JEE Mains 8 April 2019 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
08/04/2019
Evening
Page 2


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
08/04/2019
Evening
PART–A : PHYSICS
1. A cell of internal resistance r drives current
through an external resistance R. The power
delivered by the cell to the external resistance
will be maximum when :
(1) R = 1000 r (2) R = r
(3) R = 2r (4) R = 0.001 r
Answer (2)
Sol. For maximum power in external resistance,
Internal resistance =  External resistance
? R = r
2. A body of mass m
1
 moving with an unknown
velocity of 
1
ˆ
vi , undergoes a collinear collision
with a body of mass m
2
 moving with a velocity
2
ˆ
vi . After collision, m
1
 and m
2
 move with
velocities of 
3
ˆ
vi and 
4
ˆ
vi , respectively.
If m
2
 = 0.5 m
1
 and v
3
 = 0.5 v
1
, then v
1
 is :
(1)
2
4
v
v–
4
(2) v
4
 – v
2
(3) v
4
 + v
2
(4)
2
4
v
v–
2
Answer (2)
Sol. m
1
v
1
 + m
2
v
2
 = m
1
v
3
 + m
2
v
4
m
1
v
1
 + 0.5m
1
v
2
 = 0.5m
1
v
1
 + 0.5m
1
v
4
v
1
 = v
4
 – v
2
3. In a line of sight radio communication, a
distance of about 50 km is kept between the
transmitting and receiving antennas. If the
height of the receiving antenna is 70 m, then
the minimum height of the transmitting antenna
should be :
(Radius of the Earth = 6.4 × 10
6 
m).
(1) 20 m (2) 51 m
(3) 32 m (4) 40 m
Answer (3)
Sol.
3
ERE
270 R 2 h R 5010 ?? ? ? ? ? ?
Putting R
E
 = 6.4 × 10
6
 m and solving we get
h
R
 = 32 m
4. The electric field in a region is given by
??
ˆ
EAxBi ??

, where E is in NC
–1
 and x is in
metres. The values of constants are A = 20 SI
unit and B = 10 SI unit. If the potential at
x = 1 is V
1
 and that at x = – 5 is V
2
, then
V
1
 – V
2
 is :
(1) 180 V (2) –520 V
(3) 320 V (4) –48 V
Answer (1)
Sol.
??
dV –E dr – Ax B dx ?? ? ?
  
??
1
2
V
1
V–5
dV – Ax B dx ??
??
1
2
12
–5
x
V–V –A –Bx
2
??
?
??
??
??
AA
– –B 25 B –5
22
??? ?
?? ?
??? ?
??? ?
= 12A – 6B = 240 – 60 = 180 V
5. A parallel plate capacitor has 1 ?F capacitance.
One of its two plates is given +2 ?C charge and
the other plate, +4 ?C charge. The potential
difference developed across the capacitor is :
(1) 5 V (2) 1 V
(3) 3 V (4) 2 V
Answer (2)
Sol.
q1C
V
C1F
?
??
?
–
–
–
–
–
–
+
+
+
+
+
+
3C ?? 3C ??
–1 C ??
–1 C ? ?
= 1 V
6. The given diagram shows four processes i.e.,
isochoric, isobaric, isothermal and adiabatic.
The correct assignment of the processes, in
the same order is given by :
P
d
c
b
a
V
(1) a d c b (2) a d b c
(3) d a b c (4) d a c b
Answer (3)
Page 3


