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### Physics Champion Contest-3

Welcome to the Physics Champion Contest!!!!!!!!!!!

We present the First Physics Champion Contest

Winner will get Heat and thermodynamics package free.!!!!!!!!!!!!!!!!!!!!!
Procedure
1 This contest is for all the students studying in 10th to 12th ,students preparing Engineering and Medical examination
2.There are 5 multiple choice questions.You have pick the correct choice ,send to us in Format given below
3.You will be choosen for lucky draw only when all the question are correct.
4. Winner will be choosen from the lucky draw from All those candidate who give all write answer
5. You have to send the answer and the information to the following mail address in following format

Subject:Physics Champion Contest-III

Body:

Name:
Father's Name:

1.
2.
3.
4.
5.

Thanks

7 This contest is valid from 26 June 2009 to 26 Sep 2009.Winner will be announced on the site on 30 Sep 2009.

Multiple choice questions with one answer only
1. The potential energy (in joule) of a body of mass 2 Kg moving in the x-y plane is given by
U=6x+8y
Where the position coordinates x and y are measured in meters.If the body is at rest at point (6m,4m) at time t=0,it will cross the yaxis at time t equal to
a. 2s
b. 1s
c. 3s
4. 4s

2.Find the electric field inside the sphere which carries a charge density proportional to the distance from the origin
ρ = kr
a. ρ/ε0
b. ρr/ε0
c. ρr20
d. none of the above

3.A conductor of nonuniform curvature, the charge
a. is distributed uniformly over its volume
b. is distributed uniformly over surface
c. has the greatest concentration on the part of greatest curvature
d. has the greatest concentration on the part of least curvature

4.Two particle X and Y travel along the x and y axis respectively with velocities
v1=2i m/s
v2=3j m/s
At t=0 they are at X(-3,0) and Y(0,-3)

Find the vector which represent the position of Y relative to X as function of the time t
a.(3t-3)j+(3-2t)i
b.(2t-3)i + (3-3t)j
c. -3i-3j
d. 3i+3j

5. Six particles situated at the corner of a regular hexagon of side 1 m move at a constant speed 2 m/s.Each particle maintians a direction towards the particle at the next corner.Calculate the time the particle will take to meet each other.
a. 2
b. 1/2
c. 1
d . 3/2

### One Dimensional Motion Tour

This tour is given to give the feel of the whole chapters plus the type of questions and ways to tackle tehm.This is quite beneficial for anybody studing One dimensional Motion.

Description:
One dimensional Motion is the study of the motion along an straight line.Complete Study material has been provided at the below link
Study Material

Most Important Points to remember about One dimension Motion
1. Distance and displacement are not the same thing.Distance is scalar while displacement is a vector quantity.Distance is never zero in a round trip while the displacement will be zero.Example When a person moves in a cirle and return to it normal position,it displacement is zero while the distance tarvelled is the circumference of the circle
2. Speed is calculated over distance while velocity over displacment.So Average speed in a round trip will never be zero while average velocity will be zero
3. Magnitude of Instantanous velocity is equal to instantanous speed
4. When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal
5. Acceleration is defined as change in velocity per unit time.A body moving with constant speed but with varying direction will have accleration as the velocity is changing.

Most Effective way to solve a one dimensional motion problem

1. First visualize the question
2. Take starting point as origin as and take one direction as positive and other as negative.This is required as we will be dealing with Vector quantities
3. Write down what is given in the question and what is required
4. If it is uniform motion ,then you can utilize the One dimensional motion equation.It the motion is varying and relation is given for that motion.Then we can utilize one dimensional motion derivative equation to find out the solutions
5. Calculate as require

Formula's are
v=u+at
s=ut+(1/2)at2
v2=u2+2as
x=∫vdt
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx

Example -1
A bus start at Station A from rest with uniform acceleration 2m/sec2.Bus moves along a straight line
1.Find the distance moved by the bus in 10 sec?
2.At what time,it velocity becomes 20m/sec?
3 How much time it will take to cover a distance of 1.6km

Solution:
Now first step to attempt such question is to visualize the whole process.Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Intial velocity=0
Acceleration=2m/sec2

