web sites
Measurements
Download
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-13-05.pdf
--------------------------
JEE 2007 Paper I
--------------------------------------------.
Q.I. Some physical quantities are given in column I and some possible SI units in which these
quantities may be expressed are given in column II. Match the physical quantities in column I
with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4 ×
4 matrix given in the ORS.
Column I
==================================================
A.
GM-e M-s ------ ---[ P. (volt) (coulomb) (metre)]
G – universal gravitational constant,
M-e – mass of the earth
M-s – Mass of the Sun
==================================================
B.
3RT/M --------------[Q. (kilogram)(metre^3)(second^-2)]
R – universal gas constant
T – absolute temperature
M – molar mass
==================================================
C.
F^2/(B^2Q^2)------------------------------[R. (metre^2)(second^-2)]
F – force
Q – charge
B – magnetic field
=================================================
D
(GM-e)/R-e ------------[S. (farad)(volt^2)(kg^-1)]
G – universal gravitational constant,
M-e – mass of the earth
R-e – radius of the earth
=================================================
Answer
(A) -- (P), (Q)
(B) -- (R), (S)
(C) -- (R) , S
(D) -- R, S
----------------------------------------------------------------
Q.II. Column I gives certain situations in which a straight metallic wire of resistance R is used and column II gives some resulting effects. Match the statement in column I with the statements in column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.
================================
Column I -----------------------------Column II
(A to D) -----------------------------(P to S)
==========================================================
(A) A charged capacitor is ---(P) A constant current flows
connected to------------------through the wire
the ends of the wire
===========================================================
(B)The wire is moved ----------(Q) thermal energy
perpendicular to its length -------generated in the wire
length with a constant
velocity in a uniform magnetic field
perpendicular to the plane of motion
===========================================================
(C)
The wire is placed in a -----------(R) A constant potential
constant electric field ---------- difference develops
that has direction-----------------between the ends of
along the length of the wire-------the wire
============================================================
(D) A battery of constant ----------(S) Charges of constant
emf is connected to the --------------- magnitude appear
ends of the wire-----------------------at the ends of the wire
============================================================
Solution
(A) -- (Q)
(B) -- (R), (S),
(C) -- (S)
(D) -- (P), (Q) , (R)
-----------------------------------------------------
Q. III.
Some laws/ processes are given in Column I. Match these with physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.
---Column I ------------------------Column II
A. Transition between -----------P.Characteristic X-rays
two atomic energy----------------
levles
B. Electron emission-------------Q. Photoelectric effect
from a material------------------
C. Mosley’s law------------------R. Hydrogen spectrum
D. Change of photon--------------S. β – decay
energy into kinetic--------------
energy of electrons--------------
===================
Solution
(A) -- (P), (R)
(B) ® (Q), (S),
(C) ® (P)
(D) ® (Q)
-----------------------------------
Q.4 JEE 2007 paper II
Column I gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II and indicate your answer by darkening appropriate bubbles
in the 4 × 4 matrix given in the ORS.
--------Column I--------------Column II
(A) Bimetallic strip-----(P)Radiation from a hot body
(B) Steam engine-------- (Q)Energy conversion
(C) Incandescent lamp----(R) Melting
(D) Electric fuse--------(S) Thermal expansion of solids
Solution:
(A) – (S)
(B) – (Q)
(C) – (P), (Q)
(D) – (R), (Q)
(A) Bimetallic strip works on the principle of thermal expansion of solids.
(B) Steam engine converts steam energy into mechanical energy.
(C) Incandescent lamp works on the principle of radiation of body due to temperature.
(D) Electric fuse is a safety device which does not allow current more than its permissible value.
----------------------------
Q5. JEE 2007 Paper II
Column I describes some situations in which a small object moves. Column II describes some characteristics of these motions. Match the situations in Column I with the characteristics in Column II and indicate your answer by darkening
appropriate bubbles in the 4 × 4 matrix given in the ORS.
Column I is give first totally. and then Column II is given next.
A. The object moves on the x-axis under a conservative force in
such a way that its “speed” and “position” satisfy
v =c1*SQRT(c2-x^2), where c1 and c2 are positive constants.
B. The object moves on the x-axis in such a way that its velocity
and its displacement from the origin satisfy kx v − = , where k
is a positive constant.
C. The object is attached to one end of a mass less spring of a
given spring constant. The other end of the spring is attached
to the ceiling on an elevator. Initially everything is at rest. The
elevator starts going upwards with a constant acceleration a.
The motion of the object is observed from the elevator during
the period it maintains this acceleration.
D.The object is projected from the earth’s surface vertically
upwards with a speed SQRT(GM-e/R-e) , where, M-e is the mass
of the earth and R-e is the radius of the earth. Neglect forces
from objects other than the earth.
Column II
(P) The object executes a simple harmonic
motion.
(Q) The object does not change its direction
(R) The kinetic energy of the object keeps on
decreasing
(S) The object can change its direction only once.
Solution:
(A) – (P)
(B) – (Q), (R)
(C) – (P)
(D) – (Q), (R)
--------------------------------------------
Wednesday, October 24, 2007
Tuesday, October 23, 2007
Study guide H C Verma JEE Physics Ch. 1 INTRODUCTION TO PHYSICS
H C Verma Book Sections
1.1 What is Physics
1.2 Physics and Mathematics
1.3 Units
1.4 Definitions of base units
1.5 Dimension
1.6 Uses of dimension
1.7 Order of magnitude
1.8 The structure of world
Study Plan
Study plan basically consists of 10 days of study and problems solving. Additional 10 days is for revision and for solving some more problems from the book, if you could not complete them during the first 10 days. Each day spend at least one hour for this subject for the main chapter and may be 15 minutes to 30 minutes for revising the earlier chapter. Thus you have to allot at least three hours and forty five minutes every day for your self study. It will be good when you are able to combine your self study with the lecture that is going in the school and coaching class. That way you will be able to complete self study in lesser time and get the benefit of reinforcement.
Day 1
Study
1.1 What is Physics
1.2 Physics and Mathematics
1.3 Units
1.4 Definitions of base units
1.5 Dimension
Do example 1.1
Do worked out examples 1 and 2
Day 2
Study
1.6 Uses of dimension
Do example 1.2
Do worked out examples 3,4,5,6
Day 3
Study
1.7 Order of magnitude
1.8 The structure of world
Do worked out examples 7 to 10
Day 4
Revise the chapter
Answer Objective I and II
Day 5
Solve Exercises 1 to 5
Day 6
Solve Exercises 6 to 10
Day 7
Solve Exercises 11 to 15
Day 8
Solve Exercises 16 to 19
Day 9 to 20
Keep revising principles, and have a look at difficult problems. If you come across any other text book, answer problems in that textbook on these topics.
By the end of these 20 days, you must be very thorough with this chapter. You are ready to write examination on this chapter at any time from today.
1.1 What is Physics
1.2 Physics and Mathematics
1.3 Units
1.4 Definitions of base units
1.5 Dimension
1.6 Uses of dimension
1.7 Order of magnitude
1.8 The structure of world
Study Plan
Study plan basically consists of 10 days of study and problems solving. Additional 10 days is for revision and for solving some more problems from the book, if you could not complete them during the first 10 days. Each day spend at least one hour for this subject for the main chapter and may be 15 minutes to 30 minutes for revising the earlier chapter. Thus you have to allot at least three hours and forty five minutes every day for your self study. It will be good when you are able to combine your self study with the lecture that is going in the school and coaching class. That way you will be able to complete self study in lesser time and get the benefit of reinforcement.
Day 1
Study
1.1 What is Physics
1.2 Physics and Mathematics
1.3 Units
1.4 Definitions of base units
1.5 Dimension
Do example 1.1
Do worked out examples 1 and 2
Day 2
Study
1.6 Uses of dimension
Do example 1.2
Do worked out examples 3,4,5,6
Day 3
Study
1.7 Order of magnitude
1.8 The structure of world
Do worked out examples 7 to 10
Day 4
Revise the chapter
Answer Objective I and II
Day 5
Solve Exercises 1 to 5
Day 6
Solve Exercises 6 to 10
Day 7
Solve Exercises 11 to 15
Day 8
Solve Exercises 16 to 19
Day 9 to 20
Keep revising principles, and have a look at difficult problems. If you come across any other text book, answer problems in that textbook on these topics.
By the end of these 20 days, you must be very thorough with this chapter. You are ready to write examination on this chapter at any time from today.
Study guide H C Verma JEE Physics Ch. 2 PHYSICS AND MATHEMATICS
HCV Chapter Sections
2.1 Vectors and scalars
2.2 Equality of vectors
2.3 Addition of vectors
2.4 Multiplication of a vector by a number
2.5 Subtraction of vectors
2.6 Resolution of vectors
2.7 DCT product or scalar product of two vectors
2.8 Cross product or vector product of two vectors
2.9 Differential calculus: dy/dx as rate measure
2.10 Maxima and Minima
2.11 Integral calculus
2.12 Significant digits
2.13 Significant digits in calculations
2.14 Errors in measurements
Study Plan
Day 1
2.1 Vectors and scalars
2.2 Equality of vectors
2.3 Addition of vectors
Example 2.1
2.4 Multiplication of a vector by a number
2.5 Subtraction of vectors
Example 2.2
Day 2
2.6 Resolution of vectors
Example 2.3
2.7 DOT product or scalar product of two vectors
Example 2.4
Worked out examples 1 to 4
Day 3
2.8 Cross product or vector product of two vectors
Example 2.5
Worked out example 5 to 10.
Day 4
2.9 Differential calculus: dy/dx as rate measure
Ex. 2.6, 2.7
2.10 Maxima and Minima
Ex.2.8
Day 5
2.11 Integral calculus
Ex. 2.9
Worked out examples 11 to 13
Day 6
2.12 Significant digits
2.13 Significant digits in calculations
Ex. 2.10 to 2.12
Worked out examples 14 to 17
Day 7
2.14 Errors in measurements
Ex. 2.13
Worked example 18
Exercise Problems: 1 to 4
Day 8
Objective I
Exercise Problems 5 to 8
Day 9
Objective II
Exercise Problems 9 to 12
Day 10
Exercise Problems 13 to 20
Day 11 Your are entering into revision 10 days
Exercise problems 21,22
Day 12
Exercise problems 23,24
Day 13
Exercise problems 25,26
Day 14
Exercise problems 27,28
Day 15
Exercise problems 29,30
Day 16
Exercise problems 31,32
Day 17
Exercise problems 33,34
Day 18
Exercise problems 35
Day 19
Attempt Questions for short answer 1 to 7
Day 20
Exercise problems
Attempt Questions for short answer 8 to 14
Web sites
lecture notes - Vectors
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-16-05.pdf
2.1 Vectors and scalars
2.2 Equality of vectors
2.3 Addition of vectors
2.4 Multiplication of a vector by a number
2.5 Subtraction of vectors
2.6 Resolution of vectors
2.7 DCT product or scalar product of two vectors
2.8 Cross product or vector product of two vectors
2.9 Differential calculus: dy/dx as rate measure
2.10 Maxima and Minima
2.11 Integral calculus
2.12 Significant digits
2.13 Significant digits in calculations
2.14 Errors in measurements
Study Plan
Day 1
2.1 Vectors and scalars
2.2 Equality of vectors
2.3 Addition of vectors
Example 2.1
2.4 Multiplication of a vector by a number
2.5 Subtraction of vectors
Example 2.2
Day 2
2.6 Resolution of vectors
Example 2.3
2.7 DOT product or scalar product of two vectors
Example 2.4
Worked out examples 1 to 4
Day 3
2.8 Cross product or vector product of two vectors
Example 2.5
Worked out example 5 to 10.
Day 4
2.9 Differential calculus: dy/dx as rate measure
Ex. 2.6, 2.7
2.10 Maxima and Minima
Ex.2.8
Day 5
2.11 Integral calculus
Ex. 2.9
Worked out examples 11 to 13
Day 6
2.12 Significant digits
2.13 Significant digits in calculations
Ex. 2.10 to 2.12
Worked out examples 14 to 17
Day 7
2.14 Errors in measurements
Ex. 2.13
Worked example 18
Exercise Problems: 1 to 4
Day 8
Objective I
Exercise Problems 5 to 8
Day 9
Objective II
Exercise Problems 9 to 12
Day 10
Exercise Problems 13 to 20
Day 11 Your are entering into revision 10 days
Exercise problems 21,22
Day 12
Exercise problems 23,24
Day 13
Exercise problems 25,26
Day 14
Exercise problems 27,28
Day 15
Exercise problems 29,30
Day 16
Exercise problems 31,32
Day 17
Exercise problems 33,34
Day 18
Exercise problems 35
Day 19
Attempt Questions for short answer 1 to 7
Day 20
Exercise problems
Attempt Questions for short answer 8 to 14
Web sites
lecture notes - Vectors
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-16-05.pdf
Study guide H C Verma JEE Physics Ch. 3 REST AND MOTION - KINEMATICS
Inertial and uniformly accelerated frames of reference;
-----------
HCV Chapter Sections
3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average accleration and instantaneous aceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame
-------
Study Plan
Day 1
3.1 Rest and Motion
3.2 Distance and displacement
Ex. 3.1
3.3 average speed and instantaneous speed
Ex. 3.2,3.3
Worked out example 1
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-2011.html
Day 2
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
Exercises: 1 to 5
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-ch3.html
Day 3
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3 to 6
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/iit-jee-physics-study-diary-ch3-rest.html
Day 4
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Day 5
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Day 6
Formula Revision
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
WOE 7 to 10, and 13,15, 19,20
Day 7
Exercises 6 to 15
Day 8
Exercises 16 to 25
Day 9
Exercises 26 to 35
Day 10
Exercises 36 to 45
Revision Period
Day 11
Exercises 46 to 50
Day 12
Exercises 51,52
Objective I
Day 13
Exercises
Objective II
Day 14
Exercises
Question for short answer
Day 15
Concept review
http://iit-jee-physics.blogspot.com/2008/05/concept-review-chapter-3-rest-and.html
Day 16
Formula review
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
Day 17 to 20
Test paper problems 10 per day can be solved.
Inertial and accelerated frames of reference.
Section 3.9 covers material related to it. But it is not using the terms inertial frame of reference.
------
Past JEE question
1982
A particle is moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
a) zero
b) [1/Sqrt(2)]m/s^2 towards northwest
c) 1/2 m/s^2 towards northwest
d) 1/2 m/s^2 towards north
Answer b.
You can find acceleration in the direction of west and check that velocity become zero in east direction in 5 seconds and also velocity in north direction becomes 5 m/s.
----------------------
Audio visual lectures
http://www.curriki.org/nroc/Introductory_Physics_1/lesson01/Container.html
Motion in two dimensions (plane)
http://www.curriki.org/nroc/Introductory_Physics_1/lesson02/Container.html
web sites
one dimensional motion
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-15-05.pdf
-----------
HCV Chapter Sections
3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average accleration and instantaneous aceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame
-------
Study Plan
Day 1
3.1 Rest and Motion
3.2 Distance and displacement
Ex. 3.1
3.3 average speed and instantaneous speed
Ex. 3.2,3.3
Worked out example 1
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-2011.html
Day 2
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
Exercises: 1 to 5
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-ch3.html
Day 3
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3 to 6
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/iit-jee-physics-study-diary-ch3-rest.html
Day 4
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Day 5
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Day 6
Formula Revision
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
WOE 7 to 10, and 13,15, 19,20
Day 7
Exercises 6 to 15
Day 8
Exercises 16 to 25
Day 9
Exercises 26 to 35
Day 10
Exercises 36 to 45
Revision Period
Day 11
Exercises 46 to 50
Day 12
Exercises 51,52
Objective I
Day 13
Exercises
Objective II
Day 14
Exercises
Question for short answer
Day 15
Concept review
http://iit-jee-physics.blogspot.com/2008/05/concept-review-chapter-3-rest-and.html
Day 16
Formula review
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
Day 17 to 20
Test paper problems 10 per day can be solved.
Inertial and accelerated frames of reference.
Section 3.9 covers material related to it. But it is not using the terms inertial frame of reference.
------
Past JEE question
1982
A particle is moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
a) zero
b) [1/Sqrt(2)]m/s^2 towards northwest
c) 1/2 m/s^2 towards northwest
d) 1/2 m/s^2 towards north
Answer b.
You can find acceleration in the direction of west and check that velocity become zero in east direction in 5 seconds and also velocity in north direction becomes 5 m/s.
----------------------
Audio visual lectures
http://www.curriki.org/nroc/Introductory_Physics_1/lesson01/Container.html
Motion in two dimensions (plane)
http://www.curriki.org/nroc/Introductory_Physics_1/lesson02/Container.html
web sites
one dimensional motion
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-15-05.pdf
Study guide H C Verma JEE Physics Ch. 5 NEWTON'S LAWS OF MOTION
Newton’s laws of motion;
--------------
Main Sections
5.1 Newton's First law
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia
---------------------
Subsections under each main section
5.1 First law of motion
Inertial frames other than earth
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia
------------
Study Plan
Day 1
5.1 First law of motion
Example 5.1
W.O.E. 1 and 2
http://iit-jee-physics.blogspot.com/2008/08/newtons-laws-of-motion-study-plan.html
Day 2
5.2 Newton's second law
5.3 Working with Newton's laws
W.O.E. 3 and 4
Points to be noted
http://iit-jee-physics.blogspot.com/2008/09/newtons-laws-of-motion-study-plan.html
Day 3
5.4 Newton's third law of motion
W.O.E. 5,6
Day 4
5.5 Pseudo forces
W.O.E. 7,8
Day 5
5.6 The Horse and the cart
5.7 Inertia
W.O.E. 9,10,11
Day 6
Chapter 5
Objective I
Day 7
Chapter 5
Objective II
Day 8
Chapter 5
Exercises 1 to 5
Day 9
Chapter 5
Exercises 6 to 10
Day 10
Chapter 5
Exercises 11 to 15
Day 11
Chapter 5 Revision
Exercises 16 to 18
Day 12
Chapter 5 Revision
Exercises 19 to 21
Day 13
Chapter 5 Revision
Exercises 22 to 24
Day 14
Chapter 5 Revision
Exercises 25 to 27
Day 15
Chapter 5 Revision
Exercises 28 to 30
Day 16
Chapter 5 Revision
Exercises 31 to 33
Day 17
Chapter 5 Revision
Exercises 34 to 36
Day 18
Chapter 5 Revision
Exercises 37 to 39
Day 19
Chapter 5 Revision
Exercises 40 to 42
Day 20
Chapter 5 Revision
Attempt Questions for short Answer 1 to 17
Audiovisual lecture
Newton's first law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson03/Container.html
Newton's second law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson04/Container.html
Newton's thrid law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson05/Container.html
Newton's laws - applications
http://www.curriki.org/nroc/Introductory_Physics_1/lesson06/Container.html
Websites
plasma4.sr.unh.edu/ng/phys407/phys407-6-23-05.pdf
-----------------
JEE Question 2007 Paper II
Statement - 1
A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.
