Vector Differentiation 1

Here in this post we will revise our concept of Vector Calculas (differentiation of vectors). This mathematical tool would help us in expressing certain basic ideas with a great convenience while studying electrodynamics.

Consider a vector function f(u) such that
f(u) = fx(u)i + fy(u)j + fz(u)k
where fx, fy(u) and fz(u) are scalar functions of u and are components of vector f(u) along x, y, and z directions.If we want to find the derivative of f(u) with respect to u we will have to proceed in the similar manner we used to do with ordinary derivatives thus

where df(u)/du is also a vecor.
Thus in cartesian coordinated derivative of vector f(u) is given by

When we talk about fields then in this case a particular scalar or vector quantity is defined not just at a point in the space but it is defined continously throughout some region in space or maybe the entire region in the space. Now a scalar field φ(x,y,z) assocites a scalar with each point in the region of space under consideration and a vector field f(x,y,z) associates a vector with each point.
In electrodynamics we will come across the cases where variation in scalar and vector fields from one point to is continous and is also differentiable in the particular region of space under consideration.

Consider a scalar field φ(x,y,z). This function depends on three variables. Now how would we find the derivative of such functions. If we infinitesimal change dx, dy and dz along x, y, and z axis simultaneously then total differential Dφ of function φ(x,y,z) is given as

above expression comes from our previous knowledge of partial differentiation.
If we closely examine above equation this could be a result of dot product of two vectors thus,


is gradient of φ(x,y,z) and gradient of a scalar function is a vector quantity as it is the multiplication of a vector by a scalar.
Thus we see that gradient of any scalar field has both magnitude and direction. Again consider the function φ(x,y,z) then from ordinary calculas any change in this function as discussed above is given by

dφ=(φ)•(dr) = |φ| |dr|cosθ
FRom this we see that dφ(x,y,z) will be maximum when cosθ=1 which would be the case when dr would be parallel to φ. Thus function φ changes maximally when one moves in the direction same as that of gradient. So we can say that the direction of φ is along the greatest increase of φ and the mahnitude of |φ| gives the slope along that direction.
CONCLUSION: The gradient φ points in the direction of the maximum increase of function φ(x,y,z) and the magnitude |φ| gives the slope or rate of increase along the maximal direction.

While discussing gradient of a scalar function we find that gradient of any function is given by


where the term in parentheses is called "del"

Del is an vector derivative or vector operator and this operator acts on everything to its right in an expression, until the end of the expression or a closing bracket is reached. There are three ways in which can act or operate on a scalar or vector function
1. On a scalar function φ : φ (the gradient);
2. On a vector function f, via the dot product: • v (the divergence);
3. On a vector function f, via the cross product: x v (the curl).
Out of these three ways of operation of on any function we have already discussed gradient of a scalar function.

In this post we learned about scalar and vector fields, gradient of scalar fields and operator. In the next post we'll lern more about vector differential calculas i.e, in particular we'll discuss divergance and curl of vector fields.
Yes, your comments and querries are heartly invited please do give us a reaponce and comment on our work so that we could further improve it and your responce is more then enough to encourage us.

Vector Algebra 2

In this post we'll lern Vector algebra in component form.
Component of any vector is the projection of that vector along the three coordinate axis X, Y, Z.

In component form addition of two vectors is
C = (Ax+ Bx)i + (Ay+ By)j + (Ay+ By)k
A = (Ax, Ay, Az) and B = (Bx, By, Bz)
Thus in component form resultant vector C becomes,
Cx = Ax+ Bx
Cy = Ay+ By
Cz = Az+ Bz

In component form subtraction of two vectors is
D = (Ax- Bx)i + (Ay- By)j + (Ay- By)k
A = (Ax, Ay, Az) and B = (Bx, By, Bz)
Thus in component form resultant vector D becomes,
Dx = Ax - Bx
Dy = Ay- By
Dz = Az- Bz

NOTE:- Two vectors add or subtract like components.

A.B = (Axi + Ayj + Azk) . (Bxi + Byj + Bzk)
= AxBx + AyBy + AzBz.
Thus for calculating the dot product of two vectors, first multiply like components, and then add.


A x B = (Axi + Ayj + Azk) x (Bxi + Byj + Bzk)
= (AyBz - AzBy)i + (AzBx - AxBz)j + ( AxBy - AyBx)k.

