What are black holes

According to special theory of relativity, the maximum velocity that can exist is the velocity of light
c=3X108 m/s
Now we also Know that escape velocity which is the velocity needed for an object to become essentially free of the gravitational effect of another object is given
v = 2 GM R
or
v 2 = 2 GM R
We may speculate on the mass and radius of a steller body that has an escape velocity of c.
Then
c 2 = 2 GM R
or
R = 2 GM c 2
This quantity R is called the Schwarzschild Radius and is usually designated by RS
Substituting the values of G and c,we have
RS=1.485 X10-2 M
The above equation gives us the relation between M and RS. It states that A body of Mass M in kg and radius RS in m or smaller produces such a strong gravtitional at its surface that no particle on its surface can escape.This even applied to electromagnetic radiation ( photons) including light.
So Even light cannot escape from such planet or body. That is the reason such body bodies are termed as Black holes
The most common way for a black hole to form is probably in a supernova, an exploding star. When a star with about 25 times the mass of the Sun ends its life, it explodes. The outer part of the star screams outward at high speed, but the inner part of the star, its core, collapses down. If there is enough mass, the gravity of the collapsing core will compress it so much that it can become a black hole. When it’s all over, the black hole will have a few times the mass of the Sun. This is called a “stellar-mass black hole”, what many astronomers think of as a “regular” black hole.

Fictitous Force and Free falling Elevator


A non inertial Frame of Refrence is the frame of refrence which is having acceleration(a0) Example Free Falling elevator
Newtons law are valid in inertial frame of refrence only. We cannot apply them as it is in non inertial frame of refrence as it left out the acceleration of the object because of the frame of refrence.
Let me provide simple example to illustrate this.
You are moving in car which is accelerating. From your frame of refrence,a block on the road would be accelerating in opposite direction.
Now if we apply newton's law, we will find acceleration to be zero as no net force is acting. But the block has acceleration from car frame of refrence.
So to fix this issue,we will add one fictitous quantity in the Newton law equation like
F+F0=Ma
Where F0 is the fictitous force and it is in opposite direction of accleration frame
F0=-Ma0
Where ais the acceleration of Non inertial frame of refrence
Now let me further illustrate this with example of free falling elevator
A free falling elevator is an non inertial frame of refrence.The acceleration of the elevator as defined from the observer on the ground will be
g=-gk
Where k is the unit vector along vertical axis in the upward direction
Let us take an unattached object of mass M in the elevator.
From the frame of refrence of free falling elevator,Two forces are acting on the mass
Fictitous Force due to non inertail frame of refrence
Ffic=MgIt is opposite to the acceleration of the elevator
Gravitional Force on the Mass
FG=-Mgk
So the total Force on the body from elevator frame of refrence
FT=Ffic+FG=0
Which mean body is unacclerated in the elevator frame of refrence. The body will appear to be suspended in the air ,if it has no initial relative velocity to the elevator
Few more important things about Fictitous force
  • Fictitous force are called pseudo force.
  • It simple represent the effect of the acceleration of the non inertial frame of refrence
  • They dont have any physical significance
  • Whenever we work the problem in non inertial frame of refrence,always make sure to include fictitous force in the newton law

Different type of Force and their origin



A) Normal Force: When we put a book on a table ,the molecules of the book exert downward forces on the molecules of the table.The molecules composing the upper layer of the tabletop move downward until the repulsion of the molecules below balances the forces applied by the book.
This compression always occur but it is too slight to notice
The following things are good to notice about normal force
a) For a body resting on a surface,Normal force is equal and opposite to the resultant of all the other forces which acts on the body in the direction perpendicular to surface
b) Weight and Normal force are not action reaction pair as they act on same body whereas action -reaction acts on different body

B) Tension: Tension usually comes under picture for string and rope. The string can be think of small short sections interacting by contact forces
Each section pulls the section to either side of it and by Newton third law,it is pulled by the adjacent section. The magnitude of the forces acting between the section is called Tension
The Following things are good to notice about tension
a) If the tension is uniform,the net string force on each small section is zero and section remains at rest
b) if the string is accelerating ,the tension generally varies along the string
c) If there is external force on the string then also tension varies along the string

C) Friction Force: Friction force generally comes into picture when one body moves or tries to move along the surface of the second body
Friction always opposes the motion which would occur in its absence. Friction occurs becuase of interatomic forces at the actual area of contact on atomic scale.
a) When we try to push a box on the table, if we push the box gently ,the box remains at rest. Force of friction assumes a value equal to the force applied by us. The force of friction cannot increase indefinately. If we push the box hard,the box start to slide

D) Viscous Forces: A body moving through liquid or gas is retarted by the force of viscosity exerted on it by fluid. It arised because a body moving through the medium exerts forces which set the nearby fluid into motion.By newton third law,the fluid exerts a reaction force on the body
The Following things are good to notice about viscosity
a)Viscous forces are proportional to velocity
b) It always retard the motion

Polarization by reflection


  • This simple method of obtaining plane polarized light by reflection was discovered by malus in 1808
  • We found that when a beam of light is reflected from the surface of a transparent medium like glass or water, the reflected light is partially polarized and degree of the polarization varies with angle of incidence
  • The percentage of polarized light is greatest in reflected beam when light beam is incident on the transparent medium with an incident angle equal to the angle of polarization
  • For ordinary glass with refractive index =1.52 ,angle of polarization is 57.50
  • Figure below shows the polarization of light by reflection