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
08/04/2019
Evening
PART–A : PHYSICS
1. A cell of internal resistance r drives current
through an external resistance R. The power
delivered by the cell to the external resistance
will be maximum when :
(1) R = 1000 r (2) R = r
(3) R = 2r (4) R = 0.001 r
Answer (2)
Sol. For maximum power in external resistance,
Internal resistance =  External resistance
? R = r
2. A body of mass m
1
 moving with an unknown
velocity of 
1
ˆ
vi , undergoes a collinear collision
with a body of mass m
2
 moving with a velocity
2
ˆ
vi . After collision, m
1
 and m
2
 move with
velocities of 
3
ˆ
vi and 
4
ˆ
vi , respectively.
If m
2
 = 0.5 m
1
 and v
3
 = 0.5 v
1
, then v
1
 is :
(1)
2
4
v
v–
4
(2) v
4
 – v
2
(3) v
4
 + v
2
(4)
2
4
v
v–
2
Answer (2)
Sol. m
1
v
1
 + m
2
v
2
 = m
1
v
3
 + m
2
v
4
m
1
v
1
 + 0.5m
1
v
2
 = 0.5m
1
v
1
 + 0.5m
1
v
4
v
1
 = v
4
 – v
2
3. In a line of sight radio communication, a
distance of about 50 km is kept between the
transmitting and receiving antennas. If the
height of the receiving antenna is 70 m, then
the minimum height of the transmitting antenna
should be :
(Radius of the Earth = 6.4 × 10
6 
m).
(1) 20 m (2) 51 m
(3) 32 m (4) 40 m
Answer (3)
Sol.
3
ERE
270 R 2 h R 5010 ?? ? ? ? ? ?
Putting R
E
 = 6.4 × 10
6
 m and solving we get
h
R
 = 32 m
4. The electric field in a region is given by
??
ˆ
EAxBi ??

, where E is in NC
–1
 and x is in
metres. The values of constants are A = 20 SI
unit and B = 10 SI unit. If the potential at
x = 1 is V
1
 and that at x = – 5 is V
2
, then
V
1
 – V
2
 is :
(1) 180 V (2) –520 V
(3) 320 V (4) –48 V
Answer (1)
Sol.
??
dV –E dr – Ax B dx ?? ? ?
  
??
1
2
V
1
V–5
dV – Ax B dx ??
??
1
2
12
–5
x
V–V –A –Bx
2
??
?
??
??
??
AA
– –B 25 B –5
22
??? ?
?? ?
??? ?
??? ?
= 12A – 6B = 240 – 60 = 180 V
5. A parallel plate capacitor has 1 ?F capacitance.
One of its two plates is given +2 ?C charge and
the other plate, +4 ?C charge. The potential
difference developed across the capacitor is :
(1) 5 V (2) 1 V
(3) 3 V (4) 2 V
Answer (2)
Sol.
q1C
V
C1F
?
??
?
–
–
–
–
–
–
+
+
+
+
+
+
3C ?? 3C ??
–1 C ??
–1 C ? ?
= 1 V
6. The given diagram shows four processes i.e.,
isochoric, isobaric, isothermal and adiabatic.
The correct assignment of the processes, in
the same order is given by :
P
d
c
b
a
V
(1) a d c b (2) a d b c
(3) d a b c (4) d a c b
Answer (3)
Sol. Between the isothermal and the adiabatic
processes, P-V graph for adiabatic is steeper.
7. Two magnetic dipoles X and Y are placed at a
separation d, with their axes perpendicular to
each other. The dipole moment of Y is twice
that of X. A particle of charge q is passing
through their mid-point P, at angle ? = 45° with
the horizontal line, as shown in figure. What
would be the magnitude of force on the particle
at that instant? (d is much larger than the
dimensions of the dipole)
d
S N
S
N
?
P
X
(M)
Y
(2 M)
(1) 0
(2)
? ??
?
??
?
????
??
??
0
3
M
2 qv
4
d
2
(3)
? ??
?
??
?
????
??
??
0
3
2M
qv
4
d
2
(4)
? ??
?
??
?
????
??
??
0
3
M
qv
4
d
2
Answer (1)
Sol. ??
m F qVB ??
 
x y
BB B ??
 
SN
S
N
X 
M 
B 
y
Y
B 
x
d
2
d
2
 
Since M
y
 = 2M
x
?
xy
BB ?

net
B

 is parallel to V

?
F0 ?

