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Vaah terminator,, tune toh forum expert ko bhi nahi baksha!! nice..

Yups ,, now answering seriously,, the question stated by u,, itself states dat the spheres are kept at a wide distance,, their interaction will be minimum , so while calculating their individual capacitances, the interaction of the other wont be considered ,, or as earlier quoted dat ,, it can be considered dat the other capacitor lies at infinite distance,, their individual capacitances become $4\pi {\epsilon}_{0} {R}_{1}$ and $4\pi {\epsilon}_{0} {R}_{2}$ ..

What was hard in this?/ But the question would have been tricky if u would have asked,, that what will be capacitance of the system ,, if they are joined by any conductor(or medium made conducting),, then after calculating individual capacitances ,, the connection would be like two parallel capacitors ,,, so simply add them..

I would add that,, yes if distance would have been finite,,, then one has to consider induction and calculation of interaction energy would have been done by conformal mapping....

maza aaya... rate me quick..

Mozingo method is used to convert carbonyl compound into its dithio-acetal or ketal using ethanedithiol in the presence of a Lewis acid.. This obtained dithio-acetal is hydrogenlysed over Raney-Ni.

suppose that ${(R)}_{2}C=O$ is the carbonyl comp. to be treated ,, then ,,let it react with ethane dithiol first that is :: $HSC{H}_{2}-C{H}_{2}SH$

in presence of lewis acid(mostly ${Et}_{2} {O}^{+}- B{{F}_{3}}^{-}$ ) at room temperature,, to form dithio-acetal accompanied by the release of water molecule

R S---CH₂

\ / |

C |

/ \ |

R S---CH₂

the above given type of compound will be formed,, which when hydrogenlysed with raney nickel,, is reduced by breaking the bonds of the ring,, i.e, S--C bonds and Hydrogen is attached there,, thereby giving back ethane dithiol and giving ${(R)}_{2}-C{H}_{2}$ ...

is it wrong?/

if then pls. correct it..

well c..

From first bag,, the white ball can be selected in a/(a+b) ways .. And black in b/(a+b)... Let the white be selected ,, then placed in second bag.. so second bag contains (a+b+1) for the moment,, If white is selected then it has (a+1) white balls,,, and if black is selected than it has,, (b+1) black balls[and a white] from total...

so probability of gettin white from this second bag is given by

$\frac{a}{a+b} X \frac{a+1}{a+b+1} + \frac{b}{a+b} X \frac{a}{a+b+1} = \frac{a}{a+b}$

thus selection of white from every consecutive bag (in filling process) is given by a/(a+b)..

For the kth bag,, the selected bag can go through four cases::

1) that it has taken white from previous bag, and given white to next bag..;

2)that it has taken white from previous bag, and given black to next bag..;

3)that it has taken black from previous bag, and given white to next bag..; and

4)that it has taken black from previous bag, and given black to next bag............

For first case,, the bag would have taken a white ball from previous bag with prob. a/(a+b) and given white to following bag with a/(a+b) prob. the cases occur one after other..Now while selecting white ball,, the total no. of balls is (a+b) and no. of white balls is "a"...

so prob. of selection becomes::: $\frac{a}{a+b}X \frac{a}{a+b}X \frac{a}{a+b}$

since three of them occur simultaneously..

Following similar approach,, we get it for

2) $\frac{a}{a+b}X \frac{b}{a+b}X \frac{a+1}{a+b}$

3) $\frac{b}{a+b}X\frac{a}{a+b}X \frac{a-1}{a+b}$

4) $\frac{b}{a+b}X \frac{b}{a+b}X \frac{a}{a+b}$

Adding all these possibilities we get..

$\frac{a}{{(a+b)}^{3}}X ({a}^{2} + {b}^{2})$

bas yaar,, i could think till here!!!

abe pagal,,, that is in orbits placed in divisions of atmosphere of earth,, like ionosphere or exosphere,, where gravity is considered..

And pls. dont say that atmosphere is in space...

If u r saying so then satellites will be rotating anywhere around the earth...

Whenever one talks of space ,,, its always gravity free..

Its an ideal concept,, like point object...

Still, I knew u will create havoc,, so i wrote that,,"in space,, u cant consider gravity, and if u do so,, then according to ur statement u r making net force zero.... "

Y d hell r u fighting,, u were wrong in 1 sentence,, y arguing on it,,, and if u consider space to be filled of ur gravity,, still ur statement remains wrong,, because ur statement makes net force zero(refer it),, completely against question's statement stating:: "FIRST OF ALL..............IS THIS POSSIBLE........THAT A NET FORCE IS ACTIN ON THE ROCKET WHEN IT IS HAVIN ZERO ACCELERATION??????"

vaise y gettin so much screwed up,, actually Rocket-science is not in IIT syllabus!! Chill.. Dont make me waste ma posts!!

boss this is basic,, and many dont knw it.. or forgotten it.. I'll suggest do read this article.. gr8 work...

