HC Verma Concepts Of Physics Exercise Solutions Of Chapter 1 (Introduction To Physics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 1 - Introduction To Physics:

EXERCISE

1. Find the dimensions of

(a) Linear momentum, (b) frequency and (c) pressure.

Sol: 

(a) Dimensionally,

Linear momentum = mass * velocity 
And velocity = displacement/time

Dimension of velocity, [v] = L/T = LT^-1

Hence, [linear momentum] = MLT ^-1.

(b) Dimensionally,

Frequency = 1/time period

[Frequency] = 1/T =T ^-1.

(c) Dimensionally,

Pressure = force/area

[Pressure] = [force]/[area] 
Or, [Pressure] = MLT^-2 / L² = ML^-1T ^-2.

2. Find the dimensions of

(a) Angular speed ω, (b) angular acceleration α, (c) torque Γ and (d) moment of inertia.

 

Some of the equations involving these quantities are

The symbols have standard meanings. 

Sol:
(a) [ω] = [θ]/[t] = 1/T = T ^-1  
∵ [θ] = [arc]/[radius] = L/L = 1

(b) [α] = [ω]/[t] = T ^-2.

(c) [Γ] = [F] [r] =ML²T^-2 = ML²T ^-2.

(d) [I] = [m] [r²] = ML².

3. Find the dimensions of

(a) Electric field E, (b) magnetic field B and (c) magnetic permeability µ₀.

The relevant equation are F = qE, F = qvB, and B = µ₀I/2πa; Where F is force, q is charge, v is speed, I is current, and a is distance.

Sol: 

(a) F = qE or, E = F/q

[E] = [F]/[q] = MLT ^-2/IT = MLT^-3I^-1.
∵ [q] = IT

(b) F = qvB or, B = F/qv

[B] = [F] / [q][v] = MLT^-2/(IT LT^-1) = MT^-2I^-1.

(c) B = µ₀I/2πa or, µ₀ = B 2πa/I

[µ₀] = [B] [a]/I = MLT^-2I^-1 L/I = MLT^-2I^-2.

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 charge and I is current.

Sol: 

(a) p = q.d or, [p] = [q]*[d] = ITL = LTI.

(b) M = IA or, [M] = I*[A] = L²I.

5. Find the dimensions of Planck’s constant h from the equation E = hν where E is the energy and ν is the frequency.

Sol: 

Dimensionally,

→ Energy = mass* (velocity)²
Or, [E] = ML²T^-2 and [ν] = T^-1

Given, E = hν or, h = E/ν

[h] = ML²T^-2/ T^-1 = ML²T^-1.

6. Find the dimensions of

(a) The specific heat capacity c, (b) the coefficient of linear expansion α and (c) the gas constant R. Some of the equations involving these quantities are Q = mc (T₂ – T₁), l_t = l₀[1+α(T₂ – T₁)] and PV = nRT.

Sol:

(a) Q = mc (T₂ – T₁) 
Or, c = Q/ {m (T₂ – T₁)}

Where Q is heat energy and [Q] = ML²T^-2; [T] = K (Kelvin)

[c] = [Q]/ [m] [T] = ML²T^-2/MK = L²T^-2K^-1.

(b) l_t = l₀[1 + α (T₂ – T₁)] 
Or, α = (l_t/l₀ – 1)/( T₂ – T₁)

Or, [α] = 1/K = K^-1.

(c) PV = nRT
Or, R = PV/nT

[R] = [P][V]/[n][T] 
Or, [R] = ML^-1T^-2L³/(mol)K

Or, [R] = ML²T^-2K^-1(mol)^-1.

7. Taking force, length and time to be fundamental quantities find the dimensions of (a) Density, (b) pressure, (c) momentum, and (d) energy.
Sol:
Taking force, length and time as fundamental quantities
(a) Dimensionally, density = mass/volume 

Or, density = (force/acceleration)/volume

Or, [density] = F/ {(LT^-2) (L³)} = FL^-4T².

(b) Dimensionally, pressure = Force/area = FL^-2.

(c) Dimensionally, momentum = mass * velocity

Or, [momentum] = (FL^-1T²) (LT^-1)

Or, [momentum] = FT.

(d) Dimensionally, Energy = Force * distance

Or, [Energy] = FL.           

8. Suppose the acceleration due to gravity at a place is 10 m/s². Find its value in cm/(minute)². 

Sol:

1 m/s² = (1 m) (1 s)^-2  
And, 1 cm/min² = (1 cm) (1 min)^-2

                 

Or, 1 m/s² =36 * 10⁴ cm/min² 
Or, 10 m/s² = 36 * 10⁵ cm/min².

9. The average speed of a snail is 0.020 miles/hour and that of leopard is 70 miles/hour. Convert these speeds in SI units.

Sol:

1 mile/hour = (1 mile) (1 hour)^-1  
And, 1 m/s = (1 m) (1 s)^-1

Or, 1 mile/hour = 0.45 m/s

Therefore, the average speed of snail in SI unit = 0.45 * 0.02 = 0.009 m/s.
And, the average speed of Leopard in SI unit = 0.45 * 70 = 31.5 m/s.

10. The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data: Specific gravity of mercury = 13.6, Density of water = 10³ kg/m³, g = 9.8 m/s² at Calcutta. Pressure = hρg in usual symbols.

Sol:                                           

Given: h = 75 cm; S.G. of mercury = 13.6; Density of water = 10³ kg/m³; g = 9.8 m/s²

→ Pressure, P = h ρ_hg g

Or, P = 0.75 * 13.6 * 10³ * 9.8

Or, P = 10 *10⁴ N/m²   [In SI Unit]

Dimension of pressure = ML^-1T^-2

So,    1 N/m² = (1 kg) (1m)^-1 (1 s)^-1 
And, 1 dyne/cm² = (1 g) (1 cm)^-1 (1 s)^-1

              

Or, 1 N/m² = 10 dyne/cm² 
Therefore, pressure in CGS is = 10 * 10⁵ = 10⁶ dyne/cm².