8. A damped harmonic oscillator has a frequency
of 5 oscillations per second. The amplitude
drops to half its value for every 10 oscillations.
The time it will take to drop to 
1
1000
 of the
original amplitude is close to:
(1) 100 s
(2) 10 s
(3) 50 s
(4) 20 s
Answer (4)
Sol. Time for 10 oscillations = 
10
2s
5
?
A = A
0
 e
–kt
2k
1
e
2
?
?
? In2 = 2k
10
–3
 = e
–kt
? 3In10 = kt
3In10 3In10
t2
kIn2
?? ?
2.3
620s
0.69
?? ?
9. If Surface tension (S), Moment of Inertia (I) and
Planck’s constant (h), were to be taken as the
fundamental units, the dimensional formula for
linear momentum would be:
(1) S
1/2
I
3/2
h
–1
(2) S
3/2
I
1/2
h
0
(3) S
1/2
I
1/2
h
–1
(4) S
1/2
I
1/2
h
0
Answer (4)
Sol. [p] = MLT
–1
 = [I
x
 h
y
 S
z
]
= M
x
L
2x
 (ML
2
T
–1
)
y
 (MT
–2
)
z
= M
x + y + z
 L
2x + 2y
 T
–y – 2z
x + y + z = 1
2(x + y) = 1 ?
11
xy z
22
?? ? ?
y + 2z = 1 ? y = 0     
1
x
2
? ?
? ?pIS ?
Page 4


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
08/04/2019
Evening
PART–A : PHYSICS
1. A cell of internal resistance r drives current
through an external resistance R. The power
delivered by the cell to the external resistance
will be maximum when :
(1) R = 1000 r (2) R = r
(3) R = 2r (4) R = 0.001 r
Answer (2)
Sol. For maximum power in external resistance,
Internal resistance =  External resistance
? R = r
2. A body of mass m
1
 moving with an unknown
velocity of 
1
ˆ
vi , undergoes a collinear collision
with a body of mass m
2
 moving with a velocity
2
ˆ
vi . After collision, m
1
 and m
2
 move with
velocities of 
3
ˆ
vi and 
4
ˆ
vi , respectively.
If m
2
 = 0.5 m
1
 and v
3
 = 0.5 v
1
, then v
1
 is :
(1)
2
4
v
v–
4
(2) v
4
 – v
2
(3) v
4
 + v
2
(4)
2
4
v
v–
2
Answer (2)
Sol. m
1
v
1
 + m
2
v
2
 = m
1
v
3
 + m
2
v
4
m
1
v
1
 + 0.5m
1
v
2
 = 0.5m
1
v
1
 + 0.5m
1
v
4
v
1
 = v
4
 – v
2
3. In a line of sight radio communication, a
distance of about 50 km is kept between the
transmitting and receiving antennas. If the
height of the receiving antenna is 70 m, then
the minimum height of the transmitting antenna
should be :
(Radius of the Earth = 6.4 × 10
6 
m).
(1) 20 m (2) 51 m
(3) 32 m (4) 40 m
Answer (3)
Sol.
3
ERE
270 R 2 h R 5010 ?? ? ? ? ? ?
Putting R
E
 = 6.4 × 10
6
 m and solving we get
h
R
 = 32 m
4. The electric field in a region is given by
??
ˆ
EAxBi ??

, where E is in NC
–1
 and x is in
metres. The values of constants are A = 20 SI
unit and B = 10 SI unit. If the potential at
x = 1 is V
1
 and that at x = – 5 is V
2
, then
V
1
 – V
2
 is :
(1) 180 V (2) –520 V
(3) 320 V (4) –48 V
Answer (1)
Sol.
??
dV –E dr – Ax B dx ?? ? ?
  
??
1
2
V
1
V–5
dV – Ax B dx ??
??
1
2
12
–5
x
V–V –A –Bx
2
??
?
??
??
??
AA
– –B 25 B –5
22
??? ?
?? ?
??? ?
??? ?
= 12A – 6B = 240 – 60 = 180 V
5. A parallel plate capacitor has 1 ?F capacitance.
One of its two plates is given +2 ?C charge and
the other plate, +4 ?C charge. The potential
difference developed across the capacitor is :
(1) 5 V (2) 1 V
(3) 3 V (4) 2 V
Answer (2)
Sol.
q1C
V
C1F
?
??
?
–
–
–
–
–
–
+
+
+
+
+
+
3C ?? 3C ??
–1 C ??
–1 C ? ?
= 1 V
6. The given diagram shows four processes i.e.,
isochoric, isobaric, isothermal and adiabatic.
The correct assignment of the processes, in
the same order is given by :
P
d
c
b
a
V
(1) a d c b (2) a d b c
(3) d a b c (4) d a c b
Answer (3)
Sol. Between the isothermal and the adiabatic
processes, P-V graph for adiabatic is steeper.
7. Two magnetic dipoles X and Y are placed at a
separation d, with their axes perpendicular to
each other. The dipole moment of Y is twice
that of X. A particle of charge q is passing
through their mid-point P, at angle ? = 45° with
the horizontal line, as shown in figure. What
would be the magnitude of force on the particle
at that instant? (d is much larger than the
dimensions of the dipole)
d
S N
S
N
?
P
X
(M)
Y
(2 M)
(1) 0
(2)
? ??
?
??
?
????
??
??
0
3
M
2 qv
4
d
2
(3)
? ??
?
??
?
????
??
??
0
3
2M
qv
4
d
2
(4)
? ??
?
??
?
????
??
??
0
3
M
qv
4
d
2
Answer (1)
Sol. ??
m F qVB ??
 
