Courses

# Chapter 1: Introduction to Physics - HC Verma Solution, Physics Class 11 Notes | EduRev

## Class 12 : Chapter 1: Introduction to Physics - HC Verma Solution, Physics Class 11 Notes | EduRev

``` Page 1

1.1
SOLUTIONS TO CONCEPTS
CHAPTER – 1
1. a) Linear momentum : mv = [MLT
–1
]
b) Frequency :
T
1
= [M
0
L
0
T
–1
]
c) Pressure :
] L [
] MLT [
Area
Force
2
2 ?
? = [ML
–1
T
–2
]
2. a) Angular speed ? = ?/t = [M
0
L
0
T
–1
]
b) Angular acceleration ? = ? ?
?
?
T
T L M
t
2 0 0
[M
0
L
0
T
–2
]
c) Torque ? = F r = [MLT
–2
] [L] = [ML
2
T
–2
]
d) Moment of inertia = Mr
2
= [M] [L
2
] = [ML
2
T
0
]
3. a) Electric field E = F/q = ] I MLT [
] IT [
MLT
1 3
2
? ?
?
?
b) Magnetic field B = ] I MT [
] LT ][ IT [
MLT
qv
F
1 2
1
2
? ?
?
?
? ?
c) Magnetic permeability ?
0
= ] I MLT [
] I [
] L [ ] I MT
I
a 2 B
2 2
1 2
? ?
? ?
?
?
?
? ?
4. a) Electric dipole moment P = qI = [IT] × [L] = [LTI]
b) Magnetic dipole moment M = IA = [I] [L
2
] [L
2
I]
5. E = h ? where E = energy and ? = frequency.
h = ] T ML [
] T [
] T ML [ E
1 2
1
2 2
?
?
?
?
?
6. a) Specific heat capacity = C = ] K T L [
] K ][ M [
] T ML [
T m
Q
1 2 2
2 2
? ?
?
? ?
?
b) Coefficient of linear expansion = ? = ] K [
] R ][ L [
] L [
T L
L L
1
0
2 1 ?
? ?
?
?
c) Gas constant = R = ] ) mol ( K T ML [
] K )][ mol [(
] L ][ T ML [
nT
PV
1 1 2 2
3 2 1
? ? ?
? ?
? ? ?
7. Taking force, length and time as fundamental quantity
a) Density = ] T FL [
T L
F
] L [
] LT / F [
Volume
leration) force/acce (
V
m
2 4
2 4 2
2
?
?
?
? ? ? ?
b) Pressure = F/A = F/L
2
= [FL
–2
]
c) Momentum = mv (Force / acceleration) × Velocity = [F / LT
–2
] × [LT
–1
] = [FT]
d) Energy =
2 2
) velocity (
on accelerati
Force
mv
2
1
? ?
= ] FL [ ] T L [
] LT
F
] LT [
LT
F
2 2
2
2 1
2
? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
8. g =
2
sec
metre
10 = 36 ? 10
5
cm/min
2
9. The average speed of a snail is 0.02 mile/hr
Converting to S.I. units,
0.02 1.6 1000
3600
? ?
m/sec [1 mile = 1.6 km = 1600 m]  = 0.0089 ms
–1
The average speed of leopard = 70 miles/hr
In SI units = 70 miles/hour =
70 1.6 1000
3600
? ?
= 31 m/s
Page 2

1.1
SOLUTIONS TO CONCEPTS
CHAPTER – 1
1. a) Linear momentum : mv = [MLT
–1
]
b) Frequency :
T
1
= [M
0
L
0
T
–1
]
c) Pressure :
] L [
] MLT [
Area
Force
2
2 ?
? = [ML
–1
T
–2
]
2. a) Angular speed ? = ?/t = [M
0
L
0
T
–1
]
b) Angular acceleration ? = ? ?
?
?
