Explore topic-wise InterviewSolutions in .

A. 3.2 × 10¹³ Hz

D. a lower frequency and a longer wavelength than before

B. occur at the same coordinates

D. in the same direction as the spaceships at 0.95c relative to us

B. on board the explorer craft

WINDOWS is the main operating system for CAD systems.

Design and drafting using CAD softwares

Medicines are more effective in the colloidal form because of large surface area and are easily assimilated in this form.

In hcp structure, the packing fraction is 0.74.

The time taken for the concentration to change from 0.1 M to 0.025 M is  30 minutes

For loop ABCDA, 

20I + 40(I – I₁) = 80 

⇒ 60I – 40 I₁ = 80 

⇒ 3I – 2I₁ = 4 ……….. (1) 

For loop EDCFE,

-40(I – I₁) + 10I₁ = 40 

-40 I + 40I₁ + 10I₁ = 40 

50I₁ – 40I = 40 

5I₂ – 4I = 4 …………. (2) 

From (1) and (2),

On multiplying (1) × 4 and (2) × 3 we get; 

15I₁ – 8I₁ = 28 

7I₁ = 28 

I₁ = 4A 

Putting in (1), 

3I - 2× 4 = 4 

3I = 12 

I = 4A 

∴ Current through 20Ω is I = 4A 

Current through 40Ω in (I – I₁) = 4 – 4 = 0

Concept:

A set of vectors {v1, v2,…, vp} in a vector space V is said to be linearly independent if the vector equation c1v1 + c2v2 +…+ cpvp = 0 has only one trivial solution c1 = 0, c2 = 0,…, cp = 0;

The set is said to be linearly dependent if there exists weights c1, c2,…, cp not all 0, such that c1v1 + c2v2 +…+ cpvp = 0

Calculation:

Given S1 = {(7,7,7)} ⇒ v1 = (7,7,7)

S2 = (0,0,0)

A set containing zero vector is linearly dependent ⇒ S2 is linearly dependent

Let S1 = (v1) and the vector equation be av1 = 0

⇒ 7a = 0 ⇒ a = 0

Concept:

Let V and W be a vector space over F and T: V → W is a linear transformation then null space of T is given by Ker T = {v ∈ V | T(v) = 0 where 0 ∈ W}

So, the nullity of T is the dimension of the null space of T.

Range space of T is given by R(T) = {w ∈ W| T(v) = w fo v ∈ V} and Rank of T is the dimension of R(T)

Rank - Nullity Theorem:

Let T be a linear transformation from V to W i.e T: V → W and V is a finite-dimensional vector space then Rank (T) + Nullity (T) = dim V

Analysis:

Given:

T : R⁴ → R⁴

T(x, y, z, u) = (x, y, 0, 0)

Let us first find the dimension of null space (nullty)

Null space of T = kernel of T (ker T)

Ker T = { v ∈ V | T(v) = 0}

Ker T = (0, 0, 0, 0) = (x, y, 0, 0)

x = 0, y = 0

The dimension of Ker T = dimension of basis of Ker T = 2

∴ Nullity = 2

dim (V) = 4

∴ 4 = Nullity + Rank (T)

Rank (T)  = 4 – 2 = 2

\(\frac{d}{dx}\) sin(1 - 4x²) = cos(1 - 4x²) \(\frac{d}{dx}\) (1 - 4x²)

= cos(1 - 4x²) \((\frac{d}{dx}1-4\,\frac{d}{dx}x^2)\)

= cos(1 - 4x²) (0 - 8x)

= -8x cos(1 - 4x²)

The negation of the given statement is:

∃ n ∈ A, such that n + 7 ≤ 6.

OR

∃ n ∈ A, such that n + 7≯ 6.

