Messages posted by varshavallig

y do u take these things seriously.forget about the quota and put all ur efforts in securing a good rank in IIT.when u have the will power to achieve something y do u fell for such things.
RELAX!!!!!!!!!
ALL THE BEST
!!!!!!!!!!!!!!!!!!

simple
in 1mole------>2g of H is present
so in 0.4 moles-------->0.8g of H is present

in 1mole--------->32g of S is present
so in 0.4 moles--------->12.8g of S is present

hope this helps
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P/P⁰ = 2/100 = n₂/n₁

2/100 = n2/wt1/18

2/18*100 = n2/wt1

10/9 =n2 *1000/wt1

1.1 =molality

here there is an approximation of n1+n2 as n2 as given in NCERT.

rate if satisfied.

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ans is d)
the sum bcomz=1+(1+2)+(1+2+3)+(1+2+3+4)................

so the sum of 1st 3 terms is 10,which option d) satisfies.
plz correct me if m wrong
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for rotating a point(x,y) by an angle A to get a new pt (h,k)

[cosA -sinA ] [x] = [h]
[sinA cosA] [y] [k]

all r matrices
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9)e^-x =1/e^x
multiply and divide by 2 ,put t=e^2x +1 dt=2e^2x

10)e^-x=1/e^x
put e^x+1=t dt=e^x

13)i=tanx*tan^2x=tanx(sec^2x-1)
put t=tanx dt=sec^2x

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6)cotx * cot^2x=cotx(cosec^2x-1)
put t=cotx

8)take out a^2cos^2x as common from the denom,it will bcom sec^2x *1/a^2 in numerator
i=(1/a^2)sec^2x/(tan^2x+(b^2/a^2))
put t=tan^2x
apply the formula .in the formula A=b/a

1)i=sec^4x secxtanx put t=secx dt =secxtanx

2)cos2x=cos^2x - sin^2x
substitute and put t=sinx +cosx dt=cosx-sinx

3)put t=x^e+e^x dt=e(x^e-1+e^x-1)

5)tanx=sinx/cosx substitute,proceed same as 2nd
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tnx a loooooooooooootttttttttttttttt!!!!!!!!!!!!!!!!!!!!!!!

For fluorite structure, it is given that co.no. for Ca is 4 and for F is 8.

How do we rep. the coordination no. for the entire structure?

Is it 4:8 or 8:4?

reply soon, urgent

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17)put a+bcosx=1
dt = -bsinx dx
cosx=(t-a)/b

substituting,v get i= (-2/b^2) *(t-a)/t^2

splitting into 2 integrals
i=(-2/b^2)(1/t - a/t^2)

now u can integrate it
i think the first term is a bit wrong in the ans u gave.
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4)i=secxtanx(sec^4x) put secx=t dt=secxtanx

5)put cosx=t dt=(-sinx)

7)multiply and divide by 4

i=1/4 (4x^3sin(x^4)) put x^4=t dt=4x^3

8)put cotx=t dt=(-cosecx)^2

9)put tanx=t

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@ Ronaldooo
it is c-root(a^2+b^2) and c+root(a^2+b^2)
u missed the 'c'.here c is 5
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its simple ,just apply this formula:

root(c/wavelength)=a(Z-b)

for K shell b=1

now Z is inversely proportional to wavelength.so if

Z1<Z2=>lambda1-lamda(min)>lambda2-lambda(min)

so b) is right.

rate if satisfied

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

CO2 has 2 structures.

O=C=O <----->O-C# O (# means triple bond)
out of these the 1st one is more stable.
correct me if m wrong
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is the ans 5.07g????????
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put x=a * tanA
dx=a sec^2A
apply change of limits,it vill be from 0 to pie/2.
substituting ,v get I = (1/a^13) integral cos^12A.
since the power is even,

I=(1/a^13)(11*9*7*5*3*1/12*10*8*6*4*2)pie/2
I=(1/a^13)(10395/46080)pie/2

correct me if m wrong

rate me if m right
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@ BALGANESH
it is 0.2 and 0.6 moles not 2 and 6.
so the ans is 0.16 moles
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for ur 2nd Q:

let the no. of natural nos be n.
x^2 + y^2= (x+y)^2 -2xy shud be div by 5.

so x+y and xy (both )shud be div by 5.

for x+y:

the nos can be 5m+1,5m+4(1) or 5m+2,5m+3(2) or 5m and 5m(3)

P1= 2P(1) +2P(2)+P(3) can b determined only if total nos is given.

for xy:

either x or y or both x and y shud be div by 5.

P2 can b calculated.

but here both the cond for x+y and xy shud satisfy.

so only prob is that both shud be a multiple of 5.

so P(x and y shud be multiples of 5)=

=(1/[n/5])^2/n -----[] indicates greatest integer.

plz correct me if m wrong.

!!!!!!!!!!!!!!!!!!!!!

@ sunip_the_mini_mastermind
but there in the step for finding Vr ,it is diff from what v got ,so if the 1st step is wrong,obviously sum of Vr is also wrong.
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