You can enjoy 31st May as a break.
Chemistry
June 2009
3. Atomic Structure
http://iit-jee-chemistry.blogspot.com/2009/03/jee-study-guide-3-atomic-structure.html
4. Classification of Elements and Periodicity in Properties
http://iit-jee-chemistry.blogspot.com/2009/03/jee-study-guide-4-classification-of.html
Mathematics
June 2009
5. Complex numbers XI
http://iit-jee-maths.blogspot.com/2008/11/5-complex-numbers-revision-facilitator.html
6. Sequences and series XI
http://iit-jee-maths.blogspot.com/2008/11/6-sequences-and-series-revision.html
Physics
June 1 to 10
4. The forces
http://iit-jee-physics.blogspot.com/2007/10/study-guide-h-c-verma-jee-physics-ch_23.html
The force chapter is a small chapter. It is indicated in detailed study plan that you can use 7th June to 20th June to revise chapters 1 to 4. Don't waste time. Use the scheduled time to revise all the difficult problems in the four chapters. By revising difficult problems, you will understand the logic of solving those problems. Read the concepts once again. Understanding and frequent recollecting the materials will make it easy for you in the future. Try to do new problems whenever you have time. In the examination, you may not get the problems that you have solved. But solving many problems will give you the confidence and you can solve the new problems given in the examination. You should not fear the new problem. You become capable of solving new problems by solving a large variety of problems.
June 11 to 20
5. Newton’s law of motion
http://iit-jee-physics.blogspot.com/2007/10/study-guide-h-c-verma-jee-physics-ch-5.html
June 21 to 30
6. Friction
http://iit-jee-physics.blogspot.com/2007/10/study-guide-h-c-verma-jee-physics-ch-6.html
Saturday, May 30, 2009
Tuesday, May 26, 2009
IIT JEE 2009 Toppers - Planned Schedule is the Key
According to Times India Report dated 26.5.2009, Nitin Jain, the topper said planned schedule is the key to do well in any competitive examination. Listening to tutors and following it up diligently is the trick.
Friday, May 22, 2009
IIT JEE 2011 Physics Study Diary - Ch.3 Rest and Motion - Day 5
Rest and Motion
Day 5 May 25 2009 study plan
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Points to Note
The main theme of the section is expressing velocity w.r.t. one Frame into velocity w.r.t. to a different frame
If XOY is one frame called S and X'O'Y' is another frame called S' we can express velocity of a body B w.r.t. S as a combination of velocity of body w.r.t. to S' and velocity of S' w.r.t to S.
V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body w.r.t to S
V(B,S') = velocity of body w.r.t to S'
V(S',S) = velocity of S' w.r.t to S
we can rewrite above equation as
V(B,S') = V(B,S)- V(S',S)
We can interpret the above equation in terms of two bodies. Assume S', and B are two bodies. If we know velocities of two bodies with respect to a common frame (in this case S)we can find the velocity of one body with respect to the other body (V(B,S')
The above expressions for velocity were derived from the relation between position vectors of the body w.r.t. to S and S' and position vector of origin of S' with respect to origin of S.
r(B,S) = r(B,S')+r(S',S)
Differentiating the position vectors with respect to gives respective velocity
Formulas covered in the session
26. r(B,S) = r(B,S')+r(S',S)
Where
r(B,S) = Position vector
r(B,S') = Position vector
r(S',S) = Position vector
27. V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body wrt to S)
V(B,S') = velocity of body wrt to S')
V(S',S) = velocity of S' wrt to S)
we can rewrite above equation as
28. V(B,S') = V(B,S)- V(S',S)
Day 5 May 25 2009 study plan
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Points to Note
The main theme of the section is expressing velocity w.r.t. one Frame into velocity w.r.t. to a different frame
If XOY is one frame called S and X'O'Y' is another frame called S' we can express velocity of a body B w.r.t. S as a combination of velocity of body w.r.t. to S' and velocity of S' w.r.t to S.