Cross product of two vectors is itself a vector.
To calculate the cross product, form the determinantwhose first row is x, y, z, whose second row is A (in component form), and whose third row is B.


Vector product of two vectors can be made to undergo dot or cross product with any third vector.

(a) Scalar tripple product:-
For three vectors A, B, and C, their scalar triple product is defined as
A . (B x C) = B . (C x A) = C . (A x B)
obtained in cyclic permutation. If A = (Ax, Ay, Az) , B = (Bx, By, Bz) , and C = (Cx, Cy, Cz) then A . (B x C) is the volume of a parallelepiped having A, B, and C as edges and can easily obtained by finding the determinant of the 3 x 3 matrix formed by A, B, and C.

(b) Vector Triple Product:-
For vectors A, B, and C, we define the vector tiple product as
A x (B x C) = B(A . C) - C(A - B)
Note that
(A . B)C ≠ A(B . C)
(A . B)C = C(A . B).

Electric resistance and Resistivity

(A) Resistivity

  • In the previous post we derived that current density is
    j = nqvd
    where vd is the drift velocity.

  • Current density in general depend on electric field and for metals current density is nearly proportional to the electric field. (Results can be derived using theory of metallic conduction.)

  • Thus for metals ratio of E and j is constant and for a particular material its resistivity ρ is defined as the ratio of magnitude of electric field to current density,
    ρ = E/j
    This relationship is known as Ohm's law discovered by german physicist Georg Simon Ohm (1787-1854) in 1826.

  • Greater would be the resistivity of a given material greater field would be required to establish a given current density in the material or we can say that smaller would be the current density for a given field.

  • Unit of resistivity is Ωm (ohm. meter).

  • Materials having zero resistivity are known as perfect conductors and those having infinite resistivities are known as perfect insulators. Real materials lie between these two limits.

  • Metals and alloys are materials having lowest resistivities and are good conductors of electricity.

  • Insulators have resistivities many times (of the order of 1022) greater then that of metals.

  • Reciprocal of resistivity is conductivity. Unit of conductivity is (Ωm)-1.

  • Metals or good conductors of electricity have conductivity greater than that of insulators.

  • Semiconductors are those materials which have resistivities intermediate between those of metals and insulators.

    (B) Resistivity and temperature

  • Resistivity of a conductor depends on a number of factors and temperature of the metal is one such factor. As the temperature of the conductor is increased its resistivity also increases.

  • For small variations in temperature resistivity of materials is given by the relation
    ρ(T) = ρ(T0)[ 1 +α(T-T0)]
    where, ρ(T) and ρ(T0) are resistivities at temperature T and T0 respectively and α is constant for a given material which also depends on temperature to a small extent. This constant α is known as temperature coefficent of resistivity.

    (C) Resistance

  • We already know that for a conducror relation between electric field E and current density is given as
    E = ρj
    where ρ is a constant independent of E.

  • When we study electric circuits we are more interested in the total current in a conductor rather then current density j and more interested in knowing the potential difference between the ends of the conductor than in Electric field becaude current and potential difference are easier to measure then j and E.

  • Consider a conducting wire of length l and uniform crossectional area A. If V is the potential difference between both the ends of the wire then electric field inside the conductor would be
    E = V/l
    If i is the current flowing inside the wire then current density is given by
    j = i/A
    putting these values in Ohm's law ρ = E/j we get
    V = ρi (l/A)
    or , V=Ri
    where, R=ρ(l/A)
    which is known as resistance of a given conductor.

  • Unit of resistance is ohm or volt per ampere.

  • Thus how much current will flow in a wire not only depends on the potential difference between two ends of the wire but also on the resistance offered by the conductor to the flow of electric charge.

  • From the above discussion we can easily conclude that The resistance of a wire depends both on the thickness and length of the wire and also on its resistivity.

  • Thick wires have less resistance then thin ones and longer wires have more resistance then shorter ones.

  • Since the resistivity of a marerial varies with temperature, the resistance of any particular conductor also varies with temperature. For temperature ranges that are not too great, this variation is approximately a linear relationship, analogous to the one we learned for resistivity
    R(T) = R(T0)[1 + α(T - T0)]
    In this equation. R (T) is the resistance at temperature T and R(T0) is the resistance at temperature T0. The temperature coefficient of resistance α is the same constant that appears in case of resistivity.

