Chetankumar Mohane

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Viewing 10 posts - 11 through 20 (of 37 total)
    1. I wants to request to InI community to start a new single thread for “Engineering Mathematics” & “General Aptitude” sections.
      Reasons for that are:-
      1)These subjects are common for almost all disciplines.
      2)Mathematics & Apti sections are frequently neglected by students.
      3)New thread will give these topics more scope & versatility.
      4)These are the sections where we can improve our GATE score.

    1. For those who don’t know the question asked by Ganesh, Here it is:-
      Please right click the link and click on “open in new tab”

    1. @Kanika, I think now you can not change the zone! Because as far as I know only examination city can be changed & not the zone.
      As they never ever mentioned anywhere about changing the zone.
      The only way to know if this is possible or not is to contact the officials in IIT Kanpur
      You can make a call to IIT kanpur or email them, they have given contact info in their website.
      But Hurry up; the last date of accepting change in city is 30th Dec.

      Hope you will get your desired city of examination.

    1. @Ritesh,
      According to PK Nag,
      “TdS= dU+pdV this equation can be applied in all the processes undergone by closed system” is true provided that there is no other forms of major energies (like KE, PE, viscuos, Electical) involved in the process.
      You will get the answer after you can understands how actually this equation comes.
      The original equation is,
      dQ-dW=dE …………1 … (from First law of thermodynamics)
      where dQ=small change in heat
      dW=small change in work
      dE=small change in total energy
      now, dE=dU+dKE+dPE+dE+…..
      where, dU,dKE,dPE,dE are small change in internal energy, K.E., P.E., Electrical energy resp.
      we no for many processes we can neglect dKE,dPE&dE & other minor energies.
      Also we know dS=dQ/T
      & dW=pdV…. this is the work done on the system or by the system
      so equ 1 becomes,
      TdS= dU+pdV

      So the term “pdV” represents the work done on the system or by the system & it have no mean whether the process is reversible or irreversible.

      Hope you understands it. 🙂

    1. @Ganesh, Yes I agree, Heat Transfer is difficult subject. Up to conduction it is quite easy but after that is becomes more difficult because of very long & complex formulas & conditions as you said.
      But we have to remember those formulas exactly; we dont have any choice for this 🙁
      Only thing we can do is write those formulas in systematic way & try to remember them.
      Also if you solve many problems on it you are getting used to remembering those formulas.
      All the best..

    1. @Ritesh, One of my friend also joined Targate test series. I think it is good organization; keep it up. If you find any interesting questions then you can post here on InI.

      I am also going to solve previous years papers in January. Right now I am studying. Still lots of study is left. But I must practice some papers. I wish I can start practicing as early as possible. 🙂

    1. @Ganesh, I got the question paper of GATE 2005 & read that question. Still I took much time to understand the question right. May be because of I am not aware of how exactly the elevator works. I always wondered that why that counterweight(CW) is there when we have electric motor in place. Then I googled it finally understands the use of CW is to save the energy.
      Now coming back to the question;
      Here, we can see the energy in the cage is in the form of K.E. which equals to, 0.5*m*v*v
      Also K.E. in counterweight(CW) is, 0.5*m*v*v
      Even the cage & CW are moving in opposite direction, they are pulling the chain in same direction (clockwise).
      So combining the two K.E.
      (0.5*m*v*v) + (0.5*m*v*v) = m*v*v
      Now the motor should have minimum energy which could be able to oppose this combined energy. In other words, the motor should do “WORK” to oppose the K.E. which is equals to, m*v*v.
      So, minimum work the motor must generate = m*v*v
      Now as we know, Power = Work done / time
      we can say,
      Minimum power the motor must have = (m*v*v)/t

      check out the answer.

    1. @Dinesh; Thanks Dinesh. Downloaded the doc successfully. It is good presentation about forming process.
      Nice trick btw 🙂

    1. @Ritesh, Exactly, thats what I wants to say, concentrate more on our core subjects 🙂

      Ritesh I haven’t joined any test series. Did you know any good organization??

    1. @Ganesh, Frankly speaking I could not able to imagine the problem right!!! More explanation is needed 🙁
      whether they give any figure along with the question?
      You said this question is from GATE 05; please also state the stream (production, mech, civil) so that I can search this question there!!!
      Reply soon…

Viewing 10 posts - 11 through 20 (of 37 total)