Boss,,i mean eshaan, in space,, u cant consider gravity, and if u do so,, then according to ur statement u r making net force zero..

And yes,, in space(gravity free),,if gases are burnt then they will eject out to create force on the system(rocket),, and create acceleration,, constant speed itself refers zero acc...
For,, ultimator, the statement given by him,, states the hand to exert the balancing force ,, the force that balances the frictional force and hand moves with nearly const. vel.,,, again net force zero... Otherwise , if u would keep ur hand on table,, it will move without any mechanical work,,hahaha..

Gettin on serious notes,, yes mindfreak is rite... there is no net force that can create zero acc.(in this case)!!
But if machinery is such that it somehow stores burnt gases,, and if certainly burning of gases stop, and certain nozzle is opened,, due to pressure diff., the gases move out.. But still there will be many practical factors to be considered... I hope u got it.. Rate quickly!! Take care..

mahnat hai teri,,, but u wasted time dude,, still i vote 4 u!!

v hav already got d ans.. good.. its rite...

i liked d pic.. nyways wats d answer u have written

is it:::

2(1+x)-3

or $2{(1+x)}^{-3}$

m gettin it second one..

Boss this site is about iit ...

But rite now not many are online,, vacation time..Sirf mere jaise pagal milenge.. Evn m new..

It takes time to reach questions..

Nyways,, whatever i know i'll be sharing.

velocity of sound is given by: $v= \frac{1}{\sqrt{\epsilon\mu}}=\frac{1}{\sqrt{{\epsilon}_{r} {\epsilon}_{0} {\mu}_{r}{\mu}_{0}}}$

now,, this ${\epsilon}_{r}$ is relative permittivity,, and same is case for ${\mu}_{r}$ .

depending on behaviour of the material in electric and magnetic field..

For a diamagnetic material,,, ${\mu}_{r}$ varies in 0-1,,, and ${\epsilon}_{r}$ is 1+ ${\mu}_{r}$ ,, according to which,, in this time aieee paper,,

${\mu}_{r}$ is 0.5 and ${\epsilon}_{r}$ is 1.5,,,,, is the only possible set of values.. All this information is from this year NCERT (XII).

rate karne mein time mat lagana...

yaar,, is this a challenging question,, cause u wrote "any one with clear concept should try this "... Hahaha..

nyways,, just arrange ur question as,,

(3z+1)(12z+1) (6z+1)(4z+1)=2.

which gives : (36z² + 15z + 1)( 24z² + 10z +1) = 2

bas substitute (12z² + 5z) with "x" then equation becomes

(3x+1)(2x+1)=2; this equation on being solved yields x= -1, 1/6.

we obtain four solutions such that 12z² + 5z = -1,1/6

giving $z= \frac{-5 + i \sqrt{23}}{24}$ ;

$z= \frac{-5 - i \sqrt{23}}{24}$ ;

$z= \frac{-30 + \sqrt{1188}}{144}$ ; and

$z= \frac{-30 - \sqrt{1188}}{144}$ ....

shayad mere concepts clear hain..

dont 4get to rate dear!!hehe...

ohk.. but i doubt such case...

if it would have been ${|cotx|}^{sin2x}$

but yes we can consider such case also..

since its tending to infinity,, it will be a number.. and for any number "n",,

${n}^{0}={n}^{1-1}=\frac{{n}^{1}}{{n}^{1}}=1$

thus limit becomes tending to 1..

nice...

sine function follows similar nature as zero,, means,, while argument that is x in sin2x,, tends to 0+ sin2x tends to 0+ and same for 0-.....

On other hand cot x yields infinity ,,, but but,, it gives +infinity at 0+ and -infinity at 0-....refer graph,, ncert, for cotx..
thus limit does not exist!!

samjhe.. toh rate karne mein sharmana kya?/

Here we are talking about acidity and basicity,, which can be judged on the basis of +I effect. Here it is cyclic ring, and mind it ,, it is not sp2 hybridised . In such cases, ring is more symmetric +I effect causing group, since all are sp3 hybridised, but but in comp. (1), NH2 is joined to carbon, which in turn is joined to two more carbons,,unlike comp. (2) . Thus it decreases acidity and increases basicity. So (1) is more basic and (2) more acidic.
No confusions in case now.

dont forget to rate dear!!