    In the next post we'll do some worked examples related to this topic
  • Electric Current and Current Density

    - Electric current is the motion of electrically charged particles from one region to another. This motion of electric charges takes place within electric circuit which is a conducting path forming a closed loop.
    - In a conductor electric charge will flow from its one end to another if and only if both the ends of the conductor are at different electric potentials.
    - Continous flow of electric current in a conductor for a relatively long period of time can be attained using battries which could supply continous flow of charge at low potentials.

    Electric current
    - Here in this section we would discuss about the electric current in conductors.
    - When electric field inside the conductor is zero then there would be no net flow of current in the conductor because in this case electrons in the conductors moves about randomly leaving no net flow of charge in any direction and without the flow of charge there would be no net electric current.
    -To maintain a constant current in a conductor we would have to ensure that a constant and steady electric field is established inside the conductor in order to maintain a force on the mobile charges in the conductor.
    -Once the electric field is maintained inside the conductor charged particles in the conductors are now under the influence of driving force F = qE.
    - In an conductor charged particles undergoes frequent inelastic collision with fixed massive ions in the conductor and undergoes random change in the direction of motion.
    - Hence on an average charged particle moves in the direction of driving force with an average velocity known ad drift velocity.
    - Electric current is defined as the quantity of charge ΔQ flowing through cross-sectional area A in time interval Δt . Thus,
    Iav = ΔQ/Δt
    which is the average current flow in time Δt.
    - If dQ is the amount of charge flowing in infinitesimally time interval dt through a cross-sectional area of the conductor then instantaneous current I is defined as
    I = dQ/dt
    - Electric current is a scalar quantity.
    - SI unit of current is Ampere (A) where 1A is one coulomb per second.

    Drift velocity and Current density - Consider a portion of a conductor of cross-sectional area A. Also consider a small section of conductor of length Δx. Now volume of conductor under consideration is AΔx.
    - If there are n number of mobile charge carriers per unit volume then total charge in the section under consideration is
    ΔQ = (number of charge carriers in the section) x (charge per carrier)
    = (nAΔx)q .................1
    where q is the amout of charge on each carrier.
    - If charge carriers are moving with speed vd, then distance travelled by the charge carriers is Δx = vdΔt. Putting this in equation 1 we have,
    ΔQ = (nAvdΔt)q
    - Now current in the conductor is
    I = ΔQ/Δt
    I = nqvdA ....................2
    where vd is average velocity known as drift velocity as defined earlier.
    - Current density j is defined as current per unit cross-sectional area. Thus from 2
    j = I/A = nqvd
    Current is a scalar quantity but current density can also be defined too include both magnitude and direction. Thus vector current density is
    j = nqvd
    Direction of current density is same as the direction of electric field.
    - Unit of current density is A.m-2.
    - Current density tells us about how charges flow at a certain point and also about the direction of the flow at that point but current describes how charges flow throughout an extended object.

    Solved example
    In this solved example we will lern to apply the principle of current to the problems.

    Question:-Amount of charge that passes through a certain conducting wire in 4 sec is 6.5 C. Find (a) the current in the wire and (b) number of electrons that passes through the wire in 8 sec.
    Solution : -
    Problem solving strategy
    1. From the lesson learned about electric current identify the equation for calculating current when charge and time are given.
    2. Put the values of charge and time at respective place and calculate the answer.

    In this case Q = 6.5 C and t = 4 sec.
    now I = q/t = 6.5C/4 sec = 1.625 A
    which is the required answer.

    Problem solving strategy
    1. Here we have to find total charge through the wire in a given time.
    2. Total charge through the wire would be equal to number of electrons through the circuit multiplied by the charge on each electron.

    Here we have to find the number of electrons passing through the wire in 8 sec. If qe is the amount of charge on a single electron and total n number of electrons passes through the wire then
    I = nqe/t
    or, N = It/qe
    putting values of I , t and qe = 1.6 x 10-19 and calculating we get
    N = 8.125 x 1019

    Same way problems related to current density and drift velocity can be solved.

    Vector Algebra 1

     Here in this post we will go through a quick recap of vector algebra keeping in mind that reader already had detail knowledge and problem solving skills related to the topic being discussed. Here we are briefing Vector Algebra because concepts of electrostatics , electromagnetism and many more physical phenomenon can best be conveniently expressed using this tool.

