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JEE-Advanced-28-08-2022 
[Paper-2]  
 
PHYSICS 
 
Question: A particle of mass 1 kg is subjected to a force which depends on the position as 
( )
2
ˆˆ
F k xi yj kgms
-
=-+
??
 with k =1 kgs
-2
. At time t = 0, the particle’s position 
1
ˆˆ
2
2
r i jm
??
= +
??
??
?
and its velocity
1
2
ˆ ˆˆ
22 v i j k ms
p
-
??
= -+ +
??
??
?
. Let vx and vy denote the x and 
the y components of the particle’s velocity, respectively. Ignore gravity. When z = 0.5 m, the 
value of 
( )
yx
xv yv - is _____ m
2
 s
-1
. 
Answer: 3.00 
Solution: 
( ) ( )
( )
x
2
2
ma
,
sin         
m
       1
m
and cos
1
0, 
2
and 2 m / s
1
sin
2
and 2 co
1
 
n
/
s
ta
2
5
and   
2
x x x
x
x x
xx
x
x
x
xx
x
vA t
A
F
A
att x
d
v
So A
A
x
x
So a x
dt
x t rad s
= +
= =
=- =
= = -
?= +
=
=
= -
=
-
?=-
=
??
?
f
f
f
f
f
?
 
Similarly 
2
2
y y
F y ma
dy
y
dt
=-=
?=-
 
Page 2


   
JEE-Advanced-28-08-2022 
[Paper-2]  
 
PHYSICS 
 
Question: A particle of mass 1 kg is subjected to a force which depends on the position as 
( )
2
ˆˆ
F k xi yj kgms
-
=-+
??
 with k =1 kgs
-2
. At time t = 0, the particle’s position 
1
ˆˆ
2
2
r i jm
??
= +
??
??
?
and its velocity
1
2
ˆ ˆˆ
22 v i j k ms
p
-
??
= -+ +
??
??
?
. Let vx and vy denote the x and 
the y components of the particle’s velocity, respectively. Ignore gravity. When z = 0.5 m, the 
value of 
( )
yx
xv yv - is _____ m
2
 s
-1
. 
Answer: 3.00 
Solution: 
( ) ( )
( )
x
2
2
ma
,
sin         
m
       1
m
and cos
1
0, 
2
and 2 m / s
1
sin
2
and 2 co
1
 
n
/
s
ta
2
5
and   
2
x x x
x
x x
xx
x
x
x
xx
x
vA t
A
F
A
att x
d
v
So A
A
x
x
So a x
dt
x t rad s
= +
= =
=- =
= = -
?= +
=
=
= -
=
-
?=-
=
??
?
f
f
f
f
f
?
 
Similarly 
2
2
y y
F y ma
dy
y
dt
=-=
?=-
 
   
( )
, sin( )
( 1rad / s)and cos
0 2 m
and 2 m / s
So 2 sin   and 
2 cos
and 2
4
yy
yy y
y
yy
yy
y y
So y A t
v A t
att y
v
A
A
A
= +
= = +
= =
=
=
=
?= =
?f
? ? ?f
f
f
p
f
 
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
(
( )
( )
55
 So, sin 2cos 2sin cos
22
5
2 sin cos sin cos
2
10 sin
10 sin cos cos sin
y x x y y x
x y y x
xy
x y xy
xv yv t t t t
t t t t
n
?f ?f ?f ?f
?f ?f ?f ?f
ff
f f ff
- = + × + - + × +
= × + + - +× +
= -
= -
 
11 2 1
10 3
52 5 2
?? ??
= × - - × =
?? ??
?? ??
 