     
  • we can use a Polaroid as an analyzer to show that reflected light is plane polarized . we rather say that reflected light is partially plane polarized
  • the examination of transmitted light shows the variation in intensity indicating that the light is partially polarized
  • The vibrations of this plane polarized reflected light are found to be perpendicular to the plane of incidence and therefore ,the reflected light is said to be plane polarized in the plane of incidence

Kinematics Question

Question:

A ball of mass 100 g is projected vertically upward from the ground with avelocity of 49m/s. At the same time another identical ball is dropped from aheight of 98 m to fall freely along the same path as followed by the firstball. After some time ,the two balls collide and stick together and finallyfall together .

g=9.8m/s2

i is the unit vector along upward direction

1) Find the time at which balls collide in air

a) 2.1 sec

b) 1.2 sec

c)1 sec

d) 2 sec

2) Find the velocity vector of the combined mass just after collision

a) 4.8i

b) 3i

c)-4.9i

d) 4.9i

3) Find the time of the flight of masses

a) 3sec

b) 4 sec

c) 6.53 sec

d) 7 sec

Solution:

a) We will first find where and when the two balls collide. Let us assumethat the balls collide at time t after they have been set into motion. At thisinstant t when two balls collide they are at the same height h from the groundas shown below in the figure.

The height of the first ball after t seconds = 49t-0.5(9.8t2) =4.9t(20-t)

Height of second ball after t secinds = 98 - downwards distance moved by itin t seconds

=98-0.5t2=4.9(20-t2)

therefore, 4.9t(20-t)=4.9(20-t2)

or, 10t-t2=20-t2 or t=2s

b) The ball thus collides 2s after the start of their motion. Theirvelocities at this instance are

ball 1 : v1= (49-98×2)m/s = 29.4 m/s directed upwards=29.4 i

ball 2 : v2=(0+9.8×2)m/s = 19.6 m/s directed downwards=-19.6 i

If v is the velocity of the combined mass of two balls after theystick togather due to their collision then from law of conservation of momentum

200×v=100×29.4-100×19.6

v=4.9m/s upward direction =4.9 i

c) The joint mass thus moves upwards , after collision with a velocity of4.9 m/s. Its height above the ground at this instant is (consider the positionof either of the balls)

(98-0.5×9.8×22)m=78.4m

We can now find the time t' taken by this joined mass of the balls to reachthe ground. For this joined mass we have

u=4.9m/s , s=78.4m , a=-g = -9.8m/s2

-78.4=4.9t'+0.5(-9.8)t2

t'2-t'-16=0

Solving the equation for t' using formula for quadratic equations and leaving out the negative solution we get t'=4.532 s

The joint mass thus takes 4.53 s to fall to the ground. Since the balls collide 2s after they started their motion ,

the total time of flight is (2+4.53) s = 6.53 s

Question on Bolr atom model

Question :

The ionization energy of a hydrogen like Bohr atom is 4 rydbergs.

(a) What is the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state.

(b) Whet is the radius of the first orbit of this atom ? Given that Bohr radius of hydrogen atom = 5×10-11m and 1 rydberg = 2.2×10-18J.

Solution :

In terms of rydberg constant R , the energy of electron in the n'th orbit of hydrogen like atom is

E n = R Z 2 n 2

where R=2.2×10-18J. The ionization energy of the atom is

ΔE = E E 1 = R Z 2 ( 1 1 1 2 ) = R Z 2

Given that ΔE=4R. Therefore , 4R=RZ2 or, Z=2

(a) The energy of the radiation emitted when the electron jumps from the first excited state (n=2) to the ground state (n=1) is

E = E 2 E 1 = R Z 2 ( 1 2 2 1 1 2 ) = 3 R Z 2 4 = 3 R = 3 × 2.2 × 10 18 = 6.6 × 10 18 J (since Z=2)

Therefore wavelength of the radiation is given by

λ = hc E = 6.63 × 10 34 × 3 × 10 8 6.6 × 10 18 = 301 Å

(b) Radius of the first bohr orbit of the given atom is = r 1 Z = 2.5 × 10 11 m

Seeback Effect


  • Seeback effect was first discovered by Thomas John seaback
  • It stated that when two different conductor are joined to form a circuit and the two junctions are held at the different temperature then an emf is developed which results in the flow of the electric current through the circuit.Arrangement is shown as below in figure
  • Maginitude of thermo-electric emf depends upon the nature of the two metals and on the temperature difference between terminals


  • Seaback effect is reversible i.e, if the hot and cold junctions are reversed the direction of thermoelectric current is alse reversed
  • Seaback investigated thermo-electric properties of a large number of metals and arranged them in a series known as thermo-electric series or seaback series and is given as follows
    Bi,Ni,Co,Pt,Cu,Mn,Hg,Pb,Sn,Au,Ag,Zn,Cd,Fe,As,Sb,Te
  • When any two of these metals in the series is used to form a thermocouple ,the thermo emf is greater when two metals used are farther apart in the circuit
  • Figure 1 shows the thermocouple of Cu and Fe.The current in this couple flows from Cu to Fe through the hot junction
  • The thermo emf of this couple is only 1.3 milivolt for a temperature difference of 100 C between the hot and cold junction

Question on electric charge

Question
Two isolated metallic spheres of radii R and 2R are charged such that both these have the same charge densiti σ. The sphares are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger wire.
Solution
Initial charges on sphere of radius R and 2R respectively are
Q1= surface area × charge density = 4πR2σ
and Q2=4π(2R)2σ = 16πR2σ
therefore Q1+Q2=4πR2σ+16πR2σ (1)
As the charges are different, their initial potentials are also different. When the spheres are connected by long thin conducting wire, the charges are redistributed untill their potentials become equal. Let Q'1 and Q'2 be the new charges on spheres of radii R and 2R respectively, their common potential is
V = 1 4 π ε 0 . Q ' 1 R = 1 4 π ε 0 . Q ' 2 2 R
which gives
Q ' 1 = Q ' 2 2
Therefore,
Q ' 1 + Q ' 2 = 3 Q ' 2 2
From Law of conservation of charge
Q1+Q2=Q'1+Q'2
Using equation 1 and 2 we have
Q ' 2 = 40 π R 2 σ 3
The new charge density on the bigger sphere of radius R is
σ ' = Q ' 2 4 π ( 2 R ) 2 = 5 σ 6