Because
Statement - 2
For every action there is an equal an opposite reaction.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Correct choice: (B)
S1 is correct.
S2 is also correct but it does not explain S1. S1 is possible because of inertia.
-----------------------
---------
JEE Question 2007 Paper I
Two particles of mass m each are tied at the ends of a light string of
length 2a. The whole system is kept on a frictionless horizontal surface
with the string held tight so that each mass is at a distance ‘a’ from the
centre P (as shown in the figure. *figure not given in this post). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result,the particles move towards each other on the surface. The magnitude of
acceleration, when the separation between them becomes 2x, is
(A) (F/2m)(a/SQRT(a^2-x^2))
(B) (F/2m)(x/SQRT(a^2-x^2))
(C) (F/2m)(x/a)
(D) (F/2m)(SQRT(a^2-x^2)/x)
Solution: B
Imagine the string being pulled upward from midpoint, but only horizontal movement is there.Assume that the string makes with the horizontal surface is θ (theta) , and the tension in the string is T.
F = 2Tsinθ (Tension in each side of the string)
Force in the horintal direction on each particle = F' = ma = T Cosθ
So a = T Cosθ/m
T = F/2Sinθ
a = (F/2m)cotθ
In the triange formed between half of the string and horizontal surface,and the imaginary vertical line through which force is being applied, diagonal is half of the string hence a, and the horizontal side is x. The third side is SQRT(a^2-x^2)
Hence cotθ = x/SQRT(a^2-x^2)
a = (F/2m)(x/SQRT(a^2-x^2))
---------------------
--------------
Main Sections
5.1 Newton's First law
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia
---------------------
Subsections under each main section
5.1 First law of motion
Inertial frames other than earth
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia
------------
Study Plan
Day 1
5.1 First law of motion
Example 5.1
W.O.E. 1 and 2
http://iit-jee-physics.blogspot.com/2008/08/newtons-laws-of-motion-study-plan.html
Day 2
5.2 Newton's second law
5.3 Working with Newton's laws
W.O.E. 3 and 4
Points to be noted
http://iit-jee-physics.blogspot.com/2008/09/newtons-laws-of-motion-study-plan.html
Day 3
5.4 Newton's third law of motion
W.O.E. 5,6
Day 4
5.5 Pseudo forces
W.O.E. 7,8
Day 5
5.6 The Horse and the cart
5.7 Inertia
W.O.E. 9,10,11
Day 6
Chapter 5
Objective I
Day 7
Chapter 5
Objective II
Day 8
Chapter 5
Exercises 1 to 5
Day 9
Chapter 5
Exercises 6 to 10
Day 10
Chapter 5
Exercises 11 to 15
Day 11
Chapter 5 Revision
Exercises 16 to 18
Day 12
Chapter 5 Revision
Exercises 19 to 21
Day 13
Chapter 5 Revision
Exercises 22 to 24
Day 14
Chapter 5 Revision
Exercises 25 to 27
Day 15
Chapter 5 Revision
Exercises 28 to 30
Day 16
Chapter 5 Revision
Exercises 31 to 33
Day 17
Chapter 5 Revision
Exercises 34 to 36
Day 18
Chapter 5 Revision
Exercises 37 to 39
Day 19
Chapter 5 Revision
Exercises 40 to 42
Day 20
Chapter 5 Revision
Attempt Questions for short Answer 1 to 17
Audiovisual lecture
Newton's first law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson03/Container.html
Newton's second law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson04/Container.html
Newton's thrid law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson05/Container.html
Newton's laws - applications
http://www.curriki.org/nroc/Introductory_Physics_1/lesson06/Container.html
Websites
plasma4.sr.unh.edu/ng/phys407/phys407-6-23-05.pdf
-----------------
JEE Question 2007 Paper II
Statement - 1
A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.
Because
Statement - 2
For every action there is an equal an opposite reaction.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Correct choice: (B)
S1 is correct.
S2 is also correct but it does not explain S1. S1 is possible because of inertia.
-----------------------
---------
JEE Question 2007 Paper I
Two particles of mass m each are tied at the ends of a light string of
length 2a. The whole system is kept on a frictionless horizontal surface
with the string held tight so that each mass is at a distance ‘a’ from the
centre P (as shown in the figure. *figure not given in this post). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result,the particles move towards each other on the surface. The magnitude of
acceleration, when the separation between them becomes 2x, is
(A) (F/2m)(a/SQRT(a^2-x^2))
(B) (F/2m)(x/SQRT(a^2-x^2))
(C) (F/2m)(x/a)
(D) (F/2m)(SQRT(a^2-x^2)/x)
Solution: B
Imagine the string being pulled upward from midpoint, but only horizontal movement is there.Assume that the string makes with the horizontal surface is θ (theta) , and the tension in the string is T.
F = 2Tsinθ (Tension in each side of the string)
Force in the horintal direction on each particle = F' = ma = T Cosθ
So a = T Cosθ/m
T = F/2Sinθ
a = (F/2m)cotθ
In the triange formed between half of the string and horizontal surface,and the imaginary vertical line through which force is being applied, diagonal is half of the string hence a, and the horizontal side is x. The third side is SQRT(a^2-x^2)
Hence cotθ = x/SQRT(a^2-x^2)
a = (F/2m)(x/SQRT(a^2-x^2))
---------------------
Study guide H C Verma JEE Physics Ch. 6 FRICTION
JEE Syllabus
Static and dynamic friction;
----
Sections of the chapter
6.1 Friction as the component of contact force
6.2 Kinetic friction
6.3 Static friction
6.4 Laws of friction
6.5 Understandng friction at atomic level
6.6 A Laboratory method to measure friction coefficient
--------
Study Plan
Day 1
6.1 Friction as the component of contact force
Example 6.1
6.2 Kinetic friction
Example 6.2
Day 2
6.3 Static friction
Ex. 6.3
W.O.E. 1,2,3,4
Day 3
6.4 Laws of friction
W.O.E. 5 to 8
Day 4
6.5 Understandng friction at atomic level
6.6 A Laboratory method to measure friction coefficient
Ex. 6.4
W.O.E. 9,10
Day 5
Chapter 6
Objective I
Day 6
Chapter 6
Objective II
Day 7
Chapter 6
Exercises 1 to 5
Day 8
Chapter 6
Exercises 6 to 10
Day 9
Chapter 6
Exercises 11 to 15
Day 10
Chapter 6
Exercises 16 to 20
Revision Period
Day 11
Chapter 6 Revision
Exercises 21 to 23
Day 12
Chapter 6 Revision
Exercises 24 to 27
Day 13
Chapter 6 Revision
Exercises 28 to 30
Day 14
Chapter 6 Revision
Exercises 31,
Day 15
Chapter 6 Revision
Questions for Short Answer: 1 to 11
-----------
web sites
lecture notes
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-27-05.pdf
----------------
JEE question 2007 paper I
STATEMENT-1
A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30 degrees with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation.
because
STATEMENT-2
The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Correct choice: (C)
Coefficient of friction doesn’t depend on the angle of inclination of the plane.
----------------------------------------
Static and dynamic friction;
----
Sections of the chapter
6.1 Friction as the component of contact force
6.2 Kinetic friction
6.3 Static friction
6.4 Laws of friction
6.5 Understandng friction at atomic level
6.6 A Laboratory method to measure friction coefficient
--------
Study Plan
Day 1
6.1 Friction as the component of contact force
Example 6.1
6.2 Kinetic friction
Example 6.2
Day 2
6.3 Static friction
Ex. 6.3
W.O.E. 1,2,3,4
Day 3
6.4 Laws of friction
W.O.E. 5 to 8
Day 4
6.5 Understandng friction at atomic level
6.6 A Laboratory method to measure friction coefficient
Ex. 6.4
W.O.E. 9,10
Day 5
Chapter 6
Objective I
Day 6
Chapter 6
Objective II
Day 7
Chapter 6
Exercises 1 to 5
Day 8
Chapter 6
Exercises 6 to 10
Day 9
Chapter 6
Exercises 11 to 15
Day 10
Chapter 6
Exercises 16 to 20
Revision Period
Day 11
Chapter 6 Revision
Exercises 21 to 23
Day 12
Chapter 6 Revision
Exercises 24 to 27
Day 13
Chapter 6 Revision
Exercises 28 to 30
Day 14
Chapter 6 Revision
Exercises 31,
Day 15
Chapter 6 Revision
Questions for Short Answer: 1 to 11
-----------
web sites
lecture notes
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-27-05.pdf
----------------
JEE question 2007 paper I
STATEMENT-1
A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30 degrees with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation.
because
STATEMENT-2
The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Correct choice: (C)
Coefficient of friction doesn’t depend on the angle of inclination of the plane.
----------------------------------------
Study guide H C Verma JEE Physics Ch. 7 CIRCULAR MOTION
JEE Syllabus
Uniform Circular motion;
-----------
Sections of the chapter
7.1 Angular variables
7.2 Unit vectors along the radius and the tangent
7.3 Acceleration in circular motion
7.4 Dynamics of circular motion
7.5 Circular turnings and banking of roads
7.6 Centrifugal force
7.7 Effect of earth's rotation on apprarent weight
------------
Study Plan
Day 1
7.1 Angular variables
Ex. 7.1, 7.2
7.2 Unit vectors along the radius and the tangent
Day 2
7.3 Acceleration in circular motion
Ex. 7.3,7.4
7.4 Dynamics of circular motion
Ex. 7.5
W.O.E. 3,4,5,7
Day 3
7.5 Circular turnings and banking of roads
Ex. 7.6
W.O.E. 1 and 2
Day 4
7.6 Centrifugal force
w.O.E. 7, 8,9,10,
Day 5
7.7 Effect of earth's rotation on apprarent weight
Ex. 7.7
w.O.E. 11 to 13
Day 6
Chapter 7
Objective I
Day 7
Chapter 7
Objective II
Day 8
Chapter 7
Exercises 1 to 5
Day 9
Chapter 7
Exercises 6 to 10
Day 10
Chapter 7
Exercises 11 to 15
Revision period
Chapter 7 Revision
Day 11
Chapter 7 Revision
Exercises: 16 to 18
Day 12
Chapter 7 Revision
Exercises: 19 to 21
Day 13
Chapter 7 Revision
Exercises: 22 to 24
Day 14
Chapter 7 Revision
Exercises: 25 to 27
Day 15
Chapter 7 Revision
Exercises: 28 to 30
Day 16
Chapter 7 Revision
Short answer questions: 1 to 10
Day 17
Chapter 7 Revision
Try some past JEE questions
Day 18
Chapter 7 Revision
Try some past JEE questions
Day 19
Chapter 7 Revision
Try some past JEE questions
Day 20
Chapter 7 Revision
Try some past JEE questions
Concepts covered
7.1 Angular variables
Angular position
angular velocity
Angular acceleration
Linear speed
Rate of change of speed
Tangential accelaration
7.2 Unit vectors along the radius and the tangent
radial unit vector
Tangential unit vector
7.3 Acceleration in circular motion
Uniform circular motion
Centripetal acceleration
radial component of acceleration
Tangential component of acceleration
7.4 Dynamics of circular motion
Centripetal force
7.5 Circular turnings and banking of roads
When vehicles go through turnings, they travel along a nearly circular arc. That means there is centrepetal accelaration. What forces cause this acceleration? Friction fs can act towards the centre. However this may not be sufficient and the vehicle may skid.
To take care of it, the roads are banked t the turn so that outer part of the road is somewhat lifted up as compared to the inner part. therefore the normal force makes an angle θ wit hte vertical. The horizontal component of the normal force helps in providing the accelaration required.
The θ required for a speed of the vehicle of v is given by
tanθ = v²/rg
7.6 Centrifugal force
What psuedo force is required if the frame of reference rotates at a constant angular velocity ω with respect to an inertial frame?
Cetrifugal force
Coriolis force
Note: It is a common minsconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrigual force acts (or is assumed to act) becasue we describe the particle from a rotating frame which is noninertial and still use Newton's laws.
7.7 Effect of earth's rotation on apprarent weight
A plumb line stays in a direction which is different from true vertical to earth at that point. The walls of building are built by making them parallel to the plumb line and not to the true vertical.
The weight of a body is mg' and not mg and g' is less than g.
g' = g only at the poles as the poles themselved do not rotate and hence the effect of earth's rotation is not felt there.
---------
JEE Question 2007 Paper I
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2w using the entire potential energy of a spring compressed by a distance x-1 . Disc B is imparted an angular velocity by a spring having the same spring constant and compressed by a distance x-2 . Both the discs rotate in the
clockwise direction
1. The ratio x-1 /x-2 is
(A) 2
(B)1/2
(C) SQRT(2)
(D)1/SQRT(2)
Answer C
2. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t.The average frictional torque on one disc by the other during this period is
3.The loss of kinetic energy during the above process is
--------------------------
JEE question 2007 Paper II
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v^2/4g with respect to the initial position. The object is
(A) ring
(B) solid sphere
(C) hollow sphere
(D) Disc
Answer: D
-------------------------------
JEE Question Paper II 2007
Statement - 1
If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.
Because
Statement - 2
The linear momentum of an isolated system remains constant.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Right choice; D
S1 is incorrect. without external torque but with external force the centre can move.
S2 is correct.
----------------------
JEE 2006
A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be
(A) 2R/SQRT(15)
(B) R*(SQRT(2/15)
(C) 4R/SQRT(15)
(D) R/4
Answer: A
-------------------------
Uniform Circular motion;
-----------
Sections of the chapter
7.1 Angular variables
7.2 Unit vectors along the radius and the tangent
7.3 Acceleration in circular motion
7.4 Dynamics of circular motion
7.5 Circular turnings and banking of roads
7.6 Centrifugal force
7.7 Effect of earth's rotation on apprarent weight
------------
Study Plan
Day 1
7.1 Angular variables
Ex. 7.1, 7.2
7.2 Unit vectors along the radius and the tangent
Day 2
7.3 Acceleration in circular motion
Ex. 7.3,7.4
7.4 Dynamics of circular motion
Ex. 7.5
W.O.E. 3,4,5,7
Day 3
7.5 Circular turnings and banking of roads
Ex. 7.6
W.O.E. 1 and 2
Day 4
7.6 Centrifugal force
w.O.E. 7, 8,9,10,
Day 5
7.7 Effect of earth's rotation on apprarent weight
Ex. 7.7
w.O.E. 11 to 13
Day 6
Chapter 7
Objective I
Day 7
Chapter 7
Objective II
Day 8
Chapter 7
Exercises 1 to 5
Day 9
Chapter 7
Exercises 6 to 10
Day 10
Chapter 7
Exercises 11 to 15
Revision period
Chapter 7 Revision
Day 11
Chapter 7 Revision
Exercises: 16 to 18
Day 12
Chapter 7 Revision
Exercises: 19 to 21
Day 13
Chapter 7 Revision
Exercises: 22 to 24
Day 14
Chapter 7 Revision
Exercises: 25 to 27
Day 15
Chapter 7 Revision
Exercises: 28 to 30
Day 16
Chapter 7 Revision
Short answer questions: 1 to 10
Day 17
Chapter 7 Revision
Try some past JEE questions
Day 18
Chapter 7 Revision
Try some past JEE questions
Day 19
Chapter 7 Revision
Try some past JEE questions
Day 20
Chapter 7 Revision
Try some past JEE questions
Concepts covered
7.1 Angular variables
Angular position
angular velocity
Angular acceleration
Linear speed
Rate of change of speed
Tangential accelaration
7.2 Unit vectors along the radius and the tangent
radial unit vector
Tangential unit vector
7.3 Acceleration in circular motion
Uniform circular motion
Centripetal acceleration
radial component of acceleration
Tangential component of acceleration
7.4 Dynamics of circular motion
Centripetal force
7.5 Circular turnings and banking of roads
When vehicles go through turnings, they travel along a nearly circular arc. That means there is centrepetal accelaration. What forces cause this acceleration? Friction fs can act towards the centre. However this may not be sufficient and the vehicle may skid.
To take care of it, the roads are banked t the turn so that outer part of the road is somewhat lifted up as compared to the inner part. therefore the normal force makes an angle θ wit hte vertical. The horizontal component of the normal force helps in providing the accelaration required.
The θ required for a speed of the vehicle of v is given by
tanθ = v²/rg
7.6 Centrifugal force
What psuedo force is required if the frame of reference rotates at a constant angular velocity ω with respect to an inertial frame?
Cetrifugal force
Coriolis force
Note: It is a common minsconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrigual force acts (or is assumed to act) becasue we describe the particle from a rotating frame which is noninertial and still use Newton's laws.
7.7 Effect of earth's rotation on apprarent weight
A plumb line stays in a direction which is different from true vertical to earth at that point. The walls of building are built by making them parallel to the plumb line and not to the true vertical.
The weight of a body is mg' and not mg and g' is less than g.
g' = g only at the poles as the poles themselved do not rotate and hence the effect of earth's rotation is not felt there.
