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.

9. The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data: The universal constant of gravitation G = 6.67 x 10^-11 N-m²/kg², mass of the moon = 7.36 * l0²² kg, mass of the earth = 6 * 10²⁴ kg and the distance between the earth and the moon = 3.8 * 10⁵ km.

Sol:

Given: G = 6.67 x 10^-11 N-m²/kg²; M_m = 7.36 * l0²² kg; M_e = 6 * 10²⁴ kg; R = 3.8 * 10⁵ km.

We have, F = GM_eM_m/ R² 

Or, F = (6.67 x 10^-11) * (6 * 10²⁴) * (7.36 * l0²²) / (3.8 * 10⁸)² 

Or, F = 2.0 * 10²⁰ N.

10. Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

Sol:

Charge on proton = 1.6 × 10^–19 C

Electric force, F_e = [1/ (4πε_o)]*[q₁q₂/ r²]

Or, F_e = (9.0 * 10⁹) * [(1.6 × 10^–19)²/ r²]

           = 23.04 * 10^-29/r²

Mass of proton = 1.732 × 10^–27 kg

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

Or, F_g = 6.67 * 10^-11 * [(1.732 * 10^–27)²/r²]

           = 20.0 * 10^-65/r²

So, F_e/ F_g = [23.04 * 10^-29/r²]/ [20.0 * 10^-65/r²]

Or, F_e/ F_g = 1.24 * 10³⁶.

11. The average separation between the proton and the electron in a hydrogen atom in ground state is 5.3 x 10^-11 m. (a) Calculate the Coulomb force between them at this separation, (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?

Sol:

The average separation between proton and electron of Hydrogen atom is r = 5.3 * 10^–11m.

(a) Coulomb’s force, F₁ = 9 * 10⁹ * (q₁q₂)/r²

Or, F₁ = 9 * 10⁹ (1.6 x 10^-19)²/ (5.3 * 10^–11)²

Or, F₁ = 8.2 * 10^-8 N.

(b) When the average distance between proton and electron becomes 4 times that of its ground state

Coulomb’s force, F₂ = F₁/16 = 8.2 * 10^-8/16 = 5.1 * 10^-9 N.

12. The geostationary orbit of the earth is at a distance of about 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.

Sol:

The geostationary orbit of earth is at a distance of about 36000km.

We know that, g’ = GM / (R+h)²

At h = 36000 km. g’ = GM / (36000+6400)²

At earth surface, g = GM / (6400)²

So, g’/ g = 0.023

 [Taking g = 9.8 m/s² at the surface of the earth]

→ g’ = 0.023 * 9.8 = 0.22 m/s² 

A 120 kg equipment placed in a geostationary satellite will have weight

Mg’ = 0.22 * 120 = 26.79 = 27 N.

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