JEE Advanced 2019 Question Paper - 1 with Solutions | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks : 12)
? This section contains FOUR (04) questions.
? Each question has FOUR options. ONLY ONE of these four options is the correct answer.
? For each question, choose the option corresponding to the correct answer.
? Answer to each question will be evaluated according to the following marking scheme :
Full Marks : +3 If ONLY the correct option is chosen.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered)
Negative Marks : –1 In all other cases
1. A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal
rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes
with time (t) as :
T(t) = T
0
 (1 + bt
1/4
)
where b is a constant with appropriate dimension while T
0
 is a constant with dimension of temperature.
The heat capacity of the metal is :
(1) 
-
b
3
0
44
0
4P(T(t) T)
T
(2) 
0
42
0
4P(T(t) T)
T
-
b
(3) 
-
b
4
0
45
0
4P(T(t) T)
T
(4) 
-
b
2
0
43
0
4P(T(t) T)
T
Ans. (1)
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-1 TEST PAPER WITH ANSWER & SOLUTION
JEE(Advanced) 2019/Paper-1
Page 2


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks : 12)
? This section contains FOUR (04) questions.
? Each question has FOUR options. ONLY ONE of these four options is the correct answer.
? For each question, choose the option corresponding to the correct answer.
? Answer to each question will be evaluated according to the following marking scheme :
Full Marks : +3 If ONLY the correct option is chosen.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered)
Negative Marks : –1 In all other cases
1. A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal
rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes
with time (t) as :
T(t) = T
0
 (1 + bt
1/4
)
where b is a constant with appropriate dimension while T
0
 is a constant with dimension of temperature.
The heat capacity of the metal is :
(1) 
-
b
3
0
44
0
4P(T(t) T)
T
(2) 
0
42
0
4P(T(t) T)
T
-
b
(3) 
-
b
4
0
45
0
4P(T(t) T)
T
(4) 
-
b
2
0
43
0
4P(T(t) T)
T
Ans. (1)
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-1 TEST PAPER WITH ANSWER & SOLUTION
JEE(Advanced) 2019/Paper-1
Sol. P = 
dQ
dt
T
(t)
 = T
0
 
1/4
(1 t) +b
= Þ=
æö
ç÷
èø
dQ dT P
msS
dT dt dt
dt
3/4 3/4 0
0
dT 1T
T 0 . t .t
dt 44
--
b éù
= +b=
êú
ëû
S = 
3/4
0
P 4P
.t
(dT/dt) T
=
b
S = 
3/4
0
4Pt
T
éù
êú
b
ëû
= +b
1/4
0
T(t)
(1 t)
T
bt
1/4
 = 
-
-=
0
00
T(t)T T(t)
1
TT
t
3/4
 = 
æö -
ç÷
b
èø
3
0
0
T(t)T
.T
Þ S = 
[ ]
éù -
=-
êú
bbb
ëû
3
3
0
0 44
0 00
T(t)T 4P 4P
T(t)T
T .TT
2. A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential
at its surface is V
0
. A hole with a small area a4pR
2
 (a<<1) is made on the shell without affecting the
rest of the shell. Which one of the following statements is correct ?
(1) The ratio of the potential at the center of the shell to that of the point at 
1
2
R from center towards
the hole will be 
1
12
-a
-a
(2) The magnitude of electric field at the center of the shell is reduced by 
0
V
2R
a
(3) The magnitude of electric field at a point, located on a line passing through the hole and shell's center
on a distance 2R from the center of the spherical shell will be reduced by 
0
V
2R
a
(4) The potential  at the center of the shell is reduced by 2aV
0
Ans. (1)
Page 3


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks : 12)
? This section contains FOUR (04) questions.
? Each question has FOUR options. ONLY ONE of these four options is the correct answer.
? For each question, choose the option corresponding to the correct answer.
? Answer to each question will be evaluated according to the following marking scheme :
Full Marks : +3 If ONLY the correct option is chosen.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered)
Negative Marks : –1 In all other cases
1. A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal
rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes
with time (t) as :
T(t) = T
0
 (1 + bt
1/4
)
where b is a constant with appropriate dimension while T
0
 is a constant with dimension of temperature.
The heat capacity of the metal is :
(1) 
-
b
3
0
44
0
4P(T(t) T)
T
(2) 
0
42
0
4P(T(t) T)
T
-
b
(3) 
-
b
4
0
45
0
4P(T(t) T)
T
(4) 
-
b
2
0
43
0
4P(T(t) T)
T
Ans. (1)
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-1 TEST PAPER WITH ANSWER & SOLUTION
JEE(Advanced) 2019/Paper-1
Sol. P = 
dQ
dt
T
(t)
 = T
0
 