---------
JEE Question 2007 Paper I
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2w using the entire potential energy of a spring compressed by a distance x-1 . Disc B is imparted an angular velocity by a spring having the same spring constant and compressed by a distance x-2 . Both the discs rotate in the
clockwise direction
1. The ratio x-1 /x-2 is
(A) 2
(B)1/2
(C) SQRT(2)
(D)1/SQRT(2)
Answer C
2. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t.The average frictional torque on one disc by the other during this period is
3.The loss of kinetic energy during the above process is
--------------------------
JEE question 2007 Paper II
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v^2/4g with respect to the initial position. The object is
(A) ring
(B) solid sphere
(C) hollow sphere
(D) Disc
Answer: D
-------------------------------
JEE Question Paper II 2007
Statement - 1
If there is no external torque on a body about its center of mass, then the velocity of the center of mass remains constant.
Because
Statement - 2
The linear momentum of an isolated system remains constant.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Right choice; D
S1 is incorrect. without external torque but with external force the centre can move.
S2 is correct.
----------------------
JEE 2006
A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be
(A) 2R/SQRT(15)
(B) R*(SQRT(2/15)
(C) 4R/SQRT(15)
(D) R/4
Answer: A
-------------------------
Study guide H C Verma JEE Physics Ch. 8 WORK AND ENERGY
Kinetic and potential energy;
Work and power; Conservation of linear momentum and mechanical energy.
-------
Sections covered
1. Kinetic enery
2. Work and work energy theorem
3. Calculation of work done
4. Work energy theorem for a system of particles
5. Potential energy
6. Conservative and nonconservative forces
7. Definition of Potential energy and conservation of mechanical energy
8. Change in the potential energy in a rigid body motion
9. Gravitational potential energy
10. Potential energy oif a compressed or extended spring
11. Different forms of energy: Mass energy equivalence
------
Study Plan
Day 1
1. Kinetic energy
2. Work and work energy theorem
3. Calculation of work done
Day 2
Ex. 8.1 and 8.2
Worked Out Examples (WOE): 1,2,3
Day 3
4. Work energy theorem for a system of particles
5. Potential energy
6. Conservative and nonconservative forces
Day 4
Chapter 8
WOE 4,5,6
Exercises 1,2,3
Day 5
7. Definition of Potential energy and conservation of mechanical energy
Ex. 8.3
8. Change in the potential energy in a rigid body motion
9. Gravitational potential energy
Ex. 8.5
Day 6
Chapter 8
WOE: 6,7,8
Exercises 4,5,6
Day 7
8.10. Potential energy of a compressed or extended spring
Ex. 8.6, 8.7,
11. Different forms of energy: Mass energy equivalence
WOE: 7 to 10
Day 8
Chapter 8
WOE 11 to 14
Exercises: 7 to 10
Day 9
Chapter 8
Exercises: 11 to 20
Day 10
Chapter 8
Exercises: 21 to 30
Revision Period
Day 11
Chapter 8 Revision
Exercises: 31 to 35
Day 12
Chapter 8 Revision
Exercises: 36 to 40
Day 13
Chapter 8 Revision
Exercises: 41 to 45
Day 14
Chapter 8 Revision
Exercises: 46 to 50
Day 15
Chapter 8 Revision
Exercises: 51 to 55
Day 16
Chapter 8 Revision
Exercises: 56 to 60
Day 17
Chapter 8 Revision
Exercises: 61 to 64
Day 18
Chapter 8 Revision
Objective I
Day 19
Chapter 8 Revision
Objective II
Day 20
Chapter 8 Revision
Questions for Short Answer 1 to 17
--------------
Concepts covered
In the chapter initially kinetic energy is discussed. The potential energy is explained in terms of change in kinetic energy. In a system of two particles kinetic energy is decreased as the system moves from configuration 1 to configuration 2.As it comes back to configuration 1, the kinetic energy once again increases. This is due to potential energy into which the decrease in kinetic energy changes. Then potential energy and various forms of potential energy are explained.
In this context, concepts of conservative and nonconservative forces are introduced.
If the work done by a force depends only on the initial and final states and not on the path taken, it is called a conservative force. Force of gravity, Coulomb force, and the force of spring are conservative forces.
The force of friction is nonconservative because the work done by the friction is not zero in a round trip.
------
Audiovisual lectures
Work energy theorem
http://www.curriki.org/nroc/Introductory_Physics_1/lesson07/Container.html
Conservative Forces and Potential Energy
http://www.curriki.org/nroc/Introductory_Physics_1/lesson08/Container.html
Conservation of Energy
http://www.curriki.org/nroc/Introductory_Physics_1/lesson09/Container.html
Power
http://www.curriki.org/nroc/Introductory_Physics_1/lesson10/Container.html
websites
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-29-05.pdf
Work and power; Conservation of linear momentum and mechanical energy.
-------
Sections covered
1. Kinetic enery
2. Work and work energy theorem
3. Calculation of work done
4. Work energy theorem for a system of particles
5. Potential energy
6. Conservative and nonconservative forces
7. Definition of Potential energy and conservation of mechanical energy
8. Change in the potential energy in a rigid body motion
9. Gravitational potential energy
10. Potential energy oif a compressed or extended spring
11. Different forms of energy: Mass energy equivalence
------
Study Plan
Day 1
1. Kinetic energy
2. Work and work energy theorem
3. Calculation of work done
Day 2
Ex. 8.1 and 8.2
Worked Out Examples (WOE): 1,2,3
Day 3
4. Work energy theorem for a system of particles
5. Potential energy
6. Conservative and nonconservative forces
Day 4
Chapter 8
WOE 4,5,6
Exercises 1,2,3
Day 5
7. Definition of Potential energy and conservation of mechanical energy
Ex. 8.3
8. Change in the potential energy in a rigid body motion
9. Gravitational potential energy
Ex. 8.5
Day 6
Chapter 8
WOE: 6,7,8
Exercises 4,5,6
Day 7
8.10. Potential energy of a compressed or extended spring
Ex. 8.6, 8.7,
11. Different forms of energy: Mass energy equivalence
WOE: 7 to 10
Day 8
Chapter 8
WOE 11 to 14
Exercises: 7 to 10
Day 9
Chapter 8
Exercises: 11 to 20
Day 10
Chapter 8
Exercises: 21 to 30
Revision Period
Day 11
Chapter 8 Revision
Exercises: 31 to 35
Day 12
Chapter 8 Revision
Exercises: 36 to 40
Day 13
Chapter 8 Revision
Exercises: 41 to 45
Day 14
Chapter 8 Revision
Exercises: 46 to 50
Day 15
Chapter 8 Revision
Exercises: 51 to 55
Day 16
Chapter 8 Revision
Exercises: 56 to 60
Day 17
Chapter 8 Revision
Exercises: 61 to 64
Day 18
Chapter 8 Revision
Objective I
Day 19
Chapter 8 Revision
Objective II
Day 20
Chapter 8 Revision
Questions for Short Answer 1 to 17
--------------
Concepts covered
In the chapter initially kinetic energy is discussed. The potential energy is explained in terms of change in kinetic energy. In a system of two particles kinetic energy is decreased as the system moves from configuration 1 to configuration 2.As it comes back to configuration 1, the kinetic energy once again increases. This is due to potential energy into which the decrease in kinetic energy changes. Then potential energy and various forms of potential energy are explained.
In this context, concepts of conservative and nonconservative forces are introduced.
If the work done by a force depends only on the initial and final states and not on the path taken, it is called a conservative force. Force of gravity, Coulomb force, and the force of spring are conservative forces.
The force of friction is nonconservative because the work done by the friction is not zero in a round trip.
------
Audiovisual lectures
Work energy theorem
http://www.curriki.org/nroc/Introductory_Physics_1/lesson07/Container.html
Conservative Forces and Potential Energy
http://www.curriki.org/nroc/Introductory_Physics_1/lesson08/Container.html
Conservation of Energy
http://www.curriki.org/nroc/Introductory_Physics_1/lesson09/Container.html
Power
http://www.curriki.org/nroc/Introductory_Physics_1/lesson10/Container.html
websites
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-29-05.pdf
Study guide H C Verma JEE Physics Ch. 9 CENTRE OF MASS, LINEAR MOMENTUM AND MASS
Centre of mass and its motion;
--------
Sections of the chapter
1. Centre of mass
2. Centre of mass og continuous bodies
3. Motion of the Centre of mass
---------------
Study Plan
Day 1
9.1. Centre of mass
Ex. 9.1, 9.2
2. Centre of mass of continuous bodies
Worked out examples (WOE): 1 to 3
Day 2
9.3. Motion of the Centre of mass
Ex. 9.3
4. Linear momentum and its conservation principle
5. Rocket propulsion
Day 3
Chapter 9
WOE 4,5,6,7,8
Exercises 1 to 5
Day 4
9.6. Collision
Ex. 9.4
7. Elastic collision in one dimension
WOE 9 to 12
Day 5
9.8. Perfectly inelastic collision in one dimension
Ex. 9.5
9. Coefficient of restitution
10. Elastic collision in two dimensions
11. Impulse and impulsive force
WOE 13 to 14
Day 6
Chapter 9
Objective I 1 to 19
Day 7
Chapter 9
Objective II 1 to 11
Day 8
Chapter 9
WOE 15 to 20
Exercises 6 to 10
Day 9
Chapter 9
WOE 21 to 24
Exercises 11 to 15
Day 10
Chapter 9
Exercises 16 to 25
Revison Period
Day 11
Chapter 9 Revision
Exercises 26 to 30
Day 12
Chapter 9 Revision
Exercises 31 to 35
Day 13
Chapter 9 Revision
Exercises 36 to 40
Day 14
Chapter 9 Revision
Exercises 41 to 45
Day 15
Chapter 9 Revision
Exercises 46 to 50
Day 16
Chapter 9 Revision
Exercises 51 to 55
Day 17
Chapter 9 Revision
Exercises 56 to 60
Day 18
Chapter 9 Revision
Exercises 61 to 64
Day 19
Chapter 9 Revision
Questions for Short Answer 1 to 12
Day 20
Chapter 9 Revision
Questions for Short Answer 13 to 25
-----------------
Audio visual lecture
Center of Mass
www.curriki.org/nroc/Introductory_Physics_1/lesson11/Container.html
Impulse and Momentum
www.curriki.org/nroc/Introductory_Physics_1/lesson12/Container.html
Conservation of Linear Momentum, Collisions
www.curriki.org/nroc/Introductory_Physics_1/lesson13/Container.html
----------------
JEE question 2007 paper I
STATEMENT-1
In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before
the collision
because
STATEMENT-2
In an elastic collision, the linear momentum of the system is conserved
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Correct choice: (B)
------------------------------
--------
Sections of the chapter
1. Centre of mass
2. Centre of mass og continuous bodies
3. Motion of the Centre of mass
---------------
Study Plan
Day 1
9.1. Centre of mass
Ex. 9.1, 9.2
2. Centre of mass of continuous bodies
Worked out examples (WOE): 1 to 3
Day 2
9.3. Motion of the Centre of mass
Ex. 9.3
4. Linear momentum and its conservation principle
5. Rocket propulsion
Day 3
Chapter 9
WOE 4,5,6,7,8
Exercises 1 to 5
Day 4
9.6. Collision
Ex. 9.4
7. Elastic collision in one dimension
WOE 9 to 12
Day 5
9.8. Perfectly inelastic collision in one dimension
Ex. 9.5
9. Coefficient of restitution
10. Elastic collision in two dimensions
11. Impulse and impulsive force
WOE 13 to 14
Day 6
Chapter 9
Objective I 1 to 19
Day 7
Chapter 9
Objective II 1 to 11
Day 8
Chapter 9
WOE 15 to 20
Exercises 6 to 10
Day 9
Chapter 9
WOE 21 to 24
Exercises 11 to 15
Day 10
Chapter 9
Exercises 16 to 25
Revison Period
Day 11
Chapter 9 Revision
Exercises 26 to 30
Day 12
Chapter 9 Revision
Exercises 31 to 35
Day 13
Chapter 9 Revision
Exercises 36 to 40
Day 14
Chapter 9 Revision
Exercises 41 to 45
Day 15
Chapter 9 Revision
Exercises 46 to 50
Day 16
Chapter 9 Revision
Exercises 51 to 55
Day 17
Chapter 9 Revision
Exercises 56 to 60
Day 18
Chapter 9 Revision
Exercises 61 to 64
Day 19
Chapter 9 Revision
Questions for Short Answer 1 to 12
Day 20
Chapter 9 Revision
Questions for Short Answer 13 to 25
-----------------
Audio visual lecture
Center of Mass
www.curriki.org/nroc/Introductory_Physics_1/lesson11/Container.html
Impulse and Momentum
www.curriki.org/nroc/Introductory_Physics_1/lesson12/Container.html
Conservation of Linear Momentum, Collisions
www.curriki.org/nroc/Introductory_Physics_1/lesson13/Container.html
----------------
JEE question 2007 paper I
STATEMENT-1
In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before
the collision
because
STATEMENT-2
In an elastic collision, the linear momentum of the system is conserved
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Correct choice: (B)
------------------------------
Study guide H C Verma JEE Physics Ch. 10 ROTATIONAL MECHANICS
JEE syllabus
Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres;
---------
1. Rotation of a rigid body
2. Kinematics
3. Rotational dynamics
4. Torque of a force about the axis of rotation
5. Г = Iα
6. Bodies in Equilibrium
7. Bending of a Cyclist on horizontal turn
8. Angular momentum
9. L = Iα
10. Conservation of Angular Momentum
11. Angular impulse
12. Kinetic energy of a rigid body rotating about a given axis
13. Rolling
14. Calculation of Momentum of inertia
15. Two important theorems on moment of inertia
16. Combined rotation and translation
17. Rolling
18. Kinetic energy of a body in combined rotation and translation
19. Angular momentum of a body in combined rotation and translation
20. why does a rolling sphere slow down
-----------
Study Plan
Day 1
10.1. Rotation of a rigid body
2. Kinematics
Exmples 10.1,2 and 3
Day 2
10.3. Rotational dynamics
4. Torque of a force about the axis of rotation
Ex. 10.4
Worked Out Examples: 1,2
Day 3
10.5. Г = Iα
Ex. 10.5
6. Bodies in Equilibrium
7. Bending of a Cyclist on horizontal turn
WOE: 3,4
Day 4
10.8. Angular momentum
9. L = Iα
10. Conservation of Angular Momentum
Ex. 10.6
WOE: 5,6
Day 5
10.11. Angular impulse
12. Kinetic energy of a rigid body rotating about a given axis
Ex. 10.7
13. Rolling
WOE: 7,8
Day 6
10.14. Calculation of Momentum of inertia
WOE: 9,10
Day 7
10.15. Two important theorems on moment of inertia
WOE: 11,12
Day 8
10.16. Combined rotation and translation
17. Rolling
WOE: 13,14
Day 9
10.18. Kinetic energy of a body in combined rotation and translation
19. Angular momentum of a body in combined rotation and translation
20. why does a rolling sphere slow down
WOE: 15,16
Day 10
Chapter 10
WOE: 17 to 29
Day 11
Chapter 10 Revision
Exercises: 1 to 5
Day 12
Chapter 10 Revision
Exercises: 6 to 10
Day 13
Chapter 10 Revision
Exercises: 11 to 15
Day 14
Chapter 10 Revision
Exercises: 16 to 20
Day 15
Chapter 10 Revision
Exercises: 21 to 25
Day 16
Chapter 10 Revision
Exercises: 26 to 30
Day 17
Chapter 10 Revision
Exercises: 31 to 35
Day 18
Chapter 10 Revision
Exercises: 36 to 40
Day 19
Chapter 10 Revision
Exercises: 41 to 45
Day 20
Chapter 10 Revision
Exercises: 46 to 50
Still 36 problems are there in the chapter that you have to do as a special task.
---------------------
Audiovisual lecture
Uniform Circular Motion
www.curriki.org/nroc/Introductory_Physics_1/lesson14/Container.html
Torque and Rotational Statics
www.curriki.org/nroc/Introductory_Physics_1/lesson15/Container.html
http://plasma4.sr.unh.edu/ng/phys407/phys407-7-13-05.pdf
Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres;
---------
1. Rotation of a rigid body
2. Kinematics
3. Rotational dynamics
4. Torque of a force about the axis of rotation
5. Г = Iα
6. Bodies in Equilibrium
7. Bending of a Cyclist on horizontal turn
8. Angular momentum
9. L = Iα
10. Conservation of Angular Momentum
11. Angular impulse
12. Kinetic energy of a rigid body rotating about a given axis
13. Rolling
14. Calculation of Momentum of inertia
15. Two important theorems on moment of inertia
16. Combined rotation and translation
17. Rolling
18. Kinetic energy of a body in combined rotation and translation
19. Angular momentum of a body in combined rotation and translation
20. why does a rolling sphere slow down
-----------
Study Plan
Day 1
10.1. Rotation of a rigid body
2. Kinematics
Exmples 10.1,2 and 3
Day 2
10.3. Rotational dynamics
4. Torque of a force about the axis of rotation
Ex. 10.4
Worked Out Examples: 1,2
Day 3
10.5. Г = Iα
Ex. 10.5
6. Bodies in Equilibrium
7. Bending of a Cyclist on horizontal turn
WOE: 3,4
Day 4
10.8. Angular momentum
9. L = Iα
10. Conservation of Angular Momentum
Ex. 10.6
WOE: 5,6
Day 5
10.11. Angular impulse
12. Kinetic energy of a rigid body rotating about a given axis
Ex. 10.7
13. Rolling
WOE: 7,8
Day 6
10.14. Calculation of Momentum of inertia
WOE: 9,10
Day 7
10.15. Two important theorems on moment of inertia
WOE: 11,12
Day 8
10.16. Combined rotation and translation
17. Rolling
WOE: 13,14
Day 9
10.18. Kinetic energy of a body in combined rotation and translation
19. Angular momentum of a body in combined rotation and translation
20. why does a rolling sphere slow down
WOE: 15,16
Day 10
Chapter 10
WOE: 17 to 29
Day 11
Chapter 10 Revision
Exercises: 1 to 5
Day 12
Chapter 10 Revision
Exercises: 6 to 10
Day 13
Chapter 10 Revision
Exercises: 11 to 15
Day 14
Chapter 10 Revision
Exercises: 16 to 20
Day 15
Chapter 10 Revision
Exercises: 21 to 25
Day 16
Chapter 10 Revision
Exercises: 26 to 30
Day 17
Chapter 10 Revision
Exercises: 31 to 35
Day 18
Chapter 10 Revision
Exercises: 36 to 40
Day 19
Chapter 10 Revision
Exercises: 41 to 45
Day 20
Chapter 10 Revision
Exercises: 46 to 50
Still 36 problems are there in the chapter that you have to do as a special task.
