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HC Verma Solutions: Chapter 1 - Introduction to Physics

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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

1. What are the main branches of physics and how do they differ from each other?
Ans. Physics divides into mechanics (motion and forces), thermodynamics (heat and energy), optics (light behaviour), electricity and magnetism, and modern physics (atoms and nuclei). Each branch studies different natural phenomena using distinct principles and mathematical approaches. Mechanics focuses on object movement, while thermodynamics examines energy transfer and temperature effects, making them fundamentally different in scope and application.
2. Why is physics called a fundamental science and what makes it different from other sciences?
Ans. Physics is fundamental because it explains the underlying laws governing all natural phenomena, forming the foundation for chemistry, biology, and engineering. Unlike descriptive sciences, physics uses mathematical models and controlled experiments to establish universal principles. It identifies cause-and-effect relationships at the most basic level, enabling predictions about how matter and energy behave across all scales and conditions.
3. How do physicists use the scientific method to solve problems and conduct experiments?
Ans. Physicists follow observation, hypothesis formation, experimentation, data analysis, and conclusion-drawing in sequence. They design controlled experiments isolating single variables to test predictions quantitatively. This systematic approach ensures reproducibility and reliability. Mathematical analysis of experimental results either validates or refutes hypotheses, leading to refined understanding and new theories about physical phenomena.
4. What's the difference between fundamental and derived quantities in physics measurements?
Ans. Fundamental quantities (length, mass, time, temperature) are independently defined with their own units-metre, kilogram, second, kelvin. Derived quantities combine fundamental units; velocity equals length divided by time (m/s), while force equals mass times acceleration (kg⋅m/s²). Understanding this distinction helps students recognise how complex physical measurements build from basic SI unit combinations.
5. Why do physicists use approximations and models instead of studying reality exactly?
Ans. Real-world systems involve countless interacting variables making exact analysis impossible; models simplify by focusing on dominant factors. Approximations reduce mathematical complexity while preserving essential behaviour-treating air resistance as negligible for slow-moving objects, or assuming uniform gravitational fields locally. This pragmatic approach allows physicists to derive accurate predictions without requiring impossible precision in measurements or calculations.
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