Kq Kq
V
rrr
= ++
2 22
B
0
1 4a 4b 4c
V
4 bbc
éù sp sp s p
= -+
êú
pÎ
ëû
22
B
0
ab
Vc
b
éù s-
=+
êú
Î
ëû
67. Two masses m
1
 = 5kg and m
2
 = 10kg,
connected by an inextensible string over a
frictionless pulley, are moving as shown in the
figure. The coefficient of friction of horizontal
surface is 0.15. The minimum weight m that
should be put on top of m
2
 to stop the motion
is :-
m
m
2
m
1
mg
1
T
T
(1) 27.3 kg (2) 43.3 kg
(3) 10.3 kg (4) 18.3 kg
Ans. (1)
Sol.
m
m
2
m
1
mg
1
= 50N
T
T
µ(m+m )g
2
(m+m )g
2
N
50 – T  = 5 × a
T  – 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
 50 = 0.15 (m + 10) 10
5 = 
3
20
(m + 10)
100
3
 = m + 10
m = 23.3 kg
68. A particle is moving in a circular path of radius
a under the action of an attractive potential
U = –
2
k
2r
. Its total energy is :-
(1)
2
k
2a
(2) Zero
(3) 
2
3k
2a
-
(4) 
2
k
4a
-
Ans. (2)
Sol. F = 
¶
-=
¶
3
uK
rr
Since it is performing circular motion
Page 4


