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Page 1
IIT-JAM 2022
Section A
22
4 zz + =
Multiple Choice Questions
Q.1-Q.10 Carry ONE marks each.
Q1. The equation in the complex plane (where z is the complex conjugate of z )
represents
(a) Ellipse (b) Hyperbola (c) Circle of radius 2 (d) Circle of radius 4
Ans.1: (b)
Solution: & z x iy z x iy =+ =-
( ) ( )
22
2 2
4 4 z z x iy x iy
-
+ =? + + - =
22
2 xy ?- = Equation of Hyperbola
Q2. A rocket ( ) ' S moves at a speed /
2
c
ms along the positive x -axis, where c is the speed of
light. When it crosses the origin, the clocks attached to the rocket and the one with a
stationary observer ( ) S located at 0 x = are both set to zero. If S observes an event at
( ) , xt the same event occurs in the ' S frame at
(a)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
??
= -
??
??
(b)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
??
= -
??
??
(c)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
(d)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
Ans. 2: (a)
Solution: From the question one Rocket ' S moving with speed of /2 C and one observer
is standing 0 x = . For point P the coordinate at S frame is ( ) , xy and the coordinate at
' S frame is ( ) ', ' x y
Page 2
IIT-JAM 2022
Section A
22
4 zz + =
Multiple Choice Questions
Q.1-Q.10 Carry ONE marks each.
Q1. The equation in the complex plane (where z is the complex conjugate of z )
represents
(a) Ellipse (b) Hyperbola (c) Circle of radius 2 (d) Circle of radius 4
Ans.1: (b)
Solution: & z x iy z x iy =+ =-
( ) ( )
22
2 2
4 4 z z x iy x iy
-
+ =? + + - =
22
2 xy ?- = Equation of Hyperbola
Q2. A rocket ( ) ' S moves at a speed /
2
c
ms along the positive x -axis, where c is the speed of
light. When it crosses the origin, the clocks attached to the rocket and the one with a
stationary observer ( ) S located at 0 x = are both set to zero. If S observes an event at
( ) , xt the same event occurs in the ' S frame at
(a)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
??
= -
??
??
(b)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
??
= -
??
??
(c)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
(d)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
Ans. 2: (a)
Solution: From the question one Rocket ' S moving with speed of /2 C and one observer
is standing 0 x = . For point P the coordinate at S frame is ( ) , xy and the coordinate at
' S frame is ( ) ', ' x y
22
22
2
'
1 1
4
c
xt
x vt
x
vc
cc
-
-
= =
- -
;
1
'1
24
c
x xt
??
=--
??
??
;
[ ]
2
' / 2
3
x x ct = -
2
2
22 2 2
/
/
2
'
1/ 1/ 4
c
t xc
t xv c
t
vc c c
-
-
= =
--
;
1
'1
24
x
tt
c
? ?
=--
? ?
? ?
;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
[ ]
2
' /2
3
x x ct = - ;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
So option (a) is correct.
Q3. Consider a classical ideal gas of N molecules equilibrium at temperature T . Each
molecule has two energy levels - ? and ?. The mean energy of the gas is
(a) 0 (b) tanh
B
N
kT
?? ?
?
??
??
(c) tanh
B
N
kT
?? ?
-?
??
??
(d)
2
?
Ans. 3: (c)
Solution: The partition function for a single gas molecule is
1
Ze e
ße ße -
= +
Mean energy per particle,
( )
1
ln
ln
ee
Z
E
ße ße
ßß
-
??
-? +
?
??
=-=-
??
( ) ( ) ( )
1
e e
ee
ße ße
ße ße
e e
-
-
? ?
=- -+
? ?
+
ee
ee
ße ße
ße ße
e
-
-
-
= -
+
? for a classical system of N -molecules,
the mean energy is
1
tanh
ee
U NE N N
e e kT
ße ße
ße ße
ß
e
ee
-
-
??
+
= =-=-
??
??
+
??
x
Observer
S Frame
0 x =
0
t t =
Frame
Frame
S S '
/2 c
p
( ) ,: xt s
( ) ,: xt s '' '
Rocket S '
/ 2, 0 Ct =
Page 3
IIT-JAM 2022
Section A
22
4 zz + =
Multiple Choice Questions
Q.1-Q.10 Carry ONE marks each.
