Open App

Class 11 Exam > Class 11 Notes > HC Verma Solution - Introduction to Physics

HC Verma Solution - Introduction to Physics - Class 11 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 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 ( ?) ?
* * * *

Top Courses for Class 11

View all

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.

3. Can HC Verma Solution help in exam preparation?

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.

5. Is HC Verma Solution available online?

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.

Related Exams

Class 11 JEE NEET

About this Document

	4.67/5 Rating
	Nov 23, 2024 Last updated

Document Description: HC Verma Solution - Introduction to Physics for Class 11 2024 is part of Class 11 preparation. The notes and questions for HC Verma Solution - Introduction to Physics have been prepared according to the Class 11 exam syllabus. Information about HC Verma Solution - Introduction to Physics covers topics like and HC Verma Solution - Introduction to Physics Example, for Class 11 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solution - Introduction to Physics.

Introduction of HC Verma Solution - Introduction to Physics in English is available as part of our Class 11 preparation & HC Verma Solution - Introduction to Physics in Hindi for Class 11 courses. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free. Class 11: HC Verma Solution - Introduction to Physics - Class 11

Description

Full syllabus notes, lecture & questions for HC Verma Solution - Introduction to Physics - Class 11 - Class 11 | Plus excerises question with solution to help you revise complete syllabus | Best notes, free PDF download

Information about HC Verma Solution - Introduction to Physics

In this doc you can find the meaning of HC Verma Solution - Introduction to Physics defined & explained in the simplest way possible. Besides explaining types of HC Verma Solution - Introduction to Physics theory, EduRev gives you an ample number of questions to practice HC Verma Solution - Introduction to Physics tests, examples and also practice Class 11 tests.

Download as PDF

Explore Courses for Class 11 exam

Top Courses for Class 11

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Exam

pdf

Previous Year Questions with Solutions

video lectures

Sample Paper

Important questions

practice quizzes

ppt

Viva Questions

HC Verma Solution - Introduction to Physics - Class 11

Summary

Semester Notes

past year papers

Objective type Questions

HC Verma Solution - Introduction to Physics - Class 11

mock tests for examination

MCQs

Extra Questions

shortcuts and tricks

study material

Free

;

Additional Information about HC Verma Solution - Introduction to Physics for Class 11 Preparation

HC Verma Solution - Introduction to Physics Free PDF Download

The HC Verma Solution - Introduction to Physics is an invaluable resource that delves deep into the core of the Class 11 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the HC Verma Solution - Introduction to Physics now and kickstart your journey towards success in the Class 11 exam.

Importance of HC Verma Solution - Introduction to Physics

The importance of HC Verma Solution - Introduction to Physics cannot be overstated, especially for Class 11 aspirants. This document holds the key to success in the Class 11 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

HC Verma Solution - Introduction to Physics Notes

HC Verma Solution - Introduction to Physics Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to HC Verma Solution - Introduction to Physics. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, HC Verma Solution - Introduction to Physics Notes on EduRev are your ultimate resource for success.

HC Verma Solution - Introduction to Physics Class 11 Questions

The "HC Verma Solution - Introduction to Physics Class 11 Questions" guide is a valuable resource for all aspiring students preparing for the Class 11 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study HC Verma Solution - Introduction to Physics on the App

Students of Class 11 can study HC Verma Solution - Introduction to Physics alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solution - Introduction to Physics, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of HC Verma Solution - Introduction to Physics is prepared as per the latest Class 11 syllabus.

Education Revolution