2.  Six articles can be chosen from 10 articles in ¹⁰C₆ ways.

So, sample space = ¹⁰C₆

For the batch to be rejected more than 2 articles should be defective.

So, there are 2 cases ( since a max. of 4 atrciles are defective) -

(i) When 3 are defective out of 4 and 3 are non-defective out of 6.

(ii) When 4 are defective out of 4 and 2 are non-defective out of 6.

So, Req. Prob. = (⁶C₃ . ⁴C₃ / ¹⁰C₆) + (⁶C₂ . ⁴C₄ / ¹⁰C₆) = 19 / 42 (Ans)...............

3. Let S be the sample space and let E be the req. event, den

n(S) = (⁵²C₂)²

For the no. of elements in E, we frst choose a card (dat v want common) nd den from de remaining cards(i.e. 51) we choose 2 cards nd distribute dem among A and B in 2! ways.

So, n(E) - ⁵²C₁ . ⁵¹C₂ . 2!

Hence, P(E) = 50 / 663 (Ans)

yeah but only if the collison is elastic.

f'(x) = [ 2 / (x-2) ] - 2x + 4

f'(x) = 0

So, 2 / (x - 2) = 2x - 4

or, (x - 2)² = 1

or, x² - 4x + 3 = 0

or, (x - 1)(x - 3) = 0

or, x = 3 or x = 4

Also, f'(x) = (2) ( 1 - (x - 2)² ) / (x - 2) = (-2) (x - 3) (x - 4) / (x - 2)

So, function decreases in (-infinity,-3)U(4,infinity) and increases in (3,4).

So, there are 15 sigma and 4 pi bonds.

Enantiomers are stereoisomers that are mirror images of each other that are "non-superposable" (not identical), much as one's left and right hands are "the same" but opposite.

Diastereomers  are stereoisomers that are not enantiomers. Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocenters and are not mirror images of each other.

Diastereomers differ from enantiomers in that the latter are pairs of stereoisomers which differ in all stereocenters and are mirror images of each other. Enantiomers of a compound with more than one stereocenter are also diastereomers of the other stereoisomers of that compound that are not their mirror image. Diastereomers have different physical properties and different reactivity, unlike enantiomers.

Diastereomers

       D-Threose                             D-Erythrose

Enantiomers

(S)-(+)-lactic acid (left) and (R)-(–)-lactic acid (right) are nonsuperposable mirror images of each other

(i) An aqueous solution of Na₂CO₃ would have more pH. This is because it does not have any H⁺ ions to give which are dere in NaHCO₃ . So, Sodium bicarbonate is acidic and has lower pH.

Gravitational force obeys inverse square law.

I got a different answer to it.

I am getting I = (pi)² / 2 sin(a).

nyhow approach is like dis.

First apply property int x from 0 to a [ f(x) ] = int x from 0 to a [ f(a - x) ] and then apply

int x from 0 to 2a [ f(x) ] = 2 int x from 0 to a [ f(x) ] , if f(2a - x) = f(x)

U'll get,

I = (pi) int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx

Let I₁ = int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx

Now put cos(a) sin(x) = y and u'll get

I₁ = int y from 0 to cos(a) ( 1 / (1 - y) (cos²a - y²)^1/2 ).dy

Now put cos²a - y² = 1/t

U will now get, I₁ = (1/sina) int t from 1 to (1 / 1 - cos(a)) [ (cos(a)/sin²(a))² - (t - (1/sin²a))² ]^1/2

So, finally I₁ = pi / 2 sin(a)

So, I = (pi)² / 2 sin(a) [Ans]..............

I did the same way u said yagyadutt sir. But this ques. was of just 3 marks. I think they by mistake mentioned radius rather than edge length.

Let the third ticket be 30.

So, first two tickets have to be less than 30. There are 29 tickets less than 30 out of which two can be selected in ²⁹C₂ ways. Third ticket 30 can be selected in 1 way only. The remaining two are to be selected from 20 tickets numbered greater than 30 which can be selected in ²⁰C₂ ways.

Five tickets can be selected from 50 tickets in ⁵⁰C₅ ways.

So, req. probability = ²⁹C₂ X 1 X ²⁰C₂ / ⁵⁰C₂ = 551 / 15134 (Ans)

Electron gain enthalpy is the amount of energy released when an isolated gaseous atom accepts an electron to become a monovalent gaseous anion.

Cl has most negative electron gain enthalpy.

Gp. 1 has most positive electron gain enthalpy.

Configuration of C(6) = 1s² 2s² 2p²

and conf. of B(5) = 1s² 2s² 2p¹

Now first IE of C is greater as removal of one elctron is difficult bcoz it prefers to accept one e- rather. On the other hand after 1 e- is removed from each atom, 2nd e- is easier to be removed from C than B as boron attains highly stable conf. i.e. 1s² 2s² which is fully filled.

Remember : Difficult the removal, More the IE.

Glass is a supercooled liquid of very high viscosity. Due to gravity it flows down. So, the glass becomes thicker at the bottom.

As temp. inc. viscosity dec. So, with inc. in temp. glass becomes less thick at bottom.

F = Gm₁m₂/r²

So, F α 1/r²

lim x-> ∞ ((x + 5) / (x - 1))^x

= lim x ->∞ ( 1 + (6 / x - 1) )^x

= e^{lim x ->∞ (6x / x - 1)} 

= e⁶

where, li(x) :  LogIntegral[x]: logarithmic integral