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8 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

```
8 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
```
8 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.
```
8 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
```
8 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

```
8 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
```
8 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

```
8 years ago
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