Friday, April 28, 2017

Study guide H C Verma JEE Physics Ch.4 THE FORCES

The study guides provides only the main issues covered in the chapter to provide some introduction to the chapter. Some past JEE questions are provided to link the issues in the chapter to questions.

One has to read the chapter fully. Study all solved examples and solve exercise problems.

The points may be used as revision points later.

No topic in JEE syllabus. But study of the chapter is necessary as there are full chapters on each of these topics later on and this introduction is essential to understand the in depth chapters.



4.1 Introduction
4.2 Gravitational forces
4.3 Electromagnetic (EM) forces
4.4 Nuclear Forces
4.5 Weak forces
4.6 Scope of Classical physics


Study Plan

Day 1

4.1 Introduction
4.2 Gravitational forces

Session important points

Day 2
4.3 Electromagnetic (EM) forces
Ex. 4.1

Session - important points

Day 3

4.4 Nuclear Forces
4.5 Weak forces
4.6 Scope of Classical physics

Session important points

Day 4
Chapter 4 Forces
Worked out examples 1 to 3
Exercises 1 to 7

Day 5
Chapter 4 Forces
Exercises 8 top 12
Objective I and II

Day 6
Chapter 4 Forces
Questions for short answer 1 to 10
Concept Review
Formula Review

Day 7 to 10
Chapter 4 Forces Revision

Day 11 to  20
Revision of all the four chapters completed so far.

Don't waste the time available. Always use the time to revision the concepts, formulas, difficult problems and to do test paper questions. You have to sqeeze the time every day and use it in a focused way.

Master each chapter as early as possible. That will give the energy to study one more chapter. Always feel happy after you do the planned work for a day. That will relax you. The feeling of happiness after you complete the day's task will relax you and make you feel energetic to eat and sleep and if some time is there to see the TV or read a paper or talk to your parents. It is important to focus, put in the required effort and then feel happy and relax every day.

Concepts covered

Force is an interaction between two objects. Force is exerted by an object A on another object B
Force is a vector quantity.

If more than one force act on a particle, we can find the resultant force using the laws of vector addition.

Gravitational forces

Any two bodies attract each other by virtue of their masses.

The force of attraction between two point masses is

F = G m1m2/r^2

where m1 and m2 are the masses of the particles and r is the distance between them.

G is a universal constant having the value 6.67 x 10^-11 N-m²/kg²

Gravitational force on small bodies by the earth

For earth, the value of radius R and mass M are 6400 km and 6 x 10^24 kg respectively.

The quantity GM/r² is a constant and has the dimensions of acceleration. It is called acceleration due to gravity and is denoted by letter g.

g and G are different.

Electro magnetic force:

Apart from gravitational force between any two bodies, the particles may exert upon each other electromagnetic forces.

If two particles hving charges q1 and q2 are at rest with respect to the observer, the force between them has a magnitude

F = (1/4πε0)(q1q2/r^2)

Where ε0 = permittivity of air or vacuum = 8.8549 x 10^-12 C² /N-m²
The quantity (1/4πε0) = 9.0 x 10^9 N-m² /C²

q1, q2 = charges
r distance between q1 and q2

This is called coulomb force and it acts along the line joining the particles.

Force due to a spring

When a metallic wire is coiled it becomes a spring.

when a spring is stretched, it pulls the bodies attached to its ends and when compressed, it pushes the bodies attached to its ends. the force exerted by thes spring is proportional to the change in its length.

If the spring has natural length of x0 and if it is changed to x, the magnitude of force exerted by the spring will be

f = k|x-x0| = k|Δx|

If the spring is extended, the force will be directed towards its centre and it compressed, it will be directed away from its centre.

Nuclear forces

The alpha particles is a bare nucleus of Helium. It contains two protons and two neutrons. It is a stable object and once created it can remain intact until it is not made to interact with other objects.

The protons in the nucleus will repel each other due to coulomb force and try to break the nucleus. Why does the Coulomb force fail to break the nucleus.

There are forces called nucluear forces and they are exerted only if the interacting particles are protons or neutrons or both. They are largely atractive, but with a short range. They are weaker than the Coulomb force if the separation between particles is more than 10^-14 m. For separation smaller than this the nuclear force is stronger than the Coulomb force and it holds the nucleus stable.

Radioactivity, nuclear energy (fission, fusion) etc. result from nuclear force.

Weak Forces

A neutron can change into proton and simulataneously emit an electron and a particle called antinutrino.

a proton can also change into neutron and simulataneously emit a positron (and a neutrino). The forces responsible for these changes are called weak forces. The effect of this force is experienced inside protons and neutrons only.

Scope of classical physics

Physics based on Newton's Laws of motion, Newton's law of gravitation, Maxwell's electromagetism, laws of thermodynamics and the Lorentz force is called classical physics. The behaviour of all the bodies of linear sizes greater than 10^-6 m are adequately described by classical physics. Grains of sands and rain drops fall into this range as well as heavenly bodies.

But sub atomic particles like atoms, nuclei, and electrons have sizes smaller than 10^-6 m and they are explained by quantum physics.

The mechanics of particles moving at velocity equal to light are explained by relativistic mechanics formulated by Einstein in 1905.

Updated on 30 April 2017, 23 October 2007

Sunday, May 8, 2016

XI - 2.1 Scalars and Vectors - Mathematics for Physics - Video Lectures

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XI - 2.3 Addition of vectors - Video Lectures

Addition of vectors

Magnitude of av+bv = SQRT(a²+b²+ 2ab cos θ)

The angle of the resultant with av is α where
tanα = b sin θ/(a+b cos θ)

Interesting point to make note of:
Two vectors having equal magnitudes of a make an angle θ with each other. Find the magnitude and direction of the resultant(Resultant is output of the addition of two vectors.)

Magnitude = 2a cos θ/2
tan α = a sin θ/(a + a cos θ) = (2asin(θ/2)cos(θ/2))/(2acos²(θ/2))
= tan (θ/2)

Example: Two vectors are of equal magnitude of 10 units. One of them is inclined at 45° to the X-axis and the other is inclined at 75° to the X-axis. Find the magnitude and direction of the resultant with respect to X-axis.

The angle between vectors is 30°.
Hence magnitude of the resultant will be 20 cos 15°
The direction - The resultant is inclined at 60° to the X axis.

Physics Vector Addition (Algebraic)


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