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Time Travel Again !

Davide Writes: (The email is posted as such. SO, sorry for the telegraphic language and poor spelling and grammar)

“Hello there i have one more question,a crazy one:) according to Einstein we need a enormous amount of energy to push matter at a speed of light,now for me the real problem there  is that an atoms at a speed of light will break a part right?so even if we will find a way to push it,we will still failed.I was wander are we looking to the right way?is it photon, so the world of small with no mass only work? now here my crazy question,lets speculate a bit,we have two red dwarf pretty close to earth Barnard’s Star” only  “5.9 ly and Wolf 359 7.7ly a baby step compere the universe,if we can send a same kind of mission the we have done with Stardust Nasa’s comet sample and collect a piece the size of a soccer ball back to earth,we could work on it outside earth atmosphere in the space base.Since we can’t go in the first place at a speed of light let’s says that the all deal will last 150 year to go and back,(assuming that we use a new energy to go bit faster)in the mean time the space base will be pretty developed and maybe we even found a new power to push to the speed of light(i remind you that we are speculate)we could put our hands on the sample using electromagnetism so that we avoid any risks and then send it at a seed of light.What i try to say is that a single atom is weak but billions of it together is a different story,here an example, lets imagine to make a spherical shape out of Lego(the toy) and send it in to space at a great speed,it will break pretty soon,now we make another one but we glue the piece together,it will last bit longer,again another with glue and a put net around it will last even longer so instead of a Lego we use our sample,the glue in the inside is the immense mass that will have (piece of red dwarf) net outside the  immense gravitational pull ,now do you think that it will be enough to keep it together at a speed of light?regardless the energy that we need to push it?  after all even in the past we believed that going faster than a  speed of sound a plane will brake a part or something bad could happen,instead only a big BUM wen we reach it and nothing more,so maybe if we can go faster than light nothing will happen,or maybe we find a new way to travel,we can even give a name for fun “a speed of matter” what do you think about it?i know it sound like sci fi but hey physics it self is blowing mind science right?   Thank you kindly for your time:)”

I don’t like to respond now.

Responses from visitors invited

Problem from Kinematics

The speed of a train changes from36km/h to 72km/h in 10sec. calculate the distance travelled during this time

Answer:

u=36km/h = 10 m/s

v=72 km/h=20 m/s

t=10 s

S=?

use v = u+ at to find a

a = 1 m/s2

Use v2-u2=2aS to find S

S = (400-100)/2 = 150 m

Ron writes on Light and Relativity

Let me start by saying I am NOT a physicist.  That much should be painfully clear soon enough.  Please bear with me as I am not certain I have the language and in depth knowledge to explain myself adequately.

A few years ago I began experiencing parallels between the world of human behavior and physics.  At the time it seemed simple enough.  That is until I started teaching myself physics.  But at its core it seemed to make sense, the rules of life recycle themselves in different forms.  Out of simple ideas comes complexity.  Newton’s laws, thermodynamics… pascals principle, snells law, red and blue shift, wave particle duality and on and on… all seem to have their parallels in human behavior and on the surface seem to follow the same mathematical equations. Then I ran into the most famous of physics equations E=MC^2.  It makes sense in human behavior.  The energy (work) we can get out of an individual is relative to that individuals’ mass.  The more matter we attach to the individual, knowledge, life experience etc (therefore increasing their mass) the more we can get out of them in the form of energy… but then the problem of the speed of light squared.  The only known constant (light speed) is a problem in human behavior.  At least to this point I know of no known behavioral constant.

I juggled the idea of it being a relative constant.  Constant for the individual but relative as it would differ for everyone.  The speed of cognition, or thought speed, would remain the same in potential throughout life of a given individual.  Or that it was just a theoretical potential that humanity had yet to obtain.  I even juggled the idea of it being a collective ability, but all of these options cause problems with the original equation.  This has forced me to contemplate the nature of light and left me with questions I simply lack the understanding of physics to answer.

Is the speed of light truly constant or is it only constant as it relates to the big picture?  As in: our perception of time as it relates to all time that has ever existed would appear as a single point in time.  The older we get the more time seems to “fly by”  if we as humans could continue to live for 13 billion years would the perception of an hour become so perceptively small that we wouldn’t even know it has passed?   Is it therefore possible we can only understand and therefore measure the speed of light at a specific point in time, even if we try to come back and remeasure and compare the speed of light now with the speed of light fifty years from now the difference between the two would be imperceptivity small as a result of the displacement in time as it relates to the whole of time?

