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Correct option: (2) 3 × 10⁴ N/C

Explanation: 

E = CB

E = 3 × 10⁸ × 100 × 10^–6 = 3 × 10⁴ N/C

Correct option: (1) X = 1, Y = 0

Explanation:

By using concept of logic gates we will get X = 1, Y = 0

Correct answer is odd numbers

Correct option is (A) A-II, B-III, C-IV, D-I

Correct option is (A) K₃[Cu(CN)₄]

K₃[Cu(CN)₄] 

O.N. of copper is Cu⁺¹ 

Cu⁺¹ = [Ar]3d¹⁰ ⇒ Diamagnetic

Correct option: (a) Narrow band interference 

The given equation in symbolic form is written as (4D³ + 4D² + D)y = 0 and its auxiliary equation is 

4D³ + 4D² + D = 0

D(4D² + 4D + 1) = 0 D(2D + 1)² = 0

i.e D = 0, - 1/2, - 1/2

whence (C.F) = c₁e⁰ + (c₂x + c₃)e^-x/2.

Let,X+Y=14  .............(1)X-Y=14  .............(2)
From 1$2 X+Y=X-YY+Y=X-X2Y=0Y=0/2
Put value in EQ 1X+Y=14X+0=14X=14 The value of X=14 And Y=0

The definition of a linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line. An example of linear equation is y=mx + b.

(D) an equivalence relation

A = {1, 2, 3} 

R = {(1, 1), (1, 2), (1, 3) (2, 1), (2, 2), (2, 3) (3,1), (3, 2) (3, 3)}

A₁ ∪ A₂ ∪ A₃ = A and A₁ ∩ A₂ ∩ A₃ = ϕ

(A)  1 - (1/x²)

To increase the creepage distance, reduce the chances of flash over.  

In the acid-catalyzed reaction, the molecule is first protonated on oxygen and then loses the C-H proton in a second step. When the enol form reverts to the keto – since this is an equilibrium process – it picks up a deuteron instead of a proton since the solution is D₂O. Also, the D+ (or H+ if water is the solvent) is regenerated at the end (catalyst).

In the base-catalyzed reaction, the C-H proton is removed first by the base (for example hydroxide ion OH^-, OD^- ) and the proton (or D⁺ in our case) added to the oxygen atom in a second step. Also, OD- (or OH- if water is the solvent) is regenerated at the end (catalyst).

The enolate ion generated from the enol (Fig. I.6) in the base-catalyzed mechanism is nucleophilic due to: Oxygen’s small atomic radius and Formal negative charge.

Correct answer is (c) BD . CD = AD²

1) ∵ (2, 0) ∈ R₁ but 0 \(\notin\) S

∴ R₁ is not a relation on S

2) domain of R₂ = {1, 2, 4} ⊂ S

Range of R₂ = {2, 3, 4} ⊂ S

Thus, R₂ is a relation on S.

3) domain of R₃ = {1} ⊂ S

Range of R₃ = {1, 2, 3, 4} ⊂ S

Thus, R₃ is a relation on S.

4) ∵ (0, 2) ∈ R₄ but 0 \(\notin\) S

Thus, R₄ is not a relation on S.

vector equation of line joining the points having position vectors \(\vec a_1=\hat i-\hat j+4\hat k\) and \(\vec a_2=2\hat i+\hat j+2\hat k\) is

\(\vec b = \vec a_2-\vec a_1=(2\hat i+\hat j+2\hat k)-(\hat i-\hat j+4\hat k)\) 

 = \(\hat i+2\hat j-2\hat k\) 

Given that required line is passing through the point having position vector \(\vec a=\hat i-2\hat j-3\hat k\) 

\(\therefore\) Equation of required line is

\((\vec r-\vec a)+\lambda \vec b=0\)

(\(\because\) Required line is passing through \(\vec a\) and parallel to line \(\vec b\))

⇒ (\(\vec r-(\hat i-2\hat j-3\hat k)\)) + \(\lambda(\hat i+2\hat j-2\hat k)=0\)

which is vector equation of required line

Let \(\vec r = x\hat i+y\hat j+z\hat k\) 

Then (x\(\hat i\) + y\(\hat j\) + z\(\hat k\)) - (\(\hat i-2\hat j-3\hat k\)) = \(-\lambda(\hat i+2\hat j-2\hat k)\)

