Tuesday, November 13, 2012

JEE (Main)

Admission criteria to Undergraduate Engineering Programs at NITs, IIITs, Other Centrally Funded Technical Institutions, Institutions funded by participating State Governments, and other Institutions shall include the performance in the class 12/equivalent qualifying Examination and in the Joint Entrance Examination, JEE (Main). 
The Paper-1 (B. E./B. Tech.) of JEE (Main) will also be an eligibility test for the JEE (Advanced), which the candidate has to take if he/she is aspiring for admission to the undergraduate programmes offered by the IITs. 

The Application Forms of the first level exam, JEE Main 2014 have been released and will be filled online only on :
Click HERE>> 

http://jeemain.nic.in/jeemainapp/Welcome.aspx

Eligibility criteria for appearing in JEE (Advanced) - 2014:

The candidates have to first appear in Paper-1 of JEE(Main)-2014. Only 1,50,000 of the top scorers of Paper-1 of JEE(Main)-2014, including all categories, will be eligible to write JEE(Advanced) – 2014. Candidates who appeared in their qualifying examination (QE) (10+2 or equivalent) earlier than 2013 are not eligible.

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Sunday, November 4, 2012

Specific Heat Capacity Of Gases

Specific Heat Capacity Of Gases

Specific heat of a gas is numerically equal to the amount of heat necessary to raise the temperature of unit mass of gas by 1°C. In order to raise the temperature of unit mass of a gas through 1°C more heat will be required if the gas was kept at constant pressure than when it is at constant volume.
(i) Molar Specific heat capacity at constant Volume: The amount of heat required to raise the temperature of 1 mole of gas by 1 °C at constant volume is called the molar specific heat and it is represented by Cv.
            Cv = (ΔQ/mΔt)v = constant
By first law of thermodynamics ΔQ = ΔU + W
But W = 0 for isochoric process, then ΔQ = ΔU
by definition of specific heat ΔQ = nCvΔT
Where CV is the specific heat (for 1 mole of gas), then ΔU = nCvΔT
The relation ΔU = nCvΔT, is used to find the change in internal energy of the system and is valid for any process where a temperature change has taken place.
Consider two isotherms on the P-V diagram:
Process 1 –> 2 represents an isochoric process
Process 1 –> 3 represents an isobaric process
In both processes the temperature has changes from T1 to T2 as 2 and 3 lie on the same isotherm. Thus change in internal energ
             U1–>2 = U1–>3 = nCvΔT
(ii) Molar Specific heat capacity at constant Pressure: The amount of heat required to raise the temperature of 1 mol of gas by 1°C keeping its pressure constant, is called molar specific heat at constant pressure and it is represented by Cp,
             Cp = (ΔQ/mΔT)P  = constant
(iii) Relation between Cp and Cv: For an isobaric process by definition
From first law ΔQ = ΔU + W
=> nCPΔT = nCVΔT + W      [∴ ΔU = nCVΔT for any process]
For isobaric process
W = PΔV = nRΔT            nCPΔT = nCVΔT + nRΔT
=> CP = CV + R      or     CP ­- CV = R     (Meyer’s relation)
(iv) Relation between specific heat (CP and CV) and degrees of freedom: If f is the number of degree of freedom of a gas molecule then the internal energy of n moles of that gas is given by
U = f × 1/2 nRT            (from law of equipartition of energy)
(∴ U = f/2 kT = f/2 R/N T (for n moles) = nN × f/2 R/N T = n f/2 RT)
=> dU = f × 1/2 nRdT
Also dU = nCVdT, 
So, nCVdT = f × 1/2 nRdTf/2 R
but CP = CV + R
∴ CP = (f/2 f/2 + 1)R. This is the relation between specific heat ratio and degree of freedom.
(v) Adiabatic Expansion of an Ideal gas: As defined earlier in adiabatic process ΔQ = 0
For a system containing an ideal gas adiabatic process can happen in two ways
(i) if the system boundary is adiabatic
(ii) if the system boundary is diathermic but the process takes place so fast that their is no time for the transfer of the heat. For example, Propagation of sound in air.  
For an adiabatic process as shall be proved
PV? = a constant, here g = CP/ CV. ? is known as adiabatic constant
Since PV = nRT for an ideal gas
P = nRT/V so,
(nRT/V)Vg = Constant           ∴ TVg–1 = a constant
Proof: Let n moles of an ideal gas expand adiabatically by a small amount ΔV.
By first law   ΔQ = ΔU + PΔV
As ΔQ = 0    we have ΔU = –PΔV
But       U = nCvΔT
So,       nΔT = –(P/Cv)ΔV                                                                               …(i)
From the ideal gas law (after differentiating) ,          PΔV = VΔP = nRΔT
Replacing R by its equal, CP ­- CV leads to this,         nΔT = PΔV + VΔP/Cp – Cv        …(ii)
equating (1) and (2) with a little algebra leads to
            ΔP/P + (Cp/Cv) ΔV/V = 0            Cp/Cv = g
and integrating we have, ln P + g In V = a constant,
or PVg = a constant, hence proved.
Again g = Cp/Cv = Cv + R/CV = 1 + R/Cv = 1 + R/(1/2 fR) = 1 + 2/f
(vi) Work done in an adiabatic process (P1V1T1) to (P2V2T2)
For adiabatic process, W = –ΔU = –n CvΔT
From ideal gas equation P1V1 – P2V2 = nR (T1 – T2) = – nRT
     W = Cv(P1V1 – P2V2)/R(P1V1 – P2V2)/?–1 = nR(T1 – T2)/–1
(vii) The table shows the values of f, Cv, Cp and g for different gasses:
Nature of gas
Degree of freedom
f = (T+R+V)
Cv
Cp
g
Monatomic
3+0+0 = 3
3
5
 5/3
Diatomic
3+2+0 = 5
5
7
 7/5
Polyatomic (linear)
3+2+0 = 5
5
7
 7/5
Polyatomic
(non-linear)
3+2+1 = 6
3R
4R
 4/3
Note: At room temperature the energy associated with vibrational motion is negligible in comparison to translational and rotational KE.
Expressions For ΔU, W, And ΔQ For Different Process
Processes
Relation between thermodynamic variables
Work Done (W)
Heat Exchange (ΔQ)
Isothermal Process
(T constant)
  P∝1/V
W = 2.303 nRT log10 (V2/V1)
ΔQ = 2.303 nRT
log10(V2/V1)
Adiabatic Process
(No heat exchange)
PVg = constant
W = (P1V1 – P2V2)/(g–1) = nR(T1– T2)/(g-1)
ΔQ = 0
Isochoric   Process
(V = constant)
 P ∝T
W = 0
ΔQ = n CΔT*
(Use definition of Cv)
Isobaric Process
(P = Constant)
 V ∝T
W = P Δ V = P(V2 –V2)
W = nR (T2 –T1)
ΔQ = n CΔT*
(Use definition of Cp)