Q1. The equation in the complex plane (where z is the complex conjugate of z )
represents
(a) Ellipse (b) Hyperbola (c) Circle of radius 2 (d) Circle of radius 4
Ans.1: (b)
Solution: & z x iy z x iy =+ =-
( ) ( )
22
2 2
4 4 z z x iy x iy
-
+ =? + + - =
22
2 xy ?- = Equation of Hyperbola
Q2. A rocket ( ) ' S moves at a speed /
2
c
ms along the positive x -axis, where c is the speed of
light. When it crosses the origin, the clocks attached to the rocket and the one with a
stationary observer ( ) S located at 0 x = are both set to zero. If S observes an event at
( ) , xt the same event occurs in the ' S frame at
(a)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
??
= -
??
??
(b)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
??
= -
??
??
(c)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
(d)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
Ans. 2: (a)
Solution: From the question one Rocket ' S moving with speed of /2 C and one observer
is standing 0 x = . For point P the coordinate at S frame is ( ) , xy and the coordinate at
' S frame is ( ) ', ' x y
22
22
2
'
1 1
4
c
xt
x vt
x
vc
cc
-
-
= =
- -
;
1
'1
24
c
x xt
??
=--
??
??
;
[ ]
2
' / 2
3
x x ct = -
2
2
22 2 2
/
/
2
'
1/ 1/ 4
c
t xc
t xv c
t
vc c c
-
-
= =
--
;
1
'1
24
x
tt
c
? ?
=--
? ?
? ?
;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
[ ]
2
' /2
3
x x ct = - ;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
So option (a) is correct.
Q3. Consider a classical ideal gas of N molecules equilibrium at temperature T . Each
molecule has two energy levels - ? and ?. The mean energy of the gas is
(a) 0 (b) tanh
B
N
kT
?? ?
?
??
??
(c) tanh
B
N
kT
?? ?
-?
??
??
(d)
2
?
Ans. 3: (c)
Solution: The partition function for a single gas molecule is
1
Ze e
ße ße -
= +
Mean energy per particle,
( )
1
ln
ln
ee
Z
E
ße ße
ßß
-
??
-? +
?
??
=-=-
??
( ) ( ) ( )
1
e e
ee
ße ße
ße ße
e e
-
-
? ?
=- -+
? ?
+
ee
ee
ße ße
ße ße
e
-
-
-
= -
+
? for a classical system of N -molecules,
the mean energy is
1
tanh
ee
U NE N N
e e kT
ße ße
ße ße
ß
e
ee
-
-
??
+
= =-=-
??
??
+
??
x
Observer
S Frame
0 x =
0
t t =
Frame
Frame
S S '
/2 c
p
( ) ,: xt s
( ) ,: xt s '' '
Rocket S '
/ 2, 0 Ct =
Q4. At a temperature T, let ß and k denote the volume expansively and isothermal
compressibility of a gas, respectively. Then
k
ß
is equal to
(a)
V
P
T
? ??
??
?
??
(b)
T
P
V
? ??
??
?
??
(c)
V
T
P
? ??
??
?
??
(d)
P
T
V
? ??
??
?
??
Ans. 4: (a)
Solution: given
1 V
VT
ß
? ??
=
??
?
??
and
1
T
V
k
VP
- ? ??
=
??
?
??
1
1
P
T
V
VT
V k
VP
ß
? ??
??
?
??
=
- ? ??
??
?
??
1
PT
VV
V
VT P
?? ?? ??
= -
?? ??
??
?? ??
TP V
PV P
VT T
?? ? ???? ? ?
=-=
???? ? ?
?? ?
???? ? ?
Here, in the last step we made use of reciprocity theorem i.e
xx
yy
y
zx
z
z
? ? ?? ? ?? ??
= -
? ? ?? ??
? ??
??
? ? ??
Q5. The resultant of the binary subtraction 1110101 0011110 - is
(a) 1001111 (b) 1010111 (c) 1010011 (d) 1010001
Ans. 5: (b)
Solution: 2s compliment of 0011110 ?
