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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
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




c) Magnetic permeability 0 = ]IMLT[
]I[
]L[]IMT
I
a2B 22
12






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




b) Coefficient of linear expansion =  = ]K[
]R][L[
]L[
TL
LL 1
0
21 



c) Gas constant = R = ])mol(KTML[
]K)][mol[(
]L][TML[
nT
PV 1122
321


 
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




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

= ]FL[]TL[
]LT
F
]LT[
LT
F 22
2
21
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 = 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+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
MF
L
K 2/12/1

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) / (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
  )xa(
dx
22
≠ 





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)
]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
Young’s 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
Stefan’s 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 ()
* * * *

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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  =    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     c) Magnetic permeability 0 = ]IMLT[ ]I[ ]L[]IMT I a2B 22 12       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     b) Coefficient of linear expansion =  = ]K[ ]R][L[ ]L[ TL LL 1 0 21     c) Gas constant = R = ])mol(KTML[ ]K)][mol[( ]L][TML[ nT PV 1122 321     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     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  = ]FL[]TL[ ]LT F ]LT[ LT F 22 2 21 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
  • 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+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 MF L K 2/12/1 
  • 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) / (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   )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 Young’s 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 Stefan’s 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 () * * * *