Sunday, October 21, 2012

Proof of Bernoulli Equation


The Bernoulli Equation for an incompressible, steady Fluid Flow:

The Bernoulli Equation is a statement derived from conservation of energy and work-energy ideas that come from Newton's Laws of Motion.
Statement:It states that the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non–viscous fluid in steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the fluid along the length of the pipe.
This statement is based on the assumption that there is no loss of energy due to friction.
To prove Bernoulli’s theorem, we make the following assumptions:
1. The liquid is incompressible.   
2. The liquid is non–viscous. 
3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.

Proof of Bernoulli’s Theorem:
Imagine an incompressible and non–viscous liquid to be flowing through a pipe of varying cross–sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v1 and at a height h1 above the reference level (earth’s surface). It leaves the pipe with a normal velocity v2 at the narrow end B of cross–sectional area a2 and at a height h2 above the earth’s surface.
Fluid flow to the right.

We examine a fluid section of mass m traveling to the right as shown in the schematic above. The net work done in moving the fluid is
Net work done = work 2 - work 1.Eq.(1)
where F denotes a force and an x a displacement. The second term picked up its negative sign because the force and displacement are in opposite directions.
Pressure is the force exerted over the cross-sectional area, or P = F/A. Rewriting this as F = PA and substituting into Eq.(1) we find that
Net work done = (PFA)_2 - (PFA)_1.Eq.(2)
The displaced fluid volume V is the cross-sectional area A times the thickness x. This volume remains constant for an incompressible fluid, so
Volume is constant for an incompressible fluid.Eq.(3)
Using Eq.(3) in Eq.(2) we have
dW = (P_1 - P_2) V.Eq.(4)
Since work has been done, there has been a change in the mechanical energy of the fluid segment. This energy change is found with the help of the next diagram.
Diagram for derivation of Bernoulli's equation for an imcompressible unrestricted steady fluid flow.


