Messages posted by rohan2007

is the proof not clear enough ? ..

that is also quite easy .. let angle of minimum deviation be d

well we know the relations

A=r1+r2 --- - -(i)

d=i+e -A

further from (i) we get

d(r1)/di = -d(r2)/di

and we know

nsin(r1)=sin(i) -- -(ii)

nsin(r2)=sin(e) -- - (iii)

=>

ncos(r1)dr(1)/d(i)=cosi

ncos(r2)d(r2)/d(i)=cos(e)(de/di)

dividing both equations

-cos(r1)/cos(r2)=(cosi/cose)(di/de)

further from (ii) and (iii)

cos(r1) = (1-(sin(i)/n)²)^1/2

and

cos(r2) = (1-(sin(e)/n)²)^1/2

further for d to be extremum

d(d)/di=0

thus

d(i+e-A)/di=0

thus

1+de/di=0

but we have already obtained de/di

solving this equation we get

(n²- sin²i)/cos²i = (n²-sin²e)/cos²e

=>

(n²-sin²i-cos²i)/cos²i = (n²-sin²e-cos²e)/cos²e

thus we get two possibilities

n=1 or i=e

but n=1 is excluded( as n we cant manipulate)

and thus we get i=e for extremum

and we can see that the value of d obtained in this case is less than the value obtained when ray falls perpendicularly

thus IT HAS TO BE MINIMUM

please reply sir ..

i cudnt get the question

do we have to consider rays parallel to base only and different values of angle of prism

and do we have to take an isoceles prism ?

i have proved it when the case is as mentioned above ...(was quite easy so most probably it is not the case)

yes aditya ..

the equillibrium constant for the following reaction

H2O => H(+) + OH(-)

is 10^{-14 =>}[H+]=10^-7

hence the concentration of the ions is very less!!

thus it is a bad conductor

argon being a noble gas is not affected

total pressure = partial pressure of argon+ partial pressure of the remaining mixture

partial pressure of argon = nRT/V where V= 110 litres n=4 T=323

so you get the partial pressure of the remaining gas molecules = 4.678 - above quantity

let x moles be dissociated

partial pressure of remnants = (5-x +x +x)RT/V=(5+x)RT/V with same value of T and V

hence you get x

degree of dissociation= x/5

Kp=Kc(RT)

Kc= x²/(5-x)

hence you get Kp

as PV=R*(nT)

PV=constant

this implies nT=constant

infinity?

BUT the answer is given e^1/2

anirudh ,

cant we take a prism of any angle until it is mentioned that there should be an emergent ray from the other side??

yes rohit i agree with u

but definitely he has missed something in the problem

(otherwise it is a wrong problem)

if you are rerferring to the angle between the two sides of the prism then

you problem is wrong as we can make a prism of any angle!!!

please make the question a bit more clear..

i think there is a mistake in the problem /answer

when we put n=1 we get the integral as 2/3 not (1/2-1/6) =1/3

in the process you may neglect the third power of x

we know that the standard gibbs free energy G⁰ doesnot change with temperature

and further G⁰ =-RTlnK₁

and G⁰=-R(T+1)lnK₂

so TlnK₁=R(T+1)lnK₂

and further

K=[H+]²

SO Ph= -LOGK/2

SO CALCULATE

in hyperconjugation

the link between C and H atom breaks (not actually but almost)

and that pair of electrons which took part in that bond are delocalised

but in resonance any pair of free electrons are delocalised

plz make the problem little clearer..

we can only locate the position if we send a ray of light or whatever photon and if it reverses that indicates the position

but .. some of the energy of the photon gets absorbed by the electron and hence that changes its momentum

let the concentration of Z=x at equillibrium

concentration of X = 0.2-x

concentration of Y = 0.5-x/2

expression for K= [Z]²/[X]²[Y] = 100

solve for x

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