---------------------
Audiovisual lecture
Uniform Circular Motion
www.curriki.org/nroc/Introductory_Physics_1/lesson14/Container.html
Torque and Rotational Statics
www.curriki.org/nroc/Introductory_Physics_1/lesson15/Container.html
http://plasma4.sr.unh.edu/ng/phys407/phys407-7-13-05.pdf
Study guide H C Verma JEE Physics Ch. 11 GRAVITATION
JEE syllabus
Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity.
---------
Sections in H C Verma
1. Historical Introduction
2. Measurement of Gravitational Constant G
3. Gravitational Potential Energy
4. Gravitational Potential
5. Calculation of Gravitational Potential
6. Gravitational Field
7. Relation between Gravitational Potential and Field
8. Calculation of Gravitational Field
9. Variation in the Value of G
10.Planets and Satellites
11.Kepler's laws
12.Weightlessness in a Satellite
13. Escape Velocity
14. Gravitational Binding Energy
15. Blackholes
16. Inertial and Gravitational Mass
17. Possible Changes in the Law of Gravitation
---------
Study Plan
Day 1
1. Historical Introduction
Ex. 11.1
Day 2
2. Measurement of Gravitational Constant G
3. Gravitational Potential Energy
Ex. 11.2
4. Gravitational Potential
Day 3
5. Calculation of Gravitational Potential
Ex. 11.3
Worked Out Examples: 1,2
Day 4
6. Gravitational Field
Ex. 11.4
7. Relation between Gravitational Potential and Field
Ex. 11.5
WOE 3,4
Day 5
8. Calculation of Gravitational Field
Ex. 11.6, and 7
WOE 5,6
Day 6
9. Variation in the Value of G
Ex. 11.8
10.Planets and Satellites
11.9
11.Kepler's laws
12.Weightlessness in a Satellite
Day 7
13. Escape Velocity
Ex. 11.10
14. Gravitational Binding Energy
15. Blackholes
Day 8
16. Inertial and Gravitational Mass
17. Possible Changes in the Law of Gravitation
WOE 7 to 9
Day 9
WOE 10 to 13
Exercises 1 to 5
Day 10
Exercises 6 to 15
Day 11
Exercises 16 to 20
Day 12
Exercises 21 to 25
Day 13
Exercises 26 to 30
Day 14
Exercises 31 to 35
Day 15
Exercises 36 to 39
Day 16
Objective I
Day 17
Objectve II
Questions for Short Answer 1 to 5
Day 18
Questions for Short Answer 6 to 10
Day 19
Questions for Short Answer 11 to 15
Day 20
Questions for Short Answer 16 to 19
------------------
Law of gravitation
Eq (11.1) oin H C Verma
F = GMm/r²
The above equation is known as the universal law of gravitation
G = 6.67*10^-11 N-m² /kg²
Gravitational potential and field
Gravitational potential at a point is equal to the change in potential energy per unit mass, as the mass is brought from the reference point to the given point.
Gravitational field: It is assumed that a body say A, creates a gravitational field in the space around it. The field has its own existence and has energy and momentum. When another body B is placed in gravitational field of a body, this field exerts a force on it. The direction and intensity of the field is defined in terms of the force it exerts on a body placed in it.
The intensity of gravitational field vector E at a point is defined by the equation
vector E = force vector F/mass
where F is the force vector exerted by the field on a body of mass m placed in the field. The intensity of gravitational field is abbreviated as gravitational field. Its SI unit is N/kg.
By the way they are defined, intensity of gravitational field and acceleration due to gravity have equal magnitudes and directions, but they are two separate physical quantities.
Obtaining gravitational potential from gravitational field: If intensity of gravitational field E is defined in term of r the distance from the body exerting the gravitational force, its potential can be obtained by integrating e with respect to r ( ∫Edr).
If the potential is known, then its partial derivatives with respect to x, y, and z can be taken and they can be combined to get E, the intensity of gravitation field.
E = iEx + jEy + kEz
http://plasma4.sr.unh.edu/ng/phys407/phys407-7-27-05.pdf
Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity.
---------
Sections in H C Verma
1. Historical Introduction
2. Measurement of Gravitational Constant G
3. Gravitational Potential Energy
4. Gravitational Potential
5. Calculation of Gravitational Potential
6. Gravitational Field
7. Relation between Gravitational Potential and Field
8. Calculation of Gravitational Field
9. Variation in the Value of G
10.Planets and Satellites
11.Kepler's laws
12.Weightlessness in a Satellite
13. Escape Velocity
14. Gravitational Binding Energy
15. Blackholes
16. Inertial and Gravitational Mass
17. Possible Changes in the Law of Gravitation
---------
Study Plan
Day 1
1. Historical Introduction
Ex. 11.1
Day 2
2. Measurement of Gravitational Constant G
3. Gravitational Potential Energy
Ex. 11.2
4. Gravitational Potential
Day 3
5. Calculation of Gravitational Potential
Ex. 11.3
Worked Out Examples: 1,2
Day 4
6. Gravitational Field
Ex. 11.4
7. Relation between Gravitational Potential and Field
Ex. 11.5
WOE 3,4
Day 5
8. Calculation of Gravitational Field
Ex. 11.6, and 7
WOE 5,6
Day 6
9. Variation in the Value of G
Ex. 11.8
10.Planets and Satellites
11.9
11.Kepler's laws
12.Weightlessness in a Satellite
Day 7
13. Escape Velocity
Ex. 11.10
14. Gravitational Binding Energy
15. Blackholes
Day 8
16. Inertial and Gravitational Mass
17. Possible Changes in the Law of Gravitation
WOE 7 to 9
Day 9
WOE 10 to 13
Exercises 1 to 5
Day 10
Exercises 6 to 15
Day 11
Exercises 16 to 20
Day 12
Exercises 21 to 25
Day 13
Exercises 26 to 30
Day 14
Exercises 31 to 35
Day 15
Exercises 36 to 39
Day 16
Objective I
Day 17
Objectve II
Questions for Short Answer 1 to 5
Day 18
Questions for Short Answer 6 to 10
Day 19
Questions for Short Answer 11 to 15
Day 20
Questions for Short Answer 16 to 19
------------------
Law of gravitation
Eq (11.1) oin H C Verma
F = GMm/r²
The above equation is known as the universal law of gravitation
G = 6.67*10^-11 N-m² /kg²
Gravitational potential and field
Gravitational potential at a point is equal to the change in potential energy per unit mass, as the mass is brought from the reference point to the given point.
Gravitational field: It is assumed that a body say A, creates a gravitational field in the space around it. The field has its own existence and has energy and momentum. When another body B is placed in gravitational field of a body, this field exerts a force on it. The direction and intensity of the field is defined in terms of the force it exerts on a body placed in it.
The intensity of gravitational field vector E at a point is defined by the equation
vector E = force vector F/mass
where F is the force vector exerted by the field on a body of mass m placed in the field. The intensity of gravitational field is abbreviated as gravitational field. Its SI unit is N/kg.
By the way they are defined, intensity of gravitational field and acceleration due to gravity have equal magnitudes and directions, but they are two separate physical quantities.
Obtaining gravitational potential from gravitational field: If intensity of gravitational field E is defined in term of r the distance from the body exerting the gravitational force, its potential can be obtained by integrating e with respect to r ( ∫Edr).
If the potential is known, then its partial derivatives with respect to x, y, and z can be taken and they can be combined to get E, the intensity of gravitation field.
E = iEx + jEy + kEz
http://plasma4.sr.unh.edu/ng/phys407/phys407-7-27-05.pdf
Study guide H C Verma JEE Physics Ch. 12 SIMPLE HARMONIC MOTION
JEE Syllabus
Linear and angular simple harmonic motions
------------------
1. Simple harmonic motion
2. Qualitative nature of simple harmonic motion
3. Equation of motion of a simple harmonic motion
4. Terms associated with simple harmonic motion
5. Simple harmonic motion as a projection of circular motion
6. Energy conservation in Simple harmonic motion
7. Angular simple harmonic motion
8. Simple pendulum
9. Physical pendulum
10.Torsional pendulum
11. Composition of two Simple harmonic motions
12. Damped harmonic motion
13. Forced oscillation and resonance
----------
Study Plan
Day 1
1. Simple harmonic motion
Ex. 12.1
2. Qualitative nature of simple harmonic motion
Ex. 12.2
3. Equation of motion of a simple harmonic motion
Day 2
4. Terms associated with simple harmonic motion
Ex. 12.3, 12.4
5. Simple harmonic motion as a projection of circular motion
6. Energy conservation in Simple harmonic motion
Ex. 12.5
Worked Out Examples 1 and 2
Day 3
7. Angular simple harmonic motion
Ex. 12.6
8. Simple pendulum
Ex. 12.7, 12.8
WOE 3,4
Day 4
9. Physical pendulum
Ex. 12.9
10.Torsional pendulum
12.10
WOE 5,6
Day 5
11. Composition of two Simple harmonic motions
Ex. 12.11
WOE 7,8
Day 6
12. Damped harmonic motion
13. Forced oscillation and resonance
WOE 9,10
Day 7
WOE 11 to 15
Exercises 1 to 5
Day 8
WOE 16 to 20
Exercises 6 to 10
Day 9
WOE 21,22
Exercises 11 to 18
Day 10
Exercises 19 to 30
Revision Period - Half an hour per day
Day 11
Exercises 31 to 35
Day 12
Exercises 36 to 40
Day 13
Exercises 41 to 45
Day 14
Exercises 46 to 50
Day 15
Exercises 51 to 55
Day 16
Exercises 56 to 58
Day 17
Objective I
Day 18
Objective II
Day 19
Questions for short answer 1 to 8
Day 20
Questions for short answer 9 to 16
----------
Audio visual lecture
Lesson 16: Simple Harmonic Motion
www.curriki.org/nroc/Introductory_Physics_1/lesson16/Container.html
http://plasma4.sr.unh.edu/ng/phys407/phys407-7-28-05.pdf
Linear and angular simple harmonic motions
------------------
1. Simple harmonic motion
2. Qualitative nature of simple harmonic motion
3. Equation of motion of a simple harmonic motion
4. Terms associated with simple harmonic motion
5. Simple harmonic motion as a projection of circular motion
6. Energy conservation in Simple harmonic motion
7. Angular simple harmonic motion
8. Simple pendulum
9. Physical pendulum
10.Torsional pendulum
11. Composition of two Simple harmonic motions
12. Damped harmonic motion
13. Forced oscillation and resonance
----------
Study Plan
Day 1
1. Simple harmonic motion
Ex. 12.1
2. Qualitative nature of simple harmonic motion
Ex. 12.2
3. Equation of motion of a simple harmonic motion
Day 2
4. Terms associated with simple harmonic motion
Ex. 12.3, 12.4
5. Simple harmonic motion as a projection of circular motion
6. Energy conservation in Simple harmonic motion
Ex. 12.5
Worked Out Examples 1 and 2
Day 3
7. Angular simple harmonic motion
Ex. 12.6
8. Simple pendulum
Ex. 12.7, 12.8
WOE 3,4
Day 4
9. Physical pendulum
Ex. 12.9
10.Torsional pendulum
12.10
WOE 5,6
Day 5
11. Composition of two Simple harmonic motions
Ex. 12.11
WOE 7,8
Day 6
12. Damped harmonic motion
13. Forced oscillation and resonance
WOE 9,10
Day 7
WOE 11 to 15
Exercises 1 to 5
Day 8
WOE 16 to 20
Exercises 6 to 10
Day 9
WOE 21,22
Exercises 11 to 18
Day 10
Exercises 19 to 30
Revision Period - Half an hour per day
Day 11
Exercises 31 to 35
Day 12
Exercises 36 to 40
Day 13
Exercises 41 to 45
Day 14
Exercises 46 to 50
Day 15
Exercises 51 to 55
Day 16
Exercises 56 to 58
Day 17
Objective I
Day 18
Objective II
Day 19
Questions for short answer 1 to 8
Day 20
Questions for short answer 9 to 16
----------
Audio visual lecture
Lesson 16: Simple Harmonic Motion
www.curriki.org/nroc/Introductory_Physics_1/lesson16/Container.html
http://plasma4.sr.unh.edu/ng/phys407/phys407-7-28-05.pdf
IIT JEE Study Guide H C Verma JEE Physics Ch. 13 FLUID MECHANICS
JEE Syllabus
Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy and surface tension, capillary rise; Viscosity (Poiseuille’s equation excluded), Stoke’s law; Terminal velocity, Streamline flow, equation of continuity, Bernoulli’s theorem and its applications.
--------------
Sections in Chapter
1. Fluids
2. Pressure in a fluid
3. Pascal's law
4. Atmospheric pressure
5. Archimedes' Principle
6. Pressure difference and buoyant force in accelerating fluids
7. Flow of fluids
8. Steady and turbulent flow
9. Irrotational flow of an incompressible and nonviscous fluid
10. Equation of continuity
11. Bernoulli's equation
12 Application of Bernoulli's equation
-----------
Study Plan
Day 1
1. Fluids
2. Pressure in a fluid
3. Pascal's law
Day 2
4. Atmospheric pressure
5. Archimedes' Principle
Ex. 13.2
6. Pressure difference and buoyant force in accelerating fluids
Day 3
7. Flow of fluids
8. Steady and turbulent flow
9. Irrotational flow of an incompressible and nonviscous fluid
Worked Out Examples: 1,2
Day 4
10. Equation of continuity
WOE 3 to 8
Day 5
11. Bernoulli's equation
WOE 9,10
Day 6
12 Application of Bernoulli's equation
WOE 11
Exercises 1 to 5
Day 7
Exercises 6 to 15
Day 8
Exercises 16 to 25
Day 9
Exercises 26 to 35
Day 10
Objective I and II
Revision Period 30 minutes per day
Day 11
Short answer questions 1 to 8
Day 12
Short Answer Questions 9 to 16
Day 13
Concept review
Day 14
Formula review
Day 15 to 20 Revision or problems test paper books
-------------
Links for
Audiovisual lectures
Lesson 21: Hydrostatic Pressure
www.curriki.org/nroc/Introductory_Physics_1/lesson21/Container.html
Lesson 22: Buoyancy
www.curriki.org/nroc/Introductory_Physics_1/lesson227/Container.html
Lesson 23: Fluid Flow Continuity
www.curriki.org/nroc/Introductory_Physics_1/lesson23/Container.html
Lesson 24: Bernoulli's Equation
www.curriki.org/nroc/Introductory_Physics_1/lesson24/Container.html
Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy and surface tension, capillary rise; Viscosity (Poiseuille’s equation excluded), Stoke’s law; Terminal velocity, Streamline flow, equation of continuity, Bernoulli’s theorem and its applications.
--------------
Sections in Chapter
1. Fluids
2. Pressure in a fluid
3. Pascal's law
4. Atmospheric pressure
5. Archimedes' Principle
6. Pressure difference and buoyant force in accelerating fluids
7. Flow of fluids
8. Steady and turbulent flow
9. Irrotational flow of an incompressible and nonviscous fluid
10. Equation of continuity
11. Bernoulli's equation
12 Application of Bernoulli's equation
-----------
Study Plan
Day 1
1. Fluids
2. Pressure in a fluid
3. Pascal's law
Day 2
4. Atmospheric pressure
5. Archimedes' Principle
Ex. 13.2
6. Pressure difference and buoyant force in accelerating fluids
Day 3
7. Flow of fluids
8. Steady and turbulent flow
9. Irrotational flow of an incompressible and nonviscous fluid
Worked Out Examples: 1,2
Day 4
10. Equation of continuity
WOE 3 to 8
Day 5
11. Bernoulli's equation
WOE 9,10
Day 6
12 Application of Bernoulli's equation
WOE 11
Exercises 1 to 5
Day 7
Exercises 6 to 15
Day 8
Exercises 16 to 25
Day 9
Exercises 26 to 35
Day 10
Objective I and II
Revision Period 30 minutes per day
Day 11
Short answer questions 1 to 8
Day 12
Short Answer Questions 9 to 16
Day 13
Concept review
Day 14
Formula review
Day 15 to 20 Revision or problems test paper books
-------------
Links for
Audiovisual lectures
Lesson 21: Hydrostatic Pressure
www.curriki.org/nroc/Introductory_Physics_1/lesson21/Container.html
Lesson 22: Buoyancy
www.curriki.org/nroc/Introductory_Physics_1/lesson227/Container.html
Lesson 23: Fluid Flow Continuity
www.curriki.org/nroc/Introductory_Physics_1/lesson23/Container.html
Lesson 24: Bernoulli's Equation
www.curriki.org/nroc/Introductory_Physics_1/lesson24/Container.html
IIT JEE Study Guide H C Verma JEE Physics Ch. 14 SOME MECHANICAL PROPERTIES OF MATTER
JEE syllabus
Hooke’s law 14.5,
Young’s modulus 14.5, 14.8
Viscosity (Poiseuille's equation excluded) (14.15),
Stoke's law (14.17);
Terminal velocity(14.18),
-----------
Sections in the Chapter
1. Molecular structure of the material
2. Elasticity
3. Stress
4. Strain
5. Hooke's law & Young's modulus
6. Relation between longitudinal stress and strain
7. Elastic potential energy of a strained body
8 Determination of Young's modulus
9. Surface tension
10. Surface energy
11. Excess pressure inside a drop
12. Excess pressure in a soap bubble
13. Contact angle
14. Rise of a liquid in a capillary tube
15. Viscosity
16. Flow through a narrow tube: Poiseuille's equation
17. Stokes law
18. Terminal velocity
19. Measuring coefficient of viscosity by Stokes' method
20. Critical velocity and Reynold's number
----------
Study Plan
Day 1
1. Molecular structure of the material
2. Elasticity
3. Stress
Ex. 14.1
Day 2
4. Strain
5. Hooke's law & Young's modulus
Ex. 14.2
Worked Out Examples 1 to 3
Day 3
6. Relation between longitudinal stress and strain
7. Elastic potential energy of a strained body
Ex. 14.3
WOE 4
Exercises 1 to 5
Day 4
8 Determination of Young's modulus
9. Surface tension
Ex.14.4
10. Surface energy
Ex. 14.5
WOE 5,6
Day 5
WOE 7 to 10
Exercises 6 to 10
Day 6
11. Excess pressure inside a drop
12. Excess pressure in a soap bubble
13. Contact angle
14. Rise of a liquid in a capillary tube
WOE 11,12
Day 7
15. Viscosity
16. Flow through a narrow tube: Poiseuille's equation
17. Stokes law
WOE 13,14
Day 8
18. Terminal velocity
19. Measuring coefficient of viscosity by Stokes' method
20. Critical velocity and Reynold's number
WOE 15 to 17
Day 9
Exercises 11 to 20
Day 10
Exercises: 21 to 30
Revision Period: 30 minutes
Day 11
Exercises: 31,32
Objective I: 1 to 15
Day 12
Objective I: 16 to 29
Day 13
Objective II: 7
Day 14
Questions for Short Answer: 1 to 10
Day 15
Questions for Short Answer: 11 to 18
Day 16
Concept review
Day 17
Formula review
Day 18 to 20 Revision and problems from test paper guides
-------------------
Some Important concepts
Hooke's law
If the deformation is small, the stress in a body is proportional to the corresponding strain.