1100001
1
1100010
+
Thus 1110101 0011110 - is
1110101
1100010
11010111
Q6. Consider a particle trapped in a three-dimensional potential well such that
( ) , , 0 U x yz = for 0 ,0 ,0 xa y a z a = = = = = = and ( ) , , U x yz = 8 everywhere else. The
degeneracy of the
th
5 excited state is
(a) 1 (b) 3 (c) 6 (d) 9
Page 4
IIT-JAM 2022
Section A
22
4 zz + =
Multiple Choice Questions
Q.1-Q.10 Carry ONE marks each.
Q1. The equation in the complex plane (where z is the complex conjugate of z )
represents
(a) Ellipse (b) Hyperbola (c) Circle of radius 2 (d) Circle of radius 4
Ans.1: (b)
Solution: & z x iy z x iy =+ =-
( ) ( )
22
2 2
4 4 z z x iy x iy
-
+ =? + + - =
22
2 xy ?- = Equation of Hyperbola
Q2. A rocket ( ) ' S moves at a speed /
2
c
ms along the positive x -axis, where c is the speed of
light. When it crosses the origin, the clocks attached to the rocket and the one with a
stationary observer ( ) S located at 0 x = are both set to zero. If S observes an event at
( ) , xt the same event occurs in the ' S frame at
(a)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
??
= -
??
??
(b)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
??
= -
??
??
(c)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
(d)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
Ans. 2: (a)
Solution: From the question one Rocket ' S moving with speed of /2 C and one observer
is standing 0 x = . For point P the coordinate at S frame is ( ) , xy and the coordinate at
' S frame is ( ) ', ' x y
22
22
2
'
1 1
4
c
xt
x vt
x
vc
cc
-
-
= =
- -
;
1
'1
24
c
x xt
??
=--
??
??
;
[ ]
2
' / 2
3
x x ct = -
2
2
22 2 2
/
/
2
'
1/ 1/ 4
c
t xc
t xv c
t
vc c c
-
-
= =
--
;
1
'1
24
x
tt
c
? ?
=--
? ?
? ?
;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
[ ]
2
' /2
3
x x ct = - ;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
So option (a) is correct.
Q3. Consider a classical ideal gas of N molecules equilibrium at temperature T . Each
molecule has two energy levels - ? and ?. The mean energy of the gas is
(a) 0 (b) tanh
B
N
kT
?? ?
?
??
??
(c) tanh
B
N
kT
?? ?
-?
??
??
(d)
2
?
Ans. 3: (c)
Solution: The partition function for a single gas molecule is
1
Ze e
ße ße -
= +
Mean energy per particle,
( )
1
ln
ln
ee
Z
E
ße ße
ßß
-
??
-? +
?
??
=-=-
??
( ) ( ) ( )
1
e e
ee
ße ße
ße ße
e e
-
-
? ?
=- -+
? ?
+
ee
ee
ße ße
ße ße
e
-
-
-
= -
+
? for a classical system of N -molecules,
the mean energy is
1
tanh
ee
U NE N N
e e kT
ße ße
ße ße
ß
e
ee
-
-
??
+
= =-=-
??
??
+
??
x
Observer
S Frame
0 x =
0
t t =
Frame
Frame
S S '
/2 c
p
( ) ,: xt s
( ) ,: xt s '' '
Rocket S '
/ 2, 0 Ct =
Q4. At a temperature T, let ß and k denote the volume expansively and isothermal
compressibility of a gas, respectively. Then
k
ß
is equal to
(a)
V
P
T
? ??
??
?
??
(b)
T
P
V
? ??
??
?
??
(c)
V
T
P
? ??
??
?
??
(d)
P
T
V
? ??
??
?
??
Ans. 4: (a)
Solution: given
1 V
VT
ß
? ??
=
??
?
??
and
1
T
V
k
VP
- ? ??
=
??
?
??
1
1
P
T
V
VT
V k
VP
ß
? ??
??
?
??
=
- ? ??
??
?
??
1
PT
VV
V
VT P
?? ?? ??
= -
?? ??
??
?? ??
TP V
PV P
VT T
?? ? ???? ? ?
=-=
???? ? ?
?? ?
???? ? ?
Here, in the last step we made use of reciprocity theorem i.e
xx
yy
y
zx
z
z
? ? ?? ? ?? ??
= -
? ? ?? ??
? ??
??
? ? ??
