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 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 ?
radian/s
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 ? ?
farad (F)
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 ( ?) ?
* * * *

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FAQs on HC Verma Solution - Introduction to Physics - Class 11

1. What is HC Verma Solution?

Ans. HC Verma Solution is a comprehensive guide that provides solutions to the problems and exercises in the book "Concepts of Physics" written by Harish Chandra Verma. It helps students in understanding and solving complex physics problems by providing step-by-step solutions.

2. Is HC Verma Solution suitable for beginners?

Ans. Yes, HC Verma Solution is suitable for beginners as well as advanced learners. The solutions are designed in a simple and easy-to-understand manner, making it accessible for students who are new to physics. It covers the fundamental concepts and gradually progresses to more challenging problems.

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Ans. Absolutely! HC Verma Solution is widely used by students for exam preparation. It covers a wide range of topics and provides detailed solutions to the problems, which helps students in practicing and understanding the concepts thoroughly. It is highly recommended for students preparing for competitive exams like JEE, NEET, and other entrance exams.

4. Are the solutions in HC Verma Solution accurate?

Ans. Yes, the solutions in HC Verma Solution are accurate. The book has been authored by Harish Chandra Verma, who is a renowned physicist and professor. The solutions are carefully written and reviewed to ensure accuracy. However, it is always recommended to cross-verify the solutions and consult with teachers or experts if any doubts arise.

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Ans. Yes, HC Verma Solution is available online. Several websites and platforms offer the solutions in a digital format, making it easily accessible to students. Additionally, there are also mobile applications and e-book versions available for those who prefer studying on digital platforms.

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