V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body w.r.t to S
V(B,S') = velocity of body w.r.t to S'
V(S',S) = velocity of S' w.r.t to S
we can rewrite above equation as
V(B,S') = V(B,S)- V(S',S)
We can interpret the above equation in terms of two bodies. Assume S', and B are two bodies. If we know velocities of two bodies with respect to a common frame (in this case S)we can find the velocity of one body with respect to the other body (V(B,S')
The above expressions for velocity were derived from the relation between position vectors of the body w.r.t. to S and S' and position vector of origin of S' with respect to origin of S.
r(B,S) = r(B,S')+r(S',S)
Differentiating the position vectors with respect to gives respective velocity
Formulas covered in the session
26. r(B,S) = r(B,S')+r(S',S)
Where
r(B,S) = Position vector
r(B,S') = Position vector
r(S',S) = Position vector
27. V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body wrt to S)
V(B,S') = velocity of body wrt to S')
V(S',S) = velocity of S' wrt to S)
we can rewrite above equation as
28. V(B,S') = V(B,S)- V(S',S)
An Intersting Quote from Mark Twain
Yesterday I saw an interesting quote from Mark Twain
I do not allow my schooling to interfere with my education.
This may apply to JEE coaching. If you feel, you are not getting adequate inputs, you have to put in your efforts. Finally, you want a seat in the best institute not they.
So do not allow your coaching institute to spoil your education. Depend on them to a large extent, but listen to your judgement regarding your time table.
I do not allow my schooling to interfere with my education.
This may apply to JEE coaching. If you feel, you are not getting adequate inputs, you have to put in your efforts. Finally, you want a seat in the best institute not they.
So do not allow your coaching institute to spoil your education. Depend on them to a large extent, but listen to your judgement regarding your time table.
IIT JEE Physics Study Diary - Rest and Motion - Day 4
Day 4 - Study Plan
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Points to Note
Motion in a plane
Motion in plane is described by x coordinate and y coordinate, if we choose X-Y plane. You can imagine time t is on the third axis.
The position of the particle or the body can be described by x and y coordinates.
r = xi = yj
Displacement during time period t to t+Δt can be represented by Δr
Δr = Δxi = Δyj
Then Δr/Δt = (Δx/Δt)i = (Δy/Δt)j
Taking the limits as Δt tends to zero
v = dr/dt = (dx/dt)i+(dy/dt)j ... (3.15)
Hence x component of velocity is dx/dt
The x-coordinate, the x component of velocity, and the x component of acceleration are related by equations of straight line motion along X axis.
Similarly y components.
Projectile
Projectile motion is an important example of motion in a plane.
What is a projectile? When a particle is thrown obliquely near the earth's surface, it is called a projectile. It moves along a curved path. If we assume the particle is close to the earth and negligible air resistance to the motion of the particle, the acceleration of the particle will be constant. We solve projectile problems with the assumption that acceleration is constant.
Vertical motion of the projectile is the motion along Y axis and horizontal motion is motion along X axis.
Terms used in describing projectile motion
Point of projection
Angle of projection
Horizontal range
Time of flight
Maximum height reached
The motion of projectile can be discussed separately for the horizontal and vertical parts.
The origin is taken as the point of projection.
The instant the particle is projected is taken as t = 0.
X-Y plane is the plane of motion.
The horizontal line OX is taken as the X axis.
Vertical line OY is the Y axis.
Vertically upward direction is taken as positive direction of Y
Initial velocity of the particle = u
Angle between the velocity and horizontal axis = θ
ux – x-component of velocity = u cos θ
ax – x component of acceleration = 0
uy – y component of velocity = u sin θ
ay = y component of acceleration = -g
Horizontal motion of the projectile – Equations of motion
ux = u cos θ
ax = 0
vx = ux +axt = ux = u cos θ (as ax = 0)
Hence x component of the velocity remains constant.
Displacement in horizontal direction = x = uxt+1/2ax t²
As ax = 0, x = ux t = ut cos θ
Vertical motion – Equations of motion
uy = u sin θ
ay = -g
vy = uy – gt
Displacement in y direction = y = uyt – ½ gt²
vy² = uy² - 2gy
22. Time of flight of the projectile = (2u sin θ)/g
23. OB = (u²sin 2θ)/g
24. t = (u sin θ)/g
At t vertical component of velocity is zero.
