HC Verma Concepts Of Physics Exercise Solutions Of Chapter 3 (Rest and Motion:Kinematics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 3 - Rest and Motion: Kinematic:

EXERCISE

1. A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field, (a) what distance he has to walk to reach the field? (b) What is his displacement from his house to the field?

Sol:

                       

(a) The distance he has to walk to reach the field is 
                 = (50 + 40 + 20) m =110 m

(b) Magnitude of displacement vector is given by AD

AD = √ {(50 - 20)² + 40²} = √ {900 + 1600} = 50 m and

From ∆ ADE, tan θ = DE/AE = 30/40 = 3/4
                          ⇒ θ = tan^-1 (3/4)

∴ Displacement = 50 m, tan^-1 (3/4) north to east.

2. A particle starts from the origin, goes along the Z-axis to the point (20 m, 0) and then returns along the same line to the point (-20 m, 0). Find the distance and displacement of the particle during the trip. 

 Sol:

                 

Distance travelled by the particle during the trip is

= AB + BC = (20 + 40) m = 60 m

Displacement ➡ shortest distance between final and initial position.

Displacement of the particle during the trip is 
            = 20 m along –ve X direction.

3. It is 260 km from Patna to Ranchi by air and 320 km by road. An aeroplane takes 30 minutes to go from Patna to Ranchi whereas a delux bus takes 8 hours, (a) Find the average speed of the plane, (b) Find the average speed of the bus. (c) Find the average velocity of the plane, (d) Find the average velocity of the bus. 

Sol:

(a) Given: distance from Patna to Ranchi by air = 260 km and time taken by an aeroplane = 30 min = 0.5 hour.

ஃ the average speed of the plane = distance/time
              = 260/0.5 km h^-1 = 520 km h^-1  

(b) Given: distance from Patna to Ranchi by road = 320 km and time taken by delux bus = 8 hours.

ஃ the average speed of the Bus = distance/time 
             = 320/8 km h^-1 = 40 km h^-1  

(c) Displacement ➡ shortest distance between final and initial position

ஃThe average velocity of the plane = displacement/time 
             = 260/0.5 km h^-1 = 520 km h^-1 Patna to Ranchi

(d) Displacement ➡ shortest distance between final and initial position

ஃThe average velocity of the plane = displacement/time 
             = 260/8 km h^-1 = 32.5 km h^-1 Patna to Ranchi

4. When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km. (a) what is the average speed of the car during this period? (b) What is the average velocity?

Sol:

(a) The average speed of the car during this period is

= (12416 - 12352)/2 km h^-1 = 32 km h^-1

(b) The average velocity is = displacement/time

Displacement ➡ shortest distance between final and initial position

Displacement = Zero [because initial and final position is same]

ஃThe average velocity is zero.

5. An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration?

Sol:
Maximum speed (v) = 18.0 km/h 
                 = (18 * 1000)/3600 = 5.0 m s^-1.

Athelete start from zero speed, 
        So initial speed (u) = 0 m s^-1.

Time taken (t) = 2 s.

ஃ the magnitude of his average acceleration is

|A _ave| = (v - u)/t = (5 – 0)/2 = 2.5 m s^-2

6. The speed of a car as a function of time is shown in figure (3-E1). Find the distance travelled by the car in 8 seconds and its acceleration.

      

Sol:  

From the graph we get following data:

Initial velocity (u) = 0 m s^-1; final velocity (v) = 20 m s^-1; 
time taken = 8 s.

We know, a = (v - u)/t 
Or, a = (20 - 0)/8 = 2.5 m s^-2

Acceleration, a = 2.5 m s^-2   

Distance, s = ut + ½ (at²) 
Or, s = 0 * 8 + ½ * 2.5 * 8² = 80 m

7. The acceleration of a cart started at t = 0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 seconds and draw the position-time graph.
                         

Sol:

At t = 0, cart started. So initial speed (u) = 0; for 1^st 10 s acceleration, a = 5.0 ft s^-2.

Distance travelled in 1^st 10 s is

S₁ = ut + ½ at² 
    = 0 * 10 + ½ * 5 * 10² = 250 ft.

At 10 s velocity of cart, v_{10 s} = u + at = 0 + 5 * 10 = 50 ft/s; a_10-20 = 0

Distance travelled in 2^nd 10 s is

S₂ = v_{10 s} * t = 50 * 10 = 500 ft/s

For last 10 s: a = - 5.0 ft s^-2; u = 50 ft/s;

Distance travelled in 3^rd 10 s is

S₃ = ut + ½ at² 
    = 50 * 10 + ½ * (-5) * 10² 
    = 500 -250 = 250 ft

Total distance travelled,

           S = S₁ + S₂ + S₃  

              = (250 + 500 +250) ft = 1000 ft

The position-time graph 

                             

8. Figure (3-E3) shows the graph of velocity versus time for a particle going along the X-axis. Find (a) the acceleration, (b) the distance travelled in 0 to 10 s and (c) the displacement in 0 to 10 s.                                                        

Sol:

From the graph we get,

Initial velocity (u) = 2 m/s; final velocity (v) = 8 m/s; 
time taken = 10 s.

(a) acceleration, a = (v - u)/t = (8 - 2)/10 = 0.6 m/s² along +ve x axis.

(b) The distance travelled in 0 to 10 s is given by

S = ut + ½ at² 
   = 2 * 10 + ½ * 0.6 * 10² = 50 m

(c) Displacement = 50 m along +ve x axis.          

Read More

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.

Read More

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.

Read More

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. 

Read More