JEE Advanced 2019 Question Paper - 2 with Solutions | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks: 32)
? This section contains EIGHT (08) questions.
? Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct
answer(s).
? For each question, choose the option(s) corresponding to (all ) the correct answer(s)
? Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONL Y two options are chosen and both
of which are correct.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –1 In all other cases.
? For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2 marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 marks;
choosing ONLY (B) will get +1 marks;
choosing ONLY (D) will get +1 marks;
choosing no option (i.e. the question is unanswered) will get 0 marks, and
choosing any other combination of options will get –1 mark.
1. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially
at pressure P
0
, volume V
0
 and temperature T
0
. If the gas mixture is adiabatically compressed to a volume
V
0
/4, then the correct statement(s) is/are,
(Give 2
1.2
 = 2.3 ; 2
3.2
 = 9.2; R is gas constant)
(1) The final pressure of the gas mixture after compression is in between 9P
0
 and 10P
0
(2) The average kinetic energy of the gas mixture after compression is in between 18RT
0
 and 19RT
0
(3) The work |W| done during the process is 13RT
0
(4) Adiabatic constant of the gas mixture is 1.6
Ans. (1,3,4)
JEE(Advanced) 2019/Paper-2
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-2 TEST PAPER WITH ANSWER
Page 2


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks: 32)
? This section contains EIGHT (08) questions.
? Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct
answer(s).
? For each question, choose the option(s) corresponding to (all ) the correct answer(s)
? Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONL Y two options are chosen and both
of which are correct.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –1 In all other cases.
? For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2 marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 marks;
choosing ONLY (B) will get +1 marks;
choosing ONLY (D) will get +1 marks;
choosing no option (i.e. the question is unanswered) will get 0 marks, and
choosing any other combination of options will get –1 mark.
1. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially
at pressure P
0
, volume V
0
 and temperature T
0
. If the gas mixture is adiabatically compressed to a volume
V
0
/4, then the correct statement(s) is/are,
(Give 2
1.2
 = 2.3 ; 2
3.2
 = 9.2; R is gas constant)
(1) The final pressure of the gas mixture after compression is in between 9P
0
 and 10P
0
(2) The average kinetic energy of the gas mixture after compression is in between 18RT
0
 and 19RT
0
(3) The work |W| done during the process is 13RT
0
(4) Adiabatic constant of the gas mixture is 1.6
Ans. (1,3,4)
JEE(Advanced) 2019/Paper-2
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-2 TEST PAPER WITH ANSWER
Sol. n
1
 = 5 moles C
V
1
 
= 
3R
2
   P
0
V
0
T
0
n
2
 = 1 mole  C
V
2
 = 
5R
2
(C
V
)
m
 = 
12
1 V 2V
12
3R 5R
51
nC nC
22
nn6
´ +´
+
=
+
5R
3
=
g
m
 = 
( )
()
P
m
V
m
c
8
c5
=
\ Option 4 is correct
(C
P
)
m
 = 
5R 8R
R
33
+=
(1) 
8/5 0
000
V
P V P P P (4)
4
g
g
æö
= Þ=
ç÷
èø
 = 9.2 P
0
 which is between 9P
0
 and 10P
0
(2) Average K.E. = 5 × 
3
RT
2
 + 1 × 
5RT
2
= 10RT
To calculate T
000
0
0
PVV
9.2P
T 4T
=´
´
so T = 
0
9.2
T
4
Now average KE = 10 R × 9.2 
0
T
4
 = 23RT
0
(3) W = 
1 1 22
PV PV
1
-
g-
= 
0
000
V
P V 9.2P
4
3/5
-´
 
0
13RT =-
Page 3


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks: 32)
? This section contains EIGHT (08) questions.
? Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct
answer(s).
? For each question, choose the option(s) corresponding to (all ) the correct answer(s)
? Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONL Y two options are chosen and both
of which are correct.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –1 In all other cases.
? For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2 marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 marks;
choosing ONLY (B) will get +1 marks;
choosing ONLY (D) will get +1 marks;
choosing no option (i.e. the question is unanswered) will get 0 marks, and
choosing any other combination of options will get –1 mark.
1. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially
at pressure P
0
, volume V
0
 and temperature T
0
. If the gas mixture is adiabatically compressed to a volume
V
0
/4, then the correct statement(s) is/are,
(Give 2
1.2
 = 2.3 ; 2
3.2
 = 9.2; R is gas constant)
(1) The final pressure of the gas mixture after compression is in between 9P
0
 and 10P
0
(2) The average kinetic energy of the gas mixture after compression is in between 18RT
0
 and 19RT
0
(3) The work |W| done during the process is 13RT
0
(4) Adiabatic constant of the gas mixture is 1.6
Ans. (1,3,4)
JEE(Advanced) 2019/Paper-2
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-2 TEST PAPER WITH ANSWER
Sol. n
1
 = 5 moles C
V
1
 