The energy change between the initial and final positions is given by
Energy change = E_2 - E_1.Eq.(5)
Here, the the kinetic energy K = mv²/2 where m is the fluid mass and v is the speed of the fluid. The potential energy U = mgh where g is the acceleration of gravity, andh is average fluid height.
The work-energy theorem says that the net work done is equal to the change in the system energy. This can be written as
Net work = energy change.Eq.(6)
Substitution of Eq.(4) and Eq.(5) into Eq.(6) yields
Expansion of dW = dE.Eq.(7)
Dividing Eq.(7) by the fluid volume, V gives us
Expansion of dW = dE.Eq.(8)
    where
Density = mass / volume.Eq.(9)
is the fluid mass density. To complete our derivation, we reorganize Eq.(8).
Expansion of dW = dE.Eq.(10)
Finally, note that Eq.(10) is true for any two positions. Therefore,
P + mgh + mv^2/2 = Constant.Eq.(11)
Equation (11) is commonly referred to as Bernoulli's equation. Keep in mind that this expression was restricted to incompressible fluids and smooth fluid flows.

Sunday, October 7, 2012

Work, Power & Energy

Work

Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.
Let a constant force  be applied on the body such that it makes an angle θ with the horizontal and body is displaced through a distance s

By resolving force  into two components :
(i) F cos θ   in the direction of displacement of the body.
(ii) F sin θ  in the perpendicular direction of displacement of the body.
Since body is being displaced in the direction of , therefore work done by the force in displacing the body through a distance is given by
                
or             
Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
If a number of force  are acting on a body and it shifts from position vector  to position vector  then .

Unit of Work

Since work is the product of force and distance, unit of work is unit of force and unit of distance. If force is in N and distance is in metres, unit of work will be N-m. In SI system of unit, 1N-m = 1 J. The term Joule s used for one N-m work done, and letter J is used as a symbol of Joule. Hence, one Joule may be defined as the amount of work done by one Newton force when it moves one metre distance in its direction.

Energy

The energy of a body is defined as its capacity for doing work. Since energy of a body is the total quantity of work done therefore it is a scalar quantity.

Power

Power of a body is defined as the rate at which the body can do the work.
Average power 
Instantaneous power          [As ]
                                                           [As ]
i.e. power is equal to the scalar product of force with velocity.
If work done by the two bodies is same then power 
i.e. the body which perform the given work in lesser time possess more power and vice-versa. As power = work/time, any unit of power multiplied by a unit of time gives unit of work (or energy) and not power.

Work Done by Variable Force

Let a varying force P move a body by distance s as shown by force-displacement curve in Fig.. At an instance, when force P is acting and the body moves by an elemental distance ' δs', then work done at that instance
= P δ s
Hence work done by the varying force in moving the body by distance ‘s’
= Σ P δ s
and it is equal to the area under the force-displacement curve. The variation of force is in a regular fashion so that its value at any instance can be represented by an expression. Then the work done can be evaluated by suitable integration. In such a case,
 work done= ΣPs = ∫ Pds
WORK DONE BY A VARYING FORCE.JPG

Sunday, September 23, 2012

Alternating Current

Alternating Current

The basic principle of the ac generator is a direct consequence of Faraday's law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency ω a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is,