Q5. The resultant of the binary subtraction 1110101 0011110 - is
(a) 1001111 (b) 1010111 (c) 1010011 (d) 1010001
Ans. 5: (b)
Solution: 2s compliment of 0011110 ?
1100001
1
1100010
+
Thus 1110101 0011110 - is
1110101
1100010
11010111
Q6. Consider a particle trapped in a three-dimensional potential well such that
( ) , , 0 U x yz = for 0 ,0 ,0 xa y a z a = = = = = = and ( ) , , U x yz = 8 everywhere else. The
degeneracy of the
th
5 excited state is
(a) 1 (b) 3 (c) 6 (d) 9
Ans. 6: (c)
Solution: Energy for 3d Potential well is
2
2 2 22
222
2
y
x z
n
n n
E
ma b c
p
? ?
= ++
? ?
? ?
? ?
?
From question abc a = = = so,
22
222
2
2
x yz
E nnn
ma
p
?? = ++
??
?
Energy State ( , ,
xy z
nn n ) Energy Degeneracy
Ground State (1,1,1)
22 2
32ma p ?
1
1
st
Excited State (1,1,2), (1,2,1), (2,1,1)
22 2
62ma p ?
3
2
nd
Excited State (1,2,2), (2,1,2), (2,2,1)
22 2
92ma p ?
3
3
rd
Excited State (1,1,3), (1,3,1), (3,1,1)
22 2
11 2ma p ?
3
4
th
Excited State (2,2,2)
22 2
12 2ma p ?
1
5
th
Excited State (1,2,3), (2,1,3), (2,3,1),
(3,1,2), (3,2,1), (1,3,2)
22 2
14 2ma p ?
6
Q7. A particle of mass mand angular momentum L moves in space where its potential
energy is ( ) ( )
2
0 U r kr k = > and r is the radial coordinate. If the particle moves in a
circular orbit, then the radius of the orbit is
(a)
1
2
4
L
mk
? ?
? ?
? ?
(b)
1
2
4
2
L
mk
??
??
??
(c)
1
2
4
2L
mk
? ?
? ?
? ?
(d)
1
2
4
4L
mk
? ?
? ?
? ?
Ans. 7: (b)
Solution: Let the velocity of the particle is v and angular momentum L .
Method 1:-
( )
2
U r kr =
We have
Force
( )
() 2
U r
F kr
r
?
=- =-
?
We know that centripetal force
2
mv
F
r
= -
2
2
mv
kr
r
- =- ;
2
2
2kr
v
m
= ;
2
2kr
v
m
=
Page 5
IIT-JAM 2022
Section A
22
4 zz + =
Multiple Choice Questions
Q.1-Q.10 Carry ONE marks each.
Q1. The equation in the complex plane (where z is the complex conjugate of z )
represents
(a) Ellipse (b) Hyperbola (c) Circle of radius 2 (d) Circle of radius 4
Ans.1: (b)
Solution: & z x iy z x iy =+ =-
( ) ( )
22
2 2
4 4 z z x iy x iy
-
+ =? + + - =
22
2 xy ?- = Equation of Hyperbola
Q2. A rocket ( ) ' S moves at a speed /
2
c
ms along the positive x -axis, where c is the speed of
light. When it crosses the origin, the clocks attached to the rocket and the one with a
stationary observer ( ) S located at 0 x = are both set to zero. If S observes an event at
( ) , xt the same event occurs in the ' S frame at
(a)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
??
= -
??
??
(b)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
??
= -
??
??
(c)
2
'
2 3
ct
x x
??
= -
??
??
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
(d)
2
'
2 3
ct
x x
? ?
= +
? ?
? ?
and
2
'
2 3
x
t t
c
? ?
= +
? ?
? ?
Ans. 2: (a)
Solution: From the question one Rocket ' S moving with speed of /2 C and one observer
is standing 0 x = . For point P the coordinate at S frame is ( ) , xy and the coordinate at
' S frame is ( ) ', ' x y
22
22
2
'
1 1
4
c
xt
x vt
x
vc
cc
-
-
= =
- -
;
1
'1
24
c
x xt
??
=--
??
??
;
[ ]
2
' / 2
3
x x ct = -
2
2
22 2 2
/
/
2
'
1/ 1/ 4
c
t xc
t xv c
t
vc c c
-
-
= =
--
;
1
'1
24
x
tt
c
? ?
