01.solutions to-concepts
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1. The document provides dimensional analysis and units for various physical quantities. It lists 18 physical quantities along with their dimensions and standard SI units. 2. Key physical quantities included are: force, work, power, gravitational constant, angular velocity, angular momentum, moment of inertia, torque, Young's modulus, surface tension, viscosity, pressure, intensity of waves, specific heat capacity, Stefan's constant, thermal conductivity, current density, and electrical conductivity. 3. The dimensional analysis verifies the consistency of equations involving these physical quantities by ensuring the dimensions on both sides of equations are equal. This helps identify any dimensionally incorrect equations.
![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 = 緒
T
TLM
t
200
[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 = ]IMLT[
]IT[
MLT 13
2
b) Magnetic field B = ]IMT[
]LT][IT[
MLT
qv
F 12
1
2
緒
c) Magnetic permeability 0 = ]IMLT[
]I[
]L[]IMT
I
a2B 22
12
逸
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 = ]TML[
]T[
]TML[E 12
1
22
6. a) Specific heat capacity = C = ]KTL[
]K][M[
]TML[
Tm
Q 122
22
緒
b) Coefficient of linear expansion = = ]K[
]R][L[
]L[
TL
LL 1
0
21
緒
c) Gas constant = R = ])mol(KTML[
]K)][mol[(
]L][TML[
nT
PV 1122
321
緒
7. Taking force, length and time as fundamental quantity
a) Density = ]TFL[
TL
F
]L[
]LT/F[
Volume
leration)force/acce(
V
m 24
242
2
緒緒緒
b) Pressure = F/A = F/L
2
= [FL
2
]
c) Momentum = mv (Force / acceleration) Velocity = [F / LT
2
] [LT
1
] = [FT]
d) Energy = 22
)velocity(
onaccelerati
Force
mv
2
1
器
= ]FL[]TL[
]LT
F
]LT[
LT
F 22
2
21
2
緒
緒器
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](https://image.slidesharecdn.com/01-140716201102-phpapp02/85/01-solutions-to-concepts-1-320.jpg)
![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 = 104
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 = kIa
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,
Ia
= [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+bc
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
MF
L
K 2/12/1
緒緒](https://image.slidesharecdn.com/01-140716201102-phpapp02/85/01-solutions-to-concepts-2-320.jpg)
![Chapter-I
1.3
18. a) h =
rg
SCos2
LHS = [L]
Surface tension = S = F/I =
2
2MLT
[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) / (8l)
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
)xa(
dx
22
x
a
sin
a
1 1
So, the equation is dimensionally incorrect.](https://image.slidesharecdn.com/01-140716201102-phpapp02/85/01-solutions-to-concepts-3-320.jpg)
![Chapter-I
1.4
20. Important Dimensions and Units :
Physical quantity Dimension SI unit
Force (F)
]TLM[ 211
newton
Work (W)
]TLM[ 221
joule
Power (P)
]TLM[ 321
watt
Gravitational constant (G)
]TLM[ 231
N-m
2
/kg
2
Angular velocity ()
]T[ 1
radian/s
Angular momentum (L)
]TLM[ 121
kg-m
2
/s
Moment of inertia (I)
]LM[ 21
kg-m
2
Torque ()
]TLM[ 221
N-m
Youngs modulus (Y)
]TLM[ 211
N/m
2
Surface Tension (S)
]TM[ 21
N/m
Coefficient of viscosity ()
]TLM[ 111
N-s/m
2
Pressure (p)
]TLM[ 211
N/m
2
(Pascal)
Intensity of wave (I)
]TM[ 31
watt/m
2
Specific heat capacity (c)
]KTL[ 122
J/kg-K
Stefans constant ()
]KTM[ 431
watt/m
2
-k
4
Thermal conductivity (k)
]KTLM[ 1311
watt/m-K
Current density (j)
]LI[ 21
ampere/m
2
Electrical conductivity ()