 
Question: In a radioactive decay chain reaction, 
230
90
Th nucleus decays into 
214
84
Po nucleus. The 
ratio of the number of a to number of 
-
ß particles emitted in this process is_______. 
Answer: 2.00 
Solution: 
Let number of a particles are n
a
 and 
-
ß particles are n
ß
so 
( )
4 230 214
4
84 90 2
2
So 2
n
n
nn
n
n
n
a
a
ßa
ß
a
ß
= -
?=
= --
=
=
 
 
Question: Two resistances 
1
R X = ? and 
2
1 R = ?are connected to a wire AB of uniform 
resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 
mm at A to l mm at B. A galvanometer (G) connected to the center of the wire, 50 cm from 
Page 3


   
JEE-Advanced-28-08-2022 
[Paper-2]  
 
PHYSICS 
 
Question: A particle of mass 1 kg is subjected to a force which depends on the position as 
( )
2
ˆˆ
F k xi yj kgms
-
=-+
??
 with k =1 kgs
-2
. At time t = 0, the particle’s position 
1
ˆˆ
2
2
r i jm
??
= +
??
??
?
and its velocity
1
2
ˆ ˆˆ
22 v i j k ms
p
-
??
= -+ +
??
??
?
. Let vx and vy denote the x and 
the y components of the particle’s velocity, respectively. Ignore gravity. When z = 0.5 m, the 
value of 
( )
yx
xv yv - is _____ m
2
 s
-1
. 
Answer: 3.00 
Solution: 
( ) ( )
( )
x
2
2
ma
,
sin         
m
       1
m
and cos
1
0, 
2
and 2 m / s
1
sin
2
and 2 co
1
 
n
/
s
ta
2
5
and   
2
x x x
x
x x
xx
x
x
x
xx
x
vA t
A
F
A
att x
d
v
So A
A
x
x
So a x
dt
x t rad s
= +
= =
=- =
= = -
?= +
=
=
= -
=
-
?=-
=
??
?
f
f
f
f
f
?
 
Similarly 
2
2
y y
F y ma
dy
y
dt
=-=
?=-
 
   
( )
, sin( )
( 1rad / s)and cos
0 2 m
and 2 m / s
So 2 sin   and 
2 cos
and 2
4
yy
yy y
y
yy
yy
y y
So y A t
v A t
att y
v
A
A
A
= +
= = +
= =
=
=
=
?= =
?f
? ? ?f
f
f
p
f
 
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
(
( )
( )
55
 So, sin 2cos 2sin cos
22
5
2 sin cos sin cos
2
10 sin
10 sin cos cos sin
y x x y y x
x y y x
xy
x y xy
xv yv t t t t
t t t t
n
?f ?f ?f ?f
?f ?f ?f ?f
ff
f f ff
- = + × + - + × +
= × + + - +× +
= -
= -
 
11 2 1
10 3
52 5 2
?? ??
= × - - × =
?? ??
?? ??
 
 
Question: In a radioactive decay chain reaction, 
230
90
Th nucleus decays into 
214
84
Po nucleus. The 
ratio of the number of a to number of 
-
ß particles emitted in this process is_______. 
Answer: 2.00 
Solution: 
Let number of a particles are n
a
 and 
-
ß particles are n
ß
so 
( )
4 230 214
4
84 90 2
2
So 2
n
n
nn
n
n
n
a
a
ßa
ß
a
ß
= -
?=
= --
=
=
 
 
Question: Two resistances 
1
R X = ? and 
2
1 R = ?are connected to a wire AB of uniform 
resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 
mm at A to l mm at B. A galvanometer (G) connected to the center of the wire, 50 cm from 
   
each end along its axis, shows zero deflection when A and B are connected to a battery. The 
value of X is ______ 
 
Answer: 5.00 
Solution: 
Resistance of frustum shaped conductor shown is 
 
l
R
ab
?
p
= 
For the shown conductor in the diagram 
 
Thus, the resistance of left half is 
6
0.5 10
0.2 0.6
P
?
p
××
=
××
 
And the resistance of right half is 
6
0.5 10
0.6 1
Q
?
p
××
=
××
 
For Wheatstone to be balanced 
Page 4


   
JEE-Advanced-28-08-2022 
[Paper-2]  
 
