HC Verma Concepts Of Physics Exercise Solutions Of Chapter 2 (Physics and Mathematics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 2 - Physics And Mathematics:

EXERCISE

1. A vector A makes an angle of 20° and B makes an angle of 110° with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.

Sol: 

So, resultant makes an angle with X-axis is 53⁰ + 20⁰ = 73⁰.

2. Let A and B be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles 30° and 60° respectively, find the resultant.

Sol:

So, resultant makes an angle with X-axis is 15⁰ + 30⁰ = 45⁰.

Alternative method,

X component of OA = 10 cos 30° = 5√3

X component of BC = 10 cos 60° = 5 

Y component of OA = 10 sin 30° = 5

Y component of BC = 1.5 sin 60° = 5√3

R_x = x component of resultant = 5 + 5√3 = 13.66 m

R_y = y component of resultant= 5 + 5√3 = 13.66 m

So, R = Resultant = 19.32 m                                    

If it makes an angle a with positive x-axis

Tan a = x component / y component = 1 
⇒ a = tan^–1 1 = 45⁰.

3. Add vectors A, B and C each having magnitude of 100 unit and inclined to the X-axis at angles 45°, 135° and 315° respectively.

Sol: 

Vector B and C are equal in magnitude but opposite in direction. Therefore, resultant of these two vectors is zero vector. If we add zero vector with another vector, resultant will be that vector.

So, resultant of vectors A, B, and C is same as vector A.

R = 100 unit at 45⁰ with X-axis.

4. Let a = 4 i + 3 j and b = 3 i + 4 j. (a) Find the magnitudes of (a) a, (b) b, (c) a + b and (d) a - b.

Sol:

(a) | a|= √ (4² + 3²) 
Or, | a| = √ (16 + 9) = √25 = 5.

(b) | b|= √ (3² + 4²) 
Or, | b| = √ (9 + 16) = √25 = 5.

(c)  a + b = (4 i + 3 j) + (3 i + 4 j) = 7 i + 7 j

Or, | a + b |= √ (7² + 7²) = 7√2.

(d) a + b = (4 i + 3 j) – (3 i + 4 j) = i – j

Or, | a – b |= √ {1² + (– 1)²} = √2.

5. Refer to figure (2-E1). Find (a) the magnitude, (b) x and y components and (c) the angle with the X-axis of the resultant of OA, BC and DE.

Sol:

Resultant vector diagram 

Angle between vector OA and BC is 90⁰.

Therefore, resultant of OA and BC is

Angle between vector OA and R₁₂ is given by

So, angle between vector DE and R₁₂ is 90⁰ + 36.9⁰ + 30⁰ = 156.9⁰. 

(a) The magnitude of the resultant of OA, BC and DE is 

(c) angle between vector DE and R is given by

angle between vector DE and R is = 180⁰ – 37⁰ = 143⁰

Therefore, angle made by resultant with X-axis is

= (143⁰ – 90⁰) = 53⁰.

(b) X component of resultant is = 1.62*cos 53⁰ = 0.98 m.

And Y component of resultant is = 1.62*sin 53⁰ = 1.3 m.

Alternative method,

X component of OA = 2 cos 30° = √3

X component of BC = 1.5 cos 120° = – 0.75

X component of DE = 1 cos 270° = 0

Y component of OA = 2 sin 30° = 1

Y component of BC = 1.5 sin 120° = 1.3

Y component of DE = 1 sin 270° = – 1

R_x = x component of resultant = √3 – 0.75 + 0 = 0.98 m

R_y = resultant y component = 1 + 1.3 – 1 = 1.3 m

So, R = Resultant = 1.6 m

If it makes an angle a with positive x-axis  

Tan a = x component / y component = 1.32

Or, a = tan^–1 1.32.

6. Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit?

Sol:

We know, the magnitude of the resultant is given by

(a) (1)² = (3)² + (4)² + 2 * 3 * 4 * cos θ

Or, cos θ = – (24/24) = – 1

Or, θ = 180⁰.

(b) (5)² = (3)² + (4)² + 2 * 3 * 4 * cos θ

Or, cos θ = (0/24) = 0  

Or, θ = 90⁰.

(c) (7)² = (3)² + (4)² + 2 * 3 * 4 * cos θ

Or, cos θ = (24/24) = 1

Or, θ = 0⁰.

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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.

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Conversion Of Unit From MKS (SI) To (CGS)

1. Velocity:

We know, 

Velocity = distance/(time)
Or, (v = s/t)

Dimensional formula of acceleration,

[v] = L/T = LT^-1

So, 1 m/s = (1 m) (1 s)^-1

      1 cm/s = (1 cm) (1 s)^-1

Thus, (1 m/s)/(1 cm/s) = (1 m/1 cm) (1 s/1 s)^-1

Or, (1 m/s)/(1 cm/s) = (100 cm/1 cm) (1 s/1 s)^-1

Or, (1 m/s)/(1 cm/s) = (10)² (1)^-1

Or, (1 m/s)/(1 cm/s) = 10²

ஃ  1 m/s = 10² cm/s = 100 cm/s.