Tensile stress/Tensile strain = Y.
Y is a constant for a given material. This ratio is called Young's modulus for the material.
Stokes law (Ch. 14.17)
The viscous force ating on a small sphere, falling through a medium is equal to 6πηrv, where r is the radius of the sphere, v is velocity of the sphere, and η is the coefficient of viscosity of the medium.
Terminal velocity (Ch 14.18)
The viscous force on a solid moving through a fluid is proportional to its velocity (see stokes law).
When a solid is dropped in a fluid, the forces acting on it are
a. weight W acting vertically downward.
b. the viscous force F acting vertically upward and tbe buoyancy force acting vertically upward.
Weight and buoyancy forces are constant but viscous force changes with velocity.
If initially velocity is zero, viscous force is zero and the body is accelerated due to the forces W-B. As the body accelerates and velocity increases, viscous forces keeps increasing and at a certain instant it becomes equal to W-B. the net force becomes zero at that point and then onwards solid falls with constant velocity.
Hooke’s law 14.5,
Young’s modulus 14.5, 14.8
Viscosity (Poiseuille's equation excluded) (14.15),
Stoke's law (14.17);
Terminal velocity(14.18),
-----------
Sections in the Chapter
1. Molecular structure of the material
2. Elasticity
3. Stress
4. Strain
5. Hooke's law & Young's modulus
6. Relation between longitudinal stress and strain
7. Elastic potential energy of a strained body
8 Determination of Young's modulus
9. Surface tension
10. Surface energy
11. Excess pressure inside a drop
12. Excess pressure in a soap bubble
13. Contact angle
14. Rise of a liquid in a capillary tube
15. Viscosity
16. Flow through a narrow tube: Poiseuille's equation
17. Stokes law
18. Terminal velocity
19. Measuring coefficient of viscosity by Stokes' method
20. Critical velocity and Reynold's number
----------
Study Plan
Day 1
1. Molecular structure of the material
2. Elasticity
3. Stress
Ex. 14.1
Day 2
4. Strain
5. Hooke's law & Young's modulus
Ex. 14.2
Worked Out Examples 1 to 3
Day 3
6. Relation between longitudinal stress and strain
7. Elastic potential energy of a strained body
Ex. 14.3
WOE 4
Exercises 1 to 5
Day 4
8 Determination of Young's modulus
9. Surface tension
Ex.14.4
10. Surface energy
Ex. 14.5
WOE 5,6
Day 5
WOE 7 to 10
Exercises 6 to 10
Day 6
11. Excess pressure inside a drop
12. Excess pressure in a soap bubble
13. Contact angle
14. Rise of a liquid in a capillary tube
WOE 11,12
Day 7
15. Viscosity
16. Flow through a narrow tube: Poiseuille's equation
17. Stokes law
WOE 13,14
Day 8
18. Terminal velocity
19. Measuring coefficient of viscosity by Stokes' method
20. Critical velocity and Reynold's number
WOE 15 to 17
Day 9
Exercises 11 to 20
Day 10
Exercises: 21 to 30
Revision Period: 30 minutes
Day 11
Exercises: 31,32
Objective I: 1 to 15
Day 12
Objective I: 16 to 29
Day 13
Objective II: 7
Day 14
Questions for Short Answer: 1 to 10
Day 15
Questions for Short Answer: 11 to 18
Day 16
Concept review
Day 17
Formula review
Day 18 to 20 Revision and problems from test paper guides
-------------------
Some Important concepts
Hooke's law
If the deformation is small, the stress in a body is proportional to the corresponding strain.
Tensile stress/Tensile strain = Y.
Y is a constant for a given material. This ratio is called Young's modulus for the material.
Stokes law (Ch. 14.17)
The viscous force ating on a small sphere, falling through a medium is equal to 6πηrv, where r is the radius of the sphere, v is velocity of the sphere, and η is the coefficient of viscosity of the medium.
Terminal velocity (Ch 14.18)
The viscous force on a solid moving through a fluid is proportional to its velocity (see stokes law).
When a solid is dropped in a fluid, the forces acting on it are
a. weight W acting vertically downward.
b. the viscous force F acting vertically upward and tbe buoyancy force acting vertically upward.
Weight and buoyancy forces are constant but viscous force changes with velocity.
If initially velocity is zero, viscous force is zero and the body is accelerated due to the forces W-B. As the body accelerates and velocity increases, viscous forces keeps increasing and at a certain instant it becomes equal to W-B. the net force becomes zero at that point and then onwards solid falls with constant velocity.
Study guide H C Verma JEE Physics Ch. 16 SOUND WAVES
JEE syllabus
Vibration of air columns;Resonance; Beats; Speed of sound in gases; Doppler effect (in sound).
----------
Sections of the chapter
1 The nature and propagation of sound waves
2. Displacement wave and pressure wave
3. Speed of a sound wave in a material medium
4. Speed of sound in a gas
5. Effect of pressure, temperature and humidity on the speed of sound in air
6. Intensity of sound waves
7. Appearance of sound to human ear
8. Interference of sound waves
9. Standing longitudinal waves and vibrations of air columns
10. Determination of speed of sound in air
11. Beats
12. Diffraction
13 Doppler effect
14. Sonic booms
15. Musical scale
16. Accoustics of buildings
--------------
Study Plan
Day 1
1 The nature and propagation of sound waves
Ex. 16.1
2. Displacement wave and pressure wave
Ex. 16.2
Worked out examples 1, 2
Day 2
3. Speed of a sound wave in a material medium
4. Speed of sound in a gas
5. Effect of pressure, temperature and humidity on the speed of sound in air
Day 3
6. Intensity of sound waves
7. Appearance of sound to human ear
WOE 3 to 5
Day 4
8. Interference of sound waves
9. Standing longitudinal waves and vibrations of air columns
Day 5
10. Determination of speed of sound in air
11. Beats
Day 6
12. Diffraction
13 Doppler effect
14. Sonic booms
Day 7
15. Musical scale
16. Accoustics of buildings
Day 8
WOE 6 to 15
Day 9
WOE 16 to 23
Day 10
Exercises 1 to 10
Day 11
Exercises 11 to 16
Day 12
Exercises 17 to 22
Day 13
Exercises 23 to 28
Day 14
Exercises 29 to 34
Day 15
Exercises 35 to 40
Day 16
Exercises 41 to 46
Day 17
Exercises 47 to 52
Day 18
Exercises 53 to 58
Day 19
Exercises 59 to 64
Day 20
Exercises 65 to 72
special task 73 to 89
Objective I and II
Questions for short answer
-------------
concepts covered
16.11. Beats
The phenomenon of periodic variation of intensity of sound when two sound waves of slighlty different frequencies interfere, is called beats.
16.12
Bending of waves from an obstacle or an opening is called diffraction.
Diffraction effects are appreciable when the dimensions of openings or the obstacles are comparable or smaller than the wave length of the wave.
16.13 Doppler effect
If the source of sound or the observed or both, move with respect to the medium, the frequency observed may be different from the frequency of the source. This apparent change in frequency of the wave due to motion of the source or the observeer is called Doppler effect.
-----------
Audiovisual lecture
Lesson 45: Sound Waves and Doppler Shift
www.curriki.org/nroc/Introductory_Physics_2/lesson45/Container.html
--------------------
JEE Question 2007 Paper II
In the experiment to determine the speed of sound using a resonance column,
(A) prongs of the tuning fork are kept in a vertical plane
(B) prongs of the tuning fork are kept in a horizontal plane
(C) in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air
(D) in one of the two resonance observed, the length of the resonating air column is close to half of the wavelength of sound in air
Correct Choice: A
---------------------
JEE Question 2007 Paper II
Two trains A and B are moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz. The spread in the frequency (highest frequency - lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.
1. The speed of sound of the whistle is
(A) 340 m/s for passengers in A and 310 m/s for passengers in B
(B) 360 m/s for passengers in A and 310 m/s for passengers in B
(C) 310 m/s for passengers in A and 360 m/s for passengers in B
(D) 340 m/s for passengers in both the trains
Solution: D
Speed of sound is not affected by motion of source or observer. It depends only on the medium and its state of motion w.r.t. the reference frame used.
2. The spread of frequency as observed by the passengers in train B is
(A) 310 Hz
(B) 330 Hz
(C) 350 Hz
(D) 290 Hz
Solution: A
f' = f-0(340-30)/(340-20) = f-0*31/32
∆f' = 31/32∆f-0 = (31/32)*320 = 310 Hz
-----------------------
Vibration of air columns;Resonance; Beats; Speed of sound in gases; Doppler effect (in sound).
----------
Sections of the chapter
1 The nature and propagation of sound waves
2. Displacement wave and pressure wave
3. Speed of a sound wave in a material medium
4. Speed of sound in a gas
5. Effect of pressure, temperature and humidity on the speed of sound in air
6. Intensity of sound waves
7. Appearance of sound to human ear
8. Interference of sound waves
9. Standing longitudinal waves and vibrations of air columns
10. Determination of speed of sound in air
11. Beats
12. Diffraction
13 Doppler effect
14. Sonic booms
15. Musical scale
16. Accoustics of buildings
--------------
Study Plan
Day 1
1 The nature and propagation of sound waves
Ex. 16.1
2. Displacement wave and pressure wave
Ex. 16.2
Worked out examples 1, 2
Day 2
3. Speed of a sound wave in a material medium
4. Speed of sound in a gas
5. Effect of pressure, temperature and humidity on the speed of sound in air
Day 3
6. Intensity of sound waves
7. Appearance of sound to human ear
WOE 3 to 5
Day 4
8. Interference of sound waves
9. Standing longitudinal waves and vibrations of air columns
Day 5
10. Determination of speed of sound in air
11. Beats
Day 6
12. Diffraction
13 Doppler effect
14. Sonic booms
Day 7
15. Musical scale
16. Accoustics of buildings
Day 8
WOE 6 to 15
Day 9
WOE 16 to 23
Day 10
Exercises 1 to 10
Day 11
Exercises 11 to 16
Day 12
Exercises 17 to 22
Day 13
Exercises 23 to 28
Day 14
Exercises 29 to 34
Day 15
Exercises 35 to 40
Day 16
Exercises 41 to 46
Day 17
Exercises 47 to 52
Day 18
Exercises 53 to 58
Day 19
Exercises 59 to 64
Day 20
Exercises 65 to 72
special task 73 to 89
Objective I and II
Questions for short answer
-------------
concepts covered
16.11. Beats
The phenomenon of periodic variation of intensity of sound when two sound waves of slighlty different frequencies interfere, is called beats.
16.12
Bending of waves from an obstacle or an opening is called diffraction.
Diffraction effects are appreciable when the dimensions of openings or the obstacles are comparable or smaller than the wave length of the wave.
16.13 Doppler effect
If the source of sound or the observed or both, move with respect to the medium, the frequency observed may be different from the frequency of the source. This apparent change in frequency of the wave due to motion of the source or the observeer is called Doppler effect.
-----------
Audiovisual lecture
Lesson 45: Sound Waves and Doppler Shift
www.curriki.org/nroc/Introductory_Physics_2/lesson45/Container.html
--------------------
JEE Question 2007 Paper II
In the experiment to determine the speed of sound using a resonance column,
(A) prongs of the tuning fork are kept in a vertical plane
(B) prongs of the tuning fork are kept in a horizontal plane
(C) in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air
(D) in one of the two resonance observed, the length of the resonating air column is close to half of the wavelength of sound in air
Correct Choice: A
---------------------
JEE Question 2007 Paper II
Two trains A and B are moving with speeds 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz. The spread in the frequency (highest frequency - lowest frequency) is thus 320 Hz. The speed of sound in still air is 340 m/s.
1. The speed of sound of the whistle is
(A) 340 m/s for passengers in A and 310 m/s for passengers in B
(B) 360 m/s for passengers in A and 310 m/s for passengers in B
(C) 310 m/s for passengers in A and 360 m/s for passengers in B
(D) 340 m/s for passengers in both the trains
Solution: D
Speed of sound is not affected by motion of source or observer. It depends only on the medium and its state of motion w.r.t. the reference frame used.
2. The spread of frequency as observed by the passengers in train B is
(A) 310 Hz
(B) 330 Hz
(C) 350 Hz
(D) 290 Hz
Solution: A
f' = f-0(340-30)/(340-20) = f-0*31/32
∆f' = 31/32∆f-0 = (31/32)*320 = 310 Hz
-----------------------
Study guide H C Verma JEE Physics Ch. 19 OPTICAL INSTRUMENTS
JEE Syllabus
Combinations of mirrors and thin lenses; Magnification.
The syllabus given in JEE syllabus, combinations of mirrors and thin lenses and magnication indicates that this chapter is to be studied. Because in these optical instruments various mirrors and lenses are used to obtain magnification.
1. The eye
2. The apparent size
3. Simple microscope
4. Compound microscope
5. Telescope
6. Resolving power of a telescope and microscope
7. Defects of vision
Study plan
Day 1
19.1. The eye
2. The apparent size
3. Simple microscope
Day 2
19.4. Compound microscope
Worked out examples 1, 2,3
Day 3
19.5. Telescope
Worked out examples 4,5,
Day 4
19.6. Resolving power of a telescope and microscope
7. Defects of vision
Worked out examples: 6 to 10.
Day 5
Questions for short answer 1 to 6
Exercises 1 to 4
Day 6
Questions for short answer 7 to 12
Exercises 5 to 8
Day 7
Objective I 1 to 11
Exercises 9 to 12
Day 8
Objective II 1 to 5
Exercises 13 to 16
Day 9
Exercises 17 - 20
Day 10
Exercises 21 to 24
Combinations of mirrors and thin lenses; Magnification.
The syllabus given in JEE syllabus, combinations of mirrors and thin lenses and magnication indicates that this chapter is to be studied. Because in these optical instruments various mirrors and lenses are used to obtain magnification.
1. The eye
2. The apparent size
3. Simple microscope
4. Compound microscope
5. Telescope
6. Resolving power of a telescope and microscope
7. Defects of vision
Study plan
Day 1
19.1. The eye
2. The apparent size
3. Simple microscope
Day 2
19.4. Compound microscope
Worked out examples 1, 2,3
Day 3
19.5. Telescope
Worked out examples 4,5,
Day 4
19.6. Resolving power of a telescope and microscope
7. Defects of vision
Worked out examples: 6 to 10.