T
T L M
t
2 0 0
[M
0
L
0
T
–2
]
c) Torque ? = F r = [MLT
–2
] [L] = [ML
2
T
–2
]
d) Moment of inertia = Mr
2
= [M] [L
2
] = [ML
2
T
0
]
3. a) Electric field E = F/q = ] I MLT [
] IT [
MLT
1 3
2
? ?
?
?
b) Magnetic field B = ] I MT [
] LT ][ IT [
MLT
qv
F
1 2
1
2
? ?
?
?
? ?
c) Magnetic permeability ?
0
= ] I MLT [
] I [
] L [ ] I MT
I
a 2 B
2 2
1 2
? ?
? ?
?
?
?
? ?
4. a) Electric dipole moment P = qI = [IT] × [L] = [LTI]
b) Magnetic dipole moment M = IA = [I] [L
2
] [L
2
I]
5. E = h ? where E = energy and ? = frequency.
h = ] T ML [
] T [
] T ML [ E
1 2
1
2 2
?
?
?
?
?
6. a) Specific heat capacity = C = ] K T L [
] K ][ M [
] T ML [
T m
Q
1 2 2
2 2
? ?
?
? ?
?
b) Coefficient of linear expansion = ? = ] K [
] R ][ L [
] L [
T L
L L
1
0
2 1 ?
? ?
?
?
c) Gas constant = R = ] ) mol ( K T ML [
] K )][ mol [(
] L ][ T ML [
nT
PV
1 1 2 2
3 2 1
? ? ?
? ?
? ? ?
7. Taking force, length and time as fundamental quantity
a) Density = ] T FL [
T L
F
] L [
] LT / F [
Volume
leration) force/acce (
V
m
2 4
2 4 2
2
?
?
?
? ? ? ?
b) Pressure = F/A = F/L
2
= [FL
–2
]
c) Momentum = mv (Force / acceleration) × Velocity = [F / LT
–2
] × [LT
–1
] = [FT]
d) Energy =
2 2
) velocity (
on accelerati
Force
mv
2
1
? ?
= ] FL [ ] T L [
] LT
F
] LT [
LT
F
2 2
2
2 1
2
? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
8. g =
2
sec
metre
10 = 36 ? 10
5
cm/min
2
9. The average speed of a snail is 0.02 mile/hr
Converting to S.I. units,
0.02 1.6 1000
3600
? ?
m/sec [1 mile = 1.6 km = 1600 m]  = 0.0089 ms
–1
The average speed of leopard = 70 miles/hr
In SI units = 70 miles/hour =
70 1.6 1000
3600
? ?
= 31 m/s
Chapter-I
1.2
10. Height h = 75 cm, Density of mercury = 13600 kg/m
3
, g = 9.8 ms
–2
then
Pressure = hfg = 10 ? 10
4
N/m
2
(approximately)
In C.G.S. Units, P = 10 × 10
5
dyne/cm
2
11. In S.I. unit 100 watt = 100 Joule/sec
In C.G.S. Unit = 10
9
erg/sec
12. 1 micro century = 10
4
× 100 years = 10
–4
? 365 ? 24 ? 60 min
So, 100 min = 10
5
/ 52560 = 1.9 microcentury
13. Surface tension of water = 72 dyne/cm
In S.I. Unit, 72 dyne/cm = 0.072 N/m
14. K = kI
a
?
b
where k = Kinetic energy of rotating body and k = dimensionless constant
Dimensions of left side are,
K = [ML
2
T
–2
]
Dimensions of right side are,
I
a
= [ML
2
]
a
, ?
b
= [T
–1
]
b
According to principle of homogeneity of dimension,
[ML
2
T
–2
] = [ML
2
T
–2
] [T
–1
]
b
Equating the dimension of both sides,
2 = 2a and –2 = –b ? a = 1 and b = 2 ?