    A vector is a quantity that requires both a magnitude (= 0) and a direction in space it can be represented by an arrow in space for example electrostatic force, electrostatic field etc. In symbolic form we will represent vectors by bold letters. In component form vector A is written as
    A = Axi+ Ayj+Azk

    Two vectors A and B can be added together to give another resultant vector C.
    C = A + B

    Two vectors A and B can be subtracted to give another resultant vector D.
    D = A - B = A + (-B)

    When we multiply any vector A with any scalar quantity 'n' then it's direction remains unchanged and magnitude gets multiplied by 'n'. Thus,
    n(A) = nA
    Scalar multiplication of vectors is distributive i.e.,
    n(A + B) = nA +nB

    Dot product of two vectors A and B is defined as the product of the magnitudes of vectors A and B and the cosine of the angle between them when both te vectors are placed tail to tail. Dot product is represented as A.B thus,
    A.B = |A| |B| cosθ
    where θ is the angle between two vectors.
    Result of dot product of two vectors is a scalar quantity.
    Dot product is commutative : A.B = B.A
    Dot product is distributive : A . (B+C) = A.B + A.C also A.A = |A|2

    Cross product or vector product of two vectors A and B is defined as
    A x B = |A| |B| sinθ nˆ
    where nˆ is the unit vector pointing in the direction perpandicular to the plane of both A and B. Result of vector product is also a vector quantity.
    Cross product is distributive i.e., A x (B + C) = (A x B) + (A x C) but not commutative and the cross product of two parallel vectors is zero.

    In our next post we'll study vector algebra in component form and also lern about vector triple products.

    Overview of electrostatics and electricity

    Electrostatics involves electric charges namely positive and negative charges, the forces between them which is known as electric force , the field that surrounds them, and their behavior in materials. Coulumb's law is the simple relation that governs electrostatic interactions and the field around the charges is described using the concept of electric field. Coulumb's law is a inverse square law which gives the force between two charges kept at some distance (say r ) apart from each other. Like Coulumb's law, law of gravitation is also a inverse square law but gravitational interactions are only attractive in nature and electrical interactions are attractve as well as repulsive depending on the nature if interacting charges. Charges of same kind repel each other and charges different kinds , i.e. one charge positive and other negative , attract each other. One more thing electric interactions are much more stronger then gravitational interactions and gravitational force are almost negligible in comparison to the forces of electric origin. This is always true when we study the interactions of atomic and subatomic particles. But when we study objects very large in size say a person ,a planet or satellites, the net governing force in this case is gravitational force not electric.
    Now coming to the properties of electric charges we know that electric charge is quantized and it also obeys law of conservation means total charge remains conserved. In what we say electrostatic interactions electric charges are at rest in our frame of refrence. Now think what happens when we are at rest in our frame of reference and charge under consideration is moving with velocity v with respect to us such a miving charge leads to the origin of magnetism which we will discuss in later section.
    Again a question what is electricity. Electricity, deals with stationary and moving electric charges, the actions of force between these charges, and the electric and magnetic fields generated by them. Electrostatics is simply the electricity at rest. Electricity is the backbone of the modern society in which we use various instruments which depends on electric current for their functioning and without it we would not have telephones, television, household appliances and many more gadgets which are now part of our daily life. In electricity we study the motion of electric charges, or electric currents and also the voltages that produce currents and the ways to control currents.
    We have learned about what electrostatics is and what we study in it. Now we will discuss why we study electrostatics and where it finds its applications.Electric interactions are of immence importance in chemistry and biology and have many technological applications. Concepts of electricity proved to be of basic importance for studying atomic physics, nuclear physics and solid state physics. It also find importance in studying advanced level physic.

    How to prepare for Physics for IITJEE

    IITJEE is one of the toughest examination in India.SO it requires lots of systematic efforts to pass the examination.
    Here i would to try to explain what should be the strategy and plan for physics for IITJEE

    • First things is the right selection of books for the IITJEE as good books play an important role in clearing the concept in Physics.
      You can check at the following links for the list of books for the IITJEE.