How to get a job


Most of us who study hard to become a graduate want to find themselves a job. But there is lots of competition in the job market and you yourself would realize when you start looking for one. Apart from competition in the market you must be aware of the right option and job opening in the market. Before starting to look yourself a job you should first analyze yourself by this I mean that you must decide what you would enjoy the most at work because if you do not enjoy doing your work then you might fail to give the best you could give it. 
             While looking for a job engineering students might feel that lack of work experience opportunities made it much more difficult for them to get a job interview or offer in their final year even if they have all the qualifications. Most of the employers favoured students who had substantial relevant work experience. So students can go for   graduate training that are offered by large number of employers. Graduate schemes enable new recruits to settle quickly into a professional work environment, receive relevant skills development and get hands on experience, either working in a specific role within an established team, or through working on a number of assignments in different areas of the organization over the duration of the program.
              You can also search for graduate jobs over various websites available on internet and find the relevant jobs in your country and territory. For that you would have to design your C.V. carefully with true and relevant information about yourself. In your CV includes your abilities that are related to the jobs you are applying for. You can include skills that are relevant to the job field that you are interested in i.e. computer skills, software skills or language skills. You could find a lot of information about how to find graduate jobs on the internet and prepare according for interview and rest of the process. So I would like to say that if you manage to choose the right opportunity, you'll gain the experience which is of immense value and will ultimately make your way towards a really good job and a good pay check.

Root Mean square value of AC



  • We know that time average value of AC over one cycle is zero and it can be proved easily
  • Instantaneous current I and time average of AC over half cycle could be positive for one half cycle and negative for another half cycle but quantity i2 would always remain positive 
  • So time average of quantity i2 is


    This is known as the mean square current
  • The square root of mean square current is called root mean square current or rms current.
    Thus,

    thus ,the rms value of AC is .707i0 of the peak value of alternating current
  • Similarly rms value of alternating voltage or emf is

     
  • If we allow the AC current represented by i=i0sin(ωt+φ) to pass through a resistor of resistance R,the power dissipated due to flow of current would be
    P=i2R
  • Since magnitude of current changes with time ,the power dissipation in circuit also changes
  • The average Power dissipated over one complete current cycle would be


    If we pass direct current of magnitude irms through the resistor ,the power dissipate or rate of production of heat in this case would be
    P=(irms)2R
  • Thus rms value of AC is that value of steady current which would dissipate the same amount of power in a given resistance in a given tine as would gave been dissipated by alternating current
  • This is why rms value of AC is also known as virtual value of current

What is center of mass?


  • Consider a body consisting of large number of particles whose mass is equal to the total mass of all the particles. When such a body undergoes a translational motion the displacement is produced in each and every particle of the body with respect to their original position.
  • If this body is executing motion under the effect of some external forces acting on it then it has been found that there is a point in the system , where if whole mass of the system is supposed to be concentrated and the nature the motion executed by the system remains unaltered when force acting on the system is directly applied to this point. Such a point of the system is called centre of mass of the system.
  • Hence for any system Centre of mass is the point where whole mass of the system can be supposed to be concentrated and motion of the system can be defined in terms of the centre of mass.
  • Consider a stationary frame of refrance where a body of mass M is situated. This body is made up of n number of particles. Let mi be the mass and ri be the pisition vector of i'th particle of the body.
  • Let C be any point in the body whose position vector with respect to origin O of the frame of refrance is Rc and position vector of point C w.r.t. i'th particle is rci as shown below in the figure.



  • From triangle OCP
    ri=Rc+rci                               (1)
    multiplying both sides of equation 1 bt mi we get
    miri=miRc+mirci
    taking summation of above equation for n particles we get


    If for a body


    then point C is known as the centre of mass of the body.
  • Hence a point in a body w.r.t. which the sum of the product of mass of the particle and their position vector is equal to zero is equal to zero is known as centre of mass of the body.

Career opportunities in physics and engineering


Majority of students studying in senior school find physics as a more difficult subject. They find difficult to gasp various theories and formulas of physics. Physics is the most fundamental of all natural sciences and it describes how nature works using the language of mathematics. Now a question arises why should a student opt to study physics when he can choose between number of other branches of science that are much easy to study and understand. I would say before opting for physics a student needs to know why the physics is important and what career options a student may get out of studying physics at senior school.
Physics is the most basic natural science. The name physics comes from the ancient Greek word for nature. And the name fits perfectly: Physics deals with everything that occurs in nature, may it be in atoms, cars, semiconductors or outer space. This all belongs to physics.  Physics tries to explain natures by models. These are theoretical constructs, written in the language of logic, mathematics. Models can be falsified by experiments that show different outcome than expected, they cannot be verified.
To know the importance of physics in your daily life look around yourself and see how our daily life relies on technology, for example most of the electronic devices which are now important part of our life use transistors which came into existence due to research on physics of semiconductors and semiconductor devices and you can find lots of examples like this. Physics is an important subject to learn and understand if you are planning to make a career in medical sciences, engineering or technology. For example in case of medical sciences physics supplements a lot when it comes to be about branches like radiology.
In case of engineering studying and understanding physics becomes further more important as every branch of engineering be it electrical, mechanical or mechanical involves application of physics. The basic concepts we learn in physics play an important role in understanding complex scenarios in engineering. There are lots of career opportunities in engineering degree and if you want to opt engineering as an career option you can visit Online Engineering Degree for detailed information about different branches of engineering.
Further if you do not want to go for medical or engineering you can also opt to become a research scientist, teacher, lecturer or professor in physics. For this you would need to physics as a subject to study during graduation then go for post graduation and even have to go for a PhD degree in physics. So you see there are  lots of career opportunities for a student interested in physics and there are so many reasons why you should study physics.

