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# 10 years ago

Dear student,

In case 1:

Yes vapor density will remain same thruoghout the reaction course so it is correct.

In case 2:

Yes option B is very much correct

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Sagar Singh

B.Tech, IIT Delhi

10 years ago

Q-56 ..

rod will behave as chord of circle ... perpendicular distance bw rod &

center of circle can be calculated , let this distance is d ...

d2 + (L/2)2 = R2                   (by pythegorus theorem)

d2  = 4-1 = 3

d = root3

initially rod was verticle , let bottom most point of sphere  is at 0 potential then

potential of center of mass of rod is mgR  (initially , when rod is verticle)      &

potential of center of mass of rod is mg(R-d)     finally  (when rod is horizontal) ...

change in potential energy = change in KE                  (by work energy  theorem)

dKE = IW2/2

dPE = mgR - (mg(R-d) = mgd

W = [2mgd/I]1/2                  ...................1

I = Icom + md2

= ml2/12 + md2

l = 2 , d = root3 so

I = m(1/3 + 3) = 10m/3           ..................2

putting value of I in eq 1 we get

W = [6root3]1/2

option b is correct

10 years ago

dear vikas,

i got the first one yes it is correct my option is correct. but second one i'm still getting b, are you getting the same answer.

10 years ago

Q=33

differentiating this eq wrt x using leibnez rule

f1(x) = e-1 (1/2rootx)

dy/dx = 1/2erootx = 1/2ex1/2

dy = dx/2ex1/2

y = x1/2/e + c

at x = 0, y is 0

so , y = x1/2                    .......................1

or x/e = y2

this is the eq of parabola ....

option C is correct

approve my ans if u like

10 years ago

Q32=>

f(x) = exsinx           .................1

f1(x) = exsinx + excosx

f1(x) = f(x) + excosx        ..............2        (from eq 1)

f2(x) = f1(x) + excosx - exsinx     .................3

substituting excosx from 2 in 1

f2(x) = 2f1(x) - f(x) - exsinx               ..........4

substituting exsinx from eq 1 in eq 4

f2(x) = 2( f1(x) - f(x) )

f2(0) = 2( f1(0) - f(0))

in eq 1 & 2 put x = 0 , that will give

f(0) = 0

f1(0) = 1 so

f2(0) = 2

in same manner f10(0) = 2*5 = 10

option C is correct

10 years ago

C)focus of parabola is at (1,1) ...

line also passes through this point so this line is also focul chord o this parabola...

for parabola let lengths of two parts of divided by axis is l2,l1 respectively ...

l1 = a(t2+1)

l2 = a(1+1/t2)

harmonic mean is 2

k = 4

10 years ago

D) tangent to any ellipse is given

y = mx + (a2m2+b2)1/2

(y-mx)2 = a2m2 + b2               ................1

let it passes through (h,k) then

(k-mh)2 = a2m2 + b2

m2(h2-a2) - 2mkh + k2-b2 = 0

from here we get 2 slopes of tangent , let they intersect at 90o ...    (this is assumtion )

m1m2 = -1                   (multiplication of roots = -1)

k2-b2/h2-a2 = -1

k2 + h2 = a2 + b2          .....................2

x2/pi + y2/e = 1           ..............3

a2 = pi , b2 = e

now see the given point satisfies the eq 2 so we can surely say that these

tangents intersect at 90o or pi/2

k = 4

approve if u like my ans