=--
? ?
? ?
;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
[ ]
2
' /2
3
x x ct = - ;
2
'
2 3
x
t t
c
? ?
= -
? ?
? ?
So option (a) is correct.
Q3. Consider a classical ideal gas of N molecules equilibrium at temperature T . Each
molecule has two energy levels - ? and ?. The mean energy of the gas is
(a) 0 (b) tanh
B
N
kT
?? ?
?
??
??
(c) tanh
B
N
kT
?? ?
-?
??
??
(d)
2
?
Ans. 3: (c)
Solution: The partition function for a single gas molecule is
1
Ze e
ße ße -
= +
Mean energy per particle,
( )
1
ln
ln
ee
Z
E
ße ße
ßß
-
??
-? +
?
??
=-=-
??
( ) ( ) ( )
1
e e
ee
ße ße
ße ße
e e
-
-
? ?
=- -+
? ?
+
ee
ee
ße ße
ße ße
e
-
-
-
= -
+
? for a classical system of N -molecules,
the mean energy is
1
tanh
ee
U NE N N
e e kT
ße ße
ße ße
ß
e
ee
-
-
??
+
= =-=-
??
??
+
??
x
Observer
S Frame
0 x =
0
t t =
Frame
Frame
S S '
/2 c
p
( ) ,: xt s
( ) ,: xt s '' '
Rocket S '
/ 2, 0 Ct =
Q4. At a temperature T, let ß and k denote the volume expansively and isothermal
compressibility of a gas, respectively. Then
k
ß
is equal to
(a)
V
P
T
? ??
??
?
??
(b)
T
P
V
? ??
??
?
??
(c)
V
T
P
? ??
??
?
??
(d)
P
T
V
? ??
??
?
??
Ans. 4: (a)
Solution: given
1 V
VT
ß
? ??
=
??
?
??
and
1
T
V
k
VP
- ? ??
=
??
?
??
1
1
P
T
V
VT
V k
VP
ß
? ??
??
?
??
=
- ? ??
??
?
??
1
PT
VV
V
VT P
?? ?? ??
= -
?? ??
??
?? ??
TP V
PV P
VT T
?? ? ???? ? ?
=-=
???? ? ?
?? ?
???? ? ?
Here, in the last step we made use of reciprocity theorem i.e
xx
yy
y
zx
z
z
? ? ?? ? ?? ??
= -
? ? ?? ??
? ??
??
? ? ??
Q5. The resultant of the binary subtraction 1110101 0011110 - is
(a) 1001111 (b) 1010111 (c) 1010011 (d) 1010001
Ans. 5: (b)
Solution: 2s compliment of 0011110 ?
1100001
1
1100010
+
Thus 1110101 0011110 - is
1110101
1100010
11010111
Q6. Consider a particle trapped in a three-dimensional potential well such that
( ) , , 0 U x yz = for 0 ,0 ,0 xa y a z a = = = = = = and ( ) , , U x yz = 8 everywhere else. The
degeneracy of the
th
5 excited state is
(a) 1 (b) 3 (c) 6 (d) 9
Ans. 6: (c)
Solution: Energy for 3d Potential well is
2
2 2 22
222
2
y
x z
n
n n
E
ma b c
p
? ?
= ++
? ?
? ?
? ?
?
From question abc a = = = so,
22
222
2
2
x yz
E nnn
ma
p
?? = ++
??
?
Energy State ( , ,
xy z
nn n ) Energy Degeneracy
Ground State (1,1,1)
22 2
32ma p ?
1
1
st
Excited State (1,1,2), (1,2,1), (2,1,1)
22 2
62ma p ?
3
2
nd
Excited State (1,2,2), (2,1,2), (2,2,1)
22 2
92ma p ?
3
3
rd
Excited State (1,1,3), (1,3,1), (3,1,1)
22 2
11 2ma p ?
3
4
th
Excited State (2,2,2)
22 2
12 2ma p ?
1
5
th
Excited State (1,2,3), (2,1,3), (2,3,1),
(3,1,2), (3,2,1), (1,3,2)
22 2
14 2ma p ?