      IITJEE Physics Books

    • Second things is to plan your syllabus .Decide about the time for the each units.I will be soon giving u the IITJEE physics planner for theb whole year which will greatly help you in planning.
    • The next steps to stick to the schedule and start the study.First of all clear your all concepts.Read the chapter many times so that concept becomes crystal clear and you can connect the concept with our daily life.For example.Lets take Projectile Motion.A baseball in motion describes an Projectile motion.You can think of baseball example in learning the concepts.It will be good if you can prepare some notes of the chapter for fast revision.
      Check out your concept by attempting the conceptual questions.You can lots of comceptual test on the my blog.

      You can get good study Material at the following link
      IITJEE Study Material

    • Once you have cleared the concept,you are ready take the questions.First study some examples to get the feel of the concept.While going through examples,you can see and analyze the way a particulr physical concept is applied towards solving the problem. This will help you a great deal while you begin to solve problems.Try to find the best possible way to tackle the problem.Lets take the example for Projectile motion

      1.Gather all the information from the question
      2.Select a coordinate system and resolve the initial velocity vector into x and y components.
      3.Follow the techniques for solving constant-velocity problems to analyze the horizontal motion. Follow the techniques for solving constant-acceleration
      problems to analyze the vertical motion. The x and y motions share the same time of flight t.

      You can find lots of such ways and hint from the following link
      IITJEE Tips
    • Revise your material and practice questions on a systematic basis.
    • Attempts the question in previous year IITJEE questions papers.This will give you a feel of the IITJEE questions.You can find the papars and solutions in the below link
      IITJEE Previous Year Papers

    • Take Test on regular basis.It give you the feel of the examination.Give you full attempt to the test.Take online Test on the below links.
      IITJEE Test Series
    • Take some simulated IITJEE test series (not online) by the some renowned institues.This will help you in preparing for the Main Examination

    Best of luck
    Physics Expert

    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.!!!!!!!!!!!!!!!!!!!!!
    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

    TO address:
    Subject:Physics Champion Contest-III


    About Yourself
    Father's Name:



    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
    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
    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.

    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

    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

    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

    Now since it is uniform motion we can use given motion formula's in use

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

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

    2. velocity(v)=20 m/s
    So here the most suitable equation is
    or t=10 sec

    So here the most suitable equation is
    or t=40 sec

    Example -2
    A object is moving along an straight line.The motion of that object is described by
    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

    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
    Now since its motion is described by the given equation,following formula will be useful in determining the values

    1. x=? t=1sec

    Here by substituting t=1 in given equation we get the answer
    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
    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
    or a=d(a+2bt+3ct2)/dt
    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

    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

    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,

    Moment of inertia of disc Y,
    From equation (i) and (ii)

    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


    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
    T=3.14 sec

    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

    If a and b are time of ascent and descent respectively then time of flight

    Now T=2u/g

    So u=g(a+b)/2

    IITJEE Solved problem

    IITJEE solved problems are two among those problems send to us by visitors of our blog as their querries. You can also send us your querries to us to get them solved.

    Problem 1. Two masses m1 and m2 are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance x0 when the system is released from rest. Find the distance moved by the messes before they again come to rest.


    Consider blocks plus spring a system and if no external forces acts on the sysyem then centre of mass of system will remain at rest. Mean position of two SHM's would be the unstretched position.If m1 moves towards right through a distance x1 and m2 moves towards left through a distance x2before spring acquires natural length then
    x1+ x2= x0 .....................(1)
    where x1 and x2 would be the amplitudes of blocks m1 and m2 resp. Since centre of mass of system will remain same so,
    thus from 1
    x1=m2x0/(m1 + m2)
    x2=m1x0/(m1 + m2)
    to get back to rest position masses m1 and m2 would have to travel distances x1 and x2 resp. , so total distance travelled by m1 before comming to rest is
    2m2x0/(m1 + m2)
    and total distance travelled by m2 before comming to rest is,
    2m1x0/(m1 + m2)

    Problem 2. Suppose the surface charge density over a sphere of radius R depends on a polar angle θ as σ=σ0cosθ, where σ0 is a positive constant. Find the electric field strength vector inside the given sphere.
    You can visualize such a charge distribution as a result of a small relative shift of two uniformly charged balls of radius R, whose charges are equal in magnitude but opposite in sign. In such a representation inner charges cancel out each other and only charges on the surface survives having charge density σ=σ0cosθ. Charge density is maximum along +Z and -Z where the distance between the surfaces is maximum.

