Solutions to Exercises on Introduction to Physics HC Verma's Concepts of Physics Part 1 (1-4)

 Q#1

Find the dimensions of
(a) linear momentum,
(b) frequency and
(c) pressure.      

Answer:
(a) Linear momentum = mass x velocity
= mass x distance/time

p = ml/t

The dimension of mass = [M]
The dimension of velocity = [L/T] = [LT⁻¹]
Hence the dimensions of linear momentum,
[p] = [M][LT⁻¹] = [MLT⁻¹]  

(b) Frequency = 1/Time period

f = 1/T

The dimension of frequency [f] = [1/T] = [T⁻¹]

(c) Pressure = Force per unit area

P = Force/Area = Mass(acceleration)/Area

The dimension of mass = [M]
The dimension of acceleration = [LT⁻²]
The dimension of area = [L²]
Hence the dimension of pressure,
[P] = [M][LT⁻²]/[L²] =[ML⁻¹T⁻²]
           
 Q#2
Find the dimensions of
(a) angular speed ⍵,
(b) angular acceleration α
(c) Torque Г and
(d) moment of inertia I.
Some of the equations involving these quantities are

⍵ = (θ₂ – θ₁)/(t₂ – t₁), α = (⍵₂ – ⍵₁)/(t₂ – t₁),

τ = Fr and I = mr².
The symbols have standard meanings.  
  
Answer:
(a) Angular speed ⍵ = (θ₂ – θ₁)/(t₂– t₁)
The dimension of [θ₂ – θ₁] = Dimension of angle which is dimensionless
The dimension of [t₂ – t₁] = Dimension of time = [T]
Hence [⍵] = 1/[T] = [T⁻¹]  

(b) Angular acceleration α = (⍵₂ – ⍵₁)/(t₂ – t₁)          

The dimension of (⍵₂ – ⍵₁) = The dimension of angular speed = [T⁻¹]

The dimension of (t₂ – t₁) = The dimension of time = [T]

Hence the dimension of angular acceleration,

[α] = [T⁻¹]/[T] = [T⁻²]

(c) Torque = Force x Arm (distance),

i.e. τ = Fr
The dimension of force [F] = mass x acceleration = [M][LT⁻²] = [MLT⁻²]

The dimension of arm, [r] = [L]

Hence the dimension of torque,

[Г] = [MLT⁻²][L] = [ML²T⁻²]

(d) The moment of inertia, I = mr²
The dimension of mass, [m] = [M]
The dimension of radius/distance, [r] = [L]
Hence the dimension of the moment of inertia,
[I] = [M][L]² = [ML²]

Q#3
Find the dimensions of
(a) electric field E,
(b) magnetic field B and
(c) magnetic permeability µ₀.
The relevant equations are
F = qE, F = qvB, and B = µ₀I/2πa
where F is a force, q is a charge, v is speed, I is current, and a is distance.      

Answer:
(a) F = qE, hence
Electric field E = F/q
The dimension of force [F] = [MLT⁻²]
The dimension of charge, [q] = [TI]
{Since charge = current*time}
Hence the dimension of electric field,

[E] = [MLT⁻²]/[TI] = [MLT⁻³I⁻¹]

(b) Since F = qvB,

Magnetic field, B = F/qv
Now, [F] = [MLT⁻²]
[q] = [TI]
Speed, [v] = [LT⁻¹]

Hence the dimension of magnetic field,
[B] = [MLT⁻²]/[TI][LT⁻¹] = [MT⁻²I⁻¹]

(c) Given, B = µ₀I/2πa

µ₀ =2πaB/I

The dimension of distance, [a] = [L]

The dimension of the magnetic field, [B] = [MT⁻²I⁻¹]
The dimension of current, [I] = [I] and π is dimensionless. Hence the dimension of magnetic permeability,

[µ₀] = [L][MT⁻²I⁻¹]/[I] = [MLT⁻²I⁻²]  

Q#4
Find the dimensions of
(a) electric dipole moment p and
(b) magnetic dipole moment M.
The defining equations are p = q.d, and M = IA
where d is distance, A is area, q is a charge and I is current.        

Answer:
(a) The electric dipole moment, p = qd
The dimensions of charge, [q] = [TI]
The dimensions of distance, [d] = [L]  
Hence the dimensions of electric dipole moment,

[p] = [TI][L] = [LTI]

(b) The magnetic dipole moment, M = IA
The dimension of the area, [A] = [L²]
Hence the dimensions of the magnetic dipole moment, [M] = [I][L²] = [L²I]   

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