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You could actually expect a seat in NIT..........that too in first round of counselling.........even if u get a lower branch in any NIT.......u should go 4 it as there is very high probability of changing ur branch once u entered the college.
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Hi Heman, as u r from rajasthan .........hence it would be difficult 4 u to get some seat in aieee first round of counselling..........however u may get metallurgy @ nit jaipur, but not before second round of counselling..........if u want any help, i would there @ MNIT during couselling.
Good Luck!!
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Go for NIT Kurukshetra for sure................
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Go for Brilliant.............its the best..........FIITJEE is also good but the question asked in its test series especially in Maths are iirevelent (in the sense few probs. are way above JEE level)...........but FIITJEE you've an advantage of competing with more no. of students & %ile socred by you nearly matches your JEE ranks.
Good luck & cheers!!!
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let 2^x =a & 3^x =b.
then eqn. becomes
4a^2 - ab - 18b^2 = 0.
4a^2 + 8ab -9ab - 18b^2 = 0.
4a(a+2b) - 9b(a+2b) = 0.
(4a-9b)(a+2b) = 0.
but a + 2b != 0.( since sum of two exp. fn. can't be zero.)
4a=9b
4(2)^x = 9(3)^x
(3/2)^x = (3/2)^2
x=2.
good luck!!
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Welcoming all MNITinas................I'm your senior
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Yeah it has less scope than electonics.......but the course is very good & has good scope in future..........but I'm not sure of kind of placements in NIT jalandhar
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do rate me for my efforts
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Let A be the common root then, satisfying both the eqns. we get
A^3 + 3pA^2 + 3qA +r = 0.......................(1)
& A^2 + 2pA + q =0.............................(2)
reducing eqn(1) with the help of eqn.2 we get
A [ A^2 + 3pA + 3q] + r =0
=> A [ (A^2 + 2pA + q) + pA + 2q] +r =0 [Using (1)]
=> A [pA + 2q] + r = 0
=> pA^2 + 2qA + r = 0.................................(3)
solve (2) & (3) for A & A^2
put [A]^2 = A^2 to get the desired result
I hope this is enounghj
Good luck!!
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Go for NSIT for sure.............keep brach above the college in preference list.............& BITS & NSIT are almost at the same level.
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@talker
Knowing & Understanding things are different things all together.
Whatever fused_bulb said...........if could follow that with proper dicipline one could get through.
Don't take my words hard but that's true & applicable
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For EE it is 9 lac nearly.........& for civil it is 7.5 lac.........(avg,)
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x above is mole fractions of oxygen.
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as r(pure)/r(impure) = (M(impure)/M(pure))^(1/2).
Applying above formulae.....
M(impure) = 32/(1.98^2) = 8.16
since, M(impure) = M(oxygen)*x +M(ozone)*(1-x) [formuale of molecular wt. of mixture]
find x...........& hece find %age of ozone.
Good luck!!
Keep working hard!!
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Scope of chemical is not that good in pvt. colleges.............most the people goes in the field of software(more than 75%,even in NITs)........................Go for E & T for sure.........the field of telecommunication has very good scope.............
Best of luck!!
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