= 
3R
2
   P
0
V
0
T
0
n
2
 = 1 mole  C
V
2
 = 
5R
2
(C
V
)
m
 = 
12
1 V 2V
12
3R 5R
51
nC nC
22
nn6
´ +´
+
=
+
5R
3
=
g
m
 = 
( )
()
P
m
V
m
c
8
c5
=
\ Option 4 is correct
(C
P
)
m
 = 
5R 8R
R
33
+=
(1) 
8/5 0
000
V
P V P P P (4)
4
g
g
æö
= Þ=
ç÷
èø
 = 9.2 P
0
 which is between 9P
0
 and 10P
0
(2) Average K.E. = 5 × 
3
RT
2
 + 1 × 
5RT
2
= 10RT
To calculate T
000
0
0
PVV
9.2P
T 4T
=´
´
so T = 
0
9.2
T
4
Now average KE = 10 R × 9.2 
0
T
4
 = 23RT
0
(3) W = 
1 1 22
PV PV
1
-
g-
= 
0
000
V
P V 9.2P
4
3/5
-´
 
0
13RT =-
2. An electric dipole with dipole moment 
0
p
ˆˆ
(i j)
2
+
 is held fixed at the origin O in the presence of an
uniform electric field of magnitude E
0
. If the potential is constant on a circle of radius R centered at
the origin as shown in figure, then the correct statement(s) is/are:
(e
0
 is permittivity of free space, R >> dipole size)
x
R
A
y
B
45º
45º
O
(1) 
1/3
0
00
p
R
4E
æö
=
ç÷
pe
èø
(2) The magnitude of total electric field on any two points of the circle will be same
(3) Total electric field at point A is 
A0
ˆˆ
E 2E (i j) =+
r
(4) Total electric field at point B is 
B
E0 =
r
Ans. (1,4)
Sol. (1) ( )
0
P
ˆˆ
P ij
2
=+
r
E.F. at B along tangent should be zero since circle is equipotential.
So, E
0
 = 
3
K|P|
R
r
 & E
B
 = 0
So, R
3
 = 
00
0 00
PP
E 4E
æö K
=
ç÷
pÎ
èø
                  
B
E
0
3
KP
R
3
2KP
R
E
0
So R = 
1/3
0
00
P
4E
æö
ç÷
pÎ
èø
So, (1) is correct
(2) Because E
0
 is uniform & due to dipole E.F. is different at different points, so magnitude of total
E.F. will also be different at different points.
Page 4


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks: 32)
? This section contains EIGHT (08) questions.
? Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct
answer(s).
? For each question, choose the option(s) corresponding to (all ) the correct answer(s)
? Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONL Y two options are chosen and both
of which are correct.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –1 In all other cases.
? For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2 marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 marks;
choosing ONLY (B) will get +1 marks;
choosing ONLY (D) will get +1 marks;
choosing no option (i.e. the question is unanswered) will get 0 marks, and
choosing any other combination of options will get –1 mark.
1. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially
at pressure P
0
, volume V
0
 and temperature T
0
. If the gas mixture is adiabatically compressed to a volume
V
0
/4, then the correct statement(s) is/are,
(Give 2
1.2
 = 2.3 ; 2
3.2
 = 9.2; R is gas constant)
(1) The final pressure of the gas mixture after compression is in between 9P
0
 and 10P
0
(2) The average kinetic energy of the gas mixture after compression is in between 18RT
0
 and 19RT
0
(3) The work |W| done during the process is 13RT
0
(4) Adiabatic constant of the gas mixture is 1.6
Ans. (1,3,4)
JEE(Advanced) 2019/Paper-2
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-2 TEST PAPER WITH ANSWER
Sol. n
1
 = 5 moles C
V
1
 
= 
3R
2
   P
0
V
0
T
0
n
2
 = 1 mole  C
V
2
 = 
5R
2
(C
V
)
m
 = 
12
1 V 2V
12
3R 5R
51
nC nC
22
nn6
´ +´
+
=
+
5R
3
=
g
m
 = 
( )
()
P
m
V
m
c
8
c5
=
\ Option 4 is correct
(C
P
)
m
 = 
5R 8R
R
33
+=
(1) 
8/5 0
000
V
P V P P P (4)
4
g
g
æö
= Þ=
ç÷
èø
 = 9.2 P
0
 which is between 9P
0
 and 10P
0
(2) Average K.E. = 5 × 
3
RT
2
 + 1 × 
5RT
2
= 10RT
To calculate T
000
0
0
PVV
9.2P
T 4T
=´
´
so T = 
0
9.2
T
4
Now average KE = 10 R × 9.2 
0
T
4
 = 23RT
0
(3) W = 
1 1 22
PV PV
1
-
g-
= 
0
000
V
P V 9.2P
4
3/5
-´
 
