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Correct option (c) Mn⁺²

Explanation:

Mn⁺² = d⁵ = 5 unpaired electron

Correct option (d) X = Y

C_p – C_v = R (for all gases)

Correct option  (a) 60

Strength (gm/lit) = 68/22.4 x Volume strength

Correct option (a) 0.1 M NaCN

It is salt of strong base & weak acid

The IUPAC names of the following compounds are:

(i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2, 4-diol
(iii) Butane-2, 3-diol
(iv) Propane-1, 2, 3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol
(viii) 2, 6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane

Correct option (a) MF > MCl > MBr > MI

Explanation :

Alkali metals are highly electropositive and halogens are electronegative. Thus for the halides of a given alkali metal, the ionic character increases with increase in electronegativity of halogens. 

The Citizenship Act enacted by the Parliament in 1955 provides for acquisition, renunciation, termination, deprivation and determination of Indian citizenship. The Act provides for acquisition of Indian Citizenship by birth, descent, registration and naturalization.

A secret ballot is a type of vote where the voter's choices are anonymous. This is to make bribery or intimidation of voters more difficult. Secret ballots are good for many different voting systems.The voter writes only his or her choice, then places it into a sealed box. The box is emptied later for counting.

Member of the Legislative Assembly 

A Member of the Legislative Assembly (MLA) is a representative elected by the voters of an electoral district (constituency) to the legislature of State government in the Indian system of government.

Inclination angle (∅)= 30° 
Speed of centre of mass (v)= 5m/s 

(A) acceleration of the cylinder rolling up the inclined plane is 
Given by the formula, 
a = -gsin∅/(1 + k²/r²) 
Where K = radius of gyration 
Moment of inertia of cylinder = (1/2)mr²
K² = r²/2 

a = -gsin30°/(1 + 1/2) 
= -9.8/3 m/s²

Using equation of motion, 
V² = U² + 2aS 
0 = (5)² + 2(-9.8/3)×S 
S = 25×3/2×9.8 = 3.83 m 

Time taken to return the bottom = t 
S = ut + (1/2)at² 
3.83 = 0 + 1/2(9.8/3)t² 
t = 1.53 sec

Given :

Total number of lamps =10

Since at least one lamp is to be kept switched on

The total number of ways are

10C1+10C2+....10C10

⇒2¹⁰−1

⇒1024−1

⇒1023

Hence (B) is the correct answer.

Uses of convex lens are as follows:
1)Convex lens is used in microscopes and magnifying glasses to subject all the light to a specific point.
2)Convex lens is used as a camera lens in cameras as they focus light for a clean picture.
3) Convex lens is used in the correction of hypermetropia

we know that sin inverse of sin x = x

so sin inverse sin 1550 = 8.611π

Oxidation state of oxygen in H₂O₂ is -1.

Isaac Newton was the one who made Mathematical Principles of Natural Philosophy and explained his findings regarding gravity. He discovered gravity by watching an apple fall from the tree. Newton’s work promotes secularism during the Scientific Revolution because. He used his own mathematical reasoning rather than the Church‘s teachings. 

He established the three laws of motion. The most used law is the second law of motion. It states that a body at rest remains at rest unless a force is acted upon it. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. 

Solution:

H₁ =10 m, v₁ =?

Loss of energy =40% H₂ =?

On striking the floor, K.E = P.E

½ mv₁² = mg H₁

V₁ = ?(2gH₁)= ?( 2x9.8x10)

V₁= ?196= 14 m/s

As the body loses 40% of its energy, final energy E₂ =( 60/100) of initial energy

i.e. mgH₂ = (60/100)mgH₁

H₂ =3/5 H₁= (3/5)X10= 6m

Separating into half reactions: 

Oxidation half: Cu → Cu²⁺

Reduction half: NO^-₃ → NO₂ 

Balancing oxygen and hydrogen atoms: 

NO^-₃ + 2H⁺ → NO₂  + H₂O 

Balancing charge by adding electrons and making the number of electrons equal in the two half reactions: 