15. Let energy E ? M
a
C
b
where M = Mass, C = speed of light
? E = KM
a
C
b
(K = proportionality constant)
Dimension of left side
E = [ML
2
T
–2
]
Dimension of right side
M
a
= [M]
a
, [C]
b
= [LT
–1
]
b
?[ML
2
T
–2
] = [M]
a
[LT
–1
]
b
? a = 1; b = 2
So, the relation is E = KMC
2
16. Dimensional formulae of R = [ML
2
T
–3
I
–2
]
Dimensional formulae of V = [ML
2
T
3
I
–1
]
Dimensional formulae of I = [I]
?[ML
2
T
3
I
–1
] = [ML
2
T
–3
I
–2
] [I]
? V = IR
17. Frequency f = KL
a
F
b
M
c
M = Mass/unit length, L = length, F = tension (force)
Dimension of f = [T
–1
]
Dimension of right side,
L
a
= [L
a
], F
b
= [MLT
–2
]
b
, M
c
= [ML
–1
]
c
?[T
–1
] = K[L]
a
[MLT
–2
]
b
[ML
–1
]
c
M
0
L
0
T
–1
= KM
b+c
L
a+b–c
T
–2b
Equating the dimensions of both sides,
? b + c = 0 …(1)
–c + a + b = 0 …(2)
–2b = –1 …(3)
Solving the equations we get,
a = –1, b = 1/2 and c = –1/2
? So, frequency f = KL
–1
F
1/2
M
–1/2
=
M
F
L
K
M F
L
K
2 / 1 2 / 1
? ?
?
Page 3

1.1
SOLUTIONS TO CONCEPTS
CHAPTER – 1
1. a) Linear momentum : mv = [MLT
–1
]
b) Frequency :
T
1
= [M
0
L
0
T
–1
]
c) Pressure :
] L [
] MLT [
Area
Force
2
2 ?
? = [ML
–1
T
–2
]
2. a) Angular speed ? = ?/t = [M
0
L
0
T
–1
]
b) Angular acceleration ? = ? ?
?
?
T
T L M
t
2 0 0
[M
0
L
0
T
–2
]
c) Torque ? = F r = [MLT
–2
] [L] = [ML
2
T
–2
]
d) Moment of inertia = Mr
2
= [M] [L
2
] = [ML
2
T
0
]
3. a) Electric field E = F/q = ] I MLT [
] IT [
MLT
1 3
2
? ?
?
?
b) Magnetic field B = ] I MT [
] LT ][ IT [
MLT
qv
F
1 2
1
2
? ?
?
?
? ?
c) Magnetic permeability ?
0
= ] I MLT [
] I [
] L [ ] I MT
I
a 2 B
2 2
1 2
? ?
? ?
?
?
?
? ?
4. a) Electric dipole moment P = qI = [IT] × [L] = [LTI]
b) Magnetic dipole moment M = IA = [I] [L
2
] [L
2
I]
5. E = h ? where E = energy and ? = frequency.
h = ] T ML [
] T [
] T ML [ E
1 2
1
2 2
?
?
?
?
?
6. a) Specific heat capacity = C = ] K T L [
] K ][ M [
] T ML [
T m
Q
1 2 2
2 2
? ?
?
? ?
?
b) Coefficient of linear expansion = ? = ] K [
] R ][ L [
] L [
T L
L L
1
0
2 1 ?
? ?
?
?
c) Gas constant = R = ] ) mol ( K T ML [
] K )][ mol [(
] L ][ T ML [
nT
PV
1 1 2 2
3 2 1
? ? ?
? ?
? ? ?
7. Taking force, length and time as fundamental quantity
a) Density = ] T FL [
T L
F
] L [
] LT / F [
Volume
leration) force/acce (
V
m
2 4
2 4 2
2
?
?
?
? ? ? ?
b) Pressure = F/A = F/L
2
= [FL
–2
]
c) Momentum = mv (Force / acceleration) × Velocity = [F / LT
–2
] × [LT
–1
] = [FT]
d) Energy =
2 2
) velocity (
on accelerati
Force
mv
2
1
? ?
