Stress-Strain Digram

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-We all know about elasticity , plasticity and elastic materials. In this post we'll discuss about Stress-Strain digram.

-In case of solids if we go on increasing stress continually then a point is reached at which strain increases more and more rapidly and Hook's law is no longer obeyed.

-Thus, the stress at which linear relationship between stress and strain ceases to hold is referred as elastic limit of material for the stress applied.

-If the elastic limit of material is exceeded it will fail to recover its original shape or size on removal of stress and would acquire a permanent set.

-Any type of stress can be plotted against appropriate strain and the shape of resulting stress- strain digrams would have shapes, depending on the kind of material.

-Simple stress- strain digram for a bar or wire is shown below in the figure.



















(i). Portion OA is the straight line which clearly shows that stress produced is directly proportional to strain i.e., Hook's law is perfectly obeyed upto A and on removal of stress wire or bar will recover its original condition. Point A is called Proportionality limit

(ii). As soon as proportionality limit is crossed beyond point A, the strain increases more rapidly than stress and curve AB in graph shows that extension of wire in this limit is partly elastic and partly plastic and point B is the elastic limit of the material. Thus if we start decreasing load from point B the graph does not come to O via path BAO instead it traces straight line BG. So that there remains a residual strain. This is called permanent set.

(iii). If we continue to increases the stress beyond point B then for little or no increase in stress the strain increases rapidly upto point C.

(iv). Further increase of stress beyond point C produces a large increase in strain untill a point E is reached at which fracture takes place and from B to D material is said to undergo plastic flow which is irreversible.


Conclusion :

1. The wire exhibits elasticity from O to b and plasticity from b to d. If the distance between b and d is more, then the metal is ductile. If the distance between b and d is small, then metal is brittle.
2. The substances which break as soon as the stress is increased beyond elastic limit are called brittle substances eg: glass, cast iron, high carbon steel.
3. The substances which have a large plastic range are called ductile substances. Eg: copper, lead, gold, silver, iron, aluminium. Ductile materials can be drawn into wires. Malleable materials can be hammered into thin sheets. Eg: gold, silver, lead.

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System of varying mass(Rocket)

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  • While studying classical mechanics we have always considered the particle under consideration to have constant mas
  • Some times it is required to deal with the particles or system of particles in which mass is varying and motion of the rocket is one such examples
  • In a rocket fuel is burned and the exhaust gas is expelled out from the rear of the rocket
  • The force exerted by the exhaust gas on the rocket is equal and opposite to the force exerted by the rocket to expell it
  • This force exerted by the exhaust gas on the rocket propels the rocket forwards
  • The more gass is ejected from the rocket ,the mass of the rcoket decreases












  • To analyze this process let us consider a rocket being fired in upwards direction and we neglect the resistance offered by the air to the motion of the rocket and variation in the value of the acceleration due to gravity with height
  • Figure above shows a rocket of mass m at a time t after its take off moving with velocity v.Thus at time t momentum of the rocket is equal to mv.THus
    pi=mv
  • Now after a short interval of time dt,gas of total mass dm is ejected from the rocket
  • If vg represents the downward speed of the gas relative to the rocket then velocity of the gas relative to earth is
    vge=v-vg
    And its momentum equal to
    dmvge=dm(v-vg)
  • At time t+dt,the rocket and unburned fuel has mass m-dm and its moves with the speed v+dv.Thus momentum of thee rocket is
    =(m-dm)(v+dv)

  • Total momentum of the system at time t+dt is
    pf=dm(v-vg)+(m-dm)(v+dv)
    Here system constitute the ejected gas and rocket at the time t+dt
  • From Impulse momentum relation we know that change in momentum of the system is equal to the product of the resultant external force acting on the system and the time interval during which the force acts
  • Here external force on the rocket is weight -mg of the rocket ( the upward direction is taken as positive)
  • Now
    Impulse=change in momentum
    Fextdt=pf-pi
    or
    -mgdt=dm(v-vg)+(m-dm)(v+dv) - mv
    or
    -mgdt=mdv-vgdm-dmdv
    term dmdv can be dropped as this product is neglibigle in comparison of other two terms
    Thus we have







  • In equation (14) dv/dt represent the acceleration of the rocket ,so mdv/dt =resulant force on the rocket

    Therefore
    Resultant Force on rocket=Upthrust on the rocket - weight of the rocket
    where upthrust on rocket=vg (dm/dt)
  • The upthrust on rocket is proportional to both the relative velocity (vg) of the ejected gas and the mass of the gas ejected per unit time (dm/dt)
  • Again from equation (14)