Day 5
Questions for short answer 1 to 6
Exercises 1 to 4
Day 6
Questions for short answer 7 to 12
Exercises 5 to 8
Day 7
Objective I 1 to 11
Exercises 9 to 12
Day 8
Objective II 1 to 5
Exercises 13 to 16
Day 9
Exercises 17 - 20
Day 10
Exercises 21 to 24
Study guide H C Verma JEE Physics Ch. 20 DISPERSION AND SPECTRA
Deviation and dispersion of light by a prism;
Sections in the chapter
1. Dispersion
2. Dispersive power
3. Dispersion without average deviation and average deviation without dispersion
4. Spectrum
5. Kinds of spectra
6. Ultraviolet and infrared spectrum
7. Spectrometer
8. Rainbow
Study Plan
Day 1
20.1. Dispersion
2. Dispersive power
3. Dispersion without average deviation and average deviation without dispersion
Day 2
20.4. Spectrum
5. Kinds of spectra
6. Ultraviolet and infrared spectrum
Worked out examples 1 to 3
Day 3
20.7. Spectrometer
8. Rainbow
Day 4
Exercises 1 to 11
Day 5
Objective I and II
Day 6
Questions for short answer
Concept review
Day 7
Formula review
Day 8,9
Additional test paper problems
Sections in the chapter
1. Dispersion
2. Dispersive power
3. Dispersion without average deviation and average deviation without dispersion
4. Spectrum
5. Kinds of spectra
6. Ultraviolet and infrared spectrum
7. Spectrometer
8. Rainbow
Study Plan
Day 1
20.1. Dispersion
2. Dispersive power
3. Dispersion without average deviation and average deviation without dispersion
Day 2
20.4. Spectrum
5. Kinds of spectra
6. Ultraviolet and infrared spectrum
Worked out examples 1 to 3
Day 3
20.7. Spectrometer
8. Rainbow
Day 4
Exercises 1 to 11
Day 5
Objective I and II
Day 6
Questions for short answer
Concept review
Day 7
Formula review
Day 8,9
Additional test paper problems
Study guide H C Verma JEE Physics Ch. 21 SPEED OF LIGHT
No JEE topics
Sections
1. Historical introduction
2. Fizeau method
3. Foucault method
4. Michelson method
Study Plan
Day 1
1. Historical introduction
2. Fizeau method
3. Foucault method
4. Michelson method
Day 2
Exercises 1 to 3
Objective I
Objective II
Questions for Short answer
Day 3
Concept revision
Formula revision
Sections
1. Historical introduction
2. Fizeau method
3. Foucault method
4. Michelson method
Study Plan
Day 1
1. Historical introduction
2. Fizeau method
3. Foucault method
4. Michelson method
Day 2
Exercises 1 to 3
Objective I
Objective II
Questions for Short answer
Day 3
Concept revision
Formula revision
Study guide H C Verma JEE Physics Ch. 22 PHOTOMETRY
No JEE topics
Sections
1 Total radiant flux
2. Luminosity of radiant flux
3. Luminous flux: relative luminosity
4. Luminous efficiency
5. Luminous intensity or illuminating power
6. Iluminance
7. Inverse square law
8. Lambert's cosine law
9. Photometers
Sections
1 Total radiant flux
2. Luminosity of radiant flux
3. Luminous flux: relative luminosity
4. Luminous efficiency
5. Luminous intensity or illuminating power
6. Iluminance
7. Inverse square law
8. Lambert's cosine law
9. Photometers
Study guide H C Verma JEE Physics Ch. 23 HEAT AND TEMPERATURE
JEE topics
Ideal gas laws 24.7;
Thermal expansion of solids, liquids and gases 23.10;
--------------------
Sections in the chapter
23.1 Hot and Cold bodies
23.2 Zeroth law of thermodynamics
23.3 Defining scale of temperature: Mercury and resistance thermometers
23.4 Constant volume gas thermometer
23.5 Ideal gas temperature scale
23.6 Celsius temperature scale
23.7 Ideal gas equation
23.8 callender's compensated constant pressure thermometer
23.9 Adiabatic and diathermic walls
23.10 Thermal expansion
----------
Study Plan
Day 1
23.1 Hot and Cold bodies
23.2 Zeroth law of thermodynamics
23.3 Defining scale of temperature: Mercury and resistance thermometers
day 2
23.4 Constant volume gas thermometer
23.5 Ideal gas temperature scale
23.6 Celsius temperature scale
23.7 Ideal gas equation
Day 3
23.8 callender's compensated constant pressure thermometer
23.9 Adiabatic and diathermic walls
23.10 Thermal expansion
Day 4
Worked out examples 1 to 10
Day 5
WOE 11 to 17
Day 6
Exercises 1 to 10
Day 7
Exercises 11 to 20
Day 8
Exercises 21 to 30
Day 9
Exercises 31 to 34
Objective I
Day 10
Exercises
Objective II
questions for short answer 1 to 13
Revision Period
Day 11
Concept Review
Day 12
Formula Review
_____________
In this chapter, heat is defined, zeroth law of thermodynamics is given, and temparature measurement with mercury thermomters, resistance thermometer, and constant volume gas thermometer are discussed.
* The energy that is tranferred from one body to the other, without any mechanical work involved, is called heat.
* Two bodies are said to be in thermal equilibrium if no transfer of heat takes place when they are placed in contact.
* Zeroth law of thermodynamics: If two bodies A and B are in thermal equilibiurm and A and C are also in thermal equilibrium then B and C are also in thermal equilibrium.
* zeroth law allows us to introduce the concept of temperature to measure the hotness or coldness of a body.
*All bodies in thermal equilibrium are assigned equal temperature.
* To measure temperature, we can choose a substance and look for a measurable property of the substance which monotonically changes with temperature.
* Length of mercury in long capillary, resistance of a wire, pressure of gas when volume is kept constant are properties which can be used for temperature measurement.
Ideal gas laws 24.7;
Thermal expansion of solids, liquids and gases 23.10;
--------------------
Sections in the chapter
23.1 Hot and Cold bodies
23.2 Zeroth law of thermodynamics
23.3 Defining scale of temperature: Mercury and resistance thermometers
23.4 Constant volume gas thermometer
23.5 Ideal gas temperature scale
23.6 Celsius temperature scale
23.7 Ideal gas equation
23.8 callender's compensated constant pressure thermometer
23.9 Adiabatic and diathermic walls
23.10 Thermal expansion
----------
Study Plan
Day 1
23.1 Hot and Cold bodies
23.2 Zeroth law of thermodynamics
23.3 Defining scale of temperature: Mercury and resistance thermometers
day 2
23.4 Constant volume gas thermometer
23.5 Ideal gas temperature scale
23.6 Celsius temperature scale
23.7 Ideal gas equation
Day 3
23.8 callender's compensated constant pressure thermometer
23.9 Adiabatic and diathermic walls
23.10 Thermal expansion
Day 4
Worked out examples 1 to 10
Day 5
WOE 11 to 17
Day 6
Exercises 1 to 10
Day 7
Exercises 11 to 20
Day 8
Exercises 21 to 30
Day 9
Exercises 31 to 34
Objective I
Day 10
Exercises
Objective II
questions for short answer 1 to 13
Revision Period
Day 11
Concept Review
Day 12
Formula Review
_____________
In this chapter, heat is defined, zeroth law of thermodynamics is given, and temparature measurement with mercury thermomters, resistance thermometer, and constant volume gas thermometer are discussed.
* The energy that is tranferred from one body to the other, without any mechanical work involved, is called heat.
* Two bodies are said to be in thermal equilibrium if no transfer of heat takes place when they are placed in contact.
* Zeroth law of thermodynamics: If two bodies A and B are in thermal equilibiurm and A and C are also in thermal equilibrium then B and C are also in thermal equilibrium.
* zeroth law allows us to introduce the concept of temperature to measure the hotness or coldness of a body.
*All bodies in thermal equilibrium are assigned equal temperature.
* To measure temperature, we can choose a substance and look for a measurable property of the substance which monotonically changes with temperature.
* Length of mercury in long capillary, resistance of a wire, pressure of gas when volume is kept constant are properties which can be used for temperature measurement.
IIT JEE Study guide H C Verma JEE Physics Ch. 24 KINETIC THEORY OF GASES
Sections in the Chapter
1. Introduction
2. Assumptions of kinetic theory of gases
3. Calculation of pressure of an ideal gas
4. RMS Speed
5. Kinetic interpretation of temperature
6. Deductions from kinetic theory
7. Ideal gas equation
8. Maxwell's speed distribution law
9. Thermodynamic state
10. Brownian motion
11. Vapour
12. Evaporation
13. Saturated and unsaturated vapour: Vapour pressure
14. Boiling
15. Dew point
16. Humidity
17. Determination of relative humidity
18. Phase diagrams: Triple point
19. Dew and fog
-----------
This chapter contains the assumptions for developing kinetic theory of gases, derivation of an expression for pressure exerted by gas on a container, developing a concept of average of the speeds of the molecules, root-mean-square (rms) speed of molecules, derivation of translational kinetic energy of a gas, deriving the functional relation between Temperature of a gas and its rms speed of molecules, and using the kinetic theory to prove Boyle’s law, Charles’ law, Charles’ law of Pressure, Avogadro’s law, Graham’s law of diffusion, and Dalton’s law of partial pressures. The other concepts discussed are Boltzmann constant, Universal gas constant, Maxwell’s speed distribution law, description of thermodynamic state, equation of state, Brownian motion, vapour, evaporation, saturated and unsaturated vapour pressure, dew point, humidity, phase diagrams, dew and fog.
Key Points
Any sample of gas is made of molecules.
Assumptions of kinetic theory of gases :
• All gases are made up of molecules moving in al directions
• The size of the molecule is much smaller than the average separation between the molecules
• The molecules exert no force on each other or the walls of the container except during collisions
• All collisions between two molecules or the walls of the container are perfectly elastic. also the time spent during a collision is negligibly small
• The molecules obey newtons laws of motion.
• When a gas is left for a sufficient time it comes to a steady state, the density and the distribution of molecules with different velocities are independent of position ,direction ,and time. This assumption may be justified if the number of molecules is very large.
The molecular size is roughly 100 times smaller than the average separation between the molecules at 0.1atm and the room temperature
According to the kinetic theory the internal energy of the ideal gas is the same as the total translational kinetic energy of its molecules .
vtr is the rms speed of the molecules at 273.16k and hence is a constant for the given gas .
The absolute temperature of the given gas is proportional to the square of the rms speed of its molecules
The absolute temperature of the given sample of the gas is proportional to the total translational kinetic energy of its molecules .
We find that for a different kinds of gases, it is not the rms speed but the average kinetic energy of individual molecules that has a fixed value at a given temperature. The heavier molecules move with a smaller rms speed and the lighter molecules move with larger speed
Grahams law of diffusion: when two gases at same pressure and temperature are allowed to diffuse into each other,the rate of diffusion of each gas is inversely proportional to the square of the root of the density of the gas. This is known as grahams law of diffusion.
Dalton’s law of partial pressure: Daltons law of partial pressure says that the pressure exerted by a mixture of several gases equals the sum of the pressure exerted by each gas occupying the same volume as that of the mixture.
The boltzmann constant: the universal constant k is known as the boltzmann constant and its value is k=1.38*10 ^-23 J/K
The universal gas constant : R=N(a)k [n(a) =avagadros number] is another universal constant known as the universal gas constant its value is r=8.314J/mol-K.
The average speed v-bar somewhat smaller then the rms speed.
Maxwell derived an equation giving the distribution of the molecules in different speeds.
The speed v (p) at which dn/dv is maximum is called the most probable speed.
A thermodynamic state of a given sample of an ideal gas is completely described if its pressure and its volume are given
The equation relating pressure, volume and temperature of a given sample of gaseous substance is called the equation of state for that gaseous substance. For ideal gas it is pV = nRT. For a real gas, van der Waals derived the following equation: [p+(a/V^2)][V-b] = nRT.
In liquids also, molecules are in constant random motion. Such a phenomenon and motion is called Brownian motion. To observe brownian motion using normal microscope, we need to have light suspended particles in liquids.
If the temperature is sufficiently high, no amount of pressure can liquify the gas. The temperature above which this behaviour occurs is called the critical temperature of the substance.
A gas below its critical temperature is called vapour.
For water critical temperature is 374.1 degrees Celsius.
Evaporation is a process in which molecules escape slowly from the surface of a liquid due to their random motion.
The temperature at which the saturation vapour pressure is equal to the present vapour pressure is called the dew point.
Dew point is measured using Reganault’s hygrometer.
Triple points is the point at which all three phases of a substance exist in equilibrium.
Water vapour condensing on flowers, grass is termed dew.
Water vapour condensing on dust particles in air is forms thick mist called fog.
-----------------------------------
Audio visual lecture
Ideal gases
www.curriki.org/nroc/Introductory_Physics_1/lesson28/Container.html
----------------------------------
JEE questions from this chapter:
there is question from this chapter in JEE 2007 in paper II
1.
Statement-1
The total translational kinetic energy of all the molecules of an ideal gas is 1.5 times of the product of its pressure and its volume.
because
Statement-2
the molecules of a gas collide with each other and the velocities of molecules change due to the collision.
(A) Statement-1 is true and Statement-2 is true and Statement-2 is a correct explation for Statement-1.
(B) Statement-1 is true and Statement-2 is true and Statement-2 is a not correct explation for Statement-1.
(C) Statement-1 is true and Statement-2 is false.
(D) Statement-1 is false and Statement-2 is true
Correct choice: B
S2 is not a sufficient condition for S1 to be true. The reason being that the collision between molecules and that with the “walls” should be elastic for S1 to be correct.
------------------------
1. Introduction
2. Assumptions of kinetic theory of gases
3. Calculation of pressure of an ideal gas
4. RMS Speed
5. Kinetic interpretation of temperature
6. Deductions from kinetic theory
7. Ideal gas equation
8. Maxwell's speed distribution law
9. Thermodynamic state
10. Brownian motion
11. Vapour
12. Evaporation
13. Saturated and unsaturated vapour: Vapour pressure
14. Boiling
15. Dew point
16. Humidity
17. Determination of relative humidity
18. Phase diagrams: Triple point
19. Dew and fog
-----------
This chapter contains the assumptions for developing kinetic theory of gases, derivation of an expression for pressure exerted by gas on a container, developing a concept of average of the speeds of the molecules, root-mean-square (rms) speed of molecules, derivation of translational kinetic energy of a gas, deriving the functional relation between Temperature of a gas and its rms speed of molecules, and using the kinetic theory to prove Boyle’s law, Charles’ law, Charles’ law of Pressure, Avogadro’s law, Graham’s law of diffusion, and Dalton’s law of partial pressures. The other concepts discussed are Boltzmann constant, Universal gas constant, Maxwell’s speed distribution law, description of thermodynamic state, equation of state, Brownian motion, vapour, evaporation, saturated and unsaturated vapour pressure, dew point, humidity, phase diagrams, dew and fog.
Key Points
Any sample of gas is made of molecules.
Assumptions of kinetic theory of gases :
• All gases are made up of molecules moving in al directions
• The size of the molecule is much smaller than the average separation between the molecules
• The molecules exert no force on each other or the walls of the container except during collisions
• All collisions between two molecules or the walls of the container are perfectly elastic. also the time spent during a collision is negligibly small
• The molecules obey newtons laws of motion.
• When a gas is left for a sufficient time it comes to a steady state, the density and the distribution of molecules with different velocities are independent of position ,direction ,and time. This assumption may be justified if the number of molecules is very large.
The molecular size is roughly 100 times smaller than the average separation between the molecules at 0.1atm and the room temperature
According to the kinetic theory the internal energy of the ideal gas is the same as the total translational kinetic energy of its molecules .
vtr is the rms speed of the molecules at 273.16k and hence is a constant for the given gas .
The absolute temperature of the given gas is proportional to the square of the rms speed of its molecules
The absolute temperature of the given sample of the gas is proportional to the total translational kinetic energy of its molecules .
We find that for a different kinds of gases, it is not the rms speed but the average kinetic energy of individual molecules that has a fixed value at a given temperature. The heavier molecules move with a smaller rms speed and the lighter molecules move with larger speed
Grahams law of diffusion: when two gases at same pressure and temperature are allowed to diffuse into each other,the rate of diffusion of each gas is inversely proportional to the square of the root of the density of the gas. This is known as grahams law of diffusion.
Dalton’s law of partial pressure: Daltons law of partial pressure says that the pressure exerted by a mixture of several gases equals the sum of the pressure exerted by each gas occupying the same volume as that of the mixture.
The boltzmann constant: the universal constant k is known as the boltzmann constant and its value is k=1.38*10 ^-23 J/K
The universal gas constant : R=N(a)k [n(a) =avagadros number] is another universal constant known as the universal gas constant its value is r=8.314J/mol-K.
The average speed v-bar somewhat smaller then the rms speed.
Maxwell derived an equation giving the distribution of the molecules in different speeds.
The speed v (p) at which dn/dv is maximum is called the most probable speed.
A thermodynamic state of a given sample of an ideal gas is completely described if its pressure and its volume are given
The equation relating pressure, volume and temperature of a given sample of gaseous substance is called the equation of state for that gaseous substance. For ideal gas it is pV = nRT. For a real gas, van der Waals derived the following equation: [p+(a/V^2)][V-b] = nRT.
In liquids also, molecules are in constant random motion. Such a phenomenon and motion is called Brownian motion. To observe brownian motion using normal microscope, we need to have light suspended particles in liquids.
If the temperature is sufficiently high, no amount of pressure can liquify the gas. The temperature above which this behaviour occurs is called the critical temperature of the substance.
A gas below its critical temperature is called vapour.
For water critical temperature is 374.1 degrees Celsius.
Evaporation is a process in which molecules escape slowly from the surface of a liquid due to their random motion.
The temperature at which the saturation vapour pressure is equal to the present vapour pressure is called the dew point.
Dew point is measured using Reganault’s hygrometer.
Triple points is the point at which all three phases of a substance exist in equilibrium.
Water vapour condensing on flowers, grass is termed dew.
Water vapour condensing on dust particles in air is forms thick mist called fog.
-----------------------------------
Audio visual lecture
Ideal gases
www.curriki.org/nroc/Introductory_Physics_1/lesson28/Container.html
----------------------------------
JEE questions from this chapter:
there is question from this chapter in JEE 2007 in paper II
1.
Statement-1
The total translational kinetic energy of all the molecules of an ideal gas is 1.5 times of the product of its pressure and its volume.
because
Statement-2
the molecules of a gas collide with each other and the velocities of molecules change due to the collision.
(A) Statement-1 is true and Statement-2 is true and Statement-2 is a correct explation for Statement-1.
(B) Statement-1 is true and Statement-2 is true and Statement-2 is a not correct explation for Statement-1.
(C) Statement-1 is true and Statement-2 is false.
(D) Statement-1 is false and Statement-2 is true
Correct choice: B
S2 is not a sufficient condition for S1 to be true. The reason being that the collision between molecules and that with the “walls” should be elastic for S1 to be correct.
------------------------
IIT JEE Study guide H C Verma JEE Physics Ch. 25 CALORIMETRY
Ssections in the chapter
1. Heat as a form of energy
2. Units of heat
3. principles of calorimetry
4. Specific heat capacity and molar heat capacity
5. determination of specific heat capacity in a laboratory
6. Specific latent heat of fusion and vaporization
7. Measurement of specific latent heat of fusion of ice
8. Measurement of specific latent heat of vaporization of water
9. Mechanical equivalent of heat
---------------------
Study Guide
Day 1
1. Heat as a form of energy
2. Units of heat
3. principles of calorimetry
4. Specific heat capacity and molar heat capacity
Day 2
5. determination of specific heat capacity in a laboratory
6. Specific latent heat of fusion and vaporization
7. Measurement of specific latent heat of fusion of ice
Day 3
Worked out examples: 1 to 4
Exercises 1 to 6
Day 4
8. Measurement of specific latent heat of vaporization of water
9. Mechanical equivalent of heat
Day 5
WOE 5, 6
Exercises 7 to 15
Day 6
Exercises 16 to 18
Concept revision
Day 7
Formula revision
Day 8 to 10
Test paper questions around 30
-----------
This chapter covers the concept of heat as a form of energy, units of heat, principle of calorimetry, formula of the amount of heat absorbed for a given change in temperature, specific heat capacity and molar heat capacity of a substance, determination of specific heat capacity in laboratory, concepts of latent heat of fusion and vaporization, measurement of specific latent heat of fusion of ice and of vaporization of water, and mechanical equivalent of heat (Searle's Cone Method).