1/4
(1 t) +b
= Þ=
æö
ç÷
èø
dQ dT P
msS
dT dt dt
dt
3/4 3/4 0
0
dT 1T
T 0 . t .t
dt 44
--
b éù
= +b=
êú
ëû
S = 
3/4
0
P 4P
.t
(dT/dt) T
=
b
S = 
3/4
0
4Pt
T
éù
êú
b
ëû
= +b
1/4
0
T(t)
(1 t)
T
bt
1/4
 = 
-
-=
0
00
T(t)T T(t)
1
TT
t
3/4
 = 
æö -
ç÷
b
èø
3
0
0
T(t)T
.T
Þ S = 
[ ]
éù -
=-
êú
bbb
ëû
3
3
0
0 44
0 00
T(t)T 4P 4P
T(t)T
T .TT
2. A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential
at its surface is V
0
. A hole with a small area a4pR
2
 (a<<1) is made on the shell without affecting the
rest of the shell. Which one of the following statements is correct ?
(1) The ratio of the potential at the center of the shell to that of the point at 
1
2
R from center towards
the hole will be 
1
12
-a
-a
(2) The magnitude of electric field at the center of the shell is reduced by 
0
V
2R
a
(3) The magnitude of electric field at a point, located on a line passing through the hole and shell's center
on a distance 2R from the center of the spherical shell will be reduced by 
0
V
2R
a
(4) The potential  at the center of the shell is reduced by 2aV
0
Ans. (1)
Sol. Let charge on the sphere initially be Q.
\
0
kQ
V
R
=
and charge removed = aQ
(1)
=
P
C
–
R/2
and V
p
 = ( )
a
- = -a
kQ 2K Q kQ
12
R RR
V
C
 = 
kQ(1)
R
-a
\
C
p
V 1
V 12
-a
=
-a
(2) (E
C
)
initial
 = zero         
E
C
(E
C
)
final
 = 
a
2
kQ
R
Þ Electric field increases
(3) (E
P
)
initial
 = 
2
kQ
4R
              
R
P
R
P
(E
P
)
final
 = 
a
-
22
kQ kQ
4RR
DE
P
 = 
a aa
- + ==
0
2 2 22
V kQ kQ kQ kQ
4R 4R R RR
(4) (V
C
)
initial
 = 
kQ
R
C
C
(V
C
)
final
 = 
kQ(1)
R
-a
DV
C
 = 
0
kQ
()V
R
a =a
Page 4


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks : 12)
? This section contains FOUR (04) questions.
? Each question has FOUR options. ONLY ONE of these four options is the correct answer.
? For each question, choose the option corresponding to the correct answer.
? Answer to each question will be evaluated according to the following marking scheme :
Full Marks : +3 If ONLY the correct option is chosen.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered)
Negative Marks : –1 In all other cases
1. A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal
rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes
with time (t) as :
T(t) = T
0
 (1 + bt
1/4
)
where b is a constant with appropriate dimension while T
0
 is a constant with dimension of temperature.
The heat capacity of the metal is :
(1) 
-
b
3
0
44
0
4P(T(t) T)
T
(2) 
0
42
0
4P(T(t) T)
T
-
b
(3) 
-
b
4
0
45
0
4P(T(t) T)
T
(4) 
-
b
2
0
43
0
4P(T(t) T)
T
Ans. (1)
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-1 TEST PAPER WITH ANSWER & SOLUTION
JEE(Advanced) 2019/Paper-1
Sol. P = 
dQ
dt
T
(t)
 = T
0
 