    As rocket goes higher and higher ,value of the acceleration due to gravity g decreases continously .The values of vg and dm/dt parctically remains constant while fuel is being consumed but remaining mass m decreases continously .This result in continous increase in acceleration of rocket untill all the fuel is burned up
  • Now we will find the relation between the velocity at any time t and remaining mass.Again from equation (15) we have
    dv=vg (dm/m) -gdt
  • Here dm is a +ve quantity representing mass ejected in time dt.So change in mass of the rocket in time dt is -dm.So while calculating total mass change in rocket,we must change the sign of the term containing dm
    dv=-vg (dm/m) -gdt                 --(16)
  • Initially at time t=0 if the mass and velocity of the rocket are m0 and v0 respectively.After time t if m and v are mass and velocity of the rocket then integrating equation (16) within these limits






    On evaluating this integral we get
    v-v0=-vg(ln m- ln m0)-g(t-0)
    or v=v0+vgln(m0/m) -gt                (17)
  • equation(17) gives the change in velocity of the rocket in terms of exhaust speed and ration of initial amd final masses at any time t
  • The speed acquired by the rocket when the whole of the fuel is burned out is called burn-out speed of the rocket

Internal Resistance of Battery (or cell)

-The resistance offered by medium in between plates of battery (electrolytes and electrodes of the cell) to the flow of current within the battery is called internal resistance of the battery.
Internal resistance of a battery usually d branch containing batteryenoted by r and in electric circuit its representation is shown below in the figure














-Internal resistance of a battery depends on factors like seperation between plates, plate area, nature of material of plate etc. For an ideal cell r=0 , but real batteries or sourcesof emf always has same finite internal resistance.
-If P and Q are two terminals of the battery shown below in the figure






then potential difference between terminals P and Q is
VP=(VP-Vx) - (VQ - Vx) = E-Ir
let VP-VQ=V
V=E-IR
now for I=0 and V=EMF
and this potential difference V is called the terminal difference of the cell or battery and defined as the emf of the battery when no current drawn from it.
-For real battery equation(4) which gives V=E-Ir whereI is the current in the branch containing battery.
-From figure(4) potential difference across the external resistance R of the circuit would be equal to terminal potential difference of the cell. Thus
V=IR also V=E-Ir
or, IR=E-Ir
which gives
I=E/(R+r) =Net EMF/Net resistance
-From equation(4) we can calculate that when current is drawn from the battery terminal potential difference is less than the EMF of the battery.

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Resonance Column Method and End Correction

This 2007 IIT JEE physics subjective practice problem requires the student to find the end correction, or difference between the frequency of a tuning fork and the corresponding sound waves inside of a tube, given the contextual information. This type of problem has an answer between 0 and 9999, with six points awarded for a correct answer. No points are subtracted for an incorrect response, so all problems in the subjective category may be attempted without penalty.
Problem
In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode with a tuning fork is 0.30 m. When this length is changed to 1.00 m, the same tuning fork resonates with the first overtone. What is the end correction (in cm) for the column?
Solution
In the resonance column method, the pipe has one end closed and one end open. Let Δl be the end correction. Remember that the first resonance is, theoretically, one quarter of a wavelength, with the distance between resonances at half of a wavelength.
Thus, for the fundamental mode: 0.30 m + Δl = λ/4
For the first overtone: 1.00 m + Δl = 3λ/4
⇒ 3(0.30 m + Δl) = 1.00 m + Δl
⇒ Δl = 0.05 m = 5 cm.
Bio: Alexis Bonari is a freelance writer and blog junkie. She often can be found blogging about general education issues as well as information on college scholarships. In her spare time, she enjoys square-foot gardening, swimming, and avoiding her laptop.
Photo: Public Domain
URL: http://upload.wikimedia.org/wikipedia/commons/3/35/Kundt_tube.png

IITJEE/AIEEE quick links

Hi all
It is the high time that you are all set to start preparing for IITJEE/AIEEE 2011. Hera are few important links that would help you making a plan to get through IITJEE/AIEEE 2011 examination. So visit these links make your own plan that suites well with your time and start studying for getting into these high profile institutes.
1.IITJEE Pattern Analysis for Physics
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4.IITJEE Physics Syllabus
5.IITJEE Institutes Information
6.IITJEE Chemistry Syllabus
7.IITJEE Math Syllabus
8.IITJEE Books for Physics Syllabus

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Heating effect of current and thermoelectricity