CODE-C
1
61. The angular width of the central maximum in
a single slit diffraction pattern is 60°. The width
of the slit is 1 mm. The slit is illuminated by
monochromatic plane waves. If another slit of
same width is made near it, Young’s fringes can
be observed on a screen placed at a distance
50 cm from the slits. If the observed fringe
width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)
(1) 50 mm (2) 75 mm
(3) 100 mm (4) 25 mm
Ans. (4)
Sol. In diffraction
d sin 30º = l
60º
d
d
2
l=
Young's fringe width
[d' – separation between two slits]
l´
b=
D
d'
2
2
50 10 d
10
2 d'
-
-
´
=´
10
–2
 = 
62
10 50 10
2 d'
--
´´
´
d' 25m =m
62. An electron from various excited states of
hydrogen atom emit radiation to come to the ground
state. Let l
n
, l
g
 be the de Broglie wavelength of
the electron in the n
th
 state and the ground state
respectively. Let 
n
L be the wavelength of the
emitted photon in the transition from the n
th
 state
to the ground state. For large n, (A, B are
constants)
(1) 
nn
AB L » +l (2) 
22
nn
AB L » +l
(3) 
2
n
L »l (4) n 2
n
B
A L »+
l
PART C – PHYSICS
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
Ans. (4)
Sol.
l==
n
n
hh
mu 2mk
Þ k
n
 =  
l
2
2
n
h
2m
 ;  k
g
 = 
l
2
2
g
h
2m
Þ k
g
 – k
n
 = 
éù
-
êú
ll
êú
ëû
2
22
gn
h 11
2m
E
n
 = – k
n
for emitted photon
=-= -
L
n g gn
n
hc
EE KK
-
=
L
gn
n
KK
1
hc
L=
-
n
gn
hc
KK
 Þ 
L=
éù
-
êú
ll
êú
ëû
n
2
22
gn
hc
h 11
2m
L=
æö l -l
ç÷
ç÷
ll
èø
n
22
ng
22
gn
2mc
h
( )
ll
L=
l -l
22
gn
n
22
ng
2mc
h
as l
n
 µ n
l
n
 >> l
g
-
éù
ll æö
êú L=-
ç÷
l
êú èø
ëû
1
2
2
gg
n
n
2mc
1
h
éù
lll æö
êú L= ++
ç÷
ll
êú èø
ëû
2
2
ggg
n
nn
2mc
1 higher powersof
h
JEE(MAIN)-2018
2
L »+
l
n 2
n
B
A
where A = 
l
2
g
2mc
h
& B = 
l
4
g
2mc
h
63. The reading of the ammeter for a silicon diode in
the given circuit is :-
3V
200W
(1) 15 mA (2) 11.5 mA
(3) 13.5 mA (4) 0
Ans. (2)
Sol.
3V
200W
Silicon diode is in forward bias.
Hence across diode potential barrier
DV = 0.7 volts
I = 
-D-
=
V V 3 0.7
R 200
  = 
2.3
200
 = 11.5 mA
64. The density of a material in the shape of a cube
is determined by measuring three sides of the cube
and its mass. If the relative errors in measuring
the mass and length are respectively 1.5% and
1%, the maximum error in determining the density
is :-
(1) 3.5 % (2) 4.5 %
(3) 6 % (4) 2.5 %
Ans. (2)
Sol. Density = 
Mass
Volume
1d 1M 3 L
d ML
D DD
=+
= 1.5 + 3(1)
= 4.5 %
65. An electron, a proton and an alpha particle
having the same kinetic energy are moving in
circular orbits of radii r
e
, r
p
, r
a
 respectively in
a uniform magnetic field B. The relation
between r
e
, r
p
, r
a
 is:-
(1) r
e
 < r
p
 = r
a
(2) r
e
 < r
p
 < r
a
(3) r
e
 < r
a
 < r
p
(4) r
e
 > r
p
 = r
a
Ans. (1)
Sol. Radius of circular path in magnetic field is given
by  =
2Km
R
qB
where K = kinetic energy of particle
m = mass of particle
q = charge on particle
B = magnetic field intensity
R = radius of path
For electron
e
e
2Km
r
eB
= ...(i)
For proton
p
p
2Km
r
eB
= ...(ii)
For a particle
pp
2K 4m 2K m
2Km
r
q B 2eB eB
a
a
a
= ==
...(iii)
as m
e
 < m
p
   so    r
e
 < r
p
 = r
a
66. Three concentric metal shells A, B and C of
respective radii a, b and c (a < b < c) have surface
charge densities +s, –s and +s respectively. The
potential of shell B is :-
CODE-C
3
(1) 
22
0
ab
c
b
éù s-
+
êú
e
ëû
(2) 
22
0
bc
a
b
éù s-
+
êú
e
ëû
(3) 
22
0
bc
a
c
éù s-
+
êú
e
ëû
(4) 
22
0
ab
c
a
éù s-
+
êú
e
ëû
Ans. (1)
Sol.
s
–s
s
A
B
C
V
outside
 = 
Q K
r
where r is distance of point from the centre of
shell V
inside 
 = 
Q K
R
where 'R' is radius of the shell
C AB
B
b bc
Kq Kq Kq
V
rrr
= ++
2 22
B
0
1 4a 4b 4c
V
4 bbc
éù sp sp s p
= -+
êú
pÎ
ëû
22
B
0
ab
Vc
b
éù s-
=+
êú
Î
ëû
67. Two masses m
1
 = 5kg and m
2
 = 10kg,
connected by an inextensible string over a
frictionless pulley, are moving as shown in the
figure. The coefficient of friction of horizontal
surface is 0.15. The minimum weight m that
should be put on top of m
2
 to stop the motion
is :-
m
m
2
m
1
mg
1
T
T
(1) 27.3 kg (2) 43.3 kg
(3) 10.3 kg (4) 18.3 kg
Ans. (1)
Sol.
m
m
2
m
1
mg
1
= 50N
T
T
µ(m+m )g
2
(m+m )g
2
N
50 – T  = 5 × a
T  – 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
 50 = 0.15 (m + 10) 10
5 = 
3
20
(m + 10)
100
3
 = m + 10
m = 23.3 kg
68. A particle is moving in a circular path of radius
a under the action of an attractive potential
U = –
2
k
2r
. Its total energy is :-
(1)
2
k
2a
(2) Zero
(3) 
2
3k
2a
-
(4) 
2
k
4a
-
Ans. (2)
Sol. F = 
¶
-=
¶
3
uK
rr
Since it is performing circular motion
JEE(MAIN)-2018
4
F = =
2
3
mvK
rr
mv
2
 = 
2
K
r
Þ K.E. = 
1
2
mv
2
 = 
2
K
2r
Total energy = P.E. + K.E.
= 
-+
22
KK
2r 2r
 = Zero
69. A parallel plate capacitor of capacitance
90 pF is connected to a battery of emf 20V. If a
dielectric material of dielectric constant
K = 
5
3
 is inserted between the plates, the
magnitude of the induced charge will be :-
(1) 0.3 n C (2) 2.4 n C
(3) 0.9 n C (4) 1.2 n C
Ans. (4)
Sol.
K
V
–Q
+Q
Q = (kC) V
= 
5
90pF (20V)
3
æö
´
ç÷
èø
= 3000 pC
= 3nC
induced charges on dielectric
Q
ind 
 = Q 
13
1 3nC 1 1.2nC
K5
æ ö æö
- = -=
ç ÷ ç÷
è ø èø
70. A silver atom in a solid oscillates in simple
harmonic motion in some direction with a
frequency of 10
12
/sec. What is the force
constant of the bonds connecting one atom
with the other ? (Mole wt. of silver =108 and
Avagadro number = 6.02 × 10
23
 gm mole
–1
)
(1) 7.1 N/m (2) 2.2 N/m
(3) 5.5 N/m(4) 6.4 N/m
Ans. (1)
Sol. Time period of SHM is given by
T = 
p
m
2
k
frequency = 
=
p
12
1k
10
2m
where m = mass of one atom
= 
( )
-
´
´
3
23
108
10 kg
6.02 10
-
´´
2p´
23
3
1k
6.02 10
108 10
 = 10
12
On solving K = 7.1 N/m
71. It is found that if a neutron suffers an elastic
collinear collision with deuterium at rest, fractional
loss of its energy is p
d
 ; while for its similar collision
with carbon nucleus at rest, fractional loss of
energy is p
c
. The values of p
d
 and p
c
 are
respectively :
(1) (
.
28, 
.
89) (2) (0, 0)
(3) (0, 1) (4) (
.
89, 
.
28)
Ans. (4)
Sol. Let initial speed of neutron is v
0
 and kinetic energy
is K.
1
st
 