\(A = \begin{bmatrix}2 & 3 \\[0.3em]4 & 7 \\[0.3em]\end{bmatrix}\), \(B = \begin{bmatrix}1 & 3 \\[0.3em]-2 &5 \\[0.3em]\end{bmatrix}\), \(I = \begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)

Now,

2A + 3B - 5I = \( 2\begin{bmatrix}2 & 3 \\[0.3em]4 & 7 \\[0.3em]\end{bmatrix}\)+ \( 3\begin{bmatrix}1& 3 \\[0.3em]-2 & 5 \\[0.3em]\end{bmatrix}\) - \( 5\begin{bmatrix}1& 0 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)

= \( \begin{bmatrix}4+3-5 & 6+9 \\[0.3em]8-6 & 14+15-5 \\[0.3em]\end{bmatrix}\)

= \( \begin{bmatrix}2 & 15 \\[0.3em]2 &24 \\[0.3em]\end{bmatrix}\)

Eichler classified the plant kingdom into two sub-kingdoms namely Cryptogamae (absence of flowers and seeds and reproduces by spores) and Phanerogamae (presence of flowers and seeds) in the year 1833. He accepted the concept of evolution and considered the simple characters with respect to related plants.

5 carbon atoms are there in a molecule of ribulose biphosphate.

In a plant cell, Chlorophyll a pigments participates directly in the light reactions of photosynthesis.

Glycolysis is a process found in Virtually all cells.

(D) Both (B) and (C)

For an animal cell, the main advantage of aerobic cellular respiration over lactic acid fermentation is that more energy is released from each glucose molecule.

Breather is a bottle shaped steel tube, which is attached to one side of conservator to allow the air to pass in and out of the tank or conservator through the calcium chloride and silica gel, which is filled in it to absorb the moisture contained in the air. When the silica gel absorb the moisture its colour changes from blue to pink.

Perigynous flowers in which the ovary is said to be half inferior.

(i) Vapour pressure is directly proportional to temperature. 

(ii) If intermolecular force of attraction is less in the liquid, its vapour pressure will be high. 

(iii) Higher the vapour pressure, lower will be boiling point.

(iv) If atmospheric pressure is low, boiling point will be less and higher will be the vapour pressure.

Acid-base pair which differ by one photon. 

Conjugate acid of CN^- is HCN; Conjugate acid of H₂O is H₃O⁺

Solubility decreases, Due to common ion effect.

Answer is (b) 10^-8 M²

Answer is (b) 10^-8 M²

Given :- equivalent conductance = 400 Ω^-1 cm² eq^-1

conductance = 8 x 10^-5 Ω^-1 cm^-1

We know S = \(\frac{∧\times1000}{∧_{eq}}\)

Where ∧ is conductance,

∧_eq is equivalent conductance

So, S = 4 x 10^-5 x 1000/400 = 10^-4

Now, BaSO₄ ⇔ Ba⁺² + \(SO_4^{-2}\)

K_SP = [Ba⁺²] [\(SO_4^{-2}\)] = (S . S)

S² = 10^-4 x 10^-4 = 10^-8 M²

Hence, K_SP value is 10^-8 M².

(A) Both Statement I and Statement II are correct

Correct option is (A) Both Statement I and Statement II are correct.

♦ Due to pressure of double bond alkenes are highly reactive than alkanes. therefore, alkanes are less stable than alkanes.

♦ There is more energy required to break double bond than a single bond. So that, strength of double bond is greater than carbon-carbon single bond.

(3) C_NaCl(T₂ ) > C_NaCl(T₁ ) for T₂ > T₁ 

Dissolution of BaSO₄ is an endothermic reaction 50 on increasing temperature number of ions of BaSO₄ decrease so it's conduction also decrease.

The answer is (02.13)

mol of NaClO₃ = mol of O₂

mol of O₂ =PV/RT = (1 x 492/0.082 x 300) = 20 mol

mass of NaClO₃ = 20 × 106.5 = 2130 g 

100

Volume strength of H₂ O₂ at 1 atm 

273 kelvin = M × 11.2 

= 8.9 × 11.2 

= 99.68

= 100

The answer is (3)

Answer is (2) (a) and (b)

(a) Since adsorption is exothermic process, as adsorption proceeds number of active sites present over adsorbent decreases, so less heat is evolved. 

(b) Since NH₃ has higher force of attraction on adsorbent due to its polar nature (high value of 'a'). 

(c) As the adsorption increases, residual forces over surface decreases. 

(d) Since process is exothermic, on increasing temperature it shift to backward direction, so concentration of adsorbate particle decreases.