25. Maximum height reached by the projectile = (u² sin²θ)/2g
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Points to Note
Motion in a plane
Motion in plane is described by x coordinate and y coordinate, if we choose X-Y plane. You can imagine time t is on the third axis.
The position of the particle or the body can be described by x and y coordinates.
r = xi = yj
Displacement during time period t to t+Δt can be represented by Δr
Δr = Δxi = Δyj
Then Δr/Δt = (Δx/Δt)i = (Δy/Δt)j
Taking the limits as Δt tends to zero
v = dr/dt = (dx/dt)i+(dy/dt)j ... (3.15)
Hence x component of velocity is dx/dt
The x-coordinate, the x component of velocity, and the x component of acceleration are related by equations of straight line motion along X axis.
Similarly y components.
Projectile
Projectile motion is an important example of motion in a plane.
What is a projectile? When a particle is thrown obliquely near the earth's surface, it is called a projectile. It moves along a curved path. If we assume the particle is close to the earth and negligible air resistance to the motion of the particle, the acceleration of the particle will be constant. We solve projectile problems with the assumption that acceleration is constant.
Vertical motion of the projectile is the motion along Y axis and horizontal motion is motion along X axis.
Terms used in describing projectile motion
Point of projection
Angle of projection
Horizontal range
Time of flight
Maximum height reached
The motion of projectile can be discussed separately for the horizontal and vertical parts.
The origin is taken as the point of projection.
The instant the particle is projected is taken as t = 0.
X-Y plane is the plane of motion.
The horizontal line OX is taken as the X axis.
Vertical line OY is the Y axis.
Vertically upward direction is taken as positive direction of Y
Initial velocity of the particle = u
Angle between the velocity and horizontal axis = θ
ux – x-component of velocity = u cos θ
ax – x component of acceleration = 0
uy – y component of velocity = u sin θ
ay = y component of acceleration = -g
Horizontal motion of the projectile – Equations of motion
ux = u cos θ
ax = 0
vx = ux +axt = ux = u cos θ (as ax = 0)
Hence x component of the velocity remains constant.
Displacement in horizontal direction = x = uxt+1/2ax t²
As ax = 0, x = ux t = ut cos θ
Vertical motion – Equations of motion
uy = u sin θ
ay = -g
vy = uy – gt
Displacement in y direction = y = uyt – ½ gt²
vy² = uy² - 2gy
22. Time of flight of the projectile = (2u sin θ)/g
23. OB = (u²sin 2θ)/g
24. t = (u sin θ)/g
At t vertical component of velocity is zero.
25. Maximum height reached by the projectile = (u² sin²θ)/2g
Wednesday, May 20, 2009
IIT JEE 2011 Physics Study Diary - Ch.3.Rest and Motion - Session 3
Day 3 Study Plan - To be studied on 23rd May 2009
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3,4,5,6,
Points to Note
Motion in a straight line
As the motion is constrained to move on a straight line, choose the straight line in which motion is taking place as X-axis. Hence x represent the position of the particle at any time instant t. If you want you can imagine a graph between t and x but now t in on the vertical axis and x is on the horizontal axis.
Generally origin is taken at the point where the particle is situated at time t = 0.
Position of the particle at time t is given by x and also x measures displacement (not distance).
Velocity is v = dx/dt (3.9)
acceleration is a = dv/dt (3.10)
a = d²x/dt² (3.12)
Decelaration
If acceleration is negative, then it is along the negative X-axis. It is called deceleration
Motion with constant acceleration
Using integration the formulas for v velocity at any instant, x position at any instant and relation between v,u,x and a are derived in this section.
If acceleration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)
Then v = u+at (3.12)
x = distance moved in time t = ut+½at² (3.13)
Also v² = u²+2ax (3.14)
u,v, and a as well as may take negative or positive values. When u, v and a are negative it shows velocity or acceleration is in the negative X direction.
Example 3.5
a) The question asked is distance travelled. The expression for x gives only displacement. But the remark is that as the particle does not turn back it is equal to distance travelled. Be careful when initial velocity is positive and the acceleration is negative.
Example 3.6
There was a past JEE question which is based on the variable defined in the example.
