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when will the iit eml list 2010 will be published?


 

5 years ago

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Answers : (1)

										

Hi ,


As soon as the EML will be declared , you will find it on askiitians website..


 


Please feel free to post as many doubts on our disucssion forum as you can.  We are all IITians and here to help you in your IIT JEE preparation.


 


AskiitiansExpert


Amit Yadav

5 years ago

Post Your Answer

hi my name is harshit and my question is what is the minimum rank you have to get to be an aerospace engineer?
 
 
For aerospace engineerg in IITs you should get the rank in between 400 to 1000
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Avinash 6 months ago
sir my if JEE mains score is 65 in written will the person able to sit in counsiling by JOSAA
 
 
Immediately visit the official weblink of Josaa and get to know whether you may still be allotted a seat as per the counsellings, if not then try at some other renowned institutions like –...
 
Soneel Verma 6 months ago
 
Yes you can sit in counselling based on your score in OBC category I think that you can get an NIT in the 4 th round But I think josaa registrations are over!!
 
Akshit Matthew 6 months ago
which is a better option? IIT G CSE OR BITS PILANI CSE
 
 
BITS Pilani stands atop among the both but you may check in details at other institutions too in order to go for a comparative analysis, consider visiting the official weblinks of these – R ...
 
Ayushmaan Vardhan 7 months ago
 
coz it is better than iit G bt if u want the tag iit then only prefer iit G
 
navneet 7 months ago
 
both are good bt pilani is more better
 
navneet 7 months ago
how to become a topper
 
 
If you study anything by concentrating actively it will last forever.! Give your best.
  img
Komal 9 months ago
 
Doing hard work is the key to success.
 
Adarsh 8 months ago
which books should i prefer for physics and chem
 
 
For physics you can do H.C. Verma it will make your concepts clear and for more practice also do D.C.Pandey(Objective Physics for Medical Ebtrances) For chemistry you can do GRB books. You...
 
Carmel Emmanuel 5 months ago
From equilibrium class 11
 
 
For the first question Δ G o = -2.303RTlogK C logK C = Δ G o /-2.303RT = -13.8*1000/2.303*8.314*298 = -2.4186 Taking antilog we get the value of K C as 0.00382 . Please approve if my answer ...
 
Adarsh 8 months ago
 
There was a mistake in the solution of the second part of the question The correct solution is Δ G o = -2.303RTlogK C = -2.303*8.314*300*log2*10 13 = - 76403.01
 
Adarsh 8 months ago
 
For the second part of the question Δ G o = -2.303RTlogK C = -2.303*8.314*log2*10 13 = -254.6767
 
Adarsh 8 months ago
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