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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

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.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

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

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