= ] FL [ ] T L [
] LT
F
] LT [
LT
F
2 2
2
2 1
2
? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
8. g =
2
sec
metre
10 = 36 ? 10
5
cm/min
2
9. The average speed of a snail is 0.02 mile/hr
Converting to S.I. units,
0.02 1.6 1000
3600
? ?
m/sec [1 mile = 1.6 km = 1600 m]  = 0.0089 ms
–1
The average speed of leopard = 70 miles/hr
In SI units = 70 miles/hour =
70 1.6 1000
3600
? ?
= 31 m/s
Chapter-I
1.2
10. Height h = 75 cm, Density of mercury = 13600 kg/m
3
, g = 9.8 ms
–2
then
Pressure = hfg = 10 ? 10
4
N/m
2
(approximately)
In C.G.S. Units, P = 10 × 10
5
dyne/cm
2
11. In S.I. unit 100 watt = 100 Joule/sec
In C.G.S. Unit = 10
9
erg/sec
12. 1 micro century = 10
4
× 100 years = 10
–4
? 365 ? 24 ? 60 min
So, 100 min = 10
5
/ 52560 = 1.9 microcentury
13. Surface tension of water = 72 dyne/cm
In S.I. Unit, 72 dyne/cm = 0.072 N/m
14. K = kI
a
?
b
where k = Kinetic energy of rotating body and k = dimensionless constant
Dimensions of left side are,
K = [ML
2
T
–2
]
Dimensions of right side are,
I
a
= [ML
2
]
a
, ?
b
= [T
–1
]
b
According to principle of homogeneity of dimension,
[ML
2
T
–2
] = [ML
2
T
–2
] [T
–1
]
b
Equating the dimension of both sides,
2 = 2a and –2 = –b ? a = 1 and b = 2 ?
15. Let energy E ? M
a
C
b
where M = Mass, C = speed of light
? E = KM
a
C
b
(K = proportionality constant)
Dimension of left side
E = [ML
2
T
–2
]
Dimension of right side
M
a
= [M]
a
, [C]
b
= [LT
–1
]
b
?[ML
2
T
–2
] = [M]
a
[LT
–1
]
b
? a = 1; b = 2
So, the relation is E = KMC
2
16. Dimensional formulae of R = [ML
2
T
–3
I
–2
]
Dimensional formulae of V = [ML
2
T
3
I
–1
]
Dimensional formulae of I = [I]
?[ML
2
T
3
I
–1
] = [ML
2
T
–3
I
–2
] [I]
? V = IR
17. Frequency f = KL
a
F
b
M
c
M = Mass/unit length, L = length, F = tension (force)
Dimension of f = [T
–1
]
Dimension of right side,
L
a
= [L
a
], F
b
= [MLT
–2
]
b
, M
c
= [ML
–1
]
c
?[T
–1
] = K[L]
a
[MLT
–2
]
b
[ML
–1
]
c
M
0
L
0
T
–1
= KM
b+c
L
a+b–c
T
–2b
Equating the dimensions of both sides,
? b + c = 0 …(1)
–c + a + b = 0 …(2)
–2b = –1 …(3)
Solving the equations we get,
a = –1, b = 1/2 and c = –1/2
? So, frequency f = KL
–1
F
1/2
M
–1/2
=
M
F
L
K
M F
L
K
2 / 1 2 / 1
? ?
?
Chapter-I
1.3
18. a) h =
rg
SCos 2
?
?
LHS = [L]
Surface tension = S = F/I =
2
2
MLT
[MT ]
L
?
?
?
Density = ? = M/V = [ML
–3
T
0
]
Radius = r = [L], g = [LT
–2
]
RHS =
2
0 1 0
3 0 2
2Scos [MT ]
[M L T ] [L]
rg [ML T ][L][LT ]
?
? ?
?
? ? ?
?
?
LHS = RHS
So, the relation is correct
b) v =
?
p
where v = velocity
LHS = Dimension of v = [LT
–1
]
Dimension of p = F/A = [ML
–1
T
–2
]
Dimension of ? = m/V = [ML
–3
]
RHS =
1 2
2 2 1/ 2
3
p [ML T ]
[L T ]
[ML ]
? ?
