Answer laterHey there, out of all 6 semesters, ONLY my 4th semester result is awaited I have 1,2,3,5,and 6th semester result with me, since Im filling my cat form, I dont know what should I do? Shell I just put the aggregate marks of the rest results that I have or something else?
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.
the electric field inside a spherical shell of uniform surface charge density is
In a spherical shell, all the charges resides at outer surface or you can just say on the surface of the shell.
Since all the charge resides on the outer surface of the spherical shell, so according to Gauss law, electric field of a uniformly surface charge density spherical shell is zero.
how shell I caliculate my jee main entrance marks?
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.
a hollow spherical shell at outer radius R floats just submerged under the water surface. the inner radius of the shell is r.if the specific gravity of the shell material is 27/8 w.r.t water,the value of r is;
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!
Thin spherical conducting shell of radius Rhas charge Qanother charge Q centre of theshell electrostatic potential of P at distance d/2 fron the centre of the shell?
Hello Dear,
- One charge Q is present on the thin spherical shell , so the shell is totally charged Q.
- Second charge Q is placed at the center of the spherical shell.
- 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
explain the VSEPR( valence shell electron pair repulsion)theory and explain the shape of following on the basis of this theory :- CH4, NH3, H2O, SF4, ClF3 , ICL2-
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
a shell of mass m moving horizontally explodes into two equal
This is based on momentum conservation method. You can find this problem in any textbooks. Also if you want to know the physical aspects about it search in youtube or see some websites available. See previous year neet or jee questions, you will found these type of problems were asked long ago. Hope it helps.
