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I aim to pursue the integrated MS degree from IISER. I am aware that i can apply from my IIT -JEE EML rank and by direct admission . But there is a slight problem as the IIT-JEE server has not recored my application as received. Is the direct admission to IISER limited to those who pass class XII in 2010 ? I passed class XII in 2009 ( with top class marks of 92% cbse ) , but was unable to take up any academic courses for that year due to medical reasons . can you please help me out with this ? 

5 years ago


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Hi , My rank in the OBC NCL list is 15380 , what are the courses that I can get in NIT trichy or any other NITs ? Thank you
I will suggest you to visit the official website of JEE main to look for the opening and closing rank for previous year of different rounds. All the statistic is available there. Thanks
Vijay Mukati 4 months ago
To get a complete view and a detailed information regarding your stature you need to visit the official weblink of JEE Mains and you may be able to find out the exact stature of yours,...
Soneel Verma 4 months ago
Which college provides better placements? Ism dhanbad or the NIT`s? And why?
The rate of placements varies everywhere, no educational organization can commit hundred percent placement in all branches it is generally 85 to 95% and may be somewhere less than that as...
Ayushmaan Vardhan 4 months ago
It depends on the branch you take.ISM is obviously above than NIT’s but if the branch you take is low in ISM than you should take NIT.
Akshit Matthew 4 months ago
i had got 75% in pcm in 12 th cbse boards....but got only 58% in i eligible to appear for jee advanced...???
YA i knw abt it....but i was confusd abt the eligibility of my marks for jee adv......................................................
vijay sharma one month ago
my overall score in pcm is 75%....and in all subjects is 83%.........................................................
vijay sharma one month ago
but before appearing for advanced you must qualify jee-mains exam,then only your aligible to write advance.
mansi one month 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 5 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 5 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 5 months ago
sir, i have got AIR 1500 in jee advanced,Out of aeropspace or civil branch in IIT-Bombay which is better..? moreover if i want to go for MBA later on what shall be best suited for me?
aeropspace it would be a much finer field for this time and the future shobahm can u porvide me tour contact number i want to talk about your rank
grenade 5 months ago
thanks for your help....grenade
Shubham gupta 5 months ago
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