?
?
? ?
?
=
1
[LT ]
?
So, the relation is correct.
c) V = ( ?pr
4
t) / (8 ?l) ?
LHS = Dimension of V = [L
3
]
Dimension of p = [ML
–1
T
–2
], r
4
= [L
4
], t = [T]
Coefficient of viscosity = [ML
–1
T
–1
]
RHS =
4 1 2 4
1 1
pr t [ML T ][L ][T]
8 l [ML T ][L]
? ?
? ?
?
?
?
So, the relation is correct.
d) v = ) I / mgl (
2
1
?
LHS = dimension of v = [T
–1
]
RHS = ) I / mgl ( =
2
2
[M][LT ][L]
[ML ]
?
= [T
–1
]
LHS = RHS
So, the relation is correct.
19. Dimension of the left side =
2 2 2 2
dx L
(a x ) (L L )
?
? ?
? ?
= [L
0
]
Dimension of the right side = ?
?
?
?
?
?
?
x
a
sin
a
1
1
= [L
–1
]
So, the dimension of
?
? ) x a (
dx
2 2
? ?
?
?
?
?
?
?
x
a
sin
a
1
1
So, the equation is dimensionally incorrect.
Page 4

1.1
SOLUTIONS TO CONCEPTS
CHAPTER – 1
1. a) Linear momentum : mv = [MLT
–1
]
b) Frequency :
T
1
= [M
0
L
0
T
–1
]
c) Pressure :
] L [
] MLT [
Area
Force
2
2 ?
? = [ML
–1
T
–2
]
2. a) Angular speed ? = ?/t = [M
0
L
0
T
–1
]
b) Angular acceleration ? = ? ?
?
?
T
T L M
t
2 0 0
[M
0
L
0
T
–2
]
c) Torque ? = F r = [MLT
–2
] [L] = [ML
2
T
–2
]
d) Moment of inertia = Mr
2
= [M] [L
2
] = [ML
2
T
0
]
3. a) Electric field E = F/q = ] I MLT [
] IT [
MLT
1 3
2
? ?
?
?
b) Magnetic field B = ] I MT [
] LT ][ IT [
MLT
qv
F
1 2
1
2
? ?
?
?
? ?
c) Magnetic permeability ?
0
= ] I MLT [
] I [
] L [ ] I MT
I
a 2 B
2 2
1 2
? ?
? ?
?
?
?
? ?
4. a) Electric dipole moment P = qI = [IT] × [L] = [LTI]
b) Magnetic dipole moment M = IA = [I] [L
2
] [L
2
I]
5. E = h ? where E = energy and ? = frequency.
h = ] T ML [
] T [
] T ML [ E
1 2
1
2 2
?
?
?
?
?
6. a) Specific heat capacity = C = ] K T L [
] K ][ M [
] T ML [
T m
Q
1 2 2
2 2
? ?
?
? ?
?
b) Coefficient of linear expansion = ? = ] K [
] R ][ L [
] L [
T L
L L
1
0
2 1 ?
? ?
?
?
c) Gas constant = R = ] ) mol ( K T ML [
] K )][ mol [(
] L ][ T ML [
nT
PV
1 1 2 2
3 2 1
? ? ?
? ?
? ? ?
7. Taking force, length and time as fundamental quantity
a) Density = ] T FL [
T L
F
] L [
] LT / F [
Volume
leration) force/acce (
V
m
2 4
2 4 2
2
?
?
?
? ? ? ?
b) Pressure = F/A = F/L
2
= [FL
–2
]
c) Momentum = mv (Force / acceleration) × Velocity = [F / LT
–2
] × [LT
–1
] = [FT]
d) Energy =
2 2
) velocity (
on accelerati
Force
mv
2
1
? ?