1/4
(1 t) +b
= Þ=
æö
ç÷
èø
dQ dT P
msS
dT dt dt
dt
3/4 3/4 0
0
dT 1T
T 0 . t .t
dt 44
--
b éù
= +b=
êú
ëû
S = 
3/4
0
P 4P
.t
(dT/dt) T
=
b
S = 
3/4
0
4Pt
T
éù
êú
b
ëû
= +b
1/4
0
T(t)
(1 t)
T
bt
1/4
 = 
-
-=
0
00
T(t)T T(t)
1
TT
t
3/4
 = 
æö -
ç÷
b
èø
3
0
0
T(t)T
.T
Þ S = 
[ ]
éù -
=-
êú
bbb
ëû
3
3
0
0 44
0 00
T(t)T 4P 4P
T(t)T
T .TT
2. A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential
at its surface is V
0
. A hole with a small area a4pR
2
 (a<<1) is made on the shell without affecting the
rest of the shell. Which one of the following statements is correct ?
(1) The ratio of the potential at the center of the shell to that of the point at 
1
2
R from center towards
the hole will be 
1
12
-a
-a
(2) The magnitude of electric field at the center of the shell is reduced by 
0
V
2R
a
(3) The magnitude of electric field at a point, located on a line passing through the hole and shell's center
on a distance 2R from the center of the spherical shell will be reduced by 
0
V
2R
a
(4) The potential  at the center of the shell is reduced by 2aV
0
Ans. (1)
Sol. Let charge on the sphere initially be Q.
\
0
kQ
V
R
=
and charge removed = aQ
(1)
=
P
C
–
R/2
and V
p
 = ( )
a
- = -a
kQ 2K Q kQ
12
R RR
V
C
 = 
kQ(1)
R
-a
\
C
p
V 1
V 12
-a
=
-a
(2) (E
C
)
initial
 = zero         
E
C
(E
C
)
final
 = 
a
2
kQ
R
Þ Electric field increases
(3) (E
P
)
initial
 = 
2
kQ
4R
              
R
P
R
P
(E
P
)
final
 = 
a
-
22
kQ kQ
4RR
DE
P
 = 
a aa
- + ==
0
2 2 22
V kQ kQ kQ kQ
4R 4R R RR
(4) (V
C
)
initial
 = 
kQ
R
C
C
(V
C
)
final
 = 
kQ(1)
R
-a
DV
C
 = 
0
kQ
()V
R
a =a
3. Consider a spherical gaseous cloud of mass density r(r) in free space where r is the radial distance from
its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common
center with the same kinetic energy K. The force acting on the particles  is their mutual gravitational force.
If r(r) is constant in time, the particle number density n(r) = r(r)/m is :
[G is universal gravitational constant]
(1) 22
K
r mG p
(2) 22
K
6 r mG p
(3) 22
3K
r mG p
(4) 22
K
2 r mG p
Ans. (4)
Sol. Let total mass included in a sphere of radius r be M.
For a particle of mass m,
2
2
GMm mv
rr
=
Þ
GMm
2K
r
=
   Þ M = 
2Kr
Gm
\ dM= 
2Kdr
Gm
m
r
M
Þ (4pr
2
dr)r = 
2Kdr
Gm
Þr = 
2
K
2 r Gm p
\ n = 
m
r
= 22
K
2 r mG p
4. In a radioactive sample, 
40
19
K nuclei either decay into stable 
40
20
Ca nuclei with decay constant
4.5 × 10
–10
 per year or into stable 
40
18
Ar nuclei with decay constant 0.5 × 10
–10
 per year. Given that in
this sample all the stable 
40
20
Ca and 
40
18
Ar nuclei are produced by the 
40
19
K nuclei only . In time t × 10
9
 years,
if the ratio of the sum of stable 
40
20
Ca and 
40
18
Ar nuclei to the radioactive 
40
19
K nuclei is 99, the value of t
will be : [Given ln 10 = 2.3]
(1) 9.2 (2) 1.15 (3) 4.6 (4) 2.3
Ans. (1)
Page 5


PART-1 : PHYSICS
SECTION-1 : (Maximum Marks : 12)
? This section contains FOUR (04) questions.
? Each question has FOUR options. ONLY ONE of these four options is the correct answer.
? For each question, choose the option corresponding to the correct answer.
? Answer to each question will be evaluated according to the following marking scheme :
Full Marks : +3 If ONLY the correct option is chosen.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered)
Negative Marks : –1 In all other cases
1. A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal
rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes
with time (t) as :
T(t) = T
0
 (1 + bt
1/4
)
where b is a constant with appropriate dimension while T
0
 is a constant with dimension of temperature.
The heat capacity of the metal is :
(1) 
-
b
3
0
44
0
4P(T(t) T)
T
(2) 
0
42
0
4P(T(t) T)
T
-
b
(3) 
-
b
4
0
45
0
4P(T(t) T)
T
(4) 
-
b
2
0
43
0
4P(T(t) T)
T
Ans. (1)
FINAL JEE(Advanced) EXAMINATION - 2019
(Held On Monday 27
th
 MAY , 2019)
PAPER-1 TEST PAPER WITH ANSWER & SOLUTION
JEE(Advanced) 2019/Paper-1
Sol. P = 
dQ
dt
T
(t)
 = T
0
 