x y
BB B ??
 
SN
S
N
X 
M 
B 
y
Y
B 
x
d
2
d
2
 
Since M
y
 = 2M
x
?
xy
BB ?

net
B

 is parallel to V

?
F0 ?

8. A damped harmonic oscillator has a frequency
of 5 oscillations per second. The amplitude
drops to half its value for every 10 oscillations.
The time it will take to drop to 
1
1000
 of the
original amplitude is close to:
(1) 100 s
(2) 10 s
(3) 50 s
(4) 20 s
Answer (4)
Sol. Time for 10 oscillations = 
10
2s
5
?
A = A
0
 e
–kt
2k
1
e
2
?
?
? In2 = 2k
10
–3
 = e
–kt
? 3In10 = kt
3In10 3In10
t2
kIn2
?? ?
2.3
620s
0.69
?? ?
9. If Surface tension (S), Moment of Inertia (I) and
Planck’s constant (h), were to be taken as the
fundamental units, the dimensional formula for
linear momentum would be:
(1) S
1/2
I
3/2
h
–1
(2) S
3/2
I
1/2
h
0
(3) S
1/2
I
1/2
h
–1
(4) S
1/2
I
1/2
h
0
Answer (4)
Sol. [p] = MLT
–1
 = [I
x
 h
y
 S
z
]
= M
x
L
2x
 (ML
2
T
–1
)
y
 (MT
–2
)
z
= M
x + y + z
 L
2x + 2y
 T
–y – 2z
x + y + z = 1
2(x + y) = 1 ?
11
xy z
22
?? ? ?
y + 2z = 1 ? y = 0     
1
x
2
? ?
? ?pIS ?
JEE (MAIN)-2019 (Online) Phase-2
10. A circuit connected to an ac source of emf
e = e
0
sin(100t) with t in seconds, gives a phase
difference of 
?
4
 between the emf e and
current i. Which of the following circuits will
exhibit this?
(1) RL circuit with R = 1 k ? and L = 10 mH
(2) RL circuit with R = 1 k ? and L = 1 mH
(3) RC circuit with R = 1 k ? and C = 10 ?F
(4) RC circuit with R = 1 k ? and C = 1 ?F
Answer (3)
Sol. As 
4
?
?? , x
c
 = R
11. The magnetic field of an electromagnetic wave
is given by:
?
?? ? ?? ?

67 15
2
Wb
ˆˆ
B 1.6 10 cos(2 10 z 6 10 t)(2i j)
m
The associated electric field will be:
(1)
??
?? ? ?? ??
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t (i2j)
m
(2)
??
?? ? ?? ?
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t (2i j)
m
(3)
??
?? ? ?? ? ?
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t ( 2j i)
m
(4)
??
?? ? ?? ?
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t (i 2j)
m
Answer (1)
Sol. Amplitude of electric field, E = B
0
C
68
1.6 10 5 3 10
?
?? ? ??
2
4.8 10 5 V/m ??
Also EB ?

 is along 
ˆ
–k
 (the direction of
propagation)
?
27 15
V
ˆˆ
E4.810cos(210z 610 t)(i2j)
m
?? ? ?? ??
 