* the energy being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat.
* Heat is a form of enery. It is energy in transit whenever temperature differences exist. Once it is tranferred it becomes the internal energy of the receiving body.
* The amount of heat needed to increase the temperature of 1 g of water from 14.5 degrees Centigrade to 15.5 degrees Centigrade at a pressure of 1 atm is called 1 calorie.
*The calorie is now defined in terms of joule as 1 cal = 4.186 joule.
*Principle of Calorimetry: The total heat given by the hot objects equals th total heat received by the cold objects.
* Heat supplied to the body Q = ms *(change in temperature) where m = mass of the body, s = specific heat capacity of the substance.
* Units of s = Joules per Kg K or Joules per Kg-degrees Centigrade
* If the amount of substance is expressed in the number of moles
Q = nC *(change in temperature) where n is the number of moles in the sample and the constant C is called molar heat capacity.
* the quantity ms is called the heat capacity of the body. Its unit is J/K.
*The mass of water having the same heat capacity as a given body is called the water equivalent of the body.
------------------
Audiovisual lectures
Lesson 25: Mechanical Equivalent of Heat
www.curriki.org/nroc/Introductory_Physics_1/lesson25/Container.html
Lesson 26: Specific and Latent Heat
www.curriki.org/nroc/Introductory_Physics_1/lesson26/Container.html
---------------
JEE examination questions
2007 examination: No questions were asked from this chapter.
1. Heat as a form of energy
2. Units of heat
3. principles of calorimetry
4. Specific heat capacity and molar heat capacity
5. determination of specific heat capacity in a laboratory
6. Specific latent heat of fusion and vaporization
7. Measurement of specific latent heat of fusion of ice
8. Measurement of specific latent heat of vaporization of water
9. Mechanical equivalent of heat
---------------------
Study Guide
Day 1
1. Heat as a form of energy
2. Units of heat
3. principles of calorimetry
4. Specific heat capacity and molar heat capacity
Day 2
5. determination of specific heat capacity in a laboratory
6. Specific latent heat of fusion and vaporization
7. Measurement of specific latent heat of fusion of ice
Day 3
Worked out examples: 1 to 4
Exercises 1 to 6
Day 4
8. Measurement of specific latent heat of vaporization of water
9. Mechanical equivalent of heat
Day 5
WOE 5, 6
Exercises 7 to 15
Day 6
Exercises 16 to 18
Concept revision
Day 7
Formula revision
Day 8 to 10
Test paper questions around 30
-----------
This chapter covers the concept of heat as a form of energy, units of heat, principle of calorimetry, formula of the amount of heat absorbed for a given change in temperature, specific heat capacity and molar heat capacity of a substance, determination of specific heat capacity in laboratory, concepts of latent heat of fusion and vaporization, measurement of specific latent heat of fusion of ice and of vaporization of water, and mechanical equivalent of heat (Searle's Cone Method).
* the energy being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat.
* Heat is a form of enery. It is energy in transit whenever temperature differences exist. Once it is tranferred it becomes the internal energy of the receiving body.
* The amount of heat needed to increase the temperature of 1 g of water from 14.5 degrees Centigrade to 15.5 degrees Centigrade at a pressure of 1 atm is called 1 calorie.
*The calorie is now defined in terms of joule as 1 cal = 4.186 joule.
*Principle of Calorimetry: The total heat given by the hot objects equals th total heat received by the cold objects.
* Heat supplied to the body Q = ms *(change in temperature) where m = mass of the body, s = specific heat capacity of the substance.
* Units of s = Joules per Kg K or Joules per Kg-degrees Centigrade
* If the amount of substance is expressed in the number of moles
Q = nC *(change in temperature) where n is the number of moles in the sample and the constant C is called molar heat capacity.
* the quantity ms is called the heat capacity of the body. Its unit is J/K.
*The mass of water having the same heat capacity as a given body is called the water equivalent of the body.
------------------
Audiovisual lectures
Lesson 25: Mechanical Equivalent of Heat
www.curriki.org/nroc/Introductory_Physics_1/lesson25/Container.html
Lesson 26: Specific and Latent Heat
www.curriki.org/nroc/Introductory_Physics_1/lesson26/Container.html
---------------
JEE examination questions
2007 examination: No questions were asked from this chapter.
Study guide H C Verma JEE Physics Ch. 26 LAWS OF THERMODYNAMICS
JEE syllabus :
Equivalence of heat and work;
First law of thermodynamics and its applications (only for ideal gases) 26.1;
-------
Chapter sections
26.1 The first law of thermodynamics
26.2 Work done by a gas
26.3 Heat engines
26.4 The second law of thermodynamics
26.5 Reversible and irrerversible processes
26.6 entropy
27.7 Carnot engine
---------
Study plan
Day 1
26.1 The first law of thermodynamics
26.2 Work done by a gas
Day 2
26.3 Heat engines
Day 3
26.4 The second law of thermodynamics
26.5 Reversible and irrerversible processes
Day 4
26.6 entropy
27.7 Carnot engine
Day 5
Worked out examples 1 to 11
Day 6
Exercises 1 to 10
day 7
Exercises 11 to 20
Day 8
Exercises 21 to 22
Objective I
Day 9
Objective II
Questions for short answer 1 to 8
Day 10
Questions for short answer 9 to 15
Revision
Day 11
Concept review
Day 12
Formula revision
Days 13 to 20
Problems from test paper books
----------
Concepts covered
26.1 We already studied that heat is a form of energy. A system can be given energy either by supplying heat to it or by doing mechanical work on it.
Suppose in a process, an amount of ΔQ of heat is given to the gas and an amount ΔW of work is done by it.
Total energy of the gas must increase by ΔQ - ΔW. If the container along with the gas does not move (you can say that there is no systematic movement) this net energy must go into the system in the form of its internal energy.
If we denote the change in internal energy by ΔU, we can write
ΔU = ΔQ - ΔW or
ΔQ = ΔU + ΔW
The above equation is a mathematical statement of the first law of thermodynamics. The equation represents a statement of conservation of energy and is applicable to any system, however complicated.
The first law may be taken as a statement that there exists an internal energy function U that has a fixed value in a given state. Remember that internal energy is a state function.
Notation for ΔQ and ΔW
If work is done by the system, ΔW is positive. If work is done on the system ΔW is negative.
When heat is given to the system ΔQ is positive. If heat is given by the system to the surroundings ΔQ is negative.
Internal energy increases when heat is given to the system and work is done on the system.
26.2 Work done by a gas
In a cylindrical piston, if the gas expands by a small distance Δx, change in volume is ΔV which is equal to AΔx, where A is cross sectional area of cylinder or piston. As force is equal to pA work done is equal to
ΔW = p*A*Δx = p*ΔV = pΔV
Work done in expansion of from initial volume of V1 to V2 can be found by integrating pΔV and finding its definite integral value between V1 to V2.
ΔW =∫pΔV
This formual even though derived with cylindrical shape, is applicable to any shape.
Expressions for work done in various specified processes
Work done in an Isothermal process:
In an isothermal process Temperature is constant. Hence pV is constant.
For an ideal gas pV = nRT, therefore p = nRT/V
W is given by the expression nRTln[V2/V1]
Work done in an Isobaric Process
IN this process pressure is constant.
Hence W = p(V2-V1)
Work done in an Isocharic Process
As there is no change in volume of gas in this process, work done is zero.
26.5 Reversible and irrereversible processes
If the gas in thermal equilibrium all the parts of gas will be at the same temperature and the state of the gas can described by specifying its pressure, volume and temperature. If we put the container of gas on a hot stove, various parts of the gas will be at different temperatures and we cannot specify a unique temperature for the gas. The gas in not in thermodynamic equilibrium in this case.
If the process is performed in such a way that at any instant during the process the system is very nearly in thermodynamic equilibrium, the process is called quasi-static. Thus, a quasi static process is an idealized process in which all changes take place infinitely slowly. Such a process may be assumed to be reversible.
But a process can be reversible only if it satisfies two conditions. The process must be quasistatic and it should be nondissipative. This means, friction, viscosity etc. should be completely absent.
Reversible cycle: If all parts of cyclic process are reversible, it is called a reversible process.
-------------
Audiovisual lecture
Laws of thermodynamics
www.curriki.org/nroc/Introductory_Physics_1/lesson29/Container.html
Equivalence of heat and work;
First law of thermodynamics and its applications (only for ideal gases) 26.1;
-------
Chapter sections
26.1 The first law of thermodynamics
26.2 Work done by a gas
26.3 Heat engines
26.4 The second law of thermodynamics
26.5 Reversible and irrerversible processes
26.6 entropy
27.7 Carnot engine
---------
Study plan
Day 1
26.1 The first law of thermodynamics
26.2 Work done by a gas
Day 2
26.3 Heat engines
Day 3
26.4 The second law of thermodynamics
26.5 Reversible and irrerversible processes
Day 4
26.6 entropy
27.7 Carnot engine
Day 5
Worked out examples 1 to 11
Day 6
Exercises 1 to 10
day 7
Exercises 11 to 20
Day 8
Exercises 21 to 22
Objective I
Day 9
Objective II
Questions for short answer 1 to 8
Day 10
Questions for short answer 9 to 15
Revision
Day 11
Concept review
Day 12
Formula revision
Days 13 to 20
Problems from test paper books
----------
Concepts covered
26.1 We already studied that heat is a form of energy. A system can be given energy either by supplying heat to it or by doing mechanical work on it.
Suppose in a process, an amount of ΔQ of heat is given to the gas and an amount ΔW of work is done by it.
Total energy of the gas must increase by ΔQ - ΔW. If the container along with the gas does not move (you can say that there is no systematic movement) this net energy must go into the system in the form of its internal energy.
If we denote the change in internal energy by ΔU, we can write
ΔU = ΔQ - ΔW or
ΔQ = ΔU + ΔW
The above equation is a mathematical statement of the first law of thermodynamics. The equation represents a statement of conservation of energy and is applicable to any system, however complicated.
The first law may be taken as a statement that there exists an internal energy function U that has a fixed value in a given state. Remember that internal energy is a state function.
Notation for ΔQ and ΔW
If work is done by the system, ΔW is positive. If work is done on the system ΔW is negative.
When heat is given to the system ΔQ is positive. If heat is given by the system to the surroundings ΔQ is negative.
Internal energy increases when heat is given to the system and work is done on the system.
26.2 Work done by a gas
In a cylindrical piston, if the gas expands by a small distance Δx, change in volume is ΔV which is equal to AΔx, where A is cross sectional area of cylinder or piston. As force is equal to pA work done is equal to
ΔW = p*A*Δx = p*ΔV = pΔV
Work done in expansion of from initial volume of V1 to V2 can be found by integrating pΔV and finding its definite integral value between V1 to V2.
ΔW =∫pΔV
This formual even though derived with cylindrical shape, is applicable to any shape.
Expressions for work done in various specified processes
Work done in an Isothermal process:
In an isothermal process Temperature is constant. Hence pV is constant.
For an ideal gas pV = nRT, therefore p = nRT/V
W is given by the expression nRTln[V2/V1]
Work done in an Isobaric Process
IN this process pressure is constant.
Hence W = p(V2-V1)
Work done in an Isocharic Process
As there is no change in volume of gas in this process, work done is zero.
26.5 Reversible and irrereversible processes
If the gas in thermal equilibrium all the parts of gas will be at the same temperature and the state of the gas can described by specifying its pressure, volume and temperature. If we put the container of gas on a hot stove, various parts of the gas will be at different temperatures and we cannot specify a unique temperature for the gas. The gas in not in thermodynamic equilibrium in this case.
If the process is performed in such a way that at any instant during the process the system is very nearly in thermodynamic equilibrium, the process is called quasi-static. Thus, a quasi static process is an idealized process in which all changes take place infinitely slowly. Such a process may be assumed to be reversible.
But a process can be reversible only if it satisfies two conditions. The process must be quasistatic and it should be nondissipative. This means, friction, viscosity etc. should be completely absent.
Reversible cycle: If all parts of cyclic process are reversible, it is called a reversible process.
-------------
Audiovisual lecture
Laws of thermodynamics
www.curriki.org/nroc/Introductory_Physics_1/lesson29/Container.html
Study guide H C Verma JEE Physics Ch. 27 SPECIFIC HEAT CAPACITIES OF GASES
JEE syllabus
Specific heats (Cv and Cp for monoatomic and diatomic gases)27.1,27.2,27.3,27.4,
Isothermal and adiabatic processes 27.5 27.6, 27.7
27.1 Two kinds of specific heat capacities
27.2 Relation between Cp and Cv for an ideal gas
27.3 Determination of Cp of a Gas
27.4 Determination of Cv of a Gas
27.5 Isothermal and adiabatic processes
27.6 Relations between p,v, t in a reversible adiabatic process
27.7 Work done in a adiabatic process
27.8 Equipartition of energy
---------
Study Plan
27.1 Two kinds of specific heat capacities
27.2 Relation between Cp and Cv for an ideal gas
27.3 Determination of Cp of a Gas
27.4 Determination of Cv of a Gas
27.5 Isothermal and adiabatic processes
27.6 Relations between p,v, t in a reversible adiabatic process
27.7 Work done in a adiabatic process
27.8 Equipartition of energy
-------------------
Concepts covered
Day 1
27.1 Two kinds of specific heat capacities
27.2 Relation between Cp and Cv for an ideal gas
Day 2
27.3 Determination of Cp of a Gas
27.4 Determination of Cv of a Gas
Day 3
Worked out examples 1 to 5
Day 4
27.5 Isothermal and adiabatic processes
27.6 Relations between p,v, t in a reversible adiabatic process
Day 5
27.7 Work done in a adiabatic process
27.8 Equipartition of energy
Day 6
WOE 6 to 12
Day 7
Exercises 1 to 10
Day 8
Exercises 11 to 20
Day 9
exercises 21 to 30
Day 10
Exercises 31 to 35
Objective objective I
Day 11
Obejctive II
Day 12
Questions for short answer
Day 13
Concept Review
Day 14
Formula Revision
Days 15 to 20
Additional problems from test paper books
----------------
27.1
We already know that specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature.
In terms of symbols if an amount of ΔQ of heat is given to a mass of n of the substance and its temperature rises by ΔT, the specific heat capacity s is given by the equation
s = ΔQ/mΔT .......(27.1)
This definition is applicable to solids, liquids and gases.
Specific heat capacity of a gas depends on the process involved. Two specific processes are important and specific heat capacities are defined specifically for these processes.
Constant volume process and constant pressure process
When the volume of a gas of mass m is kept constant and heat is given it, the specific heat capacity calculated is called the specific heat capacity at constant volume and is denoted by symbol sv.
When the pressure of a gas of mass m is kept constant and heat is given it, the specific heat capacity calculated is called the specific heat capacity at constant pressure and is denoted by symbol sp.
The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature.
The molar heat capacity at constant volume is denoted by symbol Cv.
Cv. = ΔQ/nΔT .......(27.4)heat is given at constant volume of the gas
Where n = number of moles gas
The molar specific heat capacity at constant pressure is denoted by symbol Cp.
Cp. = ΔQ/nΔT .......(27.5) heat is given at constant pressure.
Note that quite often, the term specific heat capacity or specific heat is used for molar heat capacity also. To find out the implied meaning you have to note the units given.
The unitof specific heat capacity is J/Kg-K where as that of molar heat capacity is J/mol-K.
27.2
Specific heats (Cv and Cp for monoatomic and diatomic gases)27.1,27.2,27.3,27.4,
Isothermal and adiabatic processes 27.5 27.6, 27.7
27.1 Two kinds of specific heat capacities
27.2 Relation between Cp and Cv for an ideal gas
27.3 Determination of Cp of a Gas
27.4 Determination of Cv of a Gas
27.5 Isothermal and adiabatic processes
27.6 Relations between p,v, t in a reversible adiabatic process
27.7 Work done in a adiabatic process
27.8 Equipartition of energy
---------
Study Plan
27.1 Two kinds of specific heat capacities
27.2 Relation between Cp and Cv for an ideal gas
27.3 Determination of Cp of a Gas
27.4 Determination of Cv of a Gas
27.5 Isothermal and adiabatic processes
27.6 Relations between p,v, t in a reversible adiabatic process
27.7 Work done in a adiabatic process
27.8 Equipartition of energy
-------------------
Concepts covered
Day 1
27.1 Two kinds of specific heat capacities
27.2 Relation between Cp and Cv for an ideal gas
Day 2
27.3 Determination of Cp of a Gas
27.4 Determination of Cv of a Gas
Day 3
Worked out examples 1 to 5
Day 4
27.5 Isothermal and adiabatic processes
27.6 Relations between p,v, t in a reversible adiabatic process
Day 5
27.7 Work done in a adiabatic process
27.8 Equipartition of energy
Day 6
WOE 6 to 12
Day 7
Exercises 1 to 10
Day 8
Exercises 11 to 20
Day 9
exercises 21 to 30
Day 10
Exercises 31 to 35
Objective objective I
Day 11
Obejctive II
Day 12
Questions for short answer
Day 13
Concept Review
Day 14
Formula Revision
Days 15 to 20
Additional problems from test paper books
----------------
27.1
We already know that specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature.
In terms of symbols if an amount of ΔQ of heat is given to a mass of n of the substance and its temperature rises by ΔT, the specific heat capacity s is given by the equation
s = ΔQ/mΔT .......(27.1)
This definition is applicable to solids, liquids and gases.
Specific heat capacity of a gas depends on the process involved. Two specific processes are important and specific heat capacities are defined specifically for these processes.
Constant volume process and constant pressure process
When the volume of a gas of mass m is kept constant and heat is given it, the specific heat capacity calculated is called the specific heat capacity at constant volume and is denoted by symbol sv.
When the pressure of a gas of mass m is kept constant and heat is given it, the specific heat capacity calculated is called the specific heat capacity at constant pressure and is denoted by symbol sp.
The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature.
The molar heat capacity at constant volume is denoted by symbol Cv.