If the four dimensionality of time/space is linked shouldn’t time expand as space expands? And vice versa… if the speed of light is to remain constant as measured under such conditions isn’t light actually slowing down/speeding up over time as it relates to the whole of existing time/space?  It’s a distance displacement problem…. If points A and B are actually farther apart but light travels the same distance in the same time the “speed” may seem constant but time has actually expanded to give the light more “time” to cover that distance.  The speed of light would therefore be constant as a relation to perception and not as it relates to physical principles. Like the fact that the perception of time changes as we get older even though the actual measure of time remains the same.

How can light exist forever?  At the speed of light we theoretically freeze time for that photon but that would also require an infinite amount of energy to obtain and maintain. So even a subatomic particle with a lifespan of a nanosecond would appear to exist for all eternity but in actuality would still only exist for a nanosecond.  Because quite simply…it can’t have infinite energy and if it can I don’t understand how.  Thermodynamics: no system is a perfect system and will experience energy loss, Newton: equal and opposite reaction, if something begins it has to end to balance the equation.

Is it therefore possible that light is born of the fourth dimension… we experience it in the dimensionality of space as long as it loses its energy to the three dimensions.  A photon folds and pushes its way through space/time  The initial energy of the photon is high and generates bigger leading waves which resist the photon holding it from passing the “speed of light” as the energy dissipates/photon begins to die those folds restricts the photon less allowing it to maintain the speed of light.  We experience light because of its “ripples” in three-dimensions light dissipates as the waves of space become less folded in front of the photon.  But this would mean the photon eventually loses enough energy that it can no longer be perceived in the third dimension….  So what happens to it??  Imagine : A man running through a corn field has to exert the energy to push the stalks aside but  over time if the corn stalks are slowly spaced out even as he loses energy he can maintain his pace, because there is less impeding his path.  Our perception of light would be like being in a helicopter looking down on the field.  In the begging the field is densely packed with cornstalks and they slowly spread out until there is none.  We only know the man is there while he is running through the corn because he pushes the stalks aside and we see that movement but once there are no stalks or he lacks the energy to continue running we no longer have a way of measuring his presence.

That brings me for some reason to a theory of Dark matter.  Why?  Well what happens to photons that no longer move with enough force to be visible.  Like that guy running from a cornfield into an empty field.  How would we know he is still there?  If it was an infinite number of guys all stopping in that empty field we would know they are there by the depression their weight leaves in the field.  Or more accurately by the stones in the field rolling towards a depression we cant actually see.  The thought: An infinite number of “massless” (or perceived mass less) subatomic particles would still have infinite mass.  Infinite mass would supply more then enough gravitational pull even spread out over infinite distance to cause the continued and speeding expansion of space as more and more visible photons “die” contributing its “dark masslessness”.  Any dark matter existing within the universe would act as force… a moving invisible mass existing only on the 4th dimension pushing upon any objects in its way.  But because energy propagated internally expanding outward in all dimensions would compound on the outside as those energy’s converge.

I hope you followed those questions.  I don’t know what these ideas would do to theoretical physics but it would allow for the relativity of cognitive speed between individuals and reopen the door to “what the hell is dark matter in relation to human behavior” but that’s something different all together.

WAVE MOTION QUESTIONS AND ANSWERS FOR CLASS XI

One mark questions with answers

Q1. If a body of mass 2kg. is at restand is hit by a mass of 4kg. moving with 3m/s, find fraction of the momentumretained by the moving body assuming the collision to be elastic and head-on.

Ans1. If n = m2/m1
(n – 1)/(1 + n) is the fraction of the momentum retained by the moving body so
(n – 1)/(1 + n) = (2 – 1)/(1 + 2) = 1/3.

Q2. If mass of the moving body ismuch greater than the mass of the body at rest than what is the approximatevilocity of the moving body after head-on collision?

Ans2. When the moving mass is muchgreater than the mass at rest then after the collision the heavier mass keepson moving with the same velocity and in the same direction.

Q3. At what point the potentialenergy of a body is taken to be zero?

Ans3. The potential energy of the bodyat the surface of the earth is taken to be zero (Potential energy = mgh, whereh is the height of the body from the surface.).