]LMTI[ 3132
1
m
1
Electric dipole moment (p)
]TIL[ 111
C-m
Electric field (E)
]TILM[ 3111
V/m
Electrical potential (V)
]TILM[ 3121
volt
Electric flux ()
]LITM[ 3131
volt/m
Capacitance (C)
]LMTI[ 2142
farad (F)
Permittivity ()
]LMTI[ 3142
C
2
/N-m
2
Permeability ()
]TILM[ 3211
Newton/A
2
Magnetic dipole moment (M)
]LI[ 21
N-m/T
Magnetic flux ()
]TILM[ 2121
Weber (Wb)
Magnetic field (B)
]TIM[ 211
tesla
Inductance (L)
]TILM[ 2221
henry
Resistance (R)
]TILM[ 3221
ohm ()
* * * *](https://image.slidesharecdn.com/01-140716201102-phpapp02/85/01-solutions-to-concepts-4-320.jpg)
01.solutions to-concepts
- 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 = 緒 T TLM t 200 [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 = ]IMLT[ ]IT[ MLT 13 2 b) Magnetic field B = ]IMT[ ]LT][IT[ MLT qv F 12 1 2 緒 c) Magnetic permeability 0 = ]IMLT[ ]I[ ]L[]IMT I a2B 22 12 逸 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 = ]TML[ ]T[ ]TML[E 12 1 22 6. a) Specific heat capacity = C = ]KTL[ ]K][M[ ]TML[ Tm Q 122 22 緒 b) Coefficient of linear expansion = = ]K[ ]R][L[ ]L[ TL LL 1 0 21 緒 c) Gas constant = R = ])mol(KTML[ ]K)][mol[( ]L][TML[ nT PV 1122 321 緒 7. Taking force, length and time as fundamental quantity a) Density = ]TFL[ TL F ]L[ ]LT/F[ Volume leration)force/acce( V m 24 242 2 緒緒緒 b) Pressure = F/A = F/L 2 = [FL 2 ] c) Momentum = mv (Force / acceleration) Velocity = [F / LT 2 ] [LT 1 ] = [FT] d) Energy = 22 )velocity( onaccelerati Force mv 2 1 器 = ]FL[]TL[ ]LT F ]LT[ LT F 22 2 21 2 緒 緒器 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
- 2. 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 = 104 100 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 = kIa 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, Ia = [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+bc 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 MF L K 2/12/1 緒緒
- 3. Chapter-I 1.3 18. a) h = rg SCos2 LHS = [L] Surface tension = S = F/I = 2 2MLT [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) / (8l) 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 )xa( dx 22 x a sin a 1 1 So, the equation is dimensionally incorrect.
- 4. Chapter-I 1.4 20. Important Dimensions and Units : Physical quantity Dimension SI unit Force (F) ]TLM[ 211 newton Work (W) ]TLM[ 221 joule Power (P) ]TLM[ 321 watt Gravitational constant (G) ]TLM[ 231 N-m 2 /kg 2 Angular velocity () ]T[ 1 radian/s Angular momentum (L) ]TLM[ 121 kg-m 2 /s Moment of inertia (I) ]LM[ 21 kg-m 2 Torque () ]TLM[ 221 N-m Youngs modulus (Y) ]TLM[ 211 N/m 2 Surface Tension (S) ]TM[ 21 N/m Coefficient of viscosity () ]TLM[ 111 N-s/m 2 Pressure (p) ]TLM[ 211 N/m 2 (Pascal) Intensity of wave (I) ]TM[ 31 watt/m 2 Specific heat capacity (c) ]KTL[ 122 J/kg-K Stefans constant () ]KTM[ 431 watt/m 2 -k 4 Thermal conductivity (k) ]KTLM[ 1311 watt/m-K Current density (j) ]LI[ 21 ampere/m 2 Electrical conductivity () ]LMTI[ 3132 1 m 1 Electric dipole moment (p) ]TIL[ 111 C-m Electric field (E) ]TILM[ 3111 V/m Electrical potential (V) ]TILM[ 3121 volt Electric flux () ]LITM[ 3131 volt/m Capacitance (C) ]LMTI[ 2142 farad (F) Permittivity () ]LMTI[ 3142 C 2 /N-m 2 Permeability () ]TILM[ 3211 Newton/A 2 Magnetic dipole moment (M) ]LI[ 21 N-m/T Magnetic flux () ]TILM[ 2121 Weber (Wb) Magnetic field (B) ]TIM[ 211 tesla Inductance (L) ]TILM[ 2221 henry Resistance (R) ]TILM[ 3221 ohm () * * * *