0
13RT =-
2. An electric dipole with dipole moment 
0
p
ˆˆ
(i j)
2
+
 is held fixed at the origin O in the presence of an
uniform electric field of magnitude E
0
. If the potential is constant on a circle of radius R centered at
the origin as shown in figure, then the correct statement(s) is/are:
(e
0
 is permittivity of free space, R >> dipole size)
x
R
A
y
B
45º
45º
O
(1) 
1/3
0
00
p
R
4E
æö
=
ç÷
pe
èø
(2) The magnitude of total electric field on any two points of the circle will be same
(3) Total electric field at point A is 
A0
ˆˆ
E 2E (i j) =+
r
(4) Total electric field at point B is 
B
E0 =
r
Ans. (1,4)
Sol. (1) ( )
0
P
ˆˆ
P ij
2
=+
r
E.F. at B along tangent should be zero since circle is equipotential.
So, E
0
 = 
3
K|P|
R
r
 & E
B
 = 0
So, R
3
 = 
00
0 00
PP
E 4E
æö K
=
ç÷
pÎ
èø
                  
B
E
0
3
KP
R
3
2KP
R
E
0
So R = 
1/3
0
00
P
4E
æö
ç÷
pÎ
èø
So, (1) is correct
(2) Because E
0
 is uniform & due to dipole E.F. is different at different points, so magnitude of total
E.F. will also be different at different points.
So, (2) is incorrect
(3) ( )
0
A 3 33
2KP KP KP P
ˆˆ
E 3 ij
R RR 2
=+=+
 So, (3) is wrong
(4) E
B
 = 0
so, (4) is correct
3. A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The
rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping.
Which of the following statement(s) is/are correct, when the rod makes an angle 60º with vertical ?
[g is the acceleration due to gravity]
(1) The radial acceleration of the rod's center of mass will be 
3g
4
(2) The angular acceleration of the rod will be 
2g
L
(3) The angular speed of the rod will be 
3g
2L
(4) The normal reaction force from the floor on the rod will be 
Mg
16
Ans. (1,3,4)
Sol. We can treat contact point as hinged.
Applying work energy theorem
W
g
 = DK.E.
mg
2
2
1m
4 23
æö
=w
ç÷
èø
ll
C.M.
l/2
60º
a
t
a
r
30º
C.M.
w
a
3g
2
w=
l
radial acceleration of C.M. of rod 
2
3g
24
æö
= w=
ç÷
èø
l
Using t = I a about contact point
2
mgm
sin60º
23
=a
ll
Page 5


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks: 32)
? This section contains EIGHT (08) questions.
? Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct
answer(s).
? For each question, choose the option(s) corresponding to (all ) the correct answer(s)
? Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONL Y two options are chosen and both
of which are correct.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –1 In all other cases.
? For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then
choosing ONLY (A), (B) and (D) will get +4 marks;
choosing ONLY (A) and (B) will get +2 marks;
choosing ONLY (A) and (D) will get +2 marks;
choosing ONLY (B) and (D) will get +2 marks;
choosing ONLY (A) will get +1 marks;
choosing ONLY (B) will get +1 marks;
choosing ONLY (D) will get +1 marks;
choosing no option (i.e. the question is unanswered) will get 0 marks, and
choosing any other combination of options will get –1 mark.
1. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially
at pressure P
0
, volume V
0
 and temperature T
0
. If the gas mixture is adiabatically compressed to a volume
V
0
/4, then the correct statement(s) is/are,
(Give 2
1.2
 = 2.3 ; 2
3.2
 = 9.2; R is gas constant)
(1) The final pressure of the gas mixture after compression is in between 9P
0
 and 10P
0
(2) The average kinetic energy of the gas mixture after compression is in between 18RT
0
 and 19RT
0
(3) The work |W| done during the process is 13RT
0
(4) Adiabatic constant of the gas mixture is 1.6
Ans. (1,3,4)
JEE(Advanced) 2019/Paper-2
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-2 TEST PAPER WITH ANSWER
Sol. n
1
 = 5 moles C
V
1
 