Q4. Does the work done on a body by aforce depend upon the path followed by it?

Ans4. May or may not be. If the force isconservative then it does not depend but if it is non-conservative (friction)then it depends.

Q5. If a body hits the ground from aheight h1 and rebounds to a height h2 after havinginelastic collision with the ground then what is the coefficient ofrestitution?

Ans5. e = Ö(h2/h1)

Q6. A body hits the ground with 50m/s velocity and has inelastic collision with the ground then with whatvelocity it will rebound if the coefficient of restitution is 0.2.

Ans6. Coefficient of restitution, e = v2/v1where v1 is the velocity with which the body hits the ground and v2is the velocity of rebound.
e = 0.2 = v2/50, so v2 = 10 m/sec.

Q7. A body at rest explodes in threefragments. Is it possible that two equal parts move in mutually perpendiculardirections with the same velocity and third mass moves midway between the two?

Ans7. No, it is not possible becausemomentum before the explosion is zero and after it also the momentum must bethe same. In the given situation the third particle must go in a directionopposite to the resultant of first two parts.

Q8. If the speed of a moving vehicleis increased by 200% then how much should be the change in the retarding forceto stop the vehicle over half the previous distance?

Ans8. (1/2)mv2 = F.S where Fis the retarding force and S is the distance over which the vehicle comes torest. When ‘v’ is increased by 200% then K.E. increases by 800%. As S is halvedthen F should be made 16 times.

Q9. If 20 Joules of work is done in compressinga spring from 0 cm to 6 cm then find the work done in compressing the same from3cm to 6 cm.

Ans9. W.D = (1/2) K(x22– x12). This is the work done when the spring iscompressed from x1 to x2.
20 = (1/2)K (36 – 0)
W = (1/2)K (36 – 9)
Solving the above equations we get W = 15 J.

Two mark questions with answers

Q1. A ball is dropped from rest at aheight of 20m. If it loses 30% of its kinetic energy on striking the ground,what is the height to which it bounces? How do you account for this loss inkinetic energy?

Ans1. Suppose the ball acquires avelocity ‘v’ after falling through a height of 20m.
Because the ball is dropped from rest, hence u = 0.
Hence, v2 = u2 + 2as
= 0 + (2 x 10 x 20) = 400
So, v = 20 m/s
Kinetic energy of the ball just before hiting the ground
= (1/2)mv2 = (1/2)m(400) = 200m Joule
Because the ball loses 30% of the kinetic energy on striking the ground, hencekinetic energy retained by the ball after striking the ground = 70% of 200m J
= 140m J
The energy loss is due to the inelastic collision with the ground.

Q2. Why is no energy being consumedin planetary motion.

Ans2. A planet is a heavenly body whichrevolves round the star (the sun). The force which is responsible for circularmotion, is called centripetal force. The direction of the centripetal force isalways towards the centre. Thus, the angle between force F and displacement Sis q = 90o at every point.Work done in moving planet, W = F.S = FSCosq = FS Cos90o.So, W = 0. Hence,no energyis being consumed in planetary motion.
height=82src=”./xi%20work,%20power%20and%20energy_files/image001.gif”alt=”workpowerandenergyf06q16i.gif (1336 bytes)” v:shapes=”_x0000_i1027″>

Q3. How will you justify thathydro-electric power plant is an illustration of law of conservation of energy?

Ans3. A hydroelectric power-plant isused in generating electric energy (Power).The potential energy of water storedat a height is converted into K.E. when water is made to rush down. This fallof water is used to rotate the turbine and the coil and armature of generatoris rotated and electricity is produced . Thus, the K.E. of the fall of water isconverted into electrical form of energy. Hence the hydroelectric power-plantis an example of law of conservation of energy.

Q4. If a body is dropped from aheight of 40m then after 3 inelastic collisions with the ground to which heightthe body will rise? (given: Coefficient of restitution = 0.5)

Ans4. If the body is dropped from aheight of ‘H’ and ‘e’ is the coefficient of restitution then after ‘n’inelastic collisions with the ground the body rises to a height ‘h’ given by
h = H.e2n.
h = 40 x (1/2)2 x 3= 40 x (1/26) = 40/64 = 0.625 m

Q5. A truck and a car are moving withthe same K.E. on a straight line road. If their engines are made off at thesame time, which one of them will stop at a lesser distance.