= ] FL [ ] T L [
] LT
F
] LT [
LT
F
2 2
2
2 1
2
? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
8. g =
2
sec
metre
10 = 36 ? 10
5
cm/min
2
9. The average speed of a snail is 0.02 mile/hr
Converting to S.I. units,
0.02 1.6 1000
3600
? ?
m/sec [1 mile = 1.6 km = 1600 m]  = 0.0089 ms
–1
The average speed of leopard = 70 miles/hr
In SI units = 70 miles/hour =
70 1.6 1000
3600
? ?
= 31 m/s
Chapter-I
1.2
10. Height h = 75 cm, Density of mercury = 13600 kg/m
3
, g = 9.8 ms
–2
then
Pressure = hfg = 10 ? 10
4
N/m
2
(approximately)
In C.G.S. Units, P = 10 × 10
5
dyne/cm
2
11. In S.I. unit 100 watt = 100 Joule/sec
In C.G.S. Unit = 10
9
erg/sec
12. 1 micro century = 10
4
× 100 years = 10
–4
? 365 ? 24 ? 60 min
So, 100 min = 10
5
/ 52560 = 1.9 microcentury
13. Surface tension of water = 72 dyne/cm
In S.I. Unit, 72 dyne/cm = 0.072 N/m
14. K = kI
a
?
b
where k = Kinetic energy of rotating body and k = dimensionless constant
Dimensions of left side are,
K = [ML
2
T
–2
]
Dimensions of right side are,
I
a
= [ML
2
]
a
, ?
b
= [T
–1
]
b
According to principle of homogeneity of dimension,
[ML
2
T
–2
] = [ML
2
T
–2
] [T
–1
]
b
Equating the dimension of both sides,
2 = 2a and –2 = –b ? a = 1 and b = 2 ?
15. Let energy E ? M
a
C
b
where M = Mass, C = speed of light
? E = KM
a
C
b
(K = proportionality constant)
Dimension of left side
E = [ML
2
T
–2
]
Dimension of right side
M
a
= [M]
a
, [C]
b
= [LT
–1
]
b
?[ML
2
T
–2
] = [M]
a
[LT
–1
]
b
? a = 1; b = 2
So, the relation is E = KMC
2
16. Dimensional formulae of R = [ML
2
T
–3
I
–2
]
Dimensional formulae of V = [ML
2
T
3
I
–1
]
Dimensional formulae of I = [I]
?[ML
2
T
3
I
–1
] = [ML
2
T
–3
I
–2
] [I]
? V = IR
17. Frequency f = KL
a
F
b
M
c
M = Mass/unit length, L = length, F = tension (force)
Dimension of f = [T
–1
]
Dimension of right side,
L
a
= [L
a
], F
b
= [MLT
–2
]
b
, M
c
= [ML
–1
]
c
?[T
–1
] = K[L]
a
[MLT
–2
]
b
[ML
–1
]
c
M
0
L
0
T
–1
= KM
b+c
L
a+b–c
T
–2b
Equating the dimensions of both sides,
? b + c = 0 …(1)
–c + a + b = 0 …(2)
–2b = –1 …(3)
Solving the equations we get,
a = –1, b = 1/2 and c = –1/2
? So, frequency f = KL
–1
F
1/2
M
–1/2
=
M
F
L
K
M F
L
K
2 / 1 2 / 1
? ?
?
Chapter-I
1.3
18. a) h =
rg
SCos 2
?
?
LHS = [L]
Surface tension = S = F/I =
2
2
MLT
[MT ]
L
?
?
?
Density = ? = M/V = [ML
–3
T
0
]
Radius = r = [L], g = [LT
–2
]
RHS =
2
0 1 0
3 0 2
2Scos [MT ]
[M L T ] [L]
rg [ML T ][L][LT ]
?
? ?
?
? ? ?
?
?
LHS = RHS
So, the relation is correct
b) v =
?
p
where v = velocity
LHS = Dimension of v = [LT
–1
]
Dimension of p = F/A = [ML
–1
T
–2
]
Dimension of ? = m/V = [ML
–3
]
RHS =
1 2
2 2 1/ 2
3
p [ML T ]
[L T ]
[ML ]
? ?