Uniformly accelerated motion

Electric Field

Student with basic knowledge of electrostatics must have studied about electric field in many books with no proper definition in most of the books. But what exactly is electric field which is only said to be existed in any region in which electric force is said to be existed and this question is not the easiest one to answer. It was Michael Faraday who first referred to an electric ‘‘field of force,’’ and James Clerk Maxwell identified that field as the space around an electrified object – a space in which electric forces act. We all know that electric field and force are closely related and most basic definition of electric field is the electric force per unit area acting on the charged object.
E=F(r)/q
where this charge q is often known as test charge.
Since electric field in general is altered by the presence of test charge
E = lim Δq 0 ΔF Δq = dF dq
So, it is clear from above discussion that 

(1) E is a vector quantity with magnitude directly proportional to force and with direction given by the direction of the force on a positive test charge.
(2) E has units of newtons per coulomb (N/C)
While applying Gauss's law it is helpful to visualize electric field near charged object and the most common
approach is  to construct a visual representation of an electric field which is to use a either arrows or ‘‘field lines’' that point in the direction of the field at each point in space as shown below



  • So, electric field line is an imaginary line drawn in such a way that it's direction at any point is same as the direction of field at that point.
  • An electric field line is, in general a curve drawn in such a way that the tangent to it ateach point is the direction of net field at that point.
  • Field lines of a single position charge points radially outwards while that of a negative charge are radially inwards as shown below in the figure.

  • Field lines around the system of two positive charges gives a different picture and describe the mutual repulsion between them.

  • Field lines around a system of a positive and negative charge clearly shows the mutual attraction between them as shown below in the figure.

  • Some important general properties of field lines are
    1.Field lines start from positive charge and end on a negative charge.
    2.Field lines never cross each other if they do so then at the point of intersection there will be two direction of electric field.
    3.Electric field lines do not pass through a conductor , this shows that electric field inside a conductor is always zero.
    4.Electric field lines are continuous curves in a charge free region.

  • I would like to say that you can think of electric field as a quantity filling the space in the neighborhood of an electric charge. The electric-field concept helps us understand not only the forces between isolated stationary charged bodies but also what happens when charges move. When charges move, their motion is communicated to neighboring charged bodies in the form of a field disturbance.

    position, distance and displacement

    How to draw a free body diagram


    1. First create a mental picture of the body for which you want to write momentum balance equation.
    2. Draw rough sketch of your system showing it to be isolated from its environment.
    3. Place a dot in the center of the object and at this point all the forces are assumed to be acting upon.
    4. For every force acting on that body , draw a vector which shows size and direction of the force. each vector should start at the dot. 
    5. Label each vector based on the type of force and remember not to include numbers and calculations

    magnetic properties of matter

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    Full study material of MAGNETIC PROPERTIES OF MATTER is now available on physicscatalyst.com. To read the chapter follow this link
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    What are magnetic properties of matter


  • All substances possess magnetic properties and most general definition of magnetism defines it as a particular form of interactions originating between moving electrically charged particles.
  • Magnetic interaction relates spatially separate material objects and it is transmitted by means of magnetic field about which we have already studied .This magnetic field is important characteristics of EM form of matter.
  • We already know that source of magnetic field is a moving electric charge i.e. an electric current. On atomic scale, there are two types of macroscopic current associated with electrons.
    a) Orbital current is which electron in an atom moves about the nucleus in closed paths constituting electric currents loops
    b) Spin currents related to the internal degrees of freedom of the motion of electrons and this can only be understood through quantum mechanics.
  • Like electrons in an atom, atomic nucleus may also have magnetic properties like magnetic moment but it is fairly smaller then that of electrons.
  • Magnetic moment m is nothing but the quantitative measure of the magnetism of a particle.
  • For an elementary closed loop with a current i in it, the magnitude |m| of a magnetic moment vector equals the current times the loop area S i.e.
    |m|=iS and direction of m can be determined using right hand rule.
  • All micro structural elements of matter electrons, protons and neutrons are elementary carriers of magnetic moment and combination of these can be principle sources of magnetism
  • Thus magnetic properties are inherent to all the substances i.e. they are all magnets
  • An external magnetic field has an influence on these atomic orbital and spin currents and two basic effects of an external field are observed
    i) First is diamagnetic effect which is consequences of faraday's law of induction. According to the Lenz law’s, a magnetic field always sets up an induced current with its magnetic field direction opposite to an initial field .Therefore diamagnetic moment created by the external field is always negative related to this field
    ii) Second effect occurs if there is a resultant non zero magnetic moment in the atom i.e. there is a spin magnetic moment and orbital magnetic moment .In this case external field will attempt to orient the intrinsic atomic magnetic moment in its own direction .As a result a positive moment parallel to the field is created and this is called paramagnetic moment.
  • Because of the universality of the diamagnetic effect, all substances possess diamagnetic.
  • However, diamagnetism is by no means actually observed in all matter. This is because in many instances the diamagnetic effect is masked by the more powerful paramagnetic effect.
  • Thus in paramagnetic substances we actually always observe a difference effect produced by the prominent Para magnetism and weaker diamagnetism.
  • Polarization (quick review)