1/4
(1 t) +b
= Þ=
æö
ç÷
èø
dQ dT P
msS
dT dt dt
dt
3/4 3/4 0
0
dT 1T
T 0 . t .t
dt 44
--
b éù
= +b=
êú
ëû
S = 
3/4
0
P 4P
.t
(dT/dt) T
=
b
S = 
3/4
0
4Pt
T
éù
êú
b
ëû
= +b
1/4
0
T(t)
(1 t)
T
bt
1/4
 = 
-
-=
0
00
T(t)T T(t)
1
TT
t
3/4
 = 
æö -
ç÷
b
èø
3
0
0
T(t)T
.T
Þ S = 
[ ]
éù -
=-
êú
bbb
ëû
3
3
0
0 44
0 00
T(t)T 4P 4P
T(t)T
T .TT
2. A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential
at its surface is V
0
. A hole with a small area a4pR
2
 (a<<1) is made on the shell without affecting the
rest of the shell. Which one of the following statements is correct ?
(1) The ratio of the potential at the center of the shell to that of the point at 
1
2
R from center towards
the hole will be 
1
12
-a
-a
(2) The magnitude of electric field at the center of the shell is reduced by 
0
V
2R
a
(3) The magnitude of electric field at a point, located on a line passing through the hole and shell's center
on a distance 2R from the center of the spherical shell will be reduced by 
0
V
2R
a
(4) The potential  at the center of the shell is reduced by 2aV
0
Ans. (1)
Sol. Let charge on the sphere initially be Q.
\
0
kQ
V
R
=
and charge removed = aQ
(1)
=
P
C
–
R/2
and V
p
 = ( )
a
- = -a
kQ 2K Q kQ
12
R RR
V
C
 = 
kQ(1)
R
-a
\
C
p
V 1
V 12
-a
=
-a
(2) (E
C
)
initial
 = zero         
E
C
(E
C
)
final
 = 
a
2
kQ
R
Þ Electric field increases
(3) (E
P
)
initial
 = 
2
kQ
4R
              
R
P
R
P
(E
P
)
final
 = 
a
-
22
kQ kQ
4RR
DE
P
 = 
a aa
- + ==
0
2 2 22
V kQ kQ kQ kQ
4R 4R R RR
(4) (V
C
)
initial
 = 
kQ
R
C
C
(V
C
)
final
 = 
kQ(1)
R
-a
DV
C
 = 
0
kQ
()V
R
a =a
3. Consider a spherical gaseous cloud of mass density r(r) in free space where r is the radial distance from
its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common
center with the same kinetic energy K. The force acting on the particles  is their mutual gravitational force.
If r(r) is constant in time, the particle number density n(r) = r(r)/m is :
[G is universal gravitational constant]
(1) 22
K
r mG p
(2) 22
K
6 r mG p
(3) 22
3K
r mG p
(4) 22
K
2 r mG p
Ans. (4)
Sol. Let total mass included in a sphere of radius r be M.
For a particle of mass m,
2
2
GMm mv
rr
=
Þ
GMm
2K
r
=
   Þ M = 
2Kr
Gm
\ dM= 
2Kdr
Gm
m
r
M
Þ (4pr
2
dr)r = 
2Kdr
Gm
Þr = 
2
K
2 r Gm p
\ n = 
m
r
= 22
K
2 r mG p
4. In a radioactive sample, 
40
19
K nuclei either decay into stable 
40
20
Ca nuclei with decay constant
4.5 × 10
–10
 per year or into stable 
40
18
Ar nuclei with decay constant 0.5 × 10
–10
 per year. Given that in
this sample all the stable 
40
20
Ca and 
40
18
Ar nuclei are produced by the 
40
19
K nuclei only . In time t × 10
9
 years,
if the ratio of the sum of stable 
40
20
Ca and 
40
18
Ar nuclei to the radioactive 
40
19
K nuclei is 99, the value of t
will be : [Given ln 10 = 2.3]
(1) 9.2 (2) 1.15 (3) 4.6 (4) 2.3
Ans. (1)
Sol. Parallel radioactive decay
l
1
 = l
l
2
 = 9l
Ca
40
20
K
20
19
Ar
20
18
l = l
1
 + l
2
 = 5 × 10
–10
 per year
N = N
0
e
–lt
N
0
 – N = N
stable
N = N
radioactive
0
N
1 99
N
-=
0
N
100
N
=
t
0
N1
e
N 100
-l
==
Þ lt = 2 ln10
= 4.6
t = 9.2 × 10
9
 years
SECTION-2 : (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.