1. Heating effect of current
-In previous chapter while discussing electric energy and power ,we learned that IΔV amount of energy is lost per second when a current I flows through a potential ΔV and this energy appears in the form of heat energy
-Due to the conversion of electric energy into heat energy the conductor becomes hot .This effect is known as Joule's Heating and this heating is thermodynamics irreversible.
Cause Behind Joule's Heating:-
-Explanation behind the Joule's heating is that when a potential difference ΔV is maintained between the ends of a conductor,the free electrons in the conductor are accelerated towards the higher potential end of the conductor
-In their way electrons frequently collided withe the positive ions of the conductor due to which their velocity decreased
-This the energy electrons gained on account of acceleration is transferred to the positive lattice ions or atoms and electrons then returns to their equilibrium distribution of velocities
Thus ,lattice ions receives energy randomly at the average rate of IΔV per unit time
-Ions spends this energy by vibrating about their mean positions resulting in rise in the temperature of the conductor
-This way Joule's heating nothing but the conversion of electrical energy into heat energy
2. Thermoelectricity
-We know that currents flows in a conductor whenever there is a electric potential difference betweens the ends of the conductor If there is a temperature difference between the ends of the conductor then thermal energy flows from hotter end to the colder ends
-Thermal energy flows may also be carries by the electrons in the conductor and hence resulting the presence of electric current
-At the hotter end of the conductor electrons have slightly higher kinetic energy and hence they move faster
-So there is net flow of current towards the end of the conductor with lower temperature.Thus an electric current exists in the conductor due to the difference in the temperature of two ends of the conductor
-This phenonmenon due to which electricity is produced when two ends of the conductor are kept at different temperature is known as thermoelectricty
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Pseudo or Fictitious forces

Consider a particle P in stationary frame of refrance S in which no force is acting i.e., a=0. If any other frame of refrance S' is moving with acceleration a w.r.t. frame S, then an acceleration -a0 appears to be acting on the particle P w.r.t. S'. Therefore a force -ma0 seems to be acting on the particle P due to the accelerated motion of the frame S'. This force is known as Pseudo force or fictitious force. Therefore pseudo forces are the force which does not actually acts on the particle but seems so because of the accelerated motion of the refrance frames are. This accelerated frame of refrance S' is a non inertial frame of refrance . If due to force FR on the inertial frame of refrance , the particle seems to suffer an acceleration aR then by Newtonian mechanics FR=maR Therefore resultant force on particle in accelerated frame is
F=FR+Fa
ma=maR+(-mF0)
or,
a=aR-a0
where Fa is fictitious force. Direction of fictitious force is opposite to the accelerated motion of S' frame of refrance.

Law of conservation of linear Momentum

Law of conservation of momentum is one of the basic laws and is derived from Newton’s Laws of motion. This law always holds true even in the situations where Newton’s laws of motion fails to hold. This law states that in the absence of resultant external force acting on the system , the total linear momentum of the system always remains constant i.e.,

p=p1+p2+p3+……+pn

This is law of conservation of linear momentum for system of particles. For individual particles momentum may change but their total momentum remain unaltered in absence of external forces.

Now the law of momentum is fundamental and exact law of nature. No violation to it has ever been encountered.


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Newton's first law of motion

Hi all

What comes to your mind when you think about Newton's first law of motion . Now a days it has become an obvious statement but it was not the case when the law has been formulated. Statement of Newton's first Law of motion is

" Every body continues to be in state of rest or uniform motion untill acted upon by a net external force."

Equilibrium of the bodies is the essence of the forst law of motion.

Why we have used the word net force it is because there might be more then one forces acting on the body and net force on the body is the vector sum of all the forces acting on the body. Newton's first law of motion only gives the qualitative defination of the force that is it tells us that force is only the influence behind the moving objects but it does not tell anything about what is required to keep objects moving when they are set to motion by the application of force. In out daily life we see that bodies set to motion eventualy came to rest for example book placed on a horizontal surface is pushed , it started to move and then come to rest. What does first law of motion has to explain about this effect. What would we have to do to keep the book moving ? Can first law of motion has anything to say about it. Think about it if you find any more point to discuss let me know I'll be happy to discuss

Physics Expert
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Work , Energy and Power (Quick Recap)

1. Work done by a constant force = F.S=FS cosθ
2. Work done by a variable force = ∫F.dS
3. If force acting on the body is along the direction of displacement of the body then work done can be calculated by calculating the area enclosed between F-S curve and displacement axis.
4. Work done by conservative forces like gravitational and electrostatic forces does not depend on the path taken by the particle and only depends on the initial and final positions of the particles.
5. Work done by non conservative forces like frictional forces are path dependent.
6. Power is the rate of doing work.
       Average power = ΔW/Δt
       Instantaneous power = dW/dt = F.v
7. Slope of W vs t graph gives the power of the system and area between the P vs t curve and time axis gives work done by the system.
8. Kinetic energy = (1/2)mv2
9. Gravitational potential energy = mgh where m is the mass of the object, g is acceleration due to gravity and h is the hight of object above the referance level. PE of the object is positive when object is above the refrence level and negative when it is below the refrence level.
10. An object posess kinetic energy by virtue of its motion and it posess potential energy by virtue of its position.
11. Elastic energy of a spring is given by (1/2)kx2 where k is the elastic constant for the spring.
12. Work energy theorem states that
       Wext=Kf-Ki=ΔK
13. Law of conservation of energy states that " Total energy of the system always remains constant provided no external work is being done and energy of the system is not being converted to other forms of energy". Hence Energy can never be created or destroyed it can only be converted from one form of energy to another.

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