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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 Solutions: Chapter 1 - Introduction to Physics - Physics Class 11 - NEET

1. What is HC Verma Solutions?
Ans. HC Verma Solutions is a comprehensive guide to solving problems related to the concepts of Physics. It is authored by Harish Chandra Verma, a renowned Physicist and professor at the Indian Institute of Technology (IIT) Kanpur. The solutions provided in this book cater to a wide range of students, from those who are preparing for competitive entrance exams to those who want to strengthen their understanding of Physics.
2. Why is the first chapter of HC Verma Solutions titled "Introduction to Physics"?
Ans. The first chapter of HC Verma Solutions is titled "Introduction to Physics" as it serves as an overview of the subject. It covers the basic concepts and definitions that form the foundation of Physics. The chapter also provides an insight into the scope of Physics and its applications in our daily lives. By studying this chapter, students can develop a strong understanding of Physics and its significance.
3. What are some of the topics covered in Chapter 1 of HC Verma Solutions?
Ans. Chapter 1 of HC Verma Solutions covers a variety of topics related to Physics, such as physical quantities, units, and measurements. The chapter also covers the different branches of Physics, such as Mechanics, Optics, Thermodynamics, and Electromagnetism. Furthermore, it provides an overview of the scientific method and the importance of experimentation in Physics.
4. How can HC Verma Solutions help students prepare for Physics exams?
Ans. HC Verma Solutions is an excellent resource for students preparing for Physics exams as it provides detailed solutions to a wide range of problems related to Physics. The solutions are presented in a clear and concise manner, making it easy for students to follow and understand. Moreover, the book covers a variety of topics and concepts, making it a comprehensive guide for Physics students.
5. Is HC Verma Solutions suitable for students of all levels?
Ans. HC Verma Solutions is designed to cater to a wide range of students, from those who are just starting to learn Physics to those who are preparing for competitive entrance exams. The book is structured in a way that allows students to progress at their own pace and provides them with a solid foundation in Physics. However, it is recommended that students have a basic understanding of Mathematics before studying this book.
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