    Longitudinal wave- has the same property with respect to any plane through its line of polarization.
    Transverse waves- behaves differently in different planes. Light waves are transverse in nature and the viberation in them are at right angles to the direction in which wave is traveling.
    Plane Polarized - Since the viberations constituting the beam of light are confined only to one definite plane through the axis of the beam, the light is generally said to be plane polarized. Properties of plane polarized light beam wrt two planes, one containing the viberation and other at right angles to it

    Note that
    • Vibrations of polarized light are linear --- light is plane polarized 
    • Vibrations of polarized light are circular --- light is circularly polarized 
    • Vibrations of polarized light are elliptic --- light is elliptically polarized 
    • circular and elliptical vibrations are the resultant of two linear vibrations perpandicular to each other differing in phase by π/2. 
    Polarization by reflection
    • It is the simplest method of obtaining plane polarized light. 
    • This method was first discovered by Etiennie Louis Malus.
    • When a beam of light is reflected from the surface of a transparent medium the reflected light is partially polarized and the degree of polarization varies with the angle of incidence.
    • Percentage of polarized light in reflected beam is greatest when it is incident at an angle known as angle of polarization for the medium which is equal to 57.5 degree for glass and varies slightly with the wavelength of incident light.
    • Complete polarization is possible only with monochromatic light.
    • Reflected light is said to be plane polarized in the plane of incident.
    Brewster's Law
    • This Law states that there is a simple relation between the angle of maximum polarization and the refractive index of the medium. This relation known as Brewster's Law is given by
      μ=tan i
      where i is the angle of incidence
      and μ is the index of refraction
    • Using this law it can be showen that 'when light is incident at angle of maximum polarization the reflected ray is at right angles to the refracted ray. From Brewster's Law
      μ=tan i = (sin i)/(cos i)
      From Snell's Law
      μ = (sin i)/(sin r)
      where r is the angle of refraction
      Therefore , (sin i)/(cos i) = (sin i)/(sin r)
      (sin i)/(sin (90- i)) = (sin i)/(sin r)
      or we have
      90-i=r
      this gives i+r=90 degree
      showing that at maximum angle of polarization reflected and refracted rays are at right angles.
    • Brewster's Law is obeyed even when light is reflected at the surface of rarer medium.
    • Light reflected from both the upper and lower surfaces of a glass plate will be polarized in the plane of incidence.
    Double Refraction

    • Certain crystals split a ray of incident light into two refracted rays , one which gives the fixed image and follows all the laws of refraction and this ray is known as ordinary ray (o ray). Other ray gives an image that rotates with rotation of crystal and this ray is know as extra ordinary ray (e ray).
    • Both e and o rays are plane polarized.
    • When the crystal is rotated about the incident ray as an axis , the o-ray remains fixed but the e-ray revolves around it.
    • The index of refraction for e-ray is therefore a function of direction.
    • There is always one direction in the crystal for which there is no distinction between the o and e rays and this direction is called optic axis.
    • e ray and o ray are parallel to each other.
    • Plane of polarization of both e and o rays are at right angles to each other.
    • A class of crystals in which there is a single direction known as optic axis along which all waves are transmitted with one uniform velocities while in any other direction there are two velocities are called uni-axial crystals for example calcite and tourmaline crystals are uni-axial.
    • Crystals having two optical axis that is they have two directions of uniform velocity are bi-axial crystals for example topaz, mica etc. 
    Law of Malus
    • Intensity of incident polarized ray is equal to the sum of the intensities of two refracted rays
      Io = a2sin2θ
      and , Ie = a2cos2θ
      Io+Ie=a2=I

    Comparison between mechanical energy of mass spring system and electrical LC circuit

    Table given below compares the mechanic oscillations of mass spring system with that of electrical oscillations in an L-C circuit

    What is a magnetic field


    • We all ready know that a stationery charges gets up a electric field E in the space surrounding it and this electric field exerts a force F=q0E on the test charge q0 placed in magnetic field.
    • Similarly we can describe the intraction of moving charges that, a moving charge excert a magnetic field in the space surrounding it and this magnetic field exert a force on the moving charge.
    • Like electric field, magntic field is also a vector quantity and is represented by symbol B
    • Like electric field force which depend on the magnitude of charge and electric field, magnetic force is propotional to the magnitude of charge and the strength of magnetic field.
    • Apart from its dependence on magnitude of charge and magnetic field strength magnetic force also depends on velocity of the particle.
    • The magnitude magnetic force increase with increase in speed of charged particle.
    • Direction of magnetic force depends on direction of magnetc field B and velocity v of the chared particle.
    • The direction of magnetic force is not alonge the direction of magnetic field but direction of force is always perpendicular to direction of both magnetic field B and velocity v
    • Test charge of magnitude q0 is moving with velocity v through a point P in magnetic field B experience a deflecting force F defined by a equation
      F=qv X B 
    • As mentioned earlier this force on charged particle is perpendicular to the plane formed by v and B and its direction is determined right hand thumb rule.


    • When moving charge is positive the direction of force F is the direction of advance of hand screw whose axis is perpendicular to the plane formed by v and B.


