Problem Solving Corner

Hello Friends
We are pleased to anounce physics problem solving corner on this blog
You can send us your questions and queries ,we will try to solve them and publish them on this blog
Each month your 5 questions will be published with detailed solution.
All you need to do is mail us on following address to get the solution of your problems
mailmephysics12@rediffmail.com

Physics for IITJEE:Free Physics Ebook

We have developed few physics ebooks .The material in these books are quite good and you wont find it any where else

These Ebooks contains some good question spanning all the units.Detailed Solutions are also provided for that.
we have put in considerable work to write these e-books.
You need to mail us on following address to get that book
mailmephysics12@rediffmail.com

Physics for IITJEE:Heat and Thermodynamics Study Material

A Great offer for Students

Physics has been one of the toughest subject for the students of 11th,12th class and students preparing for Competitive examinations.Student face lots of problems in this area.They got stuck on a problem and dont find way to get out of it.They got tired of cramming solution and seek better understanding of the subject.They have to purchase lot of books to get various type of problems which involve high cost.They are quite nervous about homework and Examination.

We are proud to present our first package about the unit Heat and thermodynamics which includes complete solution for all these common faced problems.You will get good notes and lots of problems and their step by step detailed solutions at a very low cost.Besides that we also provide lot of other short stuff like examination tips,biograhies,quick recap and many other.


Heat and Thermodynamics is quite a scoring subject in 11th,12th and competition examination.This package is very useful for students preparing competitive examination like IITJEE,AIEEE,PMT,CBSE

This package includes following features

• Details Notes for all the chapter to clear your concepts
  
  
  
  
• Short Notes for quick Recap to revise at the time of any examination
  
  
  
  
• Chapter wise Questions with Detail Solutions in various format
  -Multiple choice question with only one answer
  -Multiple choice question with one or more answers
  -Linked type comphrehension
  -Subjective Questions
  
  
  
  
• Miscllaneaous Question at the end of the units with Detailed solutions in various Format
  -Assertion Reasson questions
  -Matrix Type question
  -Multiple choice question with only one answer
  -Multiple choice question with one or more answer
  -Linked type comphrehension
  -Subjective Questions
  -Fill in the blanks
  -True and False
  
  
  
  
• Difficulity Level matches that of IITJEE/AIEEE
  
  
  
  
• Biographies of all the Author of Heat And Thermodynamics
  
  
  
  
• IIT And AIEEE Heat and Thermodynamics Syllabus
  
  
  
  
• Mathematics help
  
  
  
  
• Problem solving tips
  
  
  
  
• It contains around 200 Solved problems
  
  
  
  Package Price
  The Price of Heat and thermodyanics package is Rs 100.If you want this package in CD-ROM,then it will cost Rs 50 extra including postal charges
  
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Physics for IITJEE:Optics Short Notes-1

Agenda :Summary of the basics of Optics

Reflection of light
-When light travelling in one medium is incident on a second medium some part of it is sent back to the first medium according to some definite laws and is said to be reflected.
-In practice mirrors are used to reflect light.
-The angle which the incident ray and the reflected ray make with normal to the surface are termed as Angles of incidence and of reflection respectively.

Laws of reflection:-
Law 1- The incident ray, the reflected ray and the normal to the surface , all lie in the same plane.
Law 2- The angle of incidence i is equal to the angle of reflection r.

Spherical Mirrors :-
Assuming that all the reflecting surfaces are portions of a sphere, if the inside surface reflects, the mirror is concave and if the outside surface which reflects, it is convex as shown below in the figure.
-Relation between focal length f and radius of curvature R of the mirror is
f=R/2
and both f and R are taken as positive for concave mirror and negative for convex mirror.
-Mirror formula is the relation between distance of the object (u), distance of the image (v) and the focal length (f) of the mirror and is,
(1/u)+(1/v)=(1/f)
-Mirror formula holds for a spherical mirror of small aperture only.
-The ratio of size of the image formed (by spherical mirror) to the size of the object is called magnification. Mathematically,
m=I/O
=-v/u
-Spherical abberation is the inability of a spherical mirror of large aperture to bring all the rays of wide beam of light falling

on it to focus at a single point.











More information will be given in subsequent post

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Winner of Physics Champion Contest -2

Hi All

We receive lot of entries for the contest.Thanks all for the Response

Answer to the questions are
1. c
2.a
3.b,c
4.b
5.a

The winner of the contest is Indranil Chakraborty from Mussoorie, Uttarakhand
He will soon recieve the stationary Gift worth RS 200
Congratutlation!!!!!!!!!!!!!