?
?
? ?
?
=
1
[LT ]
?
So, the relation is correct.
c) V = ( ?pr
4
t) / (8 ?l) ?
LHS = Dimension of V = [L
3
]
Dimension of p = [ML
–1
T
–2
], r
4
= [L
4
], t = [T]
Coefficient of viscosity = [ML
–1
T
–1
]
RHS =
4 1 2 4
1 1
pr t [ML T ][L ][T]
8 l [ML T ][L]
? ?
? ?
?
?
?
So, the relation is correct.
d) v = ) I / mgl (
2
1
?
LHS = dimension of v = [T
–1
]
RHS = ) I / mgl ( =
2
2
[M][LT ][L]
[ML ]
?
= [T
–1
]
LHS = RHS
So, the relation is correct.
19. Dimension of the left side =
2 2 2 2
dx L
(a x ) (L L )
?
? ?
? ?
= [L
0
]
Dimension of the right side = ?
?
?
?
?
?
?
x
a
sin
a
1
1
= [L
–1
]
So, the dimension of
?
? ) x a (
dx
2 2
? ?
?
?
?
?
?
?
x
a
sin
a
1
1
So, the equation is dimensionally incorrect.
Chapter-I
1.4
20. Important Dimensions and Units :
Physical quantity Dimension SI unit
Force (F)
] T L M [
2 1 1 ?
newton
Work (W)
] T L M [
2 2 1 ?
joule
Power (P)
] T L M [
3 2 1 ?
watt
Gravitational constant (G)
] T L M [
2 3 1 ? ?
N-m
2
/kg
2
Angular velocity ( ?) ?
] T [
1 ?
Angular momentum (L)
] T L M [
1 2 1 ?
kg-m
2
/s
Moment of inertia (I)
] L M [
2 1
kg-m
2
Torque ( ?) ?
] T L M [
2 2 1 ?
N-m
Young’s modulus (Y)
] T L M [
2 1 1 ? ?
N/m
2
Surface Tension (S)
] T M [
2 1 ?
N/m
Coefficient of viscosity ( ?) ?
] T L M [
1 1 1 ? ?
N-s/m
2
Pressure (p)
] T L M [
2 1 1 ? ?
N/m
2
(Pascal)
Intensity of wave (I)
] T M [
3 1 ?
watt/m
2
Specific heat capacity (c)
] K T L [
1 2 2 ? ?
J/kg-K
Stefan’s constant ( ?) ?
] K T M [
4 3 1 ? ?
watt/m
2
-k
4
Thermal conductivity (k)
] K T L M [
1 3 1 1 ? ?
watt/m-K
Current density (j)
] L I [
2 1 ?
ampere/m
2
Electrical conductivity ( ?) ?
] L M T I [
3 1 3 2 ? ?
?
–1
m
–1
?
Electric dipole moment (p)
] T I L [
1 1 1
C-m
Electric field (E)
] T I L M [
3 1 1 1 ? ?
V/m
Electrical potential (V)
] T I L M [
3 1 2 1 ? ?
volt
Electric flux ( ?) ?
] L I T M [
3 1 3 1 ? ?
volt/m
Capacitance (C)
] L M T I [
2 1 4 2 ? ?
Permittivity ( ?) ?
] L M T I [
3 1 4 2 ? ?
C
2
/N-m
2
Permeability ( ?) ?
] T I L M [
3 2 1 1 ? ?
Newton/A
2
Magnetic dipole moment (M)
] L I [
2 1
N-m/T
Magnetic flux ( ?) ?
] T I L M [
2 1 2 1 ? ?
Weber (Wb)
Magnetic field (B)
] T I M [
2 1 1 ? ?
tesla
Inductance (L)
] T I L M [
2 2 2 1 ? ?
henry
Resistance (R)
] T I L M [
3 2 2 1 ? ?
ohm ( ?) ?
* * * *
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Physics For JEE

188 videos|517 docs|265 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;