    • Direction of force would be opposit to the direction of advance screw for negative charge moving in same direction.
    • Magnitude of force on charged particle is
      F=q0vBsinθ
      where θ is the angle between v and B.
    • If v and B are at right angle to each other i.e. θ=90 then force acting on the particle would be maximum and is given by
      Fmax=q0vB                   ----(3)
    • When θ=180 or θ=0 i.e. v is parallel or antiparallel to B then froce acting on the particle would be zero.
    • Again from equation 2 if the velocity of the palticle in the magnetic field is zero i.e., particle is stationery in magnetic field then it does not experience any force.
    • SI unit of strength of magnetic field is tesla (T). It can be defined as follows
      B=F/qvsinθ
      for F=1N,q=1C and v=1m/s and θ=90
      1T=1NA-1m-1
      Thus if a charge of 1C when moving with velocity of 1m/s along the direction perpendicular to the magnetic field experiences a force of 1N then magnitude of field at that point is equal to 1 tesla (1T).
    • Another SI unit of magnetic field is weber/m2 Thus
      1 Wb-m-2=1T=1NA-1m-1
      In CGS system, the magnetic field is expressed in 'gauss'. And 1T= 104 gauss. Dimention formula of magnetic field (B) is [MT-2A-1]
    For full chapter visit physicscatalyst.com

    Escape Velocity


    • Escape velocity is the minimum velocity that should be given to the body to enable it to escape from the gravitational field of the earth .
    • The energy given to the body to project it with the escape velocity is called escape energy. 
    • Escape velocity of earth is 11.2Km/sec
    • Valve of escape velocity does not depend on the mass of the projected body of, instead it depends on the mass and radius of the planet from which it is being projected. 
    • There are no atmsphere on the planets where root mean square velocity is more than the escape velocity. 
    • The value of escape velocity does not depend on the angle and direction of projection instead depends on density, mass and acceleration due to gravity of the planet. 

    How to use Gauss's Law to find electric field

    We all know that Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behaviour of electrostatic field in space. Also Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux. Now we'll go through the main steps which we can employ for applying Gauss's Law

    1. First identify the symmetry properties of the charge distribution. By this we mean that the point at which the field is to be determined must lie on a surface and this surface must have enough symmetry which allows integrals involved to be evaluated properly.
    2. Determine the direction of the electric field and a surface on which the magnitude of electric field is constant. 
    3. Now choose the Gaussian surface accordingly for example if the problem has spherical symmetry then Gaussian surface would usually be spherical and for cylindrical symmetry problem Gaussian surface would be cylindrical.
    4. Calculate the flux through the Gaussian surface.
    5. Now calculate the charge enclosed inside the chosen Gaussian surface.
    6. Equate the two sides of Gauss's law in order  to find the expression for the magnitude of the electric field in that region of space.

    Online Learning Program

    Welcome Students 
    We are pleased to anounce online learning program in physics. We are targeting to provide you 1200+ questions in the form of downloadble assignment's along with full solution for each chapter. The material will be helpful for all the students in 11th,12th and 12th passed who are appearing for various competive examination.

    We are aiming to provide the clear and concise concept.The course content has been prepared to build your concepts in physics.Questions of all format has been given in the downlodable assignment.We are hopeful that you will enjoy the every part of the course material and it will be really helpful in your board examination and various competitive examinations

    We will be giving the downloadable assignment with solutions on the monthly basis as per the following schedule.



    Each chapter will consists of multiple assignment having objective and subjective quesions alomg with full detailed solutions.We expect that students will try to complete the module accordingly and solve all the assignment.Students can also mail us any doubt regarding the questions and material.

    This material comes at a cost of Rs 500.

    for more information visit this  link

    Magnetic properties of matter


  • All substances possess magnetic properties and most general definition of magnetism defines it as a particular form of interactions originating between moving electrically charged particles
  • Magnetic interaction relates spatially separate material objects and it is transmitted by means of magnetic field about which we have already studied .This magnetic field is important characteristics of EM form of matter
  • We already know that source of magnetic field is a moving electric charge i.e. an electric current. On atomic scale, there are two types of macroscopic current associated with electrons
    a) Orbital current is which electron in an atom moves about the nucleus in closed paths constituting electric currents loops
    b) Spin currents related to the internal degrees of freedom of the motion of electrons and this can only be understood through quantum mechanics

  • Like electrons in an atom, atomic nucleus may also have magnetic properties like magnetic moment but it is fairly smaller then that of electrons
  • Magnetic moment m is nothing but the quantitative measure of the magnetism of a particle
  • For an elementary closed loop with a current I in it, the magnitude |m| of a magnetic moment vector equals the current times the loop area S i.e.
    |m|=IS
    and direction of m can be determined using right hand rule
  • All micro structural elements of matter electrons, protons and neutrons are elementary carriers of magnetic moment and combination of these can be principle sources of magnetism
  • Thus magnetic properties are inherent to all the substances i.e. they are all magnets
  • An external magnetic field has an influence on these atomic orbital and spin currents and two basic effects of an external field are observed
    i) First is diamagnetic effect which is consequences of faraday's law of induction. According to the Lenz law’s, a magnetic field always sets up an induced current with its magnetic field direction opposite to an initial field .Therefore diamagnetic moment created by the external field is always negative related to this field
    ii) Second effect occurs if there is a resultant non zero magnetic moment in the atom i.e. there is a spin magnetic moment and orbital magnetic moment .In this case external field will attempt to orient the intrinsic atomic magnetic moment in its own direction .As a result a positive moment parallel to the field is created and this is called paramagnetic moment
  • Because of the universality of the diamagnetic effect, all substances possess diamagnetism
  • However, diamagnetism is by no means actually observed in all matter. This is because in many instances the diamagnetic effect is masked by the more powerful paramagnetic effect
  • Thus in paramagnetic substances we actually always observe a difference effect produced by the prominent Para magnetism and weaker diamagnetism
  • mechanics test questions

    Question 1
    A ball is thrown vertically upwards with a velocity u from top of a tower. It strikes the ground with a velocity 3u. The time taken by the ball to reach the ground is given by
    (a) u/g
    (b) 2u/g
    (c) 3u/g
    (d) 4u/g

    Question 2
    A body slides down an inclined plane of inclination . The coefficient of friction down the plane varies in the direct proportion to the distance moved down the plane(=kx). The body will move down the plane with
    (a) constant acceleration =gsin
    (b) constant acceleration =(gsin-gcos)
    (c) constant retardation =(gcos-gsin)
    (d) variable acceleration that first decreases from gsin to zero and that becomes negative.

