HC Verma Concepts Of Physics Exercise Solutions Of Chapter 26 (Laws of Thermodynamics)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 26 - Laws Of Thermodynamics:

EXERCISE

1. A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K respectively. Neglect any thermal expansion, (a) how much heat is transferred to the liquid—vessel system? (b) How much work has been done on this system? (c) How much is the increase in internal energy of the system?

Sol:

System: water + copper

Given: insulated; closed system; t_i = 15 ⁰c; t_e = 17 ⁰c; mass of copper, m_cu = 100 g = 0.1 kg; mass of water, m_w = 200 g = 0.2 kg; c_cu = 420 J/kg-K and c_w = 4200 J/kg-K.

(a) Due to insulated boundary, heat transfer to the system is zero.

(b) From the 1^st law,

We know, Q = ΔU + W = U₂ – U₁ + W

Or, W = U₁ – U₂              [Q = 0]

Or, W = (U₁ – U₂)_cu + (U₁ – U₂)_w

Or, W = m_cu c_cu (t_i – t_e) + m_w c_w (t_i – t_e)

Or, W = 0.1 * 420 (15 – 17) + 0.2 * 4200 (15 – 17)

Or, W = – 1764 J.

– Ve sign means work done on the system.

(c) The increase in internal energy of the system is ΔU = W = 1764 J.

2. Figure (26-E1) shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J/K kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. (a) how much heat is given to the liquid? (b) How much work is done on the liquid? (c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

Sol:

System: liquid + wheel + container

Given: adiabatic system; heat capacity of liquid, C_L = 4200 J/K, mass of block, m = 12 kg, change of height of block, Δh = 70 cm = 0.7 m.

(a) Heat is given to the liquid is zero, because system is adiabatic.

Q = 0

(b) Work is done on the liquid, W = change of potential energy of block

Or, W = mg (Δ h) = 12 * 10 * 0.7 = 84 J.

(c) 1^st law, Q = ΔU + W

Or, 0 = C_L (Δ t) – 84 [W = – 84 work done on the system]

Or, Δ t = 84/4200 = 0.02 ⁰c.

3. A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, (c) Find the work done in this frame by the external force holding the belt.

Sol:

Given: mass of block = 100 kg; u = 2 m/s; μ = 0.2; v = 0.

1^st law: Q = Δu + W

In this case Q = 0

Or, – Δu = W = change in K.E.

Or, Δu = – (½ mv² – ½ mu²)

Or, Δu = ½ * 100 * 2² = 200 J.

(a) The change in the internal energy of the block-belt system is 200 J.

(b) The increase in the kinetic energy of the block is 200 J.

4. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.

Sol:

Given: Q = 100 J; ΔV = 0.

1^st Law: Q = ΔU + W = ΔU + pΔV

Or, 100 = ΔU + 0 → ΔU = 100 J

So, the change in internal energy of a gas is 100 J.

5. The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Sol:

Given: Q = 0; v₁ = 200 cc = 2 * 10^-4 m³; v₂ = 50 cc = 0.5 * 10^-4; p₁ = 10 kPa; p₂ = 50 kPa.

(a) Work done during the process 1-2,

W = p₁ (v₂ – v₁) + ½ * (p₂ – p₁)(v₂ – v₁)

Or, W = 10 * (0.5 – 2)*10^-4 + ½ *40 *(0.5 – 2) * 10^-4

Or, W = – 0.0015 – 0.0030 = – 0.0045 kJ

Or, W = – 4.5 J

So, the work done by the gas is – 4.5 J.

(b) 1^st law: Q = ΔU + W

Or, 0 = ΔU + W

Or, ΔU = – W = 4.5 J.

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HC Verma Concepts Of Physics PDF Exercise Solutions Of Chapter 7 (Circular Motion)

Solutions of H.C. Verma’s Concepts of Physics chapter 7 (Circular Motion) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 7 (Circular Motion)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 7 - Circular Motion:

EXERCISE

1. Find the acceleration of the moon with respect to the earth from the following data: Distance between the earth and the moon = 3.85 * 10⁵ km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

Sol:

Given: Distance between the earth and the moon, r = 3.85 * 10⁵ km = 3.85 * 10⁸ m; T = 27.3 days = 2.36 * 10⁶ s.

