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Sreejith P Department of Physics Malabar Christian College, Calicut |
| Information about PHYSICS, JEST 2013, Question Answer Key, 17 February - Discuss the Answers and more... Free Download Question Papers and study materials.. NB : The details given here are taken from various sources. Please Verify the contents and contact us for your valuable suggestions. |
1) Poisson bracket {x^2+p,p}
ReplyDeleteans : -2x
2) Value of the series
ans : 1
3) conventional Electric and magnetic fields
ans : E along y, B along z
4) the operator {(d/dx)-x}{(d/dx)+x} is
ans : {d^2/dx^2}-x^2 + 1
5) ground state wavefunction if both e are arranged in same spin state
ans: antisymmetric state wave functions with different states (i.e combination of both states)
good all are correct....
Delete5) Total ground state energy of 4 particles in the 3 cases
ReplyDeleteans : Ef>Eb=Ec
yes its correct
Delete7)ratio of the probability that 2 particle in same state to the different state
ReplyDeleteans : boson 1, for classical particle 1/2
ya its correct
Deletein case of 5) the classical particle has h=0 and therefore ground state energy is zero (it just stays put); So the answer should be Ef>Eb>Ec
ReplyDeleteAnyway my set was R
for a diatomic ideal gas; what fraction of heat supplied available for external work= 2/7
The complex variable limit doesn't exist
for free charge to disappear time is of the order of 10^-17 secs
Probability that there are no buses in five minutes 0.19
Energy of Neutrino 236 MeV
Probability that the coin you've drawn 0.075
ratio of electrostatic energies of shell and solid sphere 5:6
maximum flat surface covered by discs pi/2*root3
Presure on the liquid sphere due to gravitation 2/3*pi*G*(rho)^2 (Rsquared-rsquared)
Drunks getting together again prob= (2n)!/4^n*(n!)^2
Parabolic and spherical orbit circular radius= 2*parabola pericentre distance
free fall time of test mass= (pi/2+1)sqrt(R^3/GM)
Viscous medium velocity sqrt(g/k)tan(sqrt(gkt))
age of rock sample 38*10^6 years
period of simple pendulum 2T/sqrt 3
mean value of x in the distribution function= 2*lambda
coordinate transformation none of the above
gyro-magnetic ratio Q/2M
electric and magnetic field n=m=1
angle ade by the string wrt vertical 7 degree
metal bullet temp increase 10 deg C
eigenvalues of operator H= sigma.a is +-modulus of a
only the first one is a possible electrostatic field, second one isn't
velocity if particle whose KE= rest energy is c/sqrt 2
I expect around 45, don't know if that's good enough. Btw, I'm in BSc final year, therefore I don't have much seats to play with either.
Best of luck to all.
sorry dear, all ans are not correct...ok....
Deletemr. Chiranjibbutun, what is the best way to prepare for the JEST exam. I am also a B.Sc. final passout. but I am confused over the preparations to be made for such exams. Jest 2013 is the first entrance that I have appeared in my life. please help me regarding the above issue
Deletevelocity of particle whose K.E.=mc2 is root3c/2..ok use this K.E.=total energy-rest mass energy...i appropriated your effort and hard work..plz give hints...to every questions which..u have done...
Deletegot it wrong. dear me, I can't even calculate. And also I got the liquid pressure prob wrong.
DeleteAnyway,Problems: The bus problem- you have to apply Poisson Distribution and set x=0;
The Ground state energy problem : take inner product of old ground state and new
maximum flat surface covered: hexagonal arrangement is the most dense by thue theorem
for free charge to disappear: mash up ohms law, eqn of contunity and gauss law and you're done
free fall time : Take energy conservation and then brute integration
Complex Variable limit: Take Re z= Imz and evaluate the limit, next take Im z=0 and evaluate the limit, they are not equal. so doesnt exist
ratio of electrostatic energy : Just integrate our good old friend u= 1/2 e0E0^2
And Ohmswasticka, I appreciate your insights very much, with the amount of insight you have, surely you'll be ranked very high. In that case, my congratulations to you. Do you have any idea when the official answer key will come out?
plz explain prob. that the coin u have drawn .075 (this result)
Deleteok thanx i will try ur hints...bt sorry i dnt have any idea abt ans keys.....gud luck...
Deleteto pass separately in part a and b?how much do we have to score in both sections pls reply?
ReplyDeletedo we have to pass separately in part a and b?
Deleteno i dnt think so....u just need a minimum marks of nearly 25 to qualify this.....
Deletebut sorry i think 25 is not enough this year. because all the friends of mine have answered about 40. and our college is a middle ranked college . I think this year 45-50 is appropriate for obtaining a rank
Deletebut i think this year 45-50 is needed. because all of frnds of mine have scored around 40 and our college is middle ranked. or may be the lower rank gives no opportunity at all
Deletebut sorry i think 25 is not enough this year. because all the friends of mine have answered about 40. and our college is a middle ranked college . I think this year 45-50 is appropriate for obtaining a rank
DeleteAnswer key is out people ....go check it on jest official site..
DeleteYou will know,when you check the answer key.Answering is not enough,answering correct is...
DeleteWhich college are youfrom,btw?
This comment has been removed by the author.
ReplyDeletePlzz give answers to following questions..
ReplyDeleteques 23 series-P,rate at which particle leave from a hole on one side of box..of area A..
ques 16 series P,probability that the particle will be in ground state after sudden expansion of walls.
ques 11 eq. of curve for which lights are reflected parallel to x-axis
Can you please say my answer is correct or not
ReplyDeletepressure on liquid sphere problem option d
answer was the one with 2 pi (rho)^2
DeleteThe answer was 2*pi*(rho)^2 something
DeleteOfficial answer key is available now on Jest website...
ReplyDeleteI've got only 15 :(
ReplyDeleteI've got only around 15 :(
ReplyDeleteAny one can suggest which books would be appropriate for topics in physics exclusively for JEST preparation?
ReplyDelete