6
Q7. A particle of mass mand angular momentum L moves in space where its potential
energy is ( ) ( )
2
0 U r kr k = > and r is the radial coordinate. If the particle moves in a
circular orbit, then the radius of the orbit is
(a)
1
2
4
L
mk
? ?
? ?
? ?
(b)
1
2
4
2
L
mk
??
??
??
(c)
1
2
4
2L
mk
? ?
? ?
? ?
(d)
1
2
4
4L
mk
? ?
? ?
? ?
Ans. 7: (b)
Solution: Let the velocity of the particle is v and angular momentum L .
Method 1:-
( )
2
U r kr =
We have
Force
( )
() 2
U r
F kr
r
?
=- =-
?
We know that centripetal force
2
mv
F
r
= -
2
2
mv
kr
r
- =- ;
2
2
2kr
v
m
= ;
2
2kr
v
m
=
2
2kr
Lm r
m
= · ;
2
22 2
2kr
Lm r
m
??
=
??
??
;
24
2 L km e =
2
4
2
L
r
km
= ;
1
2 2
2
L
r
km
??
=
??
??
So option (b) is right.
2
2
2
2
eff
L
V kr
mr
= +
Method:- 2
We know that the effective potential of the system is
;
2
3
2
eff
dV
L
kr
d mr e
-
= +
at
0
,0
eff
dV
rr
dr
= =
2
0 3
0
2
2
L
kr
mr
= ;
1/4
22
4
00
22
LL
rr
mk mk
??
= ?=
??
??
Q8. Consider a two-dimensional force field
( ) ( ) ( )
22 2 2
ˆˆ , 5 44 F x y x ay bxy x x xy y y = ++ + + +
?
If the force field is conservative, then the values of a and b are
(a) 2 a = and 4 b = (b) 2 a = and 8 b =
(c) 4 a = and 2 b = (d) 8 a = and 2 b =
Ans. 8: (b)
Solution:
22 2 2
ˆ ˆ ˆ
00
5 44 0
x y z
F
x yz
x ay bxy x xy y
? ??
?× = ? =
? ??
++ + +
??
( ) ( ) ( )
ˆˆ ˆ 00 00 8 4 2 0 x y z x y ay bx ? -- -+ + - - = 84 2 x y ay bx ?+ = +
( ) ( ) 8 42 0 bx a y ?- + - = 8, 2 b a ?= =
Q9. Consider an electrostatic field E
?
in a region of space. Identify the INCORRECT
statement.
(a) The work done in moving a charge in a closed path inside the region is zero
(b) The curl of E
?
is zero
(c) The field can be expressed as the gradient of a scalar potential
(d) The potential difference between any two points in the region is always zero
Ans. 9: (d)
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FAQs on Physics - 2022 Past Year Paper - IIT JAM Past Year Papers and Model Test Paper (All Branches)
| 1. What is the concept of angular momentum in physics? |  |
Ans. Angular momentum is a fundamental concept in physics that deals with the rotational motion of objects. It is a vector quantity that depends on an object's moment of inertia and its angular velocity. The angular momentum of an object is conserved if there are no external torques acting on it.
| 2. How is angular momentum related to an object's moment of inertia? | |
Ans. Angular momentum is directly proportional to an object's moment of inertia. The moment of inertia represents an object's resistance to changes in its rotational motion. Therefore, objects with larger moments of inertia will have larger angular momenta for the same angular velocity.
| 3. What is the relationship between torque and angular momentum? | |
Ans. Torque is the rotational equivalent of force and is responsible for changes in an object's angular momentum. According to Newton's second law for rotational motion, the torque exerted on an object is equal to the rate of change of its angular momentum. This relationship is given by the equation: Torque = (change in angular momentum)/(change in time).
| 4. Can angular momentum be negative? | |
Ans. Yes, angular momentum can be negative. The sign of angular momentum depends on the direction of rotation. If an object rotates in a counterclockwise direction, its angular momentum is positive. Conversely, if the rotation is in a clockwise direction, the angular momentum is negative.
| 5. How does conservation of angular momentum apply to real-life situations? | |
Ans. Conservation of angular momentum has several real-life applications. For example, figure skaters can increase their rotational speed by pulling their arms closer to their bodies, reducing their moment of inertia and conserving angular momentum. Similarly, divers can perform somersaults or twists by manipulating their body position to change their moment of inertia and maintain angular momentum.
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