Ans5. Given : (1/2)m1v12= (1/2)m2v22 ………….(i)
If force of brakes be the same then m1a1 = m2a2……….(ii)
If truck stops over a distance S1 then v12 =2a1S1 ……..(iii)
If car stops over a distance S2 then v22 = 2a2S2………(iv)
From (i) and (ii)
(1/2)m1v12 = (1/2)m2v22………..(v)
From (ii) and (v)
v12/a1 = v22/a2…………(vi)
From (iii), (iv) and (vi)
2a1S1/a1 = 2a2S2/a2.
S1 = S2
Hence distances covered S1 and S2 are equal.

Q6. The power of a pump motor is 4KW. How much water  in kg/minute can it raise a height of 20m? (g = 10 m/s2)

Ans6. Given power of motor P = 4KW =4000 W
If mass of water raised in one second = m kg.
Total work done in lifting water,W = mgh
Power P = W/t, but t = 1 minute = 60 sec.
4000 = mgh/60
4000 = (m x 10 x 20)/60
m = 1200 kg.

Q7. A rod of length 3m is suspendedvertically from a fixed point. It is given an angular displacement of 60oin the vertical plane. If its mass per unit length is 2 kg then find the workdone?

Ans7. Let ‘m’ be the mass of the rod and’l’ be its length then
m = 2 x 3 = 6 kg
If the rod is displaced through an angle qthen the work done on it, W = mg(l/2)(1 – Cosq).
The effective length of the rod is taken to be (l/2) because in uniformdistribution of mass the centre of mass is at the geometric centre so
W = 6 x 10 x (3/2)(1 – Cos60) =45 J.

Three mark questions with answers

Q1. When a mass m2is at rest and mass m1 moving with velocity u1 hits itelastically, show that the fraction of the momentum transferred to the mass atrest is 2n/(1 + n) where n is ratio of the masses.

Ans1. v1 = (m1 -m2)u1/(m1 + m2).
v2 = 2m1u1/(m1 + m2).
This is from the theory of conservation of momentum.
Momentum of the mass m2 after collision,
P2 = m2v2 = (2m1m2u1)/(m1+ m2)
Fraction of momentum transferred to m2.
= (2m1m2u1)/(m1 + m2)m1u1= 2m2/(m1 + m2)
= 2n/(1 + n) ….[because m2/m1 = n]

Q2. Is it possible for work to bepositive negative or zero? "Explain with example.

Ans2. Work is a scalar quantity which isgiven by the scalar product of the force applied F and the displacement S movedby body i.e.
W = F.S,
W = FS Cos q ……….(1).
If q = 0. The work is maximum.
It remains +ve for the angle q between 0oto 90o
height=45src=”./xi%20work,%20power%20and%20energy_files/image002.gif”alt=”tutor2phyworkpowerandenergyf06q2i1.gif (1062 bytes)” v:shapes=”_x0000_i1037″>
or,
q lying between 270o and360o i.e., if  the displacement is in a directionoppisite to which the force is applied.
height=59src=”./xi%20work,%20power%20and%20energy_files/image003.gif”alt=”tutor2phyworkpowerandenergyf06q2i2.gif (1097 bytes)” v:shapes=”_x0000_i1038″>
Thus work is +ve if Cos q is +ve. The work done will be -veif Cosq is -ve i.e. q lies between 90o to 270o.
If
q = 90o then Cos 90o= 0.
Hence work done W = FS Cos 90o = 0.
Thus, W may be +ve, -ve or zero

Q3. The bob of a simple pendulum isreleased from a horizontal position. If the length of the pendulum is 2m, whatis the speed with which the bob arrives at the lowermost point? Given that itdissipates 10% of its initial energy against air resistance?

Ans3. Gravitational potential energy atthe highest position
= mg x 2 Joules = 2mg Joules
Kinetic energy at lowest position
= Potential energy at the highest position – the energy dissipitated againstair resistance or friction
= [mg x 2 – (10/100) x mg x2] Joule
= mg x 18/10 J
(1/2)mv2 = mg x18/10
or, v = 1.9 ms-1.

Q4. Give the various gravitationalunits of work. Give their relations also.

