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

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All the best.

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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 = I_com + 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

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.

Q=33

 differentiating this eq wrt x using leibnez rule

f¹(x) = e^-1 (1/2rootx)

dy/dx = 1/2erootx = 1/2ex^1/2

dy = dx/2ex^1/2

y = x^1/2/e + c

at x = 0, y is 0

so , y = x^1/2                    .......................1

 or x/e = y²            

this is the eq of parabola ....

option C is correct

approve my ans if u like

Q32=>

 f(x) = e^xsinx           .................1

f¹(x) = e^xsinx + e^xcosx      

f¹(x) = f(x) + e^xcosx        ..............2        (from eq 1)

f²(x) = f¹(x) + e^xcosx - e^xsinx     .................3

substituting e^xcosx from 2 in 1

f²(x) = 2f¹(x) - f(x) - e^xsinx               ..........4

substituting e^xsinx 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

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

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 90^o ...    (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 90^o or pi/2

k = 4

approve if u like my ans