Now since it is uniform motion we can use given motion formula's in use
v=u+at
s=ut+(1/2)at2
v2=u2+2as

1. distance (s)=?
time(t)=10 sec

So here the most suitable equation is
s=ut+(1/2)at2
Substituting given values
s=(0)10 +(1/2)2(10)2=100 m

2. velocity(v)=20 m/s
time(t)=?
So here the most suitable equation is
v=u+at
20=0+2t
or t=10 sec

3.distance(s)=1.6km=1600m
t=?
So here the most suitable equation is
s=ut+(1/2)at2
1600=(1/2)(2)t2
or t=40 sec

Example -2
A object is moving along an straight line.The motion of that object is described by
x=at+bt2+ct3
where a,b,c are constants and x is in meters and t is in sec.
1. Find the displacement at t=1 sec
2. Find the velocity at t=0 and t=1 sec
3. Find the acceleration at t=0 and t=1 sec

Solution:
Now first step to attempt such question is to visualize the whole process.Here the object is moving along a straight line and its motion is described by the given equation
Now we
x=at+bt2+ct3
Now since its motion is described by the given equation,following formula will be useful in determining the values
x=∫vdt
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx

1. x=? t=1sec

Here by substituting t=1 in given equation we get the answer
x=a(1)+b(1)2+c(1)3
x=a+b+c m

2.v=? t=0,v=? t=1
Here we are having the displacement equation,so first we need to find out the velocity equation
So here the most suitable formula is
v=(dx/dt)
or v=d(at+bt2+ct3)/dt
or v=a+2bt+3ct2
Substituing t=0 we get
v=a m/s
Substituing t=1 we get
v=a+2b+3c m/s

3 a=? t=0,a=? t=1
Now we are having the velocity equation,we need to first find the acceleration equation.
So here the most suitable formula is
a=(dv/dt)
or a=d(a+2bt+3ct2)/dt
a=2b+6ct
Substituing t=0 we get
a=2b m/s2
Substituing t=1 we get
a=2b+6c m/s2

First check your concepts for this chapter by attempting the questions given at the link.Solutions are also provided there.
Conceptual Test

Once you are comfortable with the concept,you are ready to go for full length questions of various types.Please follow the below link one by one .Give your full try and check out the solutions.

Objective Test
Subjective Test
Graphical Questions Test

Once you are comfortable with these questions,I will suggest to revise all the concepts once again so that these concepts becomes embeded in your mind.

### Numericals for Class XII with detaileds Solutions

Question1:
A charge Q is divided in two parts q and Q - q separated by a distance R. If force between the two charges is maximum, find the relationship between q & Q.

Solution: F=K q(Q-q)/r2
for max. &min.dF/dq=0 , q=Q/2

Question2:A circular disc X of radius R is made from an iron pole of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. then the relation between the moment of inertia Ix and Iy is
(A) Iy=32Ix
(B) Iy=16Ix
(C) Iy=64Ix
(D) None of these

Solution2:
If t is the thickness and R is the radius of the disc, then mass = πR2
ρ = density of the material of the disc.
Moment of inertia of disc X,

Ix=(1/2)πR4
…(i)
Moment of inertia of disc Y,
Iy=32πR4
…(ii)
From equation (i) and (ii)
Iy=64Ix

Question3:
On a planet a freely falling body takes 2 sec when it is dropped from a height of 8 m, the time period of simple pendulum of length 1 m on that planet is

Solution3:

On a planet, if a body dropped initial velocity (u = 0) from a height h and takes time t to reach the ground
then h=(1/2)gpt2
or gp=4m/s2
Using
T=2π(L/g)1/2
T=3.14 sec

Question4:
A car is moving with uniform velocity on a rough horizontal road. Therefore, according to Newton's first law of motion
(a) No force is being applied by its engine
(b) An acceleration is being produced in the car
(c) The kinetic energy of the car is increasing
(d) force is surely being applied by its engine

Solution4: d Since, force needed to overcome frictional force

Question 5:
A particle is projected upwards. The times corresponding to height h while ascending and while descending are a and b respectively.Find the velocity of projection

Solution:
If a and b are time of ascent and descent respectively then time of flight
T=a+b

Now T=2u/g

So u=g(a+b)/2