    Question 3
    A escalator is moving downwards with a uniform speed u. A man of mass m is running upwards on it at a uniform speed v. If the height of escalator is h, the work done by man in going up the escalator is
    (a) zero
    (b) mgh
    (c) mghu/(v-u)
    (d) mghv/(v-u)

    Question 4
    A particle moves in a circular orbit with a uniform angular speed. However , the plane of the circular orbit is itself rotating at a constant angular speed. We may then say that
    (a) the angular velocity as well as angular acceleration of particle are both constant.
    (b) neither the angular velocity nor the angular acceleration of the particle are constant.
    (c) the angular velocity of the particle varies but its angular acceleration is constant.
    (d) the angular velocity of the particle remains constant but the angular acceleration varies.
     
    Question 5
    Two objects of mass m and 4m are at rest at infinite separation. They move towards each other under mutual gravitational attraction. Then at separation r, which one of the following is true.
    (a) the total energy of the system is not zero
    (b) the force between them is not zero
    (c) the center of mass of the system is at rest
    (d) all the above are true

    Answer
    1. b
    2. d
    3. d
    4. c
    5. d

    AIEEE test questions


    Question 1

    A body of density enters a tank of water of density ' after falling through height h. The maximum depth to which it sinks in water is









    Question 2.
    A charged particle q is shot towards another charged particle Q which is fixed, with speed v. It approaches Q up to a closest distance r and then returns. If q were given a speed 2v , the distance of approach would be
    (a) r
    (b) 2r
    (c) r/2
    (d) r/4

    Question 3.
    The Young’s modulus of a perfectly rigid body is
    (a)  zero
    (b)  unity
    (c)  infinity
    (d)  may have any finite non zero value

    Question 4.
    When an EM wave enters an ionised layer of earth atmosphere present in ionosphere
    (a)  the electron cloud will not oscillate in the electric field of the wave
    (b) the electron cloud will oscillate in the electric field of wave in  phase of sinusoidal EM wave
    (c) the electron cloud will oscillate in the electric field of wave in the opposite phase of sinusoidal EM wave
    (d) the electron cloud will oscillate in the electric field of wave with a phase retardation of 90 degree for a sinusoidal EM wave

    Question 5.
    Two bodies are charged by rubbing one against the other. During the process one becomes positively charged while other becomes negatively charged. then
    (a) mass of each body remains unchanged
    (b) mass of each body changes marginally
    (c) mass of each body changes slightly and hence the total mass
    (d) mass of each body changes slightly but total mass remains the same

    Answers
    1. (a)
    2. (d)
    3. (c)
    4. (d)



    For solutions visit link

    Elasticity question

    Question:
    A load of 31.4 kg is suspended from a wire of radius 10-3m and density 9103Kg/m3. Calculate the change in the temperature of the wire if 75% of the work done is converted into heat. The Young’s modulus and the heat capacity of the material of the wire are 9.81010Nm-1and 490 J/KgK respectively.
    Solution
    For solution visit this Link

    How to solve problems involving time dependent acceleration

    Here i tried to sum up the steps to solve problems involving time dependent acceleration. We can easily calculate the position of the particle/object
    (1) Consider that acceleration a is some function of time such that
    a=f(t)              
    this equation can also be written in terms of velocity i.e.
    a=dv/dt=f(t)                                        .....1
    (2) Now we integrate equation 1 as a function of time
    dv=f(t)dt
    v=∫dv=∫f(t)dt + C                ......2
    here C is some constant of integration and can be calculated from the initial conditions.
    (3) Now we know that v=dx/dt , using this in the solution of equation 2 we can easily find position of the particle.

    Derivation of Coulumb's Law from Gauss's Law



    • Coulumb's law can be derived from Gauss's law.
    • Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure.



    • Field of a positive charge is in radially outward direction everywhere and magnitude of electric field intensity is same for all points at a distance r from the charge.
    • We can assume Gaussian surface to be a sphere of radius r enclosing the charge q.
    • From Gauss's law

      since E is constant at all points on the surface therefore,

      surface area of the sphere is A=4πr2
      thus,
       
    • Now force acting on point charge q' at distance r from point charge q is

      This is nothing but the mathematical statement of Coulumb's law.
    For full notes visit physicscatalyst.com

    How to choose a physics book for competitive exams

    When you start preparing for competitive exams like PMT, IITJEE, AIEEEi am sure most of you face this question  "From which book should I study Physics?". There are lots of books available in the market in Physics for preparation of exams like  IITJEE/AIEEE/PMT and it could be quite confusing choosing from numerous available options.
    I would advise you all to choose one good book for example NCERT book or concepts of physics by HC Verma for reading text and clearing your concepts in physics. Use only one book for studying a given topic as different books have different approach towards the topic and you could easily get confused. Start a topic or a chapter by reading it first from NCERT book and if you fail to understand then refer another help book. Solve all problems given in NCERT as they cover range of problems from easy to tough. After that try and solve problems  asked in entrance exams from that topic. After this, you will be able to judge for yourself if reading that topic/chapter from some advanced book is necessary or not. If you feel that what you learnt from NCERT is not enough to solve questions in competitive exams then (and only then !) read that topic from some other advanced book.For this you can use "Resnik and Halliday's book" , HC Verms's Concept in physics etc. . Other books can also be used according to your personal preference.
    It's high time you start preparing for your 2012 entrance exams so all the best for your preparations. 