PHYSICS 
 
Question: A particle of mass 1 kg is subjected to a force which depends on the position as 
( )
2
ˆˆ
F k xi yj kgms
-
=-+
??
 with k =1 kgs
-2
. At time t = 0, the particle’s position 
1
ˆˆ
2
2
r i jm
??
= +
??
??
?
and its velocity
1
2
ˆ ˆˆ
22 v i j k ms
p
-
??
= -+ +
??
??
?
. Let vx and vy denote the x and 
the y components of the particle’s velocity, respectively. Ignore gravity. When z = 0.5 m, the 
value of 
( )
yx
xv yv - is _____ m
2
 s
-1
. 
Answer: 3.00 
Solution: 
( ) ( )
( )
x
2
2
ma
,
sin         
m
       1
m
and cos
1
0, 
2
and 2 m / s
1
sin
2
and 2 co
1
 
n
/
s
ta
2
5
and   
2
x x x
x
x x
xx
x
x
x
xx
x
vA t
A
F
A
att x
d
v
So A
A
x
x
So a x
dt
x t rad s
= +
= =
=- =
= = -
?= +
=
=
= -
=
-
?=-
=
??
?
f
f
f
f
f
?
 
Similarly 
2
2
y y
F y ma
dy
y
dt
=-=
?=-
 
   
( )
, sin( )
( 1rad / s)and cos
0 2 m
and 2 m / s
So 2 sin   and 
2 cos
and 2
4
yy
yy y
y
yy
yy
y y
So y A t
v A t
att y
v
A
A
A
= +
= = +
= =
=
=
=
?= =
?f
? ? ?f
f
f
p
f
 
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
(
( )
( )
55
 So, sin 2cos 2sin cos
22
5
2 sin cos sin cos
2
10 sin
10 sin cos cos sin
y x x y y x
x y y x
xy
x y xy
xv yv t t t t
t t t t
n
?f ?f ?f ?f
?f ?f ?f ?f
ff
f f ff
- = + × + - + × +
= × + + - +× +
= -
= -
 
11 2 1
10 3
52 5 2
?? ??
= × - - × =
?? ??
?? ??
 
 
Question: In a radioactive decay chain reaction, 
230
90
Th nucleus decays into 
214
84
Po nucleus. The 
ratio of the number of a to number of 
-
ß particles emitted in this process is_______. 
Answer: 2.00 
Solution: 
Let number of a particles are n
a
 and 
-
ß particles are n
ß
so 
( )
4 230 214
4
84 90 2
2
So 2
n
n
nn
n
n
n
a
a
ßa
ß
a
ß
= -
?=
= --
=
=
 
 
Question: Two resistances 
1
R X = ? and 
2
1 R = ?are connected to a wire AB of uniform 
resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 
mm at A to l mm at B. A galvanometer (G) connected to the center of the wire, 50 cm from 
   
each end along its axis, shows zero deflection when A and B are connected to a battery. The 
value of X is ______ 
 
Answer: 5.00 
Solution: 
Resistance of frustum shaped conductor shown is 
 
l
R
ab
?
p
= 
For the shown conductor in the diagram 
 
Thus, the resistance of left half is 
6
0.5 10
0.2 0.6
P
?
p
××
=
××
 
And the resistance of right half is 
6
0.5 10
0.6 1
Q
?
p
××
=
××
 
For Wheatstone to be balanced 
   
12
66
0.2 0.6 1 0.6 1
0.5 10 0.5 10
RR
PQ
X
=
×× × ×× ×
=
×× ××
pp
? ?
 
5 X ?= 
Question: In a particular system of units, a physical quantity can be expressed in terms of the 
electric charge e, electron mass ???? e, Planck’s constant h, and Coulomb’s constant 
0
1
,
4
k
pe
= 
 where 
0
e is the permittivity of vacuum. In terms of these physical constants, the dimension of 
the magnetic field is
[ ] [ ] [ ] [ ] [ ] .
e
B e m hk
a ß? d
= The value of a ß? d + + + is ______ 
Answer: 4.00 
Solution: 
[ ] [ ] [ ] [ ] [ ]
[ ] [ ]
2
1 1 2 1 3 42
e
B e m hk
M T I IT M MLT MLT I
-
- - --
=
??
? ?? ? =
? ?? ?
??
a ß? d
?d
aß
 