12. In the circuit shown, a four-wire potentiometer
is made of a 400 cm long wire, which extends
between A and B. The resistance per unit
length of the potentiometer wire is r = 0.01 ?/
cm. If an ideal voltmeter is connected as
shown with jockey J at 50 cm from end A, the
expected reading of the voltmeter will be:
1 ?
1.5 V, 1.5 V
0.5 0.5 ?? ??
100 cm
50 cm
V
J A
B
(1) 0.75 V (2) 0.50 V
(3) 0.20 V (4) 0.25 V
Answer (4)
Sol. Resistance of potentiometer wire, R
p
 = 400 × 0.01
     = 4 ?
?
3
l0.5A
6
??
? Reading of voltmeter = I R
AJ
= 0.5 × 50 × 0.01
= 0.25 V
13. A nucleus A, with a finite de-Broglie wavelength
?
A
, undergoes spontaneous fission into two
nuclei B and C of equal mass. B flies in the
same direction as that of A, while C flies in the
opposite direction with a velocity equal to half
of that of B. The de-Broglie wavelength ?
B
 and
?
C
 of B and C are respectively:
(1) 2 ?
A
,??
A
(2) ?
A
, ?
A
2
?
(3)
A
A
,
2
?
?
(4) ?
A
,
 
2 ?
A
Answer (3)
Sol.
A
A
h
mV
??
Conservation of linear momentum
?
A
mmVmV
mV V .
222 4
?? ?
?
A
4h
mV
??
?
A
V
V
4
?
A
B
h2h
m
mV 2
V
2
?
?? ? ?
cA
h4h
mV
mV
.
22
?? ? ??
Page 5


Time : 3 hrs. M.M. : 360
Answers & Solutions
for for for for for
JEE (MAIN)-2019 (Online) Phase-2
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 90 questions. The maximum marks are 360.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 30 questions in each part of equal weightage.
4. Each question is allotted 4 (four) marks for each correct response. ¼ (one-fourth) marks will be deducted
for indicating incorrect response of each question. No deduction from the total score will be made if no
response is indicated for an item in the answer sheet.
5. There is only one correct response for each question.
(Physics, Chemistry and Mathematics)
08/04/2019
Evening
PART–A : PHYSICS
1. A cell of internal resistance r drives current
through an external resistance R. The power
delivered by the cell to the external resistance
will be maximum when :
(1) R = 1000 r (2) R = r
(3) R = 2r (4) R = 0.001 r
Answer (2)
Sol. For maximum power in external resistance,
Internal resistance =  External resistance
? R = r
2. A body of mass m
1
 moving with an unknown
velocity of 
1
ˆ
vi , undergoes a collinear collision
with a body of mass m
2
 moving with a velocity
2
ˆ
vi . After collision, m
1
 and m
2
 move with
velocities of 
3
ˆ
vi and 
4
ˆ
vi , respectively.
If m
2
 = 0.5 m
1
 and v
3
 = 0.5 v
1
, then v
1
 is :
(1)
2
4
v
v–
4
(2) v
4
 – v
2
(3) v
4
 + v
2
(4)
2
4
v
v–
2
Answer (2)
Sol. m
1
v
1
 + m
2
v
2
 = m
1
v
3
 + m
2
v
4
m
1
v
1
 + 0.5m
1
v
2
 = 0.5m
1
v
1
 + 0.5m
1
v
4
v
1
 = v
4
 – v
2
3. In a line of sight radio communication, a
distance of about 50 km is kept between the
transmitting and receiving antennas. If the
height of the receiving antenna is 70 m, then
the minimum height of the transmitting antenna
should be :
(Radius of the Earth = 6.4 × 10
6 
m).
(1) 20 m (2) 51 m
(3) 32 m (4) 40 m
Answer (3)
Sol.
3
ERE
270 R 2 h R 5010 ?? ? ? ? ? ?
Putting R
E
 = 6.4 × 10
6
 m and solving we get
h
R
 = 32 m
4. The electric field in a region is given by
??
ˆ
EAxBi ??

, where E is in NC
–1
 and x is in
metres. The values of constants are A = 20 SI
unit and B = 10 SI unit. If the potential at
x = 1 is V
1
 and that at x = – 5 is V
2
, then
V
1
 – V
2
 is :
(1) 180 V (2) –520 V
(3) 320 V (4) –48 V
Answer (1)
Sol.
??
dV –E dr – Ax B dx ?? ? ?
  
