Yes , you should definitely go for JEE Advanced 2022. You might get a better college than NIT AP .There is no harm in trying for a better college. Moreover there is no risk for you as you are already in a NIT. So, I personally feel it's worth taking a shot .

Hello Applicant

Considering the existence of Bohr's atomic model as the distance from nucleus increases, energy difference between two successive Bohr orbit tends to decrease, that is why there is no sharp differentiation between two successive Bohr orbits after a certain distance without considering energy. Hence the Answer would be 'decreasing' due the the above explanation.

Hope it helps!!!

Hi there,

If the shell has n=2 quantum number value for then there will be 4 orbitals. These orbitals contains 8 electric each

So there will be one 2s orbital and three 2p orbitals in the atom.

I hope this explanation will help you revert if you face any kind of doubt.

Thanks!!

Hi. For students who are in final year, they need to upload marks only till the 4th semester. For students who are pass outs, they need to upload marks up to 6th semester but since your 4th semester marks are not with you, you can go on putting the aggregate marks , missing out on the 4th semester. They will only be checking your marks till the interview period.

Hello aspirant,

The percentile you got in JEE Mains can be used to calculate your marks. Since you haven't mentioned your percentile or your rank, I am providing you with a link to help you with the same:
https://engineering.careers360.com/articles/jee-main-marks-vs-percentile

Hope this helps you.
In case of queries, feel free to revert back in the comments.

Hello,

Solution for your question is:

4/3 π (R^3- r^3)ρmg   = 4/3 π R ^3 ρwg

1- (r/ R) ^3 =8 / 27

r /R  =( 19/27)^1/3= 19^1/3 /3

=0.88 =8/9

Hope it helps

Feel free to ask if you have  any questions

Good luck!

Hello Dear,

1. One charge Q is present on the thin spherical shell , so the shell is totally charged Q.
2. Second charge Q is placed at the center of the spherical shell.
3. Now for potential at point P we have

• Let V1 is potential at P due to Q on spherical shell and let V2 is potential at P due to center charge, let V is total potential at P. So now

V=V1+V2

V=0 + 9*10^9*Q*2/d

V=18*10^9*Q/d

Hello student,

As per valence shell electron pair repulsion theory the shape of

• ch4 is tetrahedral
• NH3 e shape is trigonal pyramidalpyramidal
• The shape of sf4 is trigonal bipyramidal
• the shape of H2O is tetrahedral
• The shape of clf3 is trigonal bipyramidal.

Hope this will help you