So, 1 ß? d + + =  …(i) 
23 0 += ?d   …(ii) 
42 a? d -- =-  …(iii) 
21 ad -=-   …(iv) 
On solving 
So, 4 a ß? d + + + = 
Question: Consider a configuration of n identical units, each consisting of three layers. The 
first layer is a column of air of height 
1
,
3
h cm = and the second and third layers are of equal 
thickness 
3 1
,
2
d cm
-
= and refractive indices 
12
3
and 3,
2
= = µµ respectively.  A light 
source O is placed on the top of the first unit, as shown in the figure. A ray of light from O is 
incident on the second layer of the first unit at an angle of ???? = 60° to the normal. For a specific 
value of n, the ray of light emerges from the bottom of the configuration at a distance 
8
,
3
l cm = as shown in the figure. The value of n is ____. 
Page 5


   
JEE-Advanced-28-08-2022 
[Paper-2]  
 
PHYSICS 
 
Question: A particle of mass 1 kg is subjected to a force which depends on the position as 
( )
2
ˆˆ
F k xi yj kgms
-
=-+
??
 with k =1 kgs
-2
. At time t = 0, the particle’s position 
1
ˆˆ
2
2
r i jm
??
= +
??
??
?
and its velocity
1
2
ˆ ˆˆ
22 v i j k ms
p
-
??
= -+ +
??
??
?
. Let vx and vy denote the x and 
the y components of the particle’s velocity, respectively. Ignore gravity. When z = 0.5 m, the 
value of 
( )
yx
xv yv - is _____ m
2
 s
-1
. 
Answer: 3.00 
Solution: 
( ) ( )
( )
x
2
2
ma
,
sin         
m
       1
m
and cos
1
0, 
2
and 2 m / s
1
sin
2
and 2 co
1
 
n
/
s
ta
2
5
and   
2
x x x
x
x x
xx
x
x
x
xx
x
vA t
A
F
A
att x
d
v
So A
A
x
x
So a x
dt
x t rad s
= +
= =
=- =
= = -
?= +
=
=
= -
=
-
?=-
=
??
?
f
f
f
f
f
?
 
Similarly 
2
2
y y
F y ma
dy
y
dt
=-=
?=-
 
   
( )
, sin( )
( 1rad / s)and cos
0 2 m
and 2 m / s
So 2 sin   and 
2 cos
and 2
4
yy
yy y
y
yy
yy
y y
So y A t
v A t
att y
v
A
A
A
= +
= = +
= =
=
=
=
?= =
?f
? ? ?f
f
f
p
f
 
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
(
( )
( )
55
 So, sin 2cos 2sin cos
22
5
2 sin cos sin cos
2
10 sin
10 sin cos cos sin
y x x y y x
x y y x
xy
x y xy
xv yv t t t t
t t t t
n
?f ?f ?f ?f
?f ?f ?f ?f
ff
f f ff
- = + × + - + × +
= × + + - +× +
= -
= -
 
11 2 1
10 3
52 5 2
?? ??
= × - - × =
?? ??
?? ??
 
 
Question: In a radioactive decay chain reaction, 
230
90
Th nucleus decays into 
214
84
Po nucleus. The 
ratio of the number of a to number of 
-
ß particles emitted in this process is_______. 
Answer: 2.00 
Solution: 
Let number of a particles are n
a
 and 
-
ß particles are n
ß
so 
( )
4 230 214
4
84 90 2
2
So 2
n
n
nn
n
n
n
a
a
ßa
ß
a
ß
= -
?=
= --
=
=
 
 
Question: Two resistances 
1
R X = ? and 
2
1 R = ?are connected to a wire AB of uniform 
resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 
mm at A to l mm at B. A galvanometer (G) connected to the center of the wire, 50 cm from 
   
each end along its axis, shows zero deflection when A and B are connected to a battery. The 
value of X is ______ 
 
Answer: 5.00 
Solution: 
Resistance of frustum shaped conductor shown is 
 
l
R
ab
?
p
= 
For the shown conductor in the diagram 
 
Thus, the resistance of left half is 
6
0.5 10
0.2 0.6
P
?
p
××
=
××
 
And the resistance of right half is 
6
0.5 10
0.6 1
Q
?
p
××
=
××
 
For Wheatstone to be balanced 
   
12
66
0.2 0.6 1 0.6 1
0.5 10 0.5 10
RR
PQ
X
=
×× × ×× ×
=
×× ××
pp
? ?
 