Cu → Cu²⁺ + 2e^- 

2NO^-₃  + 4H⁺ + 2e^- → 2NO + 2H₂O 

Adding the two half reactions to achieve the overall reaction: 

Cu + 2NO₃ + 4H⁺→ Cu²⁺ + 2NO₂ + 2H₂O 

1. Al³⁺ + 3e^- → Al

2. MnO^2-₄ → MnO^-₄  + e^- 

3. K → K⁺ + e^- 

4. Fe²⁺ → Fe³⁺ + e^- 

1. Na₂PO_4, Oxidation state of P = +6

2. H₃P₂O₇, Oxidation state of P = +5

3. PH₃, Oxidation state of P = -3

4. H₃PO₄, Oxidation state of P = +5

Oxidation number is a net charge which an atom has or appears to have when the other atoms from the molecule are removed as ions assuming that the shared pair of electrons is with more electronegative atom.

• Oxidation number of alkali metals is +1. 
• Oxidation number of alkaline earth metals is +2. 
• Oxidation number of H is +1 except in metallic hydrides.

1. The colour of the solution changes to blue. Silver is deposited on the copper rod.

2. Cu(s) +2AgNO₃ (aq) → Cu(NO₃)₂ (aq) + 2Ag(s) 

3. Oxidised species – Cu Reduced species – Ag⁺

H₂SO₄

(2 × +1) + x+ (4 × -2) = 0 

+2 + x – 8 = 0 

x – 6 = 0 

x = +6

Cr₂O₇^2-

2x + 7 ×-2 = -2 

2x = -2 + 14 

2x = 12 

∴ x = +6

NO₃^-

x + 3 × -2 = -1 

x = -1 + 5 

x = +4

(a) CH₄ ;

x + (1 × 4) = 0 

x + 4 = 0 

x = -4 

Oxidation number of C in CH₄ is -4.

CH₃Cl: 

x + (3 × 1) + -1 = 0 

x + 3 – 1= 0 

x + 2 = 0 

x = -2 

Oxidation number of C in CH₃Cl is -2.

(b) Zero

Al(OH)₃ \(\leftrightharpoons\) Al⁺³_(aq)⁺ + 3OH^-(aq)

using law of mass action

K = \(\frac{[Al^{+3}][OH^-]^3}{[Al(OH^-)_3]}\) 

⇒ K = equilibrium constant

\(\because\) [al(OH)_3] = constant

\(\therefore\) Rearranging above equation.

K x [Al(OH)₃] = [Al⁺³] [OH^-]³

^⇒ K_sp = [Al⁺³] [OH^-]³

where K_sp = [S] [3S]³

= s x 27S³

K_sp = 27s⁴

We have given, 

K_sp = 1 x 10^-36

\(\therefore\) S⁴ = \(\frac1{27}\times10^{-36}\)

= 3.7 x 10^-38

S⁴ = 370 x 10^-40

S = 4.39 x 10^-10 g/L

Hence, solubility of Al(OH)₃ will be 4.39 x 10^-10 g/L

4x² + 20 ax + 25a² - 36b² = 0

⇒ (2x)² + 2 \(\times\) 2x \(\times\) 5a + (5a)² = 36b²

⇒ (2x + 5a)² = (6b)² (\(\because\) a² + 2ab + b² = (a + b)²)

⇒ 2x + 5a = 6b or 2x + 5a = -6b

⇒ 2x = 6b - 5a or 2x = -6b - 5a

⇒ x = \(\frac{6b-5a}2\) or x = \(\frac{-6b-5a}2=\frac{-1}2\)(6b + 5a)

Hence, root of the given equation are

\(\frac{6b-5a}2\) or \(\frac{-1}2\) (6b + 5a)

x² - x - 6 and x² + 3x - 18

\(\because\) x² - x - 6 = 0

⇒  x² - 3x + 2x - 6 = 0

⇒ x(x - 3) + 2(x - 3) = 0

⇒ (x - 3) (x + 2) = 0

Hence, (x - 3) and (x + 2) are factors of x² - x - 6.