Freely falling bodies
In this case take the Y axis as the straight line on which the particle or body is moving.
You can take height above the ground as +y and work out the problems.
You can take the starting position of the body as the origin and work out the problem.
The choice may be yours or some choice may be more appropriate in case of some problems, be clear of the formula that you have to use depending on the choice you made.
g is approximately equal to 9.8 m/s², but for convenience in many problems it is given as 10m/s².
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3,4,5,6,
Points to Note
Motion in a straight line
As the motion is constrained to move on a straight line, choose the straight line in which motion is taking place as X-axis. Hence x represent the position of the particle at any time instant t. If you want you can imagine a graph between t and x but now t in on the vertical axis and x is on the horizontal axis.
Generally origin is taken at the point where the particle is situated at time t = 0.
Position of the particle at time t is given by x and also x measures displacement (not distance).
Velocity is v = dx/dt (3.9)
acceleration is a = dv/dt (3.10)
a = d²x/dt² (3.12)
Decelaration
If acceleration is negative, then it is along the negative X-axis. It is called deceleration
Motion with constant acceleration
Using integration the formulas for v velocity at any instant, x position at any instant and relation between v,u,x and a are derived in this section.
If acceleration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)
Then v = u+at (3.12)
x = distance moved in time t = ut+½at² (3.13)
Also v² = u²+2ax (3.14)
u,v, and a as well as may take negative or positive values. When u, v and a are negative it shows velocity or acceleration is in the negative X direction.
Example 3.5
a) The question asked is distance travelled. The expression for x gives only displacement. But the remark is that as the particle does not turn back it is equal to distance travelled. Be careful when initial velocity is positive and the acceleration is negative.
Example 3.6
There was a past JEE question which is based on the variable defined in the example.
Freely falling bodies
In this case take the Y axis as the straight line on which the particle or body is moving.
You can take height above the ground as +y and work out the problems.
You can take the starting position of the body as the origin and work out the problem.
The choice may be yours or some choice may be more appropriate in case of some problems, be clear of the formula that you have to use depending on the choice you made.
g is approximately equal to 9.8 m/s², but for convenience in many problems it is given as 10m/s².
Tuesday, May 19, 2009
IIT JEE 2011 Physics Study Diary - Ch.3 Rest and Motion - Day 2
Plan for Day 2
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
WOE 3 to 4
Exercises: 1 to 5
Points to Note
Average velocity
Average speed and average velocity of a body over a specified time interval may not turnout to be same.
Example See the worked out example 2 of HC Verma's book.
The teacher made 10 rounds back and forth in the room and the total distance moved is 800 feet (10 rounds back and forth of 40 ft room). As the time taken is 50 minutes, average speed is 800/50 = 16ft/min.
But because he went out of the same door that he has entered, displacement is zero and hence average velocity is zero.
Instantaneous velocity
Average acceleration
Instantaneous acceleration
Position Vector: If we join the origin to the position of a particle by a straight line and put an arrow towards the position of the particle, we get the position vector of the particle.
If the particle moves from position A to position B, we can define position vector of A and position vector of B and OB - OA will give displacement ( a vector quantity).
Another point to note: slope of velocity-time diagram gives the instant acceleration at that point.
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
WOE 3 to 4
Exercises: 1 to 5
Points to Note
Average velocity
Average speed and average velocity of a body over a specified time interval may not turnout to be same.
Example See the worked out example 2 of HC Verma's book.
The teacher made 10 rounds back and forth in the room and the total distance moved is 800 feet (10 rounds back and forth of 40 ft room). As the time taken is 50 minutes, average speed is 800/50 = 16ft/min.
But because he went out of the same door that he has entered, displacement is zero and hence average velocity is zero.
Instantaneous velocity
Average acceleration
Instantaneous acceleration
Position Vector: If we join the origin to the position of a particle by a straight line and put an arrow towards the position of the particle, we get the position vector of the particle.
If the particle moves from position A to position B, we can define position vector of A and position vector of B and OB - OA will give displacement ( a vector quantity).
Another point to note: slope of velocity-time diagram gives the instant acceleration at that point.
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