= 
3R
2
   P
0
V
0
T
0
n
2
 = 1 mole  C
V
2
 = 
5R
2
(C
V
)
m
 = 
12
1 V 2V
12
3R 5R
51
nC nC
22
nn6
´ +´
+
=
+
5R
3
=
g
m
 = 
( )
()
P
m
V
m
c
8
c5
=
\ Option 4 is correct
(C
P
)
m
 = 
5R 8R
R
33
+=
(1) 
8/5 0
000
V
P V P P P (4)
4
g
g
æö
= Þ=
ç÷
èø
 = 9.2 P
0
 which is between 9P
0
 and 10P
0
(2) Average K.E. = 5 × 
3
RT
2
 + 1 × 
5RT
2
= 10RT
To calculate T
000
0
0
PVV
9.2P
T 4T
=´
´
so T = 
0
9.2
T
4
Now average KE = 10 R × 9.2 
0
T
4
 = 23RT
0
(3) W = 
1 1 22
PV PV
1
-
g-
= 
0
000
V
P V 9.2P
4
3/5
-´
 
0
13RT =-
2. An electric dipole with dipole moment 
0
p
ˆˆ
(i j)
2
+
 is held fixed at the origin O in the presence of an
uniform electric field of magnitude E
0
. If the potential is constant on a circle of radius R centered at
the origin as shown in figure, then the correct statement(s) is/are:
(e
0
 is permittivity of free space, R >> dipole size)
x
R
A
y
B
45º
45º
O
(1) 
1/3
0
00
p
R
4E
æö
=
ç÷
pe
èø
(2) The magnitude of total electric field on any two points of the circle will be same
(3) Total electric field at point A is 
A0
ˆˆ
E 2E (i j) =+
r
(4) Total electric field at point B is 
B
E0 =
r
Ans. (1,4)
Sol. (1) ( )
0
P
ˆˆ
P ij
2
=+
r
E.F. at B along tangent should be zero since circle is equipotential.
So, E
0
 = 
3
K|P|
R
r
 & E
B
 = 0
So, R
3
 = 
00
0 00
PP
E 4E
æö K
=
ç÷
pÎ
èø
                  
B
E
0
3
KP
R
3
2KP
R
E
0
So R = 
1/3
0
00
P
4E
æö
ç÷
pÎ
èø
So, (1) is correct
(2) Because E
0
 is uniform & due to dipole E.F. is different at different points, so magnitude of total
E.F. will also be different at different points.
So, (2) is incorrect
(3) ( )
0
A 3 33
2KP KP KP P
ˆˆ
E 3 ij
R RR 2
=+=+
 So, (3) is wrong
(4) E
B
 = 0
so, (4) is correct
3. A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The
rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping.
Which of the following statement(s) is/are correct, when the rod makes an angle 60º with vertical ?
[g is the acceleration due to gravity]
(1) The radial acceleration of the rod's center of mass will be 
3g
4
(2) The angular acceleration of the rod will be 
2g
L
(3) The angular speed of the rod will be 
3g
2L
(4) The normal reaction force from the floor on the rod will be 
Mg
16
Ans. (1,3,4)
Sol. We can treat contact point as hinged.
Applying work energy theorem
W
g
 = DK.E.
mg
2
2
1m
4 23
æö
=w
ç÷
èø
ll
C.M.
l/2
60º
a
t
a
r
30º
C.M.
w
a
3g
2
w=
l
radial acceleration of C.M. of rod 
2
3g
24
æö
= w=
ç÷
èø
l
Using t = I a about contact point
2
mgm
sin60º
23
=a
ll
Þ  a = 
33
g
4l
Net vertical acceleration of C.M. of rod
a
v
 = a
r
 cos 60º + a
t
 cos 30º
= 
3g1
cos30º
422
æ öæöæö
+a
ç ÷ç÷ç÷
è øèøèø
l
= 
3g 3 3g 3
8 4 22
æö
æö
+
ç÷
ç÷
ç÷
èø
èø
l
l
= 
3g 9g 15
g
8 16 16
+=
Applying F
net
 = ma in vertical direction on rod as system
mg – N = ma
v
 = m 
15
g
16
æö
ç÷
èø
a
v
mg
N
mg
N
16
Þ=
4. A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes
elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while
the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the
piston from closed end is L = L
0
 the particle speed is v = v
0
. The piston is moved inward at a very
low speed V such that V << 
0
dL
v
L
, where dL is the infinitesimal displacement of the piston. Which
of the following statement(s) is/are correct ?
v
L
(1) The rate at which the particle strikes the piston is v/L
(2) After each collision with the piston, the particle speed increases by 2V
(3) The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from
L
0
 to 
0
1
L
2
(4) If the piston moves inward by dL, the particle speed increases by 2v
dL
L