Messages posted by krishna.gopal

Good work Vasanth

Hi I think I can't comment on this right now as I don't have HC Verma with me. But I think The way eshaan has approached is logical and if I feel it should be correct

Good work Nishant and nice try chirag. The average velocity is about 444m/s which gives time =8hrs

Right answer by Anuj and Pratikesh

Hi yahiya I think that your idea of working towards problem seems OK to me but I think that there is some mistake in your second integral. I would like to know what you have taken as dV. It should be 4pir^2dr. And hence you should not get a term containing r^1 in second expression of field.

Anyhow There is an easy way of finding this solution, by applying gauss law (But not the way DON has approached)

As its a case of spherically symetric field E*(4pir^2)=qin/epsilon_not

qin =

_{[0 ]}

^{[r ]}

*4pi*r²dr

for R/2<r<R This integral should be split in two parts

qin=

_{[0 ]}

^{[R/2 ]}

4pi*r²dr + _{[R/2 ]}

^{[r ]} 2

(1-r/R)4pi*r²dr

Integrate it to find qin and hence electric field

Pressure variation with heigh is given by

dP/dz = -dg (where dP/dz is differential and d is density and g is acc due to gravity in dg)

For air d= PM/RT

So dP/P = -(gM/RT)dz

gives P(z) =Po*exp(-gMz/(RT))

Where Po is pressure at surface = Patm

and P(z) is pressure at height z

Use this to find P(4150) and P(8300) and find difference. I have not done the exact calculation but I think it will come out to be 0.24 atm

(It can never be 2.4 and 24 atm as it must be less than 1 atm)

HI friends I would like to put my thoughts as well om this problem. What Raman sir has written is absolutly correct but i feel that there is a confusion over here. When Raman sir apply work energy theorm in his last post he is applying it on an electron (Electric force is applied on it and it gets KE). But what Vish want (As mentioned in 4th post of this thread) is to apply work energy theorm on the plate of a capacitor when we change the position from d to 2d without accilerating (Or to be more correct it should be by giving it infinitely small acceleration).

Now vish to look at your problem When two plates are kept at a distance of d from each other they are attracting each other by a force of Q^2/(2A

_o). To move it at infinitely small acceleration you must apply an equall force (or infinitely greater). If you want to apply work energy theorm the net force on plate was zero and there was no increase in its KE so work energy theorm is valid but that does not yield any result (zer = zero does not gives you any insight)

But if you find work done only by external force it is d*Q^2/(2A

_o) = Q^2(1/Cf-1/Ci) (where Ci and Cf are initial and final capacitances) which is equall to increasein energy of the capacitors. So External work done is stored in form of energy of capacitor.

Prabhat even as far as my knowledge go I think that work energy theorm can be apllied to non point objects. But if you actually have some paper which suggests that it can't be done we would be glad if you can share it with us.

And finally a kind request to everyone, please maintain the dacorum of this forum. enjoy.........

HI aman there is a mistake in your solution that you have chosen a value of g' which is an approximation if d << R. In general g'=g*r/R So force will be mgr/R

Right approach avadhesh. Varun try to solve it as avadhesh told. The answer will be R/n. You try to solve and get the answer. You may get a better insight if you make a ray diagram

Consider the sheet to be lying in yz plane and current is flowing along +z axis. The net field in the neighourhood of sheet will be have field because of sheet Bs and field because of other factors Bo. On one side of sheet field will be Bo+Bs and on other it will be Bo-Bs (Direction of Bs will be opposite on both sides)
Putting Bo+Bs = B1
and Bo-Bs= B2
Bo=(B1+B2)/2 (Here additions are vector additions)
Bs = mu*J/2 (where J is currect density)=(B1-B2)/2

Now consider a unit thin sheet of length l and width w.

It carries a current Jw and hence force on that piece is
F=iBl=JBo*lw
so pressure is JBo=(1/2mu)*(B1-B2)*(B+B2)

Sorry for delayed reply chaps but right answer is 11L/12. And Prabhat good work by you in explaining each step

Hey Arjun when force =IBLsin(theta) is calculated. You can always take l as displacement from end to and (And not the actual length of wire).

So force will be = I*B*lamda

HI Sajid as you have asked a question with three parts here i would tell you how to proceed for solution instead of giving the full solution. Try it and in case you do not get it let me know.

If sphere A was not grounded potential on this should have been K(q3/r3+q2/r2). In order to keep it at zero potential (It is grounded) A charge q1 wil come on it such that k(q1/r1+q2/r2+q3/r3)=0

This gives q1 = -3microC

Now region between A&B has a field of Kq1/r^2
Now region between B&C has a field of K(q1+q2)/r^2
Now region outside C has a field of K(q1+q2+q3)/r^2

Integrate electric field to get potentials of three spheres.
U=q1V1+q2V2+q3V3

For force on outer sphere sigma = q3/(4pir3^2)

Now force per unit area will be sigma*k( q1+q2+(q3)/2)/(r3)^2

To know why i am taking q3/2 look at concept of force per unit area on a charged conducting shell

Good work Sri. And mansi as far as reflection is concerned don't worry about reflection by construction (The basic concept remains same but construction will not be so straight forward) Remember the fact that light propogation is perpendicular to wavefront. You know direction of light propogation by ray optics and hence you can always tell the shape of wave front

Neeraj you are perfectly right. Even i can't think of anythong else.

Right answer is 6C (because capacitors are given in farads so q=CV= 3*2 = 6 C)

Yes of course you can

Adding SrCl2 in NaCl will create a non stochiometric defect where some of cation positions will be vacant and there will be Sr2+ ions on others(There numbers will be equall)

f all the bulbs are identical and all must be illuminated then that can be done in only 1 way

HI vish, I think there is a doubt. Axis of rotation is not chosen. It is the line about which all the points move in circle. In your case you can't chose one of the charge as axis of rotation

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