Now x² + 3x - 18 = 0

⇒ x² + 6x - 3x - 18 = 0

⇒ x(x + 6) - 3(x + 6) = 0

⇒ (x + 6) (x - 3) = 0

Hence, (x - 3) and (x + 6) are factors of x² + 3x - 18

\(\therefore\) common factor of x² - x - 6 and x² + 3x - 18 is x - 3

\(\therefore\) a = 3 (By comparing (x - a) with (x - 3))

\(\therefore\) x² - 3x = -2

⇒ x² - 3x + 2 = 0

⇒ (x - 2) (x - 1) = 0

⇒ (x - 2) = 0 or x - 1 = 0

⇒ x = 2 or x  = 1

And y² + 4y = 5

⇒ y² + 4y - 5 = 0

⇒ (y + 5)(y - 1) = 0

⇒ y + 5 = 0 or y - 1 = 0

⇒ y = -5 or y = 1

\(\therefore\) x = 2 or 1 and y = -5 or 1

Given function is

f(x) = (a - 11) tan x + (b - 3) lnx + (c - 4)e^x + sin x

\(\because\) tan x, ln x and e^x is unbounded function and sin x is a bounded function.

Given that f(x) is bounded function which is possible only when there is no term of tan x, ln x and e^x in f(x).

\(\therefore\) f(x) is bounded if a - 11 = 0, b - 3 = 0 and c - 4 = 0

or a = 11, b = 3 and c = 4

\(\therefore\) a + b + c = 11 + 3 + 4 = 18

Hence, a + b  + c = 18 for which given function is bounded.

We know that if n(A) = p and n(B) = q, then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9

⇒ n(A) = 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

We know that A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3, it is clear that A = {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),

(1, –1), (1, 0), and (1, 1)

Principal value rang of cos^-1x is [0, π]

\(\therefore\) Range of f(x) = 5 + cos^-1x is [ 0 + 5, 5 + π]

 = [5, 5 + π] = [5, 8.14]

Integer value in the range of f(x) = 5 + cos^-1x are 5, 6

\(\therefore\) Sum of all integer value  = 5 + 6 + 7 + 8 = 26

Domain of sin^-1x & cos^-1x is [-1, 1]

\(\therefore\) -1 \(\leq\) x \(\leq\) 1

If x = -1 or 0 or 1

then {x} = 0

Therefore, \(\frac{1+[x]}{x}\) is not defined.

\(\therefore\) x \(\neq\) -1, x \(\neq\) 0, x \(\neq\) 1

If 0 < x  < 1 then [x] = 0 and 0 < {x} < 1

Then \(\frac{1+[x]}{x}=\frac1{x}>1\)

which is not possible

because domain of cos^-1x is [-1. 1].

\(\therefore\) -1 < x < 0 then [x] = -1 and 0 < {x} < 1

\(\therefore\) \(\frac{1+[x]}{\{x\}}=\frac{1+(-1)}{\{x\}}=\frac0{\{x\}}=0\)

\(\because\) -1 < x < 0 ⇒ |x| = -x

\(\therefore\) L.H.S. = sin^-1x + cos^-1x = sin^-1(-x) + cos^-1x

 = -sin^-1x + cos^-1x (\(\because\) sin^-1(-x) = -sin^-1x)

 = \(-(\frac{\pi}2-cos^{-1}x)+cos^{-1}x\)

= \(-\frac{\pi}2+2cos^{-1}x\) 

R.H.S. = cos^-1(\(\frac{1+[x]}{\{x\}}\)) = cos^-1(0) = \(\frac{\pi}2\) 

\(\because\) L.H.S. = R.H.S.

\(\therefore\) \(-\frac{\pi}2+2cos^{-1}x = \frac{\pi}2\)

⇒ 2 cos^-1x = \(\frac{\pi}2+\frac{\pi}2=\pi\)

⇒ cos^-1x = \(\frac{\pi}2\) ⇒ x = cos(\(\frac{\pi}2\)) = 0

But -1 < x < 0

Hence, for no value of x given equality satisfies.