collision :
      
n
v
0
d n
v
1
d
v
2
m 2m m 2m
Þ
by momentum conservation
mv
0
 = mv
1
 + 2mv
2
 Þ v
1
 + 2v
2
 = v
0
by e = 1 v
2
 – v
1
 = v
0
Þ v
2
 = 
0
2v
3
 ; v
1
 = – 
0
v
3
fractional loss = 
2
2 0
0
2
0
v 11
mvm
2 23
1
mv
2
æö
-
ç÷
èø
Þ 
d
8
.
P 89
9
=»
2
nd
 collision :
 
n
v
0
c n
v
1
c
v
2
m 12m m 12m
Þ
Page 5


CODE-C
1
61. The angular width of the central maximum in
a single slit diffraction pattern is 60°. The width
of the slit is 1 mm. The slit is illuminated by
monochromatic plane waves. If another slit of
same width is made near it, Young’s fringes can
be observed on a screen placed at a distance
50 cm from the slits. If the observed fringe
width is 1 cm, what is slit separation distance?
(i.e. distance between the centres of each slit.)
(1) 50 mm (2) 75 mm
(3) 100 mm (4) 25 mm
Ans. (4)
Sol. In diffraction
d sin 30º = l
60º
d
d
2
l=
Young's fringe width
[d' – separation between two slits]
l´
b=
D
d'
2
2
50 10 d
10
2 d'
-
-
´
=´
10
–2
 = 
62
10 50 10
2 d'
--
´´
´
d' 25m =m
62. An electron from various excited states of
hydrogen atom emit radiation to come to the ground
state. Let l
n
, l
g
 be the de Broglie wavelength of
the electron in the n
th
 state and the ground state
respectively. Let 
n
L be the wavelength of the
emitted photon in the transition from the n
th
 state
to the ground state. For large n, (A, B are
constants)
(1) 
nn
AB L » +l (2) 
22
nn
AB L » +l
(3) 
2
n
L »l (4) n 2
n
B
A L »+
l
PART C – PHYSICS
JEE(MAIN) – 2018 TEST PAPER WITH SOLUTION
(HELD ON SUNDAY 08
th
 APRIL, 2018)
Ans. (4)
Sol.
l==
n
n
hh
mu 2mk
Þ k
n
 =  
l
2
2
n
h
2m
 ;  k
g
 = 
l
2
2
g
h
2m
Þ k
g
 – k
n
 = 
éù
-
êú
ll
êú
ëû
2
22
gn
h 11
2m
E
n
 = – k
n
for emitted photon
=-= -
L
n g gn
n
hc
EE KK
-
=
L
gn
n
KK
1
hc
L=
-
n
gn
hc
KK
 Þ 
L=
éù
-
êú
ll
êú
ëû
n
2
22
gn
hc
h 11
2m
L=
æö l -l
ç÷
ç÷
ll
èø
n
22
ng
22
gn
2mc
h
( )
ll
L=
l -l
22
gn
n
22
ng
2mc
h
as l
n
 µ n
l
n
 >> l
g
-
éù
ll æö
êú L=-
ç÷
l
êú èø
ëû
1
2
2
gg
n
n
2mc
1
h
éù
lll æö
êú L= ++
ç÷
ll
êú èø
ëû
2
2
ggg
n
nn
2mc
1 higher powersof
h
JEE(MAIN)-2018
2
L »+
l
n 2
n
B
A
where A = 
l
2
g
2mc
h
& B = 
l
4
g
2mc
h
63. The reading of the ammeter for a silicon diode in
the given circuit is :-
3V
200W
(1) 15 mA (2) 11.5 mA
(3) 13.5 mA (4) 0
Ans. (2)
Sol.
3V
200W
Silicon diode is in forward bias.
Hence across diode potential barrier
DV = 0.7 volts
I = 
-D-
=
V V 3 0.7
R 200
  = 
2.3
200
 = 11.5 mA
64. The density of a material in the shape of a cube
is determined by measuring three sides of the cube
and its mass. If the relative errors in measuring
the mass and length are respectively 1.5% and
1%, the maximum error in determining the density
is :-
(1) 3.5 % (2) 4.5 %
(3) 6 % (4) 2.5 %
Ans. (2)
Sol. Density = 
Mass
Volume
1d 1M 3 L
d ML
D DD
=+
= 1.5 + 3(1)
= 4.5 %
65. An electron, a proton and an alpha particle
having the same kinetic energy are moving in
circular orbits of radii r
e
, r
p
, r
a
 respectively in
a uniform magnetic field B. The relation
between r
e
, r
p
, r
a
 is:-
(1) r
e
 < r
p
 = r
a
(2) r
e
 < r
p
 < r
a
(3) r
e
 < r
a
 < r
p
(4) r
e
 > r
p
 = r
a
Ans. (1)
Sol. Radius of circular path in magnetic field is given
by  =
2Km
R
qB
where K = kinetic energy of particle