5 X ?= 
Question: In a particular system of units, a physical quantity can be expressed in terms of the 
electric charge e, electron mass ???? e, Planck’s constant h, and Coulomb’s constant 
0
1
,
4
k
pe
= 
 where 
0
e is the permittivity of vacuum. In terms of these physical constants, the dimension of 
the magnetic field is
[ ] [ ] [ ] [ ] [ ] .
e
B e m hk
a ß? d
= The value of a ß? d + + + is ______ 
Answer: 4.00 
Solution: 
[ ] [ ] [ ] [ ] [ ]
[ ] [ ]
2
1 1 2 1 3 42
e
B e m hk
M T I IT M MLT MLT I
-
- - --
=
??
? ?? ? =
? ?? ?
??
a ß? d
?d
aß
 
So, 1 ß? d + + =  …(i) 
23 0 += ?d   …(ii) 
42 a? d -- =-  …(iii) 
21 ad -=-   …(iv) 
On solving 
So, 4 a ß? d + + + = 
Question: Consider a configuration of n identical units, each consisting of three layers. The 
first layer is a column of air of height 
1
,
3
h cm = and the second and third layers are of equal 
thickness 
3 1
,
2
d cm
-
= and refractive indices 
12
3
and 3,
2
= = µµ respectively.  A light 
source O is placed on the top of the first unit, as shown in the figure. A ray of light from O is 
incident on the second layer of the first unit at an angle of ???? = 60° to the normal. For a specific 
value of n, the ray of light emerges from the bottom of the configuration at a distance 
8
,
3
l cm = as shown in the figure. The value of n is ____. 
   
 
Answer: 4.00 
Solution: 
 
o
1
11
tan 60
3 3
x x cm = = 
and, 
2
3
1 sin 60 sin
2
o
×= × ? 
o
2
2
45
xd
? ?=
? =
 
and, 
3
31
3 sin
2 2
×= × ? 
o
3
30 ? = ? 
3
3
d
x ?= 
( )
12 3
3 1
11
1
2 33
2
3
xx x
cm
-
??
? + + = + +
??
??
=
 
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FAQs on JEE Advance 2022 Question Paper - 2 with Solutions - JEE Main & Advanced Previous Year Papers

1. What is the format of JEE Advanced 2022 Question Paper?
Ans. The JEE Advanced 2022 Question Paper will consist of two papers - Paper 1 and Paper 2. Each paper will have multiple-choice questions as well as numerical answer type questions.
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Ans. The JEE Advanced 2022 Question Paper will be available on the official website of the exam conducting authority. Candidates can download the question paper from the website once it is released.
3. Are the solutions provided for JEE Advanced 2022 Question Paper accurate?
Ans. Yes, the solutions provided for the JEE Advanced 2022 Question Paper are accurate. They are prepared by subject experts and are thoroughly checked for correctness before being published.
4. Can I use a calculator while solving the JEE Advanced 2022 Question Paper?
Ans. No, the use of calculators or any other electronic devices is strictly prohibited during the JEE Advanced 2022 examination. Candidates are expected to solve the questions manually without any external aids.
5. How can I analyze my performance using the JEE Advanced 2022 Question Paper?
Ans. After attempting the JEE Advanced 2022 Question Paper, candidates can refer to the provided solutions to check their answers. They can evaluate their performance by comparing their answers with the correct solutions and calculate their estimated score. This analysis will help them identify their strengths and weaknesses in different subjects.
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