    Physics for ITTJEE/AIEEE : Radioactivity

    Law of radioactive decay: for full notes visit Radioactivity

    Introduction to radioactivity


    • Phenomenon of radioactivity was first discovered by A.H.Bacquerel in 1896 while studying fluorescence and phosphorence of compounds irradiated by visible light
    • these phosphorescent materials glow in dark after being exposed to visible light
    • while conducting experiment on uranium salts, he found that uranium salts has a capability to blacken the photographic plate kept in a dark place wrapped through a paper
    • Subsequent experiments showed that radioactivity is a nuclear phenomenon in which an unstable nucleus under goes a decay process referred as radioactive decay
    • There are three types of radioactivity decays that occur in nature .These are α decay ,β decay and γ decay.
    • We now define radioactive decay as the process by which unstable atomic nucleus looses energy by emitting ionizing particles or radiations ( α,β and γ rays) 
    • Radioactive decay of an atomic nucleus is a spontaneous process and can occur without any interaction of other particles outside the atom
    • This process of radioactive decay is random and we can not predict whether a given radioactive atom will emit radiations at a particular instant of time or not
    • Phenomenon of radioactivity is observed in heavy elements like uranium and unstable isotopes like carbon 14
    For mor information visit physicscatalyst.com

    Row reduction method and rank of a matrix

    1. Matrices are just a display of set of numbers and it does not have any value For example
    is a 2 by 3 matrix having 2 rows and 3 columns.
    Aij represents a matrix element of i’th row and jth column for example here A12=6 and A21=-2
    for rest of the document click this link

    Objective type questions(Electrostatics)

    Question 1
    Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then:
    (a) negative and distributed uniformly over the sphere
    (b) negative and appears only at the point on the sphere closest to the point-charge
    (c) negative and distributed non-uniformly over the entire surface of the sphere
    (d) zero

    Question 2
    Two similar point charges are situated at x-axis at point x=-a and x=+a . another point charge Q is placed at the origin. If Q is displaced by a small distance x on the X-axis , the change in its electric potential energy is approximately proportional to
    (a) x
    (b) x2
    (c) x3
    (d) 1/x

    Question 3
    At a point inside a charged hollow metallic sphere
    (a) the potential is zero
    (b) the electric field is zero
    (c) the potential depends on the distance of the point from the center
    (d) the electric field depends on the distance of the point from the center

    Question 4
    A hollow sphere of metal of radius 10cm is charged in such a way that the potential of its surface is 5 Volt. The potential at the center of the sphere is
    (a) 0
    (b) 5 volt
    (c) 50 Volt
    (d) equal to potential at a distance 10cm from the surface of the sphere

    Question 5
    A sphere of radius 2cm has a charge of 2 micro Coulomb while sphere of radius 5cm has charge 5 micro coulomb. The ratio of electric fields at a distance 10cm from the center of the spheres will be
    (a) 2:5
    (b) 1:1
    (c) 5:2
    (d) 4:25

    Answer:
    1. d
    2. b
    3. b
    4. b
    5. a

    Question on fluid mechanics


    Question :
    A large open top container of negligible mass and uniform area of cross section A has a small hole of cross-sectional area A/100 in its side wall near the bottom. The container is kept on the smooth horizontal floor and contains a liquid of density and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t=0 , calculate
    (i) acceleration of the container, and
    (ii) its velocity
    when 75% of the liquid has drained out

    Solution:
    For solution visit this link

    Rolling Friction

    • Consider a situation of the ring or a sphere rolling without slipping over a horizontal plane.In this case there is only one point of contact between the body and the plane

    • The frictional forces developed between two surfaces in case described above is called rolling friction

    • Rolling friction developes between two surfaces when one body rolls over the surface of another body

    • We know that it is very difficult to pull a heavy metal box on a rough surface and if we attah four metal wheel to the box it becomes easiar to move the box on the same surface

    • Thus resistance offered by the surface during rolling is relatively less than offered during sliding friction

    • This is because while rolling surfaces in contact do not rub each other

    • Rolling friction is negligible in comparision to the kinetic and static friction which are present simlutanously

    • In many parts of the machine where this type of friction is undesirable ball bearings(small steel balls) are generally kept between the rotating parts of the machines.This ways power dissipation during the motion can be reduced

    Rigid Body Rotation Question 2


    Question :
    A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligible small push is given to it. Calculate the horizontal velocity of the axis of the cylinder part of the carpet when its radius is reduced to R/2.

    For solution visit this link

    How to apply law of conservation of energy in mechanics

    We often comes across the problems in mechanics where we need to apply the law of conservation of energy where gravitational potential energy or gravity is involved . For solving such problems you can consider the following problem solving strategy,

    1. First of all define the system which includes all the interacting bodies . Now choose a zero point for gravitational potential energy according to your convenience.
    2. Select the body of interest and identify the point about which information is given in the question. Also identify the point where you want to find out asked quantity about the body of interest.
    3. Check for the possibility of the presence of non-conservative forces. If there are no non-conservative forces present then write down the energy conservation equation for the system and identify the unknown quantity asked in the question.
    4. Solve the equation for the unknown quantities asked in the question by substituting the given quantities in the equation obtained.

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