Ans4. We know that the physical workdone is given by
W = FS
(1) In S.I system,
If F = 1 kg weight or 1 kg force and S = 1m then,
W = (1 kg wt)(1m) = 1 kg m ……………(i).
Hence, one kgm is the gravitational unit of work in S.I (M.K.S) system and isdefined as the amount of work done if 1 kg force displaces a body through 1m inthe direction of the applied force.
(2) In C.G.S system,
F = 1 gmwt and S = 1cm,
W = (1 gm wt) (1 cm) = 1 gm cm ………………(ii).
Hence, one gm cm is the gravitational unit of work and is defined as the amountof work done, if 1 gm force displaces a body through 1 cm in the direction ofthe applied force.
1 gm cm = 980 ergs.
NOTE: 1 kg m = 9.8 Joules.

Q5. What is 1 e.v (electron volt)?Give its values in Joules. Give values of 1 MeV and 1 BeV also.

Ans5. One electron volt is the unit ofenergy in Atomic and Nuclear Physics.
One electron volt is the energy acquired by one elctron in moving it betweentwo point having a P.D of 1V.
Thus, 1eV = (1.6 x 10-19) C x1J/C = 1.6 x 10-19 Joules.
NOTE: The other practical units used are
1 Million electron volt = 1 MeV = 106 eV, 1 MeV = 106 x1.6 x 10-19 J, 1 MeV = 1.6 x10-13 Joules and
1 Billion eV = 109 eV, 1 BeV = 1.6 x 10-10joules.

Q6. When water is flowing through apipe then its velocity changes by 5%, find the change in the power of water?

Ans6. Power = Force xVelocity = Rate of change of momentum x velocity ={(mass/time) x velocity} xvelocity = {(adv) x v} x v =adv3 where ‘a’ is area of cross section, ‘d’ is the density of waterand ‘v’ is the velocity of flow of water.
Therefore, Power of water is directly proportional to the cube of velocity ofwater so let P = Kv3 (k is a constant and is equal  to ‘ad’.)Taking log on both sides
log P = 3log v + log k
Differentiating on both sides
DP/P = 3.Dv/v
percentage change in power,
DP/P x100 = 3 x 5%
= 15%.

Q7. The kinetic energy of rushing outwater from a dam is used in rotating a turbine. The pipe through which water isrushing is 2.4 meters and its speed is 12 m/sec. Assuming that whole of kineticenergy of the water is used in rotating the turbine, calculate the currentproduced if efficiency of the dynamo is 60% and the station transmits power at240 kV. Density of water = 103 kg/m3.

Ans7. Given that
r = radius of pipe = 1.2m, average speed of water v = 12 m/s
V = 240 kV = 240 x 103 volt, density ofwater p = 103 kg/m3.
Now, kinetic energy of rushing water per second i.e.
Power P = (1/2)(mass flowing per sec) x v2
= (1/2)pr2(l/t) rv2
= (1/2)
pr2 rv3
= (1/2) x 3.14 x (1.2)2x 103 x (12)3watt
= 3.9 x 106 watt
Current in the transmissioncables
current = output power/voltage
= (60% of power P)/(240 x 1000)
= [(60/100) x 3.9 x 106]/(240x 1000) = 9.75 amp.

Five mark questions with answers

Q1. A vehicle of mass m isaccelerated from rest when a constant power P is supplied by its engine; showthat :
(a) The velocity is given as a function of time by
v = (2Pt/m)1/2
(b) The position is given as a function of time by
s = (8P/9m)1/2t3/2.
(c) What is the shape of the graph between velocity and mass of the vehicle ifother factors remain same?
(d) What is the shape of the graph between displacement and power?

Ans1. (a) Given that Power = Fv = P =constant
i.e., m x (dv/dt) x v =P [as F = ma = m x (dv/dt)]
After rearranging and integrating on both sides
òv dv = ò(P/m) x dt
(v2/2) = (P/m) x t + C1
Now as initially the body is at rest, i.e., v = 0 at t = 0, so C1= 0.
v = (2Pt/m)1/2 …………(1)
(b) By definition v = (ds/dt),
Using eq (1) above,
ds/dt = (2Pt/m)1/2
On integration we get
òds = ò(2Pt/m)1/2 dt
s = (2P/m)1/2 x (2/3) xt3/2 + C2.
Now, as at t = 0, s = 0, so, C2 = 0
s = (8P/9m)1/2 t3/2.
(c)
height=104src=”./xi%20work,%20power%20and%20energy_files/image004.gif”alt=”5a1.gif (1325 bytes)” v:shapes=”_x0000_i1046″>

(d)
height=102src=”./xi%20work,%20power%20and%20energy_files/image005.gif”alt=”5a1i.gif (1265 bytes)” v:shapes=”_x0000_i1047″>

Q2. A simple pendulum of mass m and lengthl swings back and forth up to a maximum angle q0 with the vertical. When at an angle q, what is its (a) potential energy, (b) kinetic energy, (c) speed, and(d) tension?

