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Q. A positive point charge Q is kept inside a neutral conducting shell (not in the centre). An external uniform electric field E is applied. Then

(a) Force on Q due to E is zero.

(b) Net force on Q is zero.

(c) Net force acting on Q and conducting shell considered as a system is zero.

(d) Net force acting on the shell due to E is zero.

Please answer with proper reasoning?????

Hi Wisp

I think

The speed is

v =

(GM/R)

look,

When it is orbitting,

mv²/R = GMm/R² .........(R=radius of Earth)

So,

v =

(GM/R)

Plzzzzzzzzzz rate me..........................

Please Someone help..........................................

This is a discontinuous func. and so directly putting the limits wont work out......................

Please can anyone advice me how to order Arihant books on the Internet??

I will be greatful to you.

Solved it yo

No common terms

Ok

now let the n^thterm of the first A.P. be equal to the n' ^th term of the second AP

So,

102 + ( n-1 ) 8 = 108 + ( n'-1) 12

2 +12n' = 8n

n = ( 1+6n' )/4

now, for any value of n' , :; 6n' is even.

( 6n' +1 ) is odd

So, ( 1+ 6n' )/4 is a fraction .

So, no integral value of n exists.

So, the serieses have no term in common.

Please Rate me Guys......................................................

This is my favourite article by Kushwant Singh in the Telegraph
Thanks for putting it up on the net
U deserve Salutes man

Ya but the answer given is 7

²/2 and not -7

²/2

And keep in touch throuh ur nudgebook

_{[ -2 ]}

^{[ 5 ]} cot~1(tanx) dx

Here cot~1represents cot inverse

(a) 7

/2 (b) 7

²/2 (c) 3

/2 (d)

²

Please post solution with full explanation

Thanks Kab
You have cleared my dout.
plz accept my salute
Thanks

FIITJEE is the best
But before goin through FIITJEE , goin through the ARIHANT series makes it a perfect combination for IIT-JEE

Guys please explan how tan1 to 100 is less than 2??????????????

PLEASE HELP SOMEONE

But how does the value of tan1 to 100 lie between 2??
coz see putting x=100 we get tan~1 (100)=89.4275
Please help i am not getting this
Bubun

Cool question yo
Wait if solving it takes a bit of time, I will Surely send it to u tomorow......