Angular speed, ω = 2π/T = 2.66 * 10^-6 rad/s.

Acceleration, a = ω²r = (2.66 * 10^-6)² * 3.85 * 10⁸ = 2.73 * 10^-3 m/s².

2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

Sol:

Given: Distance between the centre of earth and the particle, r = 12800/2 km = 64 * 10⁵ m; T = 24 hrs. = 86400 s.

Angular speed, ω = 2π/T = 7.27 * 10^-5 rad/s.

Acceleration, a = ω²r = (7.27 * 10^-5)² * 64 * 10⁵ = 0.0336 m/s².

3. A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is in cm/s and t in seconds. (a) Find the radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of the acceleration at t = 1 s.

Sol:  

Given: radius, r = 1.0 cm; v = 2.0t.

Speed of particle at 1 s, v_{t = 1 s} = 2 cm/s

(a) The radial acceleration of the particle at t = 1 s

→ (a_r) _{t = 1 s} = (v_{t = 1 s})²/r = 4 cm/s².

(b) The tangential acceleration of the particle at t = 1 s 
→ (a_t) _{t = 1 s} = (dv/dt) _{t = 1 s} = d(2t)/dt = 2 cm/s².

(c) The magnitude of the acceleration at t = 1 s

→ a = √ (a_r² + a_t²) = √ (4² + 2²) = √20 cm/s².

4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?

Sol:

Given: speed, v = 36 km/hr = 10 m/s; total mass, M = 150 kg; radius of a turn, r = 30 m.

To make the turn possible, horizontal force (F) must be equal to Centripetal force (F_c). Otherwise there will be skidding.

We know, Centripetal force (F_c) = mv²/r = (150 * 10²)/30 = 500 N.

Therefore, horizontal force (F) = 500 N.

5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

Sol: 

Given: speed, v = 36 km/hr = 10 m/s; total mass, M = 150 kg; radius of a turn, r = 30 m.

From the diagram

R cos θ = Mg ------------------ (1)

R sin θ = F_C

Or, R sin θ = Mv²/r ---------- (2)

Dividing equation (2) by equation (1)

We get, tan θ = v²/rg = 10²/ (30 * 10) = 1/3

Or, θ = tan^-1 (1/3).

6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking?

Sol: 

Given: speed, v = 18 km/hr = 5 m/s; radius of a turn, r = 10 m.

Angle of banking is given by,

→ tan θ = v²/rg = 5²/ (10 * 10) = ¼

Or, θ = tan^-1 (1/4).

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HC Verma Concepts Of Physics PDF Exercise Solutions Of Chapter 6 (Friction)

Solutions of H.C. Verma’s Concepts of Physics chapter 6 (Friction) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 6 (Friction)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 6 - Friction:

EXERCISE

1. A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s². What is the coefficient of kinetic friction between the block and the plane?

Sol:

Given: acceleration, a = – 0.4 m/s².

Let mass of the body be ‘m’ and F_f → frictional force.

From the free body diagram,

Vertical motion: R – mg = 0 → R = mg ……… (1)

Horizontal motion: ma = F_f = μR ……. (2)

From equation (1) and (2)

We get, μ = a/g = 4/10 = 0.4

Therefore, the coefficient of kinetic friction between the block and the plane is 0.4.

2. A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest?

Sol:

Let mass of the body be ‘m’ and deceleration be a.

F_f → frictional force; μ = 0.10.

From the free body diagram,

Vertical motion: R – mg = 0 → R = mg ……… (1)

Horizontal motion: ma = F_f = μR ……. (2)

From equation (1) and (2)

We get, a = μg = 0.10 * 10 = 1 m/s²

Initial velocity u = 10 m/s; Final velocity v = 0 m/s; a = – 1 m/s² (deceleration).