11. Express the power of a 100 watt bulb in CGS unit.

Sol:

The unit of power is watt or Nm/s

[Power] = [work]/ [time] 
Or, [Power] = (ML²T^-2)/ (T) = ML²T^-3 

So, 1 watt = (1 kg) (1 m)² (1 s)^-3 
And, 1 erg/s = (1 g) (1 cm)² (1 s)^-3   [CGS unit of power = erg/s]

                

Or, 1 watt = 10⁷ erg/s

Therefore, 100 watt = 10⁹ erg/s.

12. The normal duration of I.sc. Physics practical period in Indian colleges is 100 minutes. Express this period in microcenturies. 1 microcentury = 10^-6 * 100 years. How many microcenturies did you sleep yesterday?

Sol:                                    

1 microcentury = 10^-6 * 100 years = 10^-4 years
Or, 1 microcentury = 10^-4 * 365 * 24 * 60 minutes

Or, 1 microcentury = 52.56 minutes

Hence, 100 minutes = 100/52.56 = 1.9 microcenturies.

If you slept 6 hours yesterday, then

6 hours = (6 * 60)/52.56 = 6.85 microcenturies.

13. The surface tension of water is 72 dyne/cm. Convert it in SI unit.

Sol:

The unit of surface tension (σ) = N/m

This suggests that it has dimensions of Force/length

Thus, [σ] = [F]/[L] = MLT^-2/L = MT^-2

So, 1 dyne/cm = (1 g) (1 s)^-2

And, 1 N/m = (1 kg) (1 s)^-2

{1 dyne/cm}/{1 N/m} = (1 g/1 kg) (1s/1s)^-2

Or, {1 dyne/cm}/{1 N/m} = 1/1000

Or, 1 dyne/cm = 0.001 N/m

Or, 72 dyne/cm = 0.072 N/m.

14. The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ω. Assuming the relation to be K = k l^a ω^b where k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is (2/5)Mr². 

Sol:

Dimensionally, 

Kinetic energy = mass * (velocity)²

Or, Kinetic energy = mass * (distance/time)²

Or, Kinetic energy = ML²T^-2

[K] = [k] [I]^a [ω]^b   [k] = 1 (dimensionless constant)

Or, ML²T^-2 = (ML²)^a (T^-1)^b     [I] = ML² & [ω] = T^-1

Or, ML²T^-2 = M^aL^2aT^-b

Comparing both side, we get

Power of ‘M’ a = 1;

Power of ‘L’ 2a = 2, or, a = 1; and

Power of ‘T’ – b = – 2, or, b = 2
[Ans: a = 1; b = 2].

15. Theory of relativity reveals that mass can be converted into energy. The energy E so obtained is proportional to certain powers of mass m and the speed c of light. Guess a relation among the quantities using the method of dimensions.

Sol:

Dimensional formula of energy (E) is ML²T^-2

Let us assume that the required equation is

    E = k m^a c^b

Where, k is dimensionless constant, m is the mass and c is the speed.

Writing dimension of both sides,

→ ML²T^-2 = k M^a (LT^-1)^b = k M^a L^b T^-b

Equating the exponents,

⇒ a = 1; b = 2

Therefore, E = k m c².

16. Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm's law, product of two of these quantities equals the third. Obtain Ohm's law from dimensional analysis. Dimensional formulae for R and V are ML²I^-2T^-3 and ML²T^-3I^-1 respectively.

Sol:

Let,

Ohm's law is     I^a R^b = V

Dimensionally,[I]^a [R]^b = [V]

Or, I^a (ML²I^-2T^-3)^b = ML²T^-3I^-1

Or, M^bL^2bT^-3bI^a-2b = ML²T^-3I^-1

Equating the exponents,

b = 1;

2b = 2,

Or, b = 1;

– 3b = – 3,

Or, b = 1;

a – 2b = – 1,

Or, a = 1

Thus, Ohm’s Law is I¹R¹ = V or, V = IR.

17. The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

Sol:

Frequency, ‘f’ of vibration of a string depends on mass per unit length ‘m’, tension ‘F’ and length of string is L.

So, f = k F^a (m)^b L^c

Where k is dimensionless constant.

[f] = [F]^a [m]^b [L]^c

Or, T^-1 = (MLT^-2)^a (M)^bL^-bL^c 
Or, T^-1 = M^a+bL^a-b+cT^-2a

Equating the exponents,

a + b = 0; a – b + c = 0; – 1 = – 2a, or, a = 1/2

And, b = – 1/2; c = – 1

So, f = k F^1/2m^-1/2L^-1 

Thus, the formula for f is (k/L)[√(F/m)].

18. Test if the following equations are dimensionally correct:

Where h = height, S = surface tension, ρ = density, P = pressure, V = volume, η = coefficient of viscosity, ν = frequency and I = moment of inertia.

Sol:
[h] = L; [S] = [F]/[l] = MT^-2; [ρ] = ML^-3; [P] = ML^-1T^-2; [r] = L; [g] = LT^-2; [v] = LT^-1;[ν] = T^-1; dimensionally, I = mr², so,[I] = ML²; [V] = L^-3; [cos θ] = 1 (dimensionless constant) and for Newtonian fluid stress is proportional to rate of deformation. Stress = η * rate of deformation. [Stress] = [F]/[area] = ML^-1T^-2; dimensionally, [rate of deformation] = [velocity]/[distance] = T^-1.So, [η] = ML^-1T^-3.

19.

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