    Electric field due to positively charged sphere E+r+/3ε0
    Electric field due to positively charged sphere E-=-ρr-/3ε0
    Electric field at P
    E = E++E- =ρ(r+-r-)/3ε0
    ρa = charge per unit area =σ0

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    IITJEE Problems: Modern Physics

    Q 1. What is the Kinetic energy of the photoelectrons ejected from a tungsten surface by UV light of wavelength 1940Å ? Express the result in eV. The photoelectric threshold of tungsten is 2300Å .
    Ans. K=1 eV
    Q 2. The work function of sodium is 2 eV.
    (a) Find the maximum energy of the photoelectrons ejected when a sodium surface is illuminated with light of 3105Å.
    (b) Calculate the maximum wavelength of the light that would produce photoelectric effect.
    Ans. 2 eV , 6210Å
    Q 3. A small plate of metal (work function 1.17 eV) is placed at a distance 2 m from monochromatic light source of wavelength 4.8x10-7and power 1.0 watt. The light falls normally on plate. Find the number of photons striking the metal plate per square meter per second. If a constant uniform magnetic field of strength 10-4 tesla is applied parallel to the metal surface , find the radius of the largest circular path followed by the emitted photoelectrons.
    Ans. 4.82 x1016 , 4.0 cm
    Q 4. Calculate the wavelength of the wave associated with :
    (a) 1 MeV photon
    (b) 1 MeV proton
    Ans. 1.242x10-10 cm, 2.84x10-12 cm
    Q 5. The molecules of a certain gas at 320 K have a rms speed of .499 Km/sec. Calculate the De Brogli wavelength of these molecules. What is the gas?
    Ans. 0.25Å , O16

    IITJEE Problems : Optics

    Q 1. A concave mirror of radius of curvature 1m is placed at bottom of a tank of water. The mirror forms an image of the sun , when it is directly overhead. Calculate the distance of the images from the mirror for different depths, 80 cm and 40 cm of the water in the tank. (ans=50 cm , 47.5 cm)
    Q 2. A pin is placed 10 cm in front of convex lens of focal length 20 cm made of material of refrective index 1.5. The surface of lens farther away from the pin is silvered and has a radius of curvature 22 cm. Determine the position of the final image. Is the image real or virtual.(Ans -11 cm)
    Q 3. An object is placed 50 cm from the surface of a glass sphere of radius 12 cm along the diameter . Where will the final image be formed by reflection on both the surfaces? Refractive index of glass is 1.60 (Ans +9.5 cm)
    Q 4. A bi-convex lens of radii of curvature 30.0 cm each is placed on the surface of water so that the lower surface is only immersed . An illuminated object 40 cm deep is observed vertically through the lens and water. Find the position of the image. Refractive index of air-glass=3/2 and of glass-air=4/3 . (ans -90 cm)
    Q 5. Interference pattern is obtained with two coherent light sources of intensity ratio n. Show that in the interference pattern

    IITJEE/AIEEE 2009 Information

    IITJEE 2009 Information

    1.The IIT JEE for 2009 would be held for 12th April 2009

    2. The JEE examination would be objective type and would contain 2 papers of 3 hours each Paper-1 would be held from 9 AM to 12 noon and the Paper-2 would be held from 2 PM to 5 PM.Each paper would have separate sections for Physics,Chemistry and Mathematics and would be designed to test comprehension,reasoning and analytical ability of the candidates.

    3.This year the examination would be held for 12 IIT’s,IT-BHU Varanasi and ISMU Dhandbad.

    AIIEE 2009 information

    1. The eighth all India Engineering Entrance Examination will be held on 26th April,2009 all over India and abroad.

    2 This is for admissions in B.E/B.Tech and B.Arch/B. Planning in various national level institutes like NIT’s,IIIT’s,Deemed Universities and government funded institutions and States like Haryana and Uttaranchal

    3. This examination will be held in two parts viz B.E./B.Tech from 9:30 a.m. to 12:30 p.m. and for B.Arch./B. Planning from 1400 hrs to 1700 hrs.

    4. Materials to be brought on the day of examination Admit Card and Ball Point Pen of good quality. For Aptitude Test in Architecture, the candidates are advised to bring their own Card Board, geometry box set, pencils,erasers and color pencils or crayons.
    5. Rough work All rough work is to be done in the Test Booklet only. The candidate should NOT do any rough work or put stray mark on the Answer Sheet.


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