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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