Cv. = ΔQ/nΔT .......(27.4)heat is given at constant volume of the gas
Where n = number of moles gas
The molar specific heat capacity at constant pressure is denoted by symbol Cp.
Cp. = ΔQ/nΔT .......(27.5) heat is given at constant pressure.
Note that quite often, the term specific heat capacity or specific heat is used for molar heat capacity also. To find out the implied meaning you have to note the units given.
The unitof specific heat capacity is J/Kg-K where as that of molar heat capacity is J/mol-K.
27.2
Study guide H C Verma JEE Physics Ch. 28. HEAT TRANSFER
JEE syllabus
Blackbody radiation 28.7:
absorptive and emissive powers;
Kirchhoff’s law 28.8;
Wien’s displacement law,
Stefan’s law 28.10.
------------------
sections in the book
28.1 Thermal conduction
28.2 Series and parallel connection of rods
28.3 Measurment of thermal conductivity of a solid
28.4 Convection
28.5 Radiation
28.6 Prevost theory of exchange
28.7 Blackbody radiation
28.8 Kirchhoff’s law
28.9 Nature of thermal radiation
28.10 Stefan-Boltzmann Law
28.11 Newton's law of cooling
28.12 Detection and Measurement of Radiation
-----------------
Study Plan
Day 1
28.1 Thermal conduction
28.2 Series and parallel connection of rods
28.3 Measurment of thermal conductivity of a solid
Day 2
Worked out examples 1 to 10
Day 3
28.4 Convection
28.5 Radiation
Day 4
WOE 11 to 15
Exercises 1 to 5
Day 5
28.6 Prevost theory of exchange
28.7 Blackbody radiation
28.8 Kirchhoff’s law
Day 6
28.9 Nature of thermal radiation
28.10 Stefan-Boltzmann Law
WOE 16,17
Day 7
28.11 Newton's law of cooling
28.12 Detection and Measurement of Radiation
WOE 18 to 20
Exercises 6 to 10
Day 8
Exercises 11 to 20
Day 9
Exercises 21 to 30
Day 10
Exercises 31 to 40
Day 11
Exercises 41 to 45
Day 12
Exercises 46 to 50
Day 13
Exercises 51 to 55
Day 14
Objective I
Day 15
Objectie II
Day 16
questions for short answer 1 to 11
Day 17
Concept review
Day 18
Formula review
Day 19,20
Additional questions from test paper guides
------------------
Audiovisual lecture
Lesson 27: Heat Transfer and Thermal Expansion
www.curriki.org/nroc/Introductory_Physics_1/lesson27/Container.html
Blackbody radiation 28.7:
absorptive and emissive powers;
Kirchhoff’s law 28.8;
Wien’s displacement law,
Stefan’s law 28.10.
------------------
sections in the book
28.1 Thermal conduction
28.2 Series and parallel connection of rods
28.3 Measurment of thermal conductivity of a solid
28.4 Convection
28.5 Radiation
28.6 Prevost theory of exchange
28.7 Blackbody radiation
28.8 Kirchhoff’s law
28.9 Nature of thermal radiation
28.10 Stefan-Boltzmann Law
28.11 Newton's law of cooling
28.12 Detection and Measurement of Radiation
-----------------
Study Plan
Day 1
28.1 Thermal conduction
28.2 Series and parallel connection of rods
28.3 Measurment of thermal conductivity of a solid
Day 2
Worked out examples 1 to 10
Day 3
28.4 Convection
28.5 Radiation
Day 4
WOE 11 to 15
Exercises 1 to 5
Day 5
28.6 Prevost theory of exchange
28.7 Blackbody radiation
28.8 Kirchhoff’s law
Day 6
28.9 Nature of thermal radiation
28.10 Stefan-Boltzmann Law
WOE 16,17
Day 7
28.11 Newton's law of cooling
28.12 Detection and Measurement of Radiation
WOE 18 to 20
Exercises 6 to 10
Day 8
Exercises 11 to 20
Day 9
Exercises 21 to 30
Day 10
Exercises 31 to 40
Day 11
Exercises 41 to 45
Day 12
Exercises 46 to 50
Day 13
Exercises 51 to 55
Day 14
Objective I
Day 15
Objectie II
Day 16
questions for short answer 1 to 11
Day 17
Concept review
Day 18
Formula review
Day 19,20
Additional questions from test paper guides
------------------
Audiovisual lecture
Lesson 27: Heat Transfer and Thermal Expansion
www.curriki.org/nroc/Introductory_Physics_1/lesson27/Container.html
Study guide H C Verma JEE Physics Ch. 29 ELECTRIC FIELD AND POTENTIAL
Syllabus
Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field;
-------------
Verma - Topics
29.1 What is electric charge?
29.2 Coulomb's law
29.3 electric field
29.4 Lines of electric force
29.5 Electric potential energy
29.6 Electric potential
29.7 Electric potential due to a point charge
29.8 Relation between electric field and potential
29.9 Electric dipole
29.10 Torque on an electric dipole placed in an lectric field
29.11 Potential energy of a diple placed in a uniform electric field
29.12 Conductors, insulators nad semiconductors
29.13 The electric field inside a conductor
--------------------
Study Plan
Day 1
29.1 What is electric charge?
29.2 Coulomb's law
Day 2
29.3 electric field
Ex. 29.1, 29.2
29.4 Lines of electric force
Worked out examples 1,2
Day 3
29.5 Electric potential energy
Ex. 29.3
29.6 Electric potential
Ex. 29.4
29.7 Electric potential due to a point charge
Ex. 29.4
WOE 3 to 5
Day 4
29.8 Relation between electric field and potential
Ex. 29.5
WOE 5 to 10
Day 5
29.9 Electric dipole
29.10 Torque on an electric dipole placed in an lectric field
29.11 Potential energy of a diple placed in a uniform electric field
Day 6
29.12 Conductors, insulators nad semiconductors
29.13 The electric field inside a conductor
WOE 11 to 15
Day 7
WOE 17 to 19
Day 8
Exercises 1 to 10
Day 9
Exercises 11 to 20
Day 10
Exercises 21 to 30
Day 11
Exercises 31 to 35
Day 12
Exercises 36 to 40
Day 13
Exercises 41 to 45
Day 14
Exercises 46 to 50
Day 15
Exercises 51 to 55
Day 16
Exercises 56 to 60
Day 17
Exercises 61 to 65
Day 18
Exercises 66 to 70
Day 19
Exercises 71 to 75
Day 20
Exercises Objectve I 1 to 9 and Objective II 1 to 8
Special task: Questions for short answer, concept review and formual review
----------------------
Introduction
The credit for the discovery of electricity goes to Thales of Miletus, regarded as one of the seven wise men of Greece. Around 600 B.C., Thales discovered that when amber is rubbed with wool, it acquires the property of attracting light objects such as straw, wood shavings etc. The investigation of this phenomenon remained in abeyance for over 2000 years.
Near the end of the 16th century A.D., william Gilbert, a physician to Queen Elizabeth I, found that like amber, there are several othe substances which, when rubbed with suitable substances, exhibit the property of attracting light objects. For example, a glass rod rubbed with silk attracts light objects such as paper, cork etc. Similarly an ebonite rod when rubbed with fur attracts light objects. As this phenomenon was first observed with amber and amber is called electron in Greek, Gilbert name the phenomenon as electricity.
He proposed that a body upon rubbing becomes eletrified or charged and acquires the property of attracting light objects. Such a body is said to have acquired an electric charge.
Subsequently, it was observed that (i) two ebonite rods rubbed with fur repel each other and; (ii) two glass rods rubbed with silk repel each other, while (iii) an ebonite rod rubbed with fur attracts a glass rod rubbed with silk.
-------------------
Concepts covered
The chapter may seem to be a difficult chapter for the first reading.
29.1 What is electric charge is a simple one.
29.2 Coulomb's law - this topic is also straight forward. Charges have forces of attraction or repulsion among them.
Through experimental data, Coulomb proposed the law that gave the formual for finding the force exerted by a charged particle on the other charged particle.
F = kq1q2/r² ... (29.1)
q1, q2 are the charges on particles
r is the separation or distance between them and
k is a constant.
k in SI units is 8.98755 * 10^9 N-m²/C²
29.3 Electric Field
A charge produces an electric field in the space around it and this electric field exerts a force on any other charge placed it.
The field takes finite time to propagate. If a charge is displaced from its position, the field at a distance r will change after a time t = r/c, where c is the speed of light.
Intensity of electric field: is defined in this way. If a charge q is brought into the electric field and if it experiences an electric Vector F (Vr(F)), we define the intensity of electric field at the given point where the charge q is there as
Vr(E) = Vr(F)/q
Vr(E) = Vector of electric field
29.4 Lines of force - concept is similar to magnetic lines of force
29.5 Electric potential energy - this topic may require some time to think over and grasp.
29.6 Electric Potential - Electric field is a vector quantity. Electric potential is a scalar quantity. This concept follows from the concept of electric potential energy.
29.7 This topics gives expressesions for electric potential for a point charge and a system having number of point charges. Follows from the topic above
29.8 This section gives the relationship between the scalar quantity potential and vector quantity field. spend some time to grasp the concepts once again.
29.9 Electric dipole: You know by now dipole. Some of the chemical compounds have dipole. In this section, electric dipole moment is defined. It is a vector quantity.
Vr(p) = q*Vr(d)
p = electric dipole moment
q = charge one positive and one negative
d = distance between charges which is small. vector gives the direction from negative charge to positive charge.
There is expression for electric potential due to a dipole. If you are already comfortable with the concept of electric potential (scalar quantity) this topic will be easy for you.
Then follows discussion of electric field (vector quantity)
29.10 gives the expression for torque experienced by dipole when placed in an electric field.
29.11 discusses potential energy of the dipole whne placed in a uniform electric field.
29.12 describes why some materials are conductors, insulators and semiconductors. The conductors have large number of free electrons.
29.13 informs us that inside a conductor the electric field zero. The electrons shift to one side of the conductor and therefore once face of it becomes negatively charged and the other face positively charged.
--------------
Audiovisual lectures
Lesson 30: Electric Charges and Coulomb's Law www.curriki.org/nroc/Introductory_Physics_2/lesson30/Container.html
Lesson 31: Electric Fields
www.curriki.org/nroc/Introductory_Physics_2/lesson31/Container.html
Lesson 32: Electric Potential
www.curriki.org/nroc/Introductory_Physics_2/lesson32/Container.html
Chapter 11: Conductors and Capacitors
Lesson 33: Electrostatics with Conductors
www.curriki.org/nroc/Introductory_Physics_2/lesson33/Container.html
Websites
http://www.colorado.edu/physics/phys1120/phys1120_fa07/notes/notes/Knight25_coul_lect.pdf
http://www.colorado.edu/physics/phys2020/phys2020_fa01///notes/lecture_notes/CH16/ch16.pdf
-------------------
JEE Question 2007 Paper II
Positive and negative point charges of equal magnitude are kept at (0,0,a/2)and (0,0,-a/2) respectively. The work done by the electric field when another positive point charge is moved form (-a,0,0) to (0,a,0) is
(A) positive
(B) negative
(C) zero
(D) depends on the path connecting the initial and final positions
Correct Choice: C
-----------------------
For more questions and answers on the topics related to electricity visit
http://iit-jee-physics-atps.blogspot.com/2008/03/jee-past-objective-questions.html
Posting started on 23.3.2008
Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field;
-------------
Verma - Topics
29.1 What is electric charge?
29.2 Coulomb's law
29.3 electric field
29.4 Lines of electric force
29.5 Electric potential energy
29.6 Electric potential
29.7 Electric potential due to a point charge
29.8 Relation between electric field and potential
29.9 Electric dipole
29.10 Torque on an electric dipole placed in an lectric field
29.11 Potential energy of a diple placed in a uniform electric field
29.12 Conductors, insulators nad semiconductors
29.13 The electric field inside a conductor
--------------------
Study Plan
Day 1
29.1 What is electric charge?
29.2 Coulomb's law
Day 2
29.3 electric field
Ex. 29.1, 29.2
29.4 Lines of electric force
Worked out examples 1,2
Day 3
29.5 Electric potential energy
Ex. 29.3
29.6 Electric potential
Ex. 29.4
29.7 Electric potential due to a point charge
Ex. 29.4
WOE 3 to 5
Day 4
29.8 Relation between electric field and potential
Ex. 29.5
WOE 5 to 10
Day 5
29.9 Electric dipole
29.10 Torque on an electric dipole placed in an lectric field
29.11 Potential energy of a diple placed in a uniform electric field
Day 6
29.12 Conductors, insulators nad semiconductors
29.13 The electric field inside a conductor
WOE 11 to 15
Day 7
WOE 17 to 19
Day 8
Exercises 1 to 10
Day 9
Exercises 11 to 20
Day 10
Exercises 21 to 30
Day 11
Exercises 31 to 35
Day 12
Exercises 36 to 40
Day 13
Exercises 41 to 45
Day 14
Exercises 46 to 50
Day 15
Exercises 51 to 55
Day 16
Exercises 56 to 60
Day 17
Exercises 61 to 65
Day 18
Exercises 66 to 70
Day 19
Exercises 71 to 75
Day 20
Exercises Objectve I 1 to 9 and Objective II 1 to 8
Special task: Questions for short answer, concept review and formual review
----------------------
Introduction
The credit for the discovery of electricity goes to Thales of Miletus, regarded as one of the seven wise men of Greece. Around 600 B.C., Thales discovered that when amber is rubbed with wool, it acquires the property of attracting light objects such as straw, wood shavings etc. The investigation of this phenomenon remained in abeyance for over 2000 years.
Near the end of the 16th century A.D., william Gilbert, a physician to Queen Elizabeth I, found that like amber, there are several othe substances which, when rubbed with suitable substances, exhibit the property of attracting light objects. For example, a glass rod rubbed with silk attracts light objects such as paper, cork etc. Similarly an ebonite rod when rubbed with fur attracts light objects. As this phenomenon was first observed with amber and amber is called electron in Greek, Gilbert name the phenomenon as electricity.
He proposed that a body upon rubbing becomes eletrified or charged and acquires the property of attracting light objects. Such a body is said to have acquired an electric charge.
Subsequently, it was observed that (i) two ebonite rods rubbed with fur repel each other and; (ii) two glass rods rubbed with silk repel each other, while (iii) an ebonite rod rubbed with fur attracts a glass rod rubbed with silk.
-------------------
Concepts covered
The chapter may seem to be a difficult chapter for the first reading.
29.1 What is electric charge is a simple one.
29.2 Coulomb's law - this topic is also straight forward. Charges have forces of attraction or repulsion among them.
Through experimental data, Coulomb proposed the law that gave the formual for finding the force exerted by a charged particle on the other charged particle.
F = kq1q2/r² ... (29.1)
q1, q2 are the charges on particles
r is the separation or distance between them and
k is a constant.
k in SI units is 8.98755 * 10^9 N-m²/C²
29.3 Electric Field
A charge produces an electric field in the space around it and this electric field exerts a force on any other charge placed it.
The field takes finite time to propagate. If a charge is displaced from its position, the field at a distance r will change after a time t = r/c, where c is the speed of light.
Intensity of electric field: is defined in this way. If a charge q is brought into the electric field and if it experiences an electric Vector F (Vr(F)), we define the intensity of electric field at the given point where the charge q is there as
Vr(E) = Vr(F)/q
Vr(E) = Vector of electric field
29.4 Lines of force - concept is similar to magnetic lines of force
29.5 Electric potential energy - this topic may require some time to think over and grasp.
29.6 Electric Potential - Electric field is a vector quantity. Electric potential is a scalar quantity. This concept follows from the concept of electric potential energy.
29.7 This topics gives expressesions for electric potential for a point charge and a system having number of point charges. Follows from the topic above
29.8 This section gives the relationship between the scalar quantity potential and vector quantity field. spend some time to grasp the concepts once again.
29.9 Electric dipole: You know by now dipole. Some of the chemical compounds have dipole. In this section, electric dipole moment is defined. It is a vector quantity.
Vr(p) = q*Vr(d)
p = electric dipole moment
q = charge one positive and one negative
d = distance between charges which is small. vector gives the direction from negative charge to positive charge.
There is expression for electric potential due to a dipole. If you are already comfortable with the concept of electric potential (scalar quantity) this topic will be easy for you.
Then follows discussion of electric field (vector quantity)
29.10 gives the expression for torque experienced by dipole when placed in an electric field.
29.11 discusses potential energy of the dipole whne placed in a uniform electric field.
29.12 describes why some materials are conductors, insulators and semiconductors. The conductors have large number of free electrons.
29.13 informs us that inside a conductor the electric field zero. The electrons shift to one side of the conductor and therefore once face of it becomes negatively charged and the other face positively charged.
--------------
Audiovisual lectures
Lesson 30: Electric Charges and Coulomb's Law www.curriki.org/nroc/Introductory_Physics_2/lesson30/Container.html
Lesson 31: Electric Fields
www.curriki.org/nroc/Introductory_Physics_2/lesson31/Container.html
Lesson 32: Electric Potential
www.curriki.org/nroc/Introductory_Physics_2/lesson32/Container.html
Chapter 11: Conductors and Capacitors
Lesson 33: Electrostatics with Conductors
www.curriki.org/nroc/Introductory_Physics_2/lesson33/Container.html
Websites
http://www.colorado.edu/physics/phys1120/phys1120_fa07/notes/notes/Knight25_coul_lect.pdf
http://www.colorado.edu/physics/phys2020/phys2020_fa01///notes/lecture_notes/CH16/ch16.pdf
-------------------
JEE Question 2007 Paper II
Positive and negative point charges of equal magnitude are kept at (0,0,a/2)and (0,0,-a/2) respectively. The work done by the electric field when another positive point charge is moved form (-a,0,0) to (0,a,0) is
(A) positive
(B) negative
(C) zero
(D) depends on the path connecting the initial and final positions
Correct Choice: C
-----------------------
For more questions and answers on the topics related to electricity visit
http://iit-jee-physics-atps.blogspot.com/2008/03/jee-past-objective-questions.html
Posting started on 23.3.2008
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