??
1
2
V
1
V–5
dV – Ax B dx ??
??
1
2
12
–5
x
V–V –A –Bx
2
??
?
??
??
??
AA
– –B 25 B –5
22
??? ?
?? ?
??? ?
??? ?
= 12A – 6B = 240 – 60 = 180 V
5. A parallel plate capacitor has 1 ?F capacitance.
One of its two plates is given +2 ?C charge and
the other plate, +4 ?C charge. The potential
difference developed across the capacitor is :
(1) 5 V (2) 1 V
(3) 3 V (4) 2 V
Answer (2)
Sol.
q1C
V
C1F
?
??
?
–
–
–
–
–
–
+
+
+
+
+
+
3C ?? 3C ??
–1 C ??
–1 C ? ?
= 1 V
6. The given diagram shows four processes i.e.,
isochoric, isobaric, isothermal and adiabatic.
The correct assignment of the processes, in
the same order is given by :
P
d
c
b
a
V
(1) a d c b (2) a d b c
(3) d a b c (4) d a c b
Answer (3)
Sol. Between the isothermal and the adiabatic
processes, P-V graph for adiabatic is steeper.
7. Two magnetic dipoles X and Y are placed at a
separation d, with their axes perpendicular to
each other. The dipole moment of Y is twice
that of X. A particle of charge q is passing
through their mid-point P, at angle ? = 45° with
the horizontal line, as shown in figure. What
would be the magnitude of force on the particle
at that instant? (d is much larger than the
dimensions of the dipole)
d
S N
S
N
?
P
X
(M)
Y
(2 M)
(1) 0
(2)
? ??
?
??
?
????
??
??
0
3
M
2 qv
4
d
2
(3)
? ??
?
??
?
????
??
??
0
3
2M
qv
4
d
2
(4)
? ??
?
??
?
????
??
??
0
3
M
qv
4
d
2
Answer (1)
Sol. ??
m F qVB ??
 
x y
BB B ??
 
SN
S
N
X 
M 
B 
y
Y
B 
x
d
2
d
2
 
Since M
y
 = 2M
x
?
xy
BB ?

net
B

 is parallel to V

?
F0 ?

8. A damped harmonic oscillator has a frequency
of 5 oscillations per second. The amplitude
drops to half its value for every 10 oscillations.
The time it will take to drop to 
1
1000
 of the
original amplitude is close to:
(1) 100 s
(2) 10 s
(3) 50 s
(4) 20 s
Answer (4)
Sol. Time for 10 oscillations = 
10
2s
5
?
A = A
0
 e
–kt
2k
1
e
2
?
?
? In2 = 2k
10
–3
 = e
–kt
? 3In10 = kt
3In10 3In10
t2
kIn2
?? ?
2.3
620s
0.69
?? ?
9. If Surface tension (S), Moment of Inertia (I) and
Planck’s constant (h), were to be taken as the
fundamental units, the dimensional formula for
linear momentum would be:
(1) S
1/2
I
3/2
h
–1
(2) S
3/2
I
1/2
h
0
(3) S
1/2
I
1/2
h
–1
(4) S
1/2
I
1/2
h
0
Answer (4)
Sol. [p] = MLT
–1
 = [I
x
 h
y
 S
z
]
= M
x
L
2x
 (ML
2
T
–1
)
y
 (MT
–2
)
z
= M
x + y + z
 L
2x + 2y
 T
–y – 2z
x + y + z = 1
2(x + y) = 1 ?
11
xy z
22
?? ? ?
y + 2z = 1 ? y = 0     
1
x
2
? ?
? ?pIS ?
JEE (MAIN)-2019 (Online) Phase-2
10. A circuit connected to an ac source of emf
e = e
0
sin(100t) with t in seconds, gives a phase
difference of 
?
4
 between the emf e and
current i. Which of the following circuits will
exhibit this?
(1) RL circuit with R = 1 k ? and L = 10 mH
(2) RL circuit with R = 1 k ? and L = 1 mH
(3) RC circuit with R = 1 k ? and C = 10 ?F
(4) RC circuit with R = 1 k ? and C = 1 ?F
Answer (3)
Sol. As 
4
?
?? , x
c
 = R
11. The magnetic field of an electromagnetic wave
is given by:
?
?? ? ?? ?