We have, 2as = v² – u²

Or, s = (v² – u²)/2a = (0² – 10²)/2(– 1) = 50 m

So, it will travel 50 m before coming to rest. 

3. A block of mass m is kept on a horizontal table. If the static friction coefficient is μ, find the frictional force acting on the block. 

Sol:

Body is kept on the horizontal table.

If no force is applied, no frictional force will be there

F_s → frictional force

F_a → Applied force

From graph it can be seen that when applied force is zero, frictional force is zero.

4. A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two.

Sol:

Given: distance covered, s = 8 m; time taken, t = 2 s; initial velocity, u = 0.

We know, s = ut + ½ at²

Or, 8 = 0 * 2 + ½ * a * 2² → a = 4 m/s².

Frictional force, F_f = μR

Let, the mass of body be ‘m’.

The equation of motion of the body along perpendicular to the plane: acceleration, a = 0.

→ R – mg cos 30⁰ = 0 → R = mg cos 30⁰ ------- (1)

The equation of motion of the body along the plane: acceleration, a = 4 m/s².

→ mg sin 30⁰ – F_f = ma

Or, mg sin 30⁰ – μR = 4m ------- (2)

Solving equation (1) and (2)

We get, mg sin 30⁰ – μmg cos 30⁰ = 4m

Or, μ = (g sin 30⁰ – 4)/g cos 30⁰ = 0.11

Therefore, the coefficient of kinetic friction between the two is 0.11.

5. Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4 kg.

Sol:

Given: mass of block, m = 5 kg; Force applied, F = 4 N; μ = 0.11.

The equation of motion of the body along perpendicular to the plane: acceleration, a = 0.

→ R – mg cos 30⁰ = 0 → R = 4g cos 30⁰ ------- (1)

The equation of motion of the body along the plane: acceleration, a.

→ mg sin 30⁰ – F_f + F = ma

Or, 4g * ½ – μR + 4 = 4a ------- (2)

Solving equation (1) and (2)

We get, 4g * ½ – 4* 0.11* g cos 30⁰ + 4 = 4a

Or, a = 5.04 ~ 5 m/s²  

Now, we have u = 0; t = 2 s; a = 5 m/s²

We know, s = ut + ½ at² = 0*2 + ½ * 5 * 2²

Or, s = 10 m

Therefore, the block move in the first two seconds after starting from rest is 10 m.

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HC Verma Concepts Of Physics PDF Exercise Solutions Of Chapter 5 (Newton's Laws of Motion)

Solutions of H.C. Verma’s Concepts of Physics chapter 5 (Newton's Laws of Motion) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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

Solutions of H.C. Verma’s Concepts of Physics chapter 4 (The Forces) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF.

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

EXERCISE

Solutions of H.C. Verma’s Concepts of Physics chapter 3 (Rest and Motion:Kinematics) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF.

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

Solutions of H.C. Verma’s Concepts of Physics chapter 2 (Physics and Mathematics) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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

Solutions of H.C. Verma’s Concepts of Physics chapter 1 (Introduction to Physics) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF.

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 5 (Newton's Laws of Motion)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 5 - Newton's Laws of Motion:

EXERCISE

1. A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.

Sol:

Given: mass of block, m = 2 kg; distance travelled, s = 10 m; time taken, t = 2 s; initial velocity, u = 0; Force, F =?

Acceleration, a = F/m = F/2

We have, s = ut + ½ * a * t²

Or, 10 = 0 * 2 + ½ * (F/2) * 2² 

Or, F = 10 N.

2. A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it?

Sol:

Given: initial velocity, u = 40 km/h = 11.1 m/s; final velocity, v = 0; distance travelled, s = 4.0 m; mass of car, m = 2000 kg.

We have, v² – u² = 2as

Or, 0² – 11.1² = 2 * a * 4

Or, acceleration, a = – 15.4 m/s²

Average force must be applied = F = ma

Or, F = 2000 * 15.4 = 30800 N

Or, F = 3.08 *10⁴ N ≈ 3.1 *10⁴ N.