2. Acceleration:

We know, 

Acceleration = length/ (time) ²     (a = L/t²)

Dimensional formula of acceleration,

[a] = L/T² = LT^-2

So, 1 m/s² = (1 m) (1 s) ^-2

      1 cm/s² = (1 cm) (1 s) ^-2

Thus, (1 m/s²)/ (1 cm/s²) = (1 m/1 cm) (1 s/1 s)^-2

Or, (1 m/s²)/ (1 cm/s²) = (100 cm/1 cm) (1 s/1 s) ^-2

Or, (1 m/s²)/ (1 cm/s²) = (10)² (1)^-2

Or, (1 m/s²)/ (1 cm/s²) = 10²

ஃ   1 m/s² = 10² cm/s² = 100 cm/s².

3. Density:                 

We know, 

Density = mass/volume     (ρ = m/v)

Dimensional formula of density,

[ρ] = M/L³ 

     = ML^-3 ^-

So, 1 kg/m³ = (1 kg) (1 m^-3)

      1 g/cm³ = (1 g) (1cm^-3)

Thus, (1 kg/m³)/ (1 g/cm³) = (1 kg/1 g) (1 m/1 cm)^-3

Or, (1 kg/m³)/ (1 g/cm³) = (1000 g/1 g) (100 cm/1cm) ^-3

Or, (1 kg/m³)/ (1 g/cm³) = (10)³ (10²)^-3

Or, (1 kg/m³)/ (1 g/cm³) = 10^-3           

ஃ 1 kg/m³ = 10^-3 g/cm³ = 0.001 g/cm³.

4. Force:

We know, 

Force = mass * acceleration     (F = m*a)

Dimensional formula of Force,

[F] = [m]*[a] = M LT^-2 = MLT^-2

So, 1 Newton = (1 kg) (1 m) (1 s)^-2

      1 dyne = (1 g) (1 cm) (1 s)^-2

Thus, 1 Newton/1 dyne = (1 kg/1 g) (1 m/1 cm) (1 s/1 s)^-2

Or, 1 Newton/1 dyne = (1000 g/1 g) (100 cm/1 cm) (1 s/1 s)^-2

Or, 1 Newton/1 dyne = (10)³ (10²)  

Or, 1 Newton/1 dyne = 10⁵

ஃ 1 Newton = 10⁵ dyne.

5. Pressure:

We know, 

 Pressure = Force/ (Area)     (P = F/A)

Dimensional formula of Pressure,

[P] = [F]/ [A] = MLT^-2/L² = ML^-1T^-2

So, 1 Pascal = (1 kg) (1 m) ^-1 (1 s)^-2

1 CGS pressure = (1 g) (1 cm)^-1 (1 s)^-2

Thus, 
1 Pascal/1 CGS pressure = (1 kg/1 g) (1 m/1 cm)^-1 (1 s/1 s)^-2

= (1000 g/1 g) (100 cm/1 cm)^-1 (1 s/1 s)^-2

= (10)³ (10²)^-1

= 10

ஃ 1 Pascal = 10 CGS pressure.

6. Work & Energy:

We know, 

Work = Force * distance    (W = F * d)

Dimensional formula of work,

[W] = [F] * [d] = MLT^-2 L = ML²T^-2

So, 1 joule = (1 kg) (1 m)² (1 s)^-2

      1 erg = (1 g) (1 cm)² (1 s)^-2

Thus, 1 joule/1 erg = (1 kg/1 g) (1 m/1 cm)² (1 s/1 s)^-2

= (1000 g/1 g) (100 cm/1 cm)² (1 s/1 s)^-2

= (10)³ (10²)²

= 10⁷

ஃ 1 joule = 10⁷ erg.

Both work and energy are dimensionally identical and same unit.

Dimensionally, Kinetic Energy = mass * (velocity)²  

                = mass* (length/time)²   

Or, Potential Energy = mass * acceleration * distance

                = mass * length/ (time)² * length

Both have same dimensional formula

[E] = ML²T^-2

So, 1 joule = (1 kg) (1 m)² (1 s)^-2

      1 erg = (1 g) (1 cm)² (1 s)^-2

Thus, 1 joule/1 erg = (1 kg/1 g) (1 m/1 cm)² (1 s/1 s)^-2

= (1000 g/1 g) (100 cm/1 cm)² (1 s/1 s)^-2

= (10)³ (10²)²

= 10⁷

ஃ 1 joule = 10⁷ erg.

7. Power:

We know, 

Power = Force * velocity      (P = F * v)

Dimensional formula of Power,

[P] = [F] [v] = MLT^-2 LT^-1 = ML²T^-3

So, 1 watt (J/s) = (1 kg) (1 m)² (1 s)^-3

      1 CGS power (erg/s) = (1 g) (1 cm)² (1 s)^-3

Thus, 

1 watt /1 CGS power = (1 kg/1 g) (1 m/1 cm)² (1 s/1 s)^-3

= (1000 g/1 g) (100 cm/1 cm)² (1 s/1 s)^-3

= (10)³ (10²)²

= 10⁷

ஃ 1 watt (J/s) = 10⁷ erg/s.

Discussion - If you have any Query or Feedback comment below. 

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