            V = V0 sin ωt                                       ... (i)
The usual circuit diagram symbol for an ac source is shown in Figure.
alternatig-current
In Equation (i) V0 is the maximum output voltage of the ac generator, or the voltage amplitude and w is the angular frequency, equal to 2p times the frequency f.
            ω = 2pf
The frequency of ac in India is 50 Hz, i.e.,
            f = 50 Hz
So,       ω = 2pf » 314 rad/s
The time of one cycle is known as time period T, the number of cycles per second the frequency f.
            T = 1/f     or  T = 2π/ω
A sinusoidal current might be described as,
            i = i0 sin ωt
If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is,
alternatig-current
Thus,      one cycle = 0
Similarly, the average value of the voltage (or emf) for one cycle is zero.
              one cycle = 0
Since, these averages for the whole cycle are zero, the dc instrument will indicate zero deflection. In ac, the average value of current is defined as its average taken over half the cycle. Hence,
alternatig-current
This is sometimes simply written as, iav. Hence,
               iav = Half Cycle = 2/π i0 ≈ 0.637 i0
Similarly, Vav =  2/π V0 ≈ 0.637 V0
A dc meter can be used in an ac circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, i.e., 2/π time the maximum current i0. A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i2 and finally take the square root of that average. This procedure defines the root-mean-square current denoted as irms. Even when i is negative, i2 is always positive so irms is never zero (unless i is zero at every instant). Hence,
alternatig-current
Similarly, we get RMS value Vrms = V0/√2 ≈ 0.707 V0
The square root of the mean square value is called the virtual value and is the value give by ac instruments.
Thus, when we speak of our house hold power supply as 220 volts ac, this means that the rms voltage is 220 volts and its voltage amplitude is,
            V0 = √2Vrms = 311 volt
Form Factor
The ratio,       
rms value/average value = (V0/√2)/(2V0/π)
                                   = π/2√2 = 1.111
alternatig-current
is known as or factor,
Note:
1. The average value of sin wt, cos wt, sin2wt, cos2wt, etc. is zero because it is positive half of the time and negative rest half of the time. Thus,
alternatig-current
2. The average value of sin2 wt and cos2 wt is 1/2
alternatig-current
3. Like SHM, general expressions of current/voltage in an sinusoidal ac are,
     = i0 sin (ωt ± f),                   V = V0 sin (ωt ± f)
or i = i0 cos (ωt ± f),        and     V = V0 cos (ωt ± f)

Ray Optics - Optical Instruments

Optical Instruments

Optical instruments are devices that are used to processes light waves to enhance an image for viewing, or analyze light waves to determine one of a number of characteristic properties. The first optical instrument invented was telescope and was used for magnification of distant images, and microscopes used for magnifying very tiny images. 

Simple Microscope

It is an optical instrument used to see very small objects. It's magnifying power is given by
m = Visual angle with instrument (β)/Visual angle when object is placed at least distance of distinct vision(α)
Simple microscope
(i) It is a single convex lens of lesser focal length.
(ii) Also called magnifying glass or reading lens.
(iii) Magnification's, when final image is formed at D and ∞
Simple Microscope
     (i.e. mD and m)    mD = (1 + D/f)max
     and       m = (D/f)min
(iv) If lens is kept at a distance a from the eye then mD = 1 + D – a /f and m = D–a/f

Compound Microscope

(i) Consist of two converging lenses called objective and eye lens.
Compound Microscope
(ii) feye lens > fobjective and (diameter)eye lens > (diameter)objective
(iii) Intermediate image is real and enlarged.
(iv) Final image is magnified, virtual and inverted.
(v) uo = Distance of object from objective(o), vo = Distance of image (A'B') formed by objective from objective, ue = Distance of A'B' from eye lens, ve = Distance of final image from eye lens, fo = Focal length of objective, fe = Focal length of eye lens.
(vi) Final image is formed at D: Magnification mD = – vo/uo (1+ D/fe) and length of the microscope tube (distance between two lenses) is LD = vo + ue.
Generally object is placed very near to the principal focus of the objective hence uo = fo. The eye piece is also of small focal length and the image formed by the objective is also near to the eye piece.
So vo = LD, the length of the tube.
Hence, we can write mD = –L/do (1+ D/fe)
(vii) Final image is formed at ∞: magnification m∞ = –vo/uo.D/fe and length of tube L∞ = vo + fe
      In terms of length m∞ = (L – fo – fe)D/fo fe
(viii) For large magnification of the compound microscope, both fo and fe should be small.
(ix) If the length of the tube of microscope increases, then its magnifying power increases.
(x) The magnifying power of the compound microscope may be expressed as M = mo × me; where mo is the magnification of the objective and me is magnifying power of eye piece.

Astronomical Telescope

By astronomical telescope heavenly bodies are seen.
(i) fobjective >; feye lens and dobjective >; deye lens
Astronomical Telescope
(ii) Intermediate image is real, inverted and small.
(iii) Final image is virtual, inverted and small.
(iv) Magnification: mD = –fo/fe (1+fe/D) and m = –fo/fe
(v) Length : LD = fo+ ue and L = fo + fe

Resolving Limit and Resolving Power

Microscope: In reference to a microscope, the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and it's reciprocal is called Resolving power (RP)
Resolving Limit and Resolving Power
λ = Wavelength of light used to illuminate the object,
µ = Refractive index of the medium between object and objective,
θ = Half angle of the cone of light from the point object,
µ sin θ = Numerical aperture.
Telescope: Smallest angular separations (dθ) between two distant objects, whose images are separated in the telescope is calledresolving limit. So revolving limit dθ = 1.22λ/a and resolving power  where a = aperture of objective.