m = mass of particle
q = charge on particle
B = magnetic field intensity
R = radius of path
For electron
e
e
2Km
r
eB
= ...(i)
For proton
p
p
2Km
r
eB
= ...(ii)
For a particle
pp
2K 4m 2K m
2Km
r
q B 2eB eB
a
a
a
= ==
...(iii)
as m
e
 < m
p
   so    r
e
 < r
p
 = r
a
66. Three concentric metal shells A, B and C of
respective radii a, b and c (a < b < c) have surface
charge densities +s, –s and +s respectively. The
potential of shell B is :-
CODE-C
3
(1) 
22
0
ab
c
b
éù s-
+
êú
e
ëû
(2) 
22
0
bc
a
b
éù s-
+
êú
e
ëû
(3) 
22
0
bc
a
c
éù s-
+
êú
e
ëû
(4) 
22
0
ab
c
a
éù s-
+
êú
e
ëû
Ans. (1)
Sol.
s
–s
s
A
B
C
V
outside
 = 
Q K
r
where r is distance of point from the centre of
shell V
inside 
 = 
Q K
R
where 'R' is radius of the shell
C AB
B
b bc
Kq Kq Kq
V
rrr
= ++
2 22
B
0
1 4a 4b 4c
V
4 bbc
éù sp sp s p
= -+
êú
pÎ
ëû
22
B
0
ab
Vc
b
éù s-
=+
êú
Î
ëû
67. Two masses m
1
 = 5kg and m
2
 = 10kg,
connected by an inextensible string over a
frictionless pulley, are moving as shown in the
figure. The coefficient of friction of horizontal
surface is 0.15. The minimum weight m that
should be put on top of m
2
 to stop the motion
is :-
m
m
2
m
1
mg
1
T
T
(1) 27.3 kg (2) 43.3 kg
(3) 10.3 kg (4) 18.3 kg
Ans. (1)
Sol.
m
m
2
m
1
mg
1
= 50N
T
T
µ(m+m )g
2
(m+m )g
2
N
50 – T  = 5 × a
T  – 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
 50 = 0.15 (m + 10) 10
5 = 
3
20
(m + 10)
100
3
 = m + 10
m = 23.3 kg
68. A particle is moving in a circular path of radius
a under the action of an attractive potential
U = –
2
k
2r
. Its total energy is :-
(1)
2
k
2a
(2) Zero
(3) 
2
3k
2a
-
(4) 
2
k
4a
-
Ans. (2)
Sol. F = 
¶
-=
¶
3
uK
rr
Since it is performing circular motion
JEE(MAIN)-2018
4
F = =
2
3
mvK
rr
mv
2
 = 
2
K
r
Þ K.E. = 
1
2
mv
2
 = 
2
K
2r
Total energy = P.E. + K.E.
= 
-+
22
KK
2r 2r
 = Zero
69. A parallel plate capacitor of capacitance
90 pF is connected to a battery of emf 20V. If a
dielectric material of dielectric constant
K = 
5
3
 is inserted between the plates, the
magnitude of the induced charge will be :-
(1) 0.3 n C (2) 2.4 n C
(3) 0.9 n C (4) 1.2 n C
Ans. (4)
Sol.
K
V
–Q
+Q
Q = (kC) V
= 
5
90pF (20V)
3
æö
´
ç÷
èø
= 3000 pC
= 3nC
induced charges on dielectric
Q
ind 
 = Q 
13
1 3nC 1 1.2nC
K5
æ ö æö
- = -=
ç ÷ ç÷
è ø èø
70. A silver atom in a solid oscillates in simple
harmonic motion in some direction with a
frequency of 10
12
/sec. What is the force
constant of the bonds connecting one atom
with the other ? (Mole wt. of silver =108 and
Avagadro number = 6.02 × 10
23
 gm mole
–1
)
(1) 7.1 N/m (2) 2.2 N/m
(3) 5.5 N/m(4) 6.4 N/m
Ans. (1)
Sol. Time period of SHM is given by
T = 
p
m
2
k
frequency = 
=
p
12
1k
10
2m
where m = mass of one atom
= 
( )
-
´
´
3
23
108
10 kg
6.02 10
-
´´
2p´
23
3
1k
6.02 10
108 10
 = 10
12
On solving K = 7.1 N/m
71. It is found that if a neutron suffers an elastic
collinear collision with deuterium at rest, fractional
loss of its energy is p
d
 ; while for its similar collision
with carbon nucleus at rest, fractional loss of
energy is p
c
. The values of p
d
 and p
c
 are
respectively :
(1) (
.
28, 
.
89) (2) (0, 0)
(3) (0, 1) (4) (
.
89, 
.
28)
Ans. (4)
Sol. Let initial speed of neutron is v
0
 and kinetic energy
is K.
1
st
 