Ans2.
height=149src=”./xi%20work,%20power%20and%20energy_files/image006.gif”alt=”5a2.gif (2041 bytes)” v:shapes=”_x0000_i1048″>
Taking the reference level at the lowest point R, we have
hP = l – l cos q0 = l(1 – cos q0)
hQ = l – l cos q = l(1 – cos q)
So (a) potential energy at Q relative to R will be
PE = mghQ
PE = mgl(1 – cos
q) ………(a)
(b) PE at P = mghP = mgl(1 – cos q0)
KE at P = 1/2 x mv2 = 0
so, total mechanical energy at P = mgl(1 – cos
q0) …….(i)
Now, if KQ is the KE at Q,
then using eq. (i)
mechanical energy at Q = KQ + mgl(1 – cos
q) ……….(ii)
But by conservation of mechanical energy between P and Q
KQ + mgl(1 – cos
q) = mgl(1 -cos q0)
i.e., KQ = mgl(cos q– cos q0) ……..(b)
(c) If v is the speed at point Q, from eq. (b)
1/2 x mv2 = mgl(cos q – cos q0)
i.e., v = height=17src=”./xi%20work,%20power%20and%20energy_files/image007.gif”alt=”i2.gif (1049 bytes)” v:shapes=”_x0000_i1049″>.
(d) If ‘E’ is the energy at
Ðq, then itis equal to mgl(1 – Cosq) + (1/2)mv2.
Since the energy remains constant throughout, E = Eo.
mgl(1 – Cosq) + (1/2)mv2 = mgl(1 -Cosqo)
or mv2 = 2mgl(Cosq – Cosqo)
Therefore, tension ‘T’ at q would begiven by
T = mv2/l + mgCos
q = mg Cosq + 2mg(Cosq– Cosqo)
or T = 3mgCosq – 2 mgCosqo.

 

Q3.What do youmean by work in the language of physics? Give its absolute and gravitationalunits. Give two illustrations of zero work, negative work and positive work.

 

Ans.(Try yourself).

 

Q4.How will youfind work done by a variable force mathematically and graphically?

 

Ans.(Try yourself).

 

Q5.What do youmean by conservative and non-conservative forces? Give their importantproperties.

 

Ans.(Try yourself).

 

Q6.What do youmean by gravitational potential energy? Show that gravitational potentialenergy is independent of the path followed.

 

Ans.(Try yourself).

 

Q7.If a body iskept on the top of a rough inclined plane, find the expression for
(i) work done in bringing it down to the bottom of the plane with constantvelocity
(ii) work done in moving it up the plane with constant acceleration
(iii) work done in moving it down the plane with constant acceleration.

 

Ans.(Try yourself).

Potential

Devna asked.

find the potential at A ….a prson brings a mass of 1 kg from infinity to the pt A .initially the mass was at rest but it moves at a speed of 2 m/sas it reches A .the work done by person on mass is -3 J

A Few Problems from Kinematics

  1. A body falling freely from rest has velocity v after it falls a height h . Calculate the distance it should fall down further for its velocity to become double
  2. A particle in uniform acceleration in a straight line has a speed of v m/s at position x meter is given by √(25-16x). What is the acceleration of the particle


Average Speed

Naman Asks:

“A car travels from A to B with a speed of 30km/h and returns back to A with a speed of 50km/h.Find distance,displacement,average speed and average velocity ifdistance between A and B is 100km.”

 

Answer:

Distance 200 m

Displacement 0

Average velocity 0

Average speed = = 37.5 kmph

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Half Yearly Exam Count Down

Half Yearly Exam Count Down KVS Ernakulam RegionOctober 17th, 2017
The Half Yearly Examination for Kendriya Vidyalayas of Kerala starts from 17 October 2017
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