⇒ (x - 1)\(\hat i+(y+2)\hat j+(z+3)\hat k=-\lambda\hat i-2\lambda \hat j+2\lambda\hat k\)

⇒ x - 1 = -\(\lambda\), y + 2 = -2\(\lambda\) and z + 3 = 2\(\lambda\)

(On comparing coefficient of \(\hat i, \hat j\&\hat k\))

⇒ \(\lambda\) = \(\frac{x-1}{-1}, \lambda=\frac{y+2}{-2},\lambda=\frac{z+3}{2}\)

⇒ \(\frac{x-1}{-1}=\frac{y+2}{-2}=\frac{z+3}2\) which is cartesian equation of required line

Alternate method:

Position vector of point is \(\hat i-2\hat j-3\hat k\) 

\(\therefore\) Point is (1, -2, -3)

Hence, required line is passing through point (1, -2, -3)

Similarly, required line is parallel to the line joining points (1, -1, 4) and (2, 1, 2).

\(\therefore\) Direction ratios of line joining points (1, -1, 4) & (2, 1, 2) are \(\pm(2-1,\pm(1-(-1))),\pm(2,-4)\)

or 1, 2, -2 or -1, -2, 2.

\(\therefore\) Direction ratios of required line are 1,2,-2 or-1, -2, 2 and it is passing through point (1, -2, -3)

 \(\therefore\) Cartesian equation of required line \(\frac{x-1}1=\frac{y-(-2)}2=\frac{z-(-3)}{-2}\)

or \(\frac{x-1}{-1}=\frac{y-(-2)}{-2}=\frac{2-(-3)}2\) 

⇒ \(\frac{x-1}1=\frac{y+2}2=\frac{z+3}{-2}\)

or \(\frac{x-1}{-1}=\frac{y+2}{-2}=\frac{z+3}2\)

Correct matching is equation (2)

As we know,

The contribution of corner atom per unit cell = 1/8. and The contribution of face centres atom per unit cell =1/2

\(\therefore\) Number of P atoms per unit cell = 8 x 1/8

\(\therefore\) Number of Q atoms per unit cell = 6 x 1/2 = 3

Therefore, the formula of compound will be PQ_3.

[Hint : (i) 2XeF₂ (s) + 2H₂O (l) → 2Xe (g) + 4HF (aq + O₂ (g)
(ii) Xe + PtF₆→ Xe+[PtF₆]-]

[Hint : Dioxygenyl hexafluoroplatinate(iv).]

(i) Because oxidation number of Xe do not change during hydrolysis of XeF₆.
(ii) XeF₄ + 2H₂ →  Xe + 4HF

[Hint : F and O are most electronegative elements Kr and Xe both have low ionization enthalpies as compared to He and Ne.]

(i) The first ionization energy of xenon (1, 170 kJ mol^{– 1} ) is quite close to that of dioxygen (1,180 kJ mol^–1). (ii) The molecular diameters of xenon and dioxygen are almost identical. Based on the above similarities Barlett (who prepared O₂ ⁺ [PtF₆ ] ^– compound) suggested that since oxygen combines with PtF₆ , so xenon should also form similar compound with PtF₆ .

[Hint : Linear, 180º]

The two N atoms in N₂ are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N₂ is less reactive at room temperature.

[Hint : Xe (g) + 2F₂ (g) → XeF₄ (s)]
Ratio 1 : 5

ICl is more reactive than I₂ because I−Cl bond in ICl is weaker than I−I bond in I₂.

Balanced equation
XeF₆ + 2 H₂O → XeO₂F₂ + 4 HF
 

[Hint : Due to smaller size of oxygen than sulphur electron-electron repulsion is more in oxygen than sulphur.]

[Hint  :  H₂O molecules are stabilized by intermolecular hydrogen bonding, while H₂S by weak van der Waal’s forces.]

(b) Dropwise condensation 

(b) Since dropwise condensation offers lesser thermal resistance during phase change thus can enhance heat transfer rate between vapour and solid surface.

(c) Choke valve 

They are people of yellow complexion, oblique eyes, high chick bones, spare hair and medium height," The reference here is to Mongoloids.