collision :
      
n
v
0
d n
v
1
d
v
2
m 2m m 2m
Þ
by momentum conservation
mv
0
 = mv
1
 + 2mv
2
 Þ v
1
 + 2v
2
 = v
0
by e = 1 v
2
 – v
1
 = v
0
Þ v
2
 = 
0
2v
3
 ; v
1
 = – 
0
v
3
fractional loss = 
2
2 0
0
2
0
v 11
mvm
2 23
1
mv
2
æö
-
ç÷
èø
Þ 
d
8
.
P 89
9
=»
2
nd
 collision :
 
n
v
0
c n
v
1
c
v
2
m 12m m 12m
Þ
CODE-C
5
by momentum conservation
  
0 12
mv mv 12mv =+
Þ +=
1 20
v 12vv
by e = 1     v
2
 – v
1
 = v
0
00
21
2v 11v
v ;v
13 13
-
==
Now fraction loss of energy
2
2 0
c0
2
0
11v 11
P mvm
2 2 13
1
mv
2
æö
=-
ç÷
èø
 = 
48
0.28
169
»
72. The dipole moment of a circular loop carrying a
current I, is m and the magnetic field at the centre
of the loop is B
1
. When the dipole moment is
doubled by keeping the current constant, the
magnetic field at the centre of the loop is B
2
. The
ratio 
1
2
B
B
is :
(1) 
3
(2) 
2
(3) 
1
2
(4) 2
Ans. (2)
Sol. Dipole moment of circular loop is m
m
1
 = I.A = I.pR
2
  {R = radius of the loop}
B
1
 = 
0
I
2R
m
moment becomes double
Þ R becomes  
2R
 (keeping current
constant)
m
2
 = I.p 
2
( 2R) = 2.IpR
2
 = 2m
1
0 1
2
I B
B
2( 2R) 2
m
==
1
2
B
2
B
=
73. In a potentiometer experiment, it is found that
no current passes through the galvanometer
when the terminals of the cell are connected
across 52 cm of the potentiometer wire. If the
cell is shunted by a resistance of 5 W, a balance
is found when the cell is connected across
40 cm of the wire. Find the internal resistance
of the cell.
(1) 1.5 W (2) 2 W
(3) 2.5 W (4) 1 W
Ans. (1)
Sol. without shunting condition :
G
E ,r
s
52cm
e
p
E
s
 = 52 × x ...(1)
when balanced
where, x = potential gradient of wire.
with shunting condition
G
R
E
s
,r
40cm
e
p
On balancing
( )
- =´
+
s
s
E
E r 40x
rR
....(2)
On solving :
(1)
(2)
 Þ 
=
-
+
1 52
r
40
1
rR
 \ r = 1.5 W