HC Verma Concepts Of Physics Exercise Solutions Of Chapter 4 (The Forces )

HC Verma Concepts of Physics Solutions - Part 1, Chapter 4 - The Forces:

EXERCISE

1. The gravitational force acting on a particle of 1 g due to a similar particle is equal to 6.67 * 10^-17 N. Calculate the separation between the particles.

Sol:

Given: m₁ = m₂ = 1 g = 0.001 kg; G = 6.67 * 10^-11 N-m²/kg²; F_g = 6.67 * 10^-17 N.

We have, 

→ Gravitational force, F_g = G * (m₁m₂)/r²

Or, 6.67 * 10^-17 = 6.67 * 10^-11 (0.001*0.001)/ r² 

Or, r = 1 m.

2. Calculate the force with which you attract the earth.

Sol:

A man is standing on the surface of earth

The force acting on the man = mg ……… (1)

Assuming that, m = mass of the man = 100 kg

And g = acceleration due to gravity on the surface of earth = 9.8 m/s².

W = mg = 100 * 9.8 = 980 N = force acting on the man by earth.

From newton’s 3^rd law, the man attracts earth with 980 N.

3. At what distance should two charges, each equal to 1 C, be placed so that the force between them equals to your weight?

Sol:

Given: q₁ = q₂ = 1 C; 1/ (4πε_o) = 9.0 * 10⁹ N-m²/C²; electrostatic force, F_e = weight, W.

Let we assume that, my weight, W = 1000 N

We have,

→ F_e = [1/ (4π ε_o)]*[q₁q₂/ r²]

Or, 1000 = (9.0 * 10⁹ * 1*1)/ r² 

Or, r = √ (9 *10⁶) = 3000 m.

4. Two spherical bodies, each of mass 50 kg, are placed at a separation of 20 cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.

Sol:

Given: m₁ = m₂ = 50 kg; r = 20 cm = 0.2 m; F_g = F_e; q₁ = q₂ = q =?

According to the question,

→ F_g = F_e

Or, G * (m₁m₂)/r² = [1/ (4πε_o)]*[q₁q₂/ r²]

Or, 6.67 * 10^-11 *50² = 9.0 * 10⁹ * q² 

Or, q = 4.3 * 10^-9 C. 

5. A monkey is sitting on a tree limb. The limb exerts a normal force of 48 N and a frictional force of 20 N. Find the magnitude of the total force exerted by the limb on the monkey.

Sol:

The limb exerts a normal force 48 N and frictional force of 20 N. The magnitude of the Resultant force,

→ R = √ (N² + F_f²) 
Or, R = √ (48² + 20²) = 52 N

Therefore, the magnitude of the total force exerted by the limb on the monkey is 52 N.

6. A body builder exerts a force of 150 N against a bullworker and compresses it by 20 cm. Calculate the spring constant of the spring in the bullworker.

Sol:

Given: F = 150 N; x = 20 cm = 0.2 m; k =?

We know, F = kx

Where, F = applied force; k = spring constant; x = deflection of spring.

So, k = F/x = 150/0.2 = 750 N/m.

Therefore, the spring constant of the spring in the bullworker is 750 N/m.

7. A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.)

Sol:

Suppose the height is h.

At earth station F = GMm/R²

Where, M = mass of earth

               m = mass of satellite

               R = Radius of earth

The force on the satellite due to the earth at height, h is F/2.

Now, F_{@ h} = GMm/ (R+h)² = F/2

Or, GMm/ (R+h)² = GMm/2R²   

Or, (R+h)² = 2R²

Or, h = √(2) * R – R =0.414 R 

Or, h = 2650 km.

8. Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force if the separation is increased to 25 cm?

Sol:

Given: separation, R₁ = 20 cm = 0.2 m; F₁ = 20 N; if R₂ = 25 cm = 0.25 m, then F₂ =?

We have, F₁ = GMm/ (R₁)²

Or, 20 = GMm/0.2² → GMm = 0.8 N-m²

Now, F₂ = GMm/ (R₂)² 

Or, F₂ = 0.8 / 0.25² = 12.8 N.

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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.          

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