hey man its easy

center of cube q/epsilon0

center of edge  q/4epsilon0

at one corner  q/8epsilon0

hope i m right or i need a revision........

hey man something about the conductors should be given dimensions ,capacitance etc.

hey i m not getting ur question

is the surrounding space filled with charge?

yes i agree with rhd for ist ques.

rhd i m also from bhilai

by conservation of momentum

v₁/v₂=tan

v₁/v₂=tan 60=root 3

v₂=v₁/root 3

hope u got it.

I know the ans of 2nd question man.

wait i m giving the ans.

yes sir thats right i think

hey angel i m not getting it plz tell the soln.

I m not cofident still i m trying

PV=nRT

PV₁=n₁RT    FOR 35Cl

PV₂=n₂RT         FOR 37Cl

NOW ASSUMING THAT BOTH THE GASES HAVE SAME WT.

L₁/L₂=37/35

TELL ME WHERE I M WRONG....

hey i think its easy

lim  {(a-n)nx-tan x}/x   *   (sin nx)/x

lim  [ {(a-n)nx}/x  -(tanx/x)]  *  (sin nx/nx)*n

lim  {(a-n)n -1} * n        .........                {sinx/x=1 & tanx/x=1 as x-0}

now putting the limit=

a=n+1/n

hope u got it......

hey guys look at my method and tell me if i m wrong

x³-1 has 1 and w as its roots.

now putting these roots in the given eqn. we get

a+b+c=-6.....................................(1)

now putting w in the eqn we get

aw+b+1+cw²+3w+2=0

(3+b)+(a+3)w+cw²=0

also we know that 1+w+w²=0

therefore b+3=a+3=c............(2)

a=b=c-3

now from eqn i and 2

a+a+a=-9

a=-3,b=-3,c=0

ans=-54

yes sir u r right.

yes spidey u r right in [-1,0)  log(1+x)>x which is also seen graphically

hey I have not posted that one bcoz i cant do such a blunder some one has posted a rubbish soln. on my name.I need to find him.

sorry for such a rubbish soln. as it is on my name.

hey man it is easy just diff. both sudes as allamraju has done

a shortcut method

also keep in mind the domain of the original function.

we get 1/(1+x)<=1  {rolles theorem}

which is always posiible in the interval (-1,infinity).

hope u got it.