67 15
2
Wb
ˆˆ
B 1.6 10 cos(2 10 z 6 10 t)(2i j)
m
The associated electric field will be:
(1)
??
?? ? ?? ??
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t (i2j)
m
(2)
??
?? ? ?? ?
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t (2i j)
m
(3)
??
?? ? ?? ? ?
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t ( 2j i)
m
(4)
??
?? ? ?? ?
 
27 15
V
ˆˆ
E 4.8 10 cos 2 10 z 6 10 t (i 2j)
m
Answer (1)
Sol. Amplitude of electric field, E = B
0
C
68
1.6 10 5 3 10
?
?? ? ??
2
4.8 10 5 V/m ??
Also EB ?

 is along 
ˆ
–k
 (the direction of
propagation)
?
27 15
V
ˆˆ
E4.810cos(210z 610 t)(i2j)
m
?? ? ?? ??
 
12. In the circuit shown, a four-wire potentiometer
is made of a 400 cm long wire, which extends
between A and B. The resistance per unit
length of the potentiometer wire is r = 0.01 ?/
cm. If an ideal voltmeter is connected as
shown with jockey J at 50 cm from end A, the
expected reading of the voltmeter will be:
1 ?
1.5 V, 1.5 V
0.5 0.5 ?? ??
100 cm
50 cm
V
J A
B
(1) 0.75 V (2) 0.50 V
(3) 0.20 V (4) 0.25 V
Answer (4)
Sol. Resistance of potentiometer wire, R
p
 = 400 × 0.01
     = 4 ?
?
3
l0.5A
6
??
? Reading of voltmeter = I R
AJ
= 0.5 × 50 × 0.01
= 0.25 V
13. A nucleus A, with a finite de-Broglie wavelength
?
A
, undergoes spontaneous fission into two
nuclei B and C of equal mass. B flies in the
same direction as that of A, while C flies in the
opposite direction with a velocity equal to half
of that of B. The de-Broglie wavelength ?
B
 and
?
C
 of B and C are respectively:
(1) 2 ?
A
,??
A
(2) ?
A
, ?
A
2
?
(3)
A
A
,
2
?
?
(4) ?
A
,
 
2 ?
A
Answer (3)
Sol.
A
A
h
mV
??
Conservation of linear momentum
?
A
mmVmV
mV V .
222 4
?? ?
?
A
4h
mV
??
?
A
V
V
4
?
A
B
h2h
m
mV 2
V
2
?
?? ? ?
cA
h4h
mV
mV
.
22
?? ? ??
14. A rocket has to be launched from earth in such
a way that it never returns. If E is the minimum
energy delivered by the rocket launcher, what
should be the minimum energy that the
launcher should have if the same rocket is to
be launched from the surface of the moon?
Assume that the density of the earth and the
moon are equal and that the earth’s volume is
64 times the volume of the moon.
(1)
E
64
(2)
E
4
(3)
E
16
(4)
E
32
Answer (3)
Sol.
E
E
GM m
E
R
?
m
M
GM m
E
R
? ?
33
EM
R64R ???
? R
E
 = 4 R
M
M E
EM
M R E11
..4
EM R 64 16
?
?? ?
?
E
E
16
? ?
15. In a simple pendulum experiment for
determination of acceleration due to gravity (g),
time taken for 20 oscillations is measured by
using a watch of 1 second least count. The
mean value of time taken comes out to be
30 s. The length of pendulum is measured by
using a meter scale of least count 1 mm and
the value obtained is 55.0 cm. The percentage
error in the determination of g is close to:
(1) 6.8%
(2) 0.2%
(3) 3.5%
(4) 0.7%
Answer (1)
Sol.
2
2
II
T2 g 4
g T
?? ? ??
gI2T 0.121
100
gI T 55 30
?? ? ? ??
?? ? ? ?
??
??
? 6.8%
16. An electric dipole is formed by two equal and
opposite charges q with separation d. The
charges have same mass m. It is kept in a
uniform electric field E. If it is slightly rotated
from its equilibrium orientation, then its angular
frequency ? is:
(1)
2qE
md
(2)
qE
2
md
(3)
qE
2md
(4)
qE
md
Answer (1)
Sol.
?
d
2
d
2
-q qE
q
qE
222
22
dmdd
Eqd I 2 .
4 dt dt
??
??? ?
?
2
2
d2Eq
md
dt
??
?
?
2Eq
md
??