3. In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5 x 10⁶ m/s in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1 x 10^-31 kg.

Sol: 

Given: mass of electron, m = 9.1 * 10^-31 kg; initial velocity, u = 0; final velocity, v = 5 * 10⁶ m/s; distance travelled, s = 1 cm = 0.01 m; F =?

We have, v² – u² = 2as

Or, (5 * 10⁶)² – 0² = 2 * a * 0.01

Or, a = 12.5 * 10¹⁴

Therefore, F = ma

Or, F = (9.1 * 10^-31) * (12.5 * 10¹⁴)

Or, F = 1.1 * 10^-15 N.

4. A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s².

Sol: 

T₁ and T₂ are the tensions in the two strings.

Taking the block 2 as the system. The forces acting on the system are

(i) M₂g downwards

(ii) T₂ upwards

As the body is in equilibrium, these forces add to zero.

T₂ – M₂g = 0 or, T₂ = M₂g = 0.3 * 10 = 3 N.

Taking the block 1 as the system. The forces acting on the system are

(i) M₁g downwards

(ii) T₁ upwards

(iii) T₂ downwards

As the body is in equilibrium, these forces add to zero.

→ T₁ – T₂ – M₁g = 0 
Or, T₁ = M₁g + T₂ 
Or, T₁ = 0.2 * 10 + 3 = 5 N.

Therefore, tensions in the strings are 5 N and 3 N.

5. Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

Sol: 

Taking both the blocks as a system. Force acting on the system is F. Due to this force, let the system move with an acceleration, a.

From Newton’s law of motion:

We have,

→ Force = mass of system * acceleration

Or, F = 2m * a

Or, a = F/2m.

Now, taking first block as a system. Force acting on the system is T and acceleration of the system is F/2m. Mass of the system m.

Again, from Newton’s law of motion:

We have,

→ Force = mass of system * acceleration

Or, T = m * a

Or, T = m * (F/2m) = F/2.

So, the tension in the string joining the blocks is F/2.

6. A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds. 

Sol:

We know that,

Slope of the curve in velocity v/s time graph = acceleration

Therefore, a_{@ 2 s} = (15/3) = 5 m/s²; a_{@ 4 s} = 0 m/s²; a_{@ 6 s} = – (15/3) = – 5 m/s².

We have, Force = mass * acceleration

So, F_{@ 2s} = (50/1000) * 5

Or, F_{@ 2s} = 0.25 N along the motion.

→ F_{@ 4s} = (50/1000) * 0 = 0.

→ F_{@ 6s} = (50/1000) * (– 5)

Or, F_{@ 6s} = – 0.25 or 0.25 N opposite to the motion.

7. Two blocks A and B of mass m_A and m_B respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?

Sol:

Given: mass of two blocks A and B are m_A and m_B respectively. Force exerted by block A on Block B is F. Therefore, the force exerted on block A by block B is also F. Let, the force exerted by the experimenter on A be P and acceleration be a.

Taking block A as a system and forces acting on block A are P and F.

Therefore, m_A * a = P – F

Or, P = m_A * a + F ------------ (1)

Taking block B as a system and forces acting on block A is F.

Therefore, m_B * a = F

Or, a = F/ m_B ------------- (2)

Substituting the equation (2) in equation (1), we get

P = m_A * (F/ m_B) + F = F (1 + m_A/ m_B).

8. Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.

Sol:

Given: radius of raindrop, r = 1 mm = 0.001 m; mass of raindrop, m = 4 mg = 4*10^-6 kg; speed of raindrop = initial velocity = 30 m/s; distance covered = radius of raindrop, r = 1 mm; final velocity = 0.

We have, v² – u² = 2 a s

Or, 0² – 30² = 2 * a * 0.001

Or, a = – 450000 m/s²

Therefore, the force exerted is F = m*a

Or, F = 4 * 10^-6 * 450000 = 1.8 N.

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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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