units and dimensions

1.1

SOLUTIONS TO CONCEPTS

CHAPTER – 1

1. a) Linear momentum

: mv

= [MLT–1]

b) Frequency

:

1 = [M0L0T–1]
T

Force [MLT2 ]

–1 –2

c) Pressure : Area
[L2 ]
= [ML T ]

2. a) Angular speed = /t = [M0L0T–1]

M0L0T2 0 0 –2

b) Angular acceleration = 

t

T
[M L T ]

c) Torque = F r = [MLT–2] [L] = [ML2T–2]
d) Moment of inertia = Mr2 = [M] [L2] = [ML2T0]

3. a) Electric field E = F/q = MLT 2

[IT]
[MLT 3I1]

F MLT2

2 1

b) Magnetic field B =

qv 

[IT][LT1]

[MT I ]

c) Magnetic permeability 0 = B 2a

I
MT2I1] [L]

[I]

[MLT 2I2 ]

4. a) Electric dipole moment P = qI = [IT] × [L] = [LTI]
b) Magnetic dipole moment M = IA = [I] [L2] [L2I]

5. E = hwhere E = energy and = frequency.

h = E 



[ML2T2 ]

[T1]

[ML2T 1]

Q [ML2T2 ]

2 2 1

6. a) Specific heat capacity = C =

mT 



[M][K] [L T K ]

b) Coefficient of linear expansion = = L1 L2 

L0 T

[L]

[L][R]

[K1]

c) Gas constant = R = PV 

nT

[ML1T2 ][L3 ]

[(mol)][K]
[ML2T 2K1(mol)1]

7. Taking force, length and time as fundamental quantity

m (force/acceleration) [F / LT2 ] 
F







 4 2

a) Density =

V

Volume

[L2 ] L4T2
[FL T ]

b) Pressure = F/A = F/L2 = [FL–2]
c) Momentum = mv (Force / acceleration) × Velocity = [F / LT–2] × [LT–1] = [FT]

d) Energy =

1 mv 2
2

Force

acceleration

 (velocity )2

= F [LT 1]2F

[L2T2 ] [FL]

LT2 
LT2 ] 



8. g = 10 metre

sec 2



= 36 105 cm/min2

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/m3, g = 9.8 ms–2 then

Pressure = hfg = 10 104 N/m2 (approximately)

In C.G.S. Units, P = 10 × 105 dyne/cm2

11. In S.I. unit 100 watt = 100 Joule/sec

In C.G.S. Unit = 109 erg/sec

12. 1 micro century = 104 × 100 years = 10–4365 24 60 min

So, 100 min = 105 / 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 = [ML2T–2]
Dimensions of right side are,

Ia = [ML2]a, b = [T–1]b

According to principle of homogeneity of dimension,

[ML2T–2] = [ML2T–2] [T–1]b

Equating the dimension of both sides,

2 = 2a and –2 = –b a = 1 and b = 2

15. Let energy E MaCb where M = Mass, C = speed of light

E = KMaCb (K = proportionality constant)

Dimension of left side

E = [ML2T–2]

Dimension of right side

Ma = [M]a, [C]b = [LT–1]b

[ML2T–2] = [M]a[LT–1]b

a = 1; b = 2

So, the relation is E = KMC2

16. Dimensional formulae of R = [ML2T–3I–2]

Dimensional formulae of V = [ML2T3I–1]

Dimensional formulae of I = [I]

[ML2T3I–1] = [ML2T–3I–2] [I]

V = IR

17. Frequency f = KLaFbMc M = Mass/unit length, L = length, F = tension (force)

Dimension of f = [T–1]
Dimension of right side,

La = [La], Fb = [MLT–2]b, Mc = [ML–1]c

[T–1] = K[L]a [MLT–2]b [ML–1]c
M0L0T–1 = KMb+c La+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–1F1/2M–1/2 = K F1/2M1/2K 

L

L

F
M

Chapter-I

1.3

p


p


L

(L2L2 )

 

18. a) h = 2SCos





rg

LHS = [L]

Surface tension = S = F/I =

MLT2

L

[MT2 ]

Density = = M/V = [ML–3T0]

Radius = r = [L], g = [LT–2]

2Scos 





[MT2 ]

0 1 0

RHS =
rg 

[ML3T0 ][L][LT2 ]

[M L T ] [L]

LHS = RHS

So, the relation is correct

b) v =

where v = velocity

LHS = Dimension of v = [LT–1]

Dimension of p = F/A = [ML–1T–2]

Dimension of = m/V = [ML–3]

2 2 1/ 2

1

RHS =



[L T ]

= [LT ]

So, the relation is correct.

c) V = (pr4t) / (8l)

LHS = Dimension of V = [L3]

Dimension of p = [ML–1T–2], r4 = [L4], t = [T]

Coefficient of viscosity = [ML–1T–1]

pr 4t [ML1T2 ][L4 ][T]

RHS =

8l [ML1T1][L]

So, the relation is correct.

d) v =

LHS = dimension of v = [T–1]

–1

RHS =

(mgl / I) =

= [T ]

LHS = RHS

So, the relation is correct.

19. Dimension of the left side = 

Dimension of the right side =



= [L0]

1 sin1a = [L–1]



a

So, the dimension of  dx
≠

x 

1 sin1a 
a

So, the equation is dimensionally incorrect.

x 

[ML1T2 ]

[ML3 ]

1 (mgl / I)
2

[M][LT2 ][L]

[ML2 ]

dx

(a2x2 )

(a2x2 )

Chapter-I

1.4

20. Important Dimensions and Units :

Force (F)

Work (W)

Power (P)

Gravitational constant (G)

Angular velocity ()

Angular momentum (L)

Moment of inertia (I)

Torque ()

Young’s modulus (Y)

Surface Tension (S)

Coefficient of viscosity ()

Pressure (p)

Intensity of wave (I)

Specific heat capacity (c)

Stefan’s constant ()

Thermal conductivity (k)

Current density (j)

Electrical conductivity ()

Electric dipole moment (p)

Electric field (E)

Electrical potential (V)

Electric flux ()

Capacitance (C)

Permittivity ()

Permeability ()

Magnetic dipole moment (M)

Magnetic flux ()

Magnetic field (B)

Inductance (L)

Resistance (R)

[M1L1T2 ]

[M1L2T2 ]

[M1L2T3 ]

[M1L3T2 ]

[T1]

[M1L2T1]

[M1L2 ]

[M1L2T2 ]

[M1L1T2 ]

[M1T2 ]

[M1L1T1]

[M1L1T2 ]

[M1T3 ]

[L2T2K1]

[M1T3K4 ]

[M1L1T3K1]

[I1L2 ]

[I2T3M1L3 ]

[L1I1T1]

[M1L1I1T3 ]

[M1L2I1T3 ]

[M1T3I1L3 ]

[I2T4M1L2 ]

[I2T4M1L3 ]

[M1L1I2T3 ]

[I1L2 ]

[M1L2I1T2 ]

[M1I1T2 ]

[M1L2I2T2 ]

[M1L2I2T3 ]

newton

joule

watt

N-m2/kg2

radian/s

kg-m2/s

kg-m2

N-m

N/m2

N/m

N-s/m2

N/m2 (Pascal)

watt/m2

J/kg-K

watt/m2-k4

watt/m-K

ampere/m2

–1 m–1

C-m

V/m

volt

volt/m

farad (F)

C2/N-m2

Newton/A2

N-m/T

Weber (Wb)

tesla

henry

ohm ()

* * * *

Physical quantity

Dimension

SI unit