Thanks to all for overwhelming response!!!!!

IITJEE PHYSICS:Current Electricity Part -I

Agenda: Summary of Current Electricity

Some basic point about Electricty
-A current flows along a metal or wire when battery is connected to it.
-Current in a wire is due to free electrons moving along the wire
-The Battery has a Potential difference between its poles due to chemical changes inside the battery.The Potential difference pushes the electrons alongs the metal.The Potential diffrence between the terminal is call the EMF of the battery.
-One pole of the battery is called positive pole and another one is call the negative pole.
-Current direction is opposite to the electron flow direction.
-EMF is measured in Volt



Electric current is defined as the rate at which charge flows inside the conductor
so
I=Q/t
- Its unit is Amphere
Also I=dQ/dt

Drift speed of electrons inside the conductor
v_d=I/neA
Where I is the current,n is the number of the electron per unit volume ,e is the electronic charge and A is the area of the crossection of the conductor

Ohms Law

V=IR
where V is the Potential difference accross its ends
I is the current flow
And R is the electrical Resistance of the conductor.Electrical Resistance unit is Ohm


ELectrical Resistivity

It is defined as
ρ=RA/L

Where R is the Resistance
A is the crossectional Area
L is the length


Series Resistors

R=R₁+R₂+R₃

Parallel Resistors

(1/R)=(1/R₁)+(1/R₂)+(1/R₂)


Terminal Voltage of the Cell

V=E-Ir

where E is the EMF of the cell and r is the internal resistance of the Cell


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Physics for IITJEE:Exercise your Brain with Miscellaneous Physics Question

Question 1:
A 2 kg body moving to the right with speed 6 m/s coliides elastcally with a stationary body of mass 4 kg.
a.Find the velocity vector of each mass relative to center of mass before and after the collsion
b.Find the velocity vector of the each mass in original frame after the collision
c.How much energy was transfered to the 4 kg body.

Take i as the unit vector in the right direction

Answer
a.
Before collision
Velocity vector of mass 2 kg=4i m/s
Velocity vector of mass 4 kg=-2i m/s

After collision
Velocity vector of mass 2 kg=-4i m/s
Velocity vector of mass 4 kg=2i m/s

b
Velocity vector of mass 2 kg=-2i m/s
Velocity vector of mass 4 kg=4i m/s

c. 32 J

Hints:
Velocity of the center of mass=m₁v₁+m₂v₂/m₁+m₂
Velocity w.r.t CM=v-v_cm

Applying law of linear momentum and energy conservation,veloclities after collosion could be find out


Question 2:
A small solid cylinder of radius r rolls down from the top of sphere of Radius R without slipping.
a. find the angle made from the vertical at which the cylinder loses contact with the surface of the sphere
b.find the angular velocity of the cylinder at the moment it loses contact with the surface of the sphere

Answer
a. cos^-1(4/7)
b.(2/r)[(R+r)g/7]^1/2

hints:

Normal reaction will become zero at the point of losing contact
so centripetal force will become equal to the weight components across the radius.
Also applying law of conservation of energy ...we can find the solution


Question 3:

An gas obeys Vander waal equation

(P+a/V²)(V-b)=nRT

Find following for such a gas

A. The dimensional formula for constant a is
a.ML⁵T^-2
b.ML⁵T^-1
c.ML^-1T^-1
d. M⁰L^-0T⁰


B. Find the workdone by this gas when gas expand from V to 2V at constant temperature T
a.nRTlog_e(2V-b/V-b) -a/V
b.nRTlog_e(2V-b/V-b) +a/V
c. nRTlog_e(2V-b/V-b)
d. none of the above

C.Bulk modulus at constant pressure is defined as
K=(-1/V)(dP/dV)_T
Find the value of k for this gas
a.(1-b/V)(p-a/V²+2a/V³)
b. (1+b/V)(p-a/V²+2a/V³)
c.(1-b/V)(p-a/V²-2a/V³)
d.None of the above

Answer:
A.ML⁵T^-2
B.nRTlog_e(2V-b/V-b) -a/V



Question 4:
An electron (mass m and charge e) is moving along the axis of the ring of radius r and carrying a charge q.Find the time period of small oscillation of the electron about the center of the ring

Answer
T=4π(πε₀mr³/eq)^1/2

Hint:
Electric field at a point on the axis of the ring at the distance x from its center is
E=qx/4πε₀(r²+x²)^3/2