Refraction of Light through Plane Surfaces

Refraction of Light

The bending of the ray of light passing from one medium to the other medium is called refraction.
Refraction of Light
(i) The refraction of light takes place on going from one medium to another because the speed of light is different in the two media.
(ii) Greater the difference in the speeds of light in the two media, greater will be the amount of refraction.
(iii) A medium in which the speed of light is more is known as optically rarer medium and a medium is which the speed of light is less, is known as optically denser medium.
(iv) When a ray of light goes from a rarer medium to a denser medium, it bends towards the normal.
Refraction of Light
(v) When a ray of light goes from a denser medium to a rarer medium, it bends away from the normal.
Refraction of Light

Refractive Index

(i) Refractive index of a medium is that characteristic which decides speed of light in it.
(ii) It is a scalar, unit less and dimensionless quantity.
(iii) Absolute refractive index:
When light travels from vacuum to any transparent medium then refractive index of medium w.r.t. vacuum is called it's absolute refractive index i.e. vacuumµmedium = c/v
Absolute refractive indices for glass, water and diamond are respectively µg = 3/2 = 1.5, µw = 4/3 = 1.33 and µD = 12/5 = 2.4
(iv) Relative refractive index:
When light travels from medium (1) to medium (2) then refractive index of medium (2) w.r.t. medium (1) is called it's relative refractive index i.e. 1µ2 = µ21 = v1/v2 (where v1 and v2 are the speed of light in medium 1 and 2 respectively).
(v) When we say refractive index we mean absolute refractive index.
(vi) The minimum value of absolute refractive index is 1. For air it is very near to 1. ( 1.003)
(vii) Cauchy's equation : µ = A + B/λ2 + C/λ4 + ...         (1Red > 1Violet so mRed < mViolet)
(viii) If a light ray travels from medium (1) to medium (2), then 1µ 2 =  µ21 =  λ1/ λ2 = v1/v2
(ix) Dependence of Refractive index:
(a) Nature of the media of incidence and refraction.
(b) Colour of light or wavelength of light.
(c) Temperature of the media : Refractive index decreases with the increase in temperature.
(x) Reversibility of light and refraction through several media:
Refractive Index
(A) 1µ2 = 1/2µ1      (B)   1µ2x2µ3x3µ1 = 1 or 2µ3 = 1µ3/1µ2
Snell's law:
The ratio of sine of the angle of incidence to the angle of refraction (r) is a constant called refractive index
i.e.   sin i/sin r = µ (a constant).
For two media, Snell's law an be written as 1µ2 = µ21 = sin i/sin r or µ1sin i = µ2sin r

Refraction Through Glass

Lateral Shift:
The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges out parallel to it's incident direction i.e. the ray undergoes no deviation δ = 0 . The angle of emergence (e) is equal to the angle of incidence (i)
Refraction Through Glass
The Lateral shift of the ray is the perpendicular distance between the incident and the emergent ray, and it is given by MN = t secrsin (i – r)

Apparent Depth

When object is in denser medium and observer is in rarer medium
(i) µ = Real depth/Apparent depth = h/h'
(ii) Real > Apparent depth
Apparent Depth
(iii) Shift d = h – h' = (1 – 1/µ)h.
(iv) If a beaker contains various immiscible liquids as shown then Apparent depth of
bottom d11 + d22 + d33 + ...
Apparent Depth
Object is in rarer medium and observer is in denser medium
(i) µ = h'/h
(ii) Real depth < Apparent depth.
(iii) d = (µ = 1)h
Apparent Depth

Total Internal Reflection (TIR)

When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes 90°, this angle of incidence is called critical angle (C).
When Angle of incidence exceeds the critical angle than light ray comes back in to the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).
(i) µ = 1/sin C = cosec C where µ means; RaserµDenser
(ii) Conditions for Total Internal Reflection
Total Internal Reflection
(a) The ray must travel from denser medium to rarer medium.
(b) The angle of incidence i must be greater than critical angle C