Explore topic-wise InterviewSolutions in Current Affairs.

Insulin or Glucagon.

Answer (b) Polythene

Answer (b) Lead storage battery

Answer (d) Butane 1-ol

Answer (c) 3

Answer (d) H₂NCH₂COOH

Answer (b) 2

The hybridisation of Fe in K₄[Fe(CN)₆] is d²sp³

Answer (c) d²sp³

Boiling point of p-nitrophenol is greater than O-nitrophenol due to intermolecular and intramolecular hydrogen bond respectively. 

Hence O-nitrophenol and p-nitrophenol are separated by fractional distillation process.

Answer (d) All of these

Flux – the foreign substance added in the roasted ore to remove infusible impurities present in the ore is called flux.

Choice of the flux–

(i) For basic infusible impurities, acidic flux (SiO₂) is used

MnO₂ + SiO₂ →  MnSiO₃

(ii) For acidic infusible material, basic flux is used 

SiO₂ + CaO → CaSiO₃

Slag – The fusible material formed due to reaction between gange and flux is called slag.

Gang + Flux = Slag

SiO₂ + CaO = CaSiO₃

XeF₆ on hydrolysis not produce XeO₄ 

(a) Ethanal (CH₃CH₂OH) perform idoform test but CH₃OH does not.

(b) Ethanal (CH₃–CHO) performs Tollen’s reagent test but propanone does not. 

Central atom –  In co-ordination complex or entity the metal atom or ion to which a fixed number of molecules or ions are attached by co-ordinate bonds is called central atom (or ions).

For example, an Ni(CO)₄ the atom Ni is central atom. An complex [CO(NH₃)₆]²⁺ , CO²⁺ ion is the central ion. 

The noble gas which is not adsorbed by coconut charcoal is He

Mineral – The chemical found in earth crust having high percentage of any one element is called mineral.

Ore – The mineral from which metal can be extracted easily and economically is called ore.

All ores are minerals but all minerals are not ore. 

Ligands – The neutral molecules or ions linked directly to the central atom/cation in the co-ordination entity having ability to donate ions prir of electrons to the central metal atom/cation are known as ligands. 

According to henry's law- "The partial pressure applied by any gas on a liquid surface is directly proportional to its mole fraction present in a liquid solvent.

P = K_n.X

K_H = henry's constant

X = mole fraction of solute.

∴ 1.5 atm = 1.83 x 10² atm \(\times\) x

x = \(\frac{1.5}{1.83\times10^2}\)

 = 0.0082

mole fraction of HCHO = 0.0082 = \(\frac{n_{CHO}}{n_{H_2o+n_{HCHO}}}\)

 ∵ nH₂O >> n_HCHO

we can neglect n_HCHO in denominator

∴ 0.0082 = \(\frac{n_{HCHO}}{n_{H_2O}}\)

0.0082 = \(\frac{n_{HCHO}}{\frac{W}{18}}\)

or = \(\frac{n_{HCHO}}w\times1000\) = 0.0082 x 18

⇒ molality = 0.0082 x 18

As we know for very dilute solution the molality and molarity are approximately equal.

∴ molarity of HCHO = 0.15 M

(i) Aldehydes are more reactive than ketone due to less steric effect. In ketones two bulkier alkyl group hinder the approach of nucleophile to the carbonyl atom. 

(ii) The carboxylic carbon is less electrophilic than carbonyl carbon because of resonance stabilised structure. 

(iii) This is due to more extensive association of carboxylic acid molecules through intermolecular H-bonding. Most of the carboxylic acid exists as dimer in vapour phase.

Let the number be x.

Then, x + (1/x) = 13/6 

=> (x² + 1)/x = 13/6 

=> 6x² – 13x + 6 = 0 

 => 6x² – 9x – 4x + 6 = 0 

=> (3x – 2) (2x – 3) = 0 

x = 2/3 or x = 3/2 

Hence the required number is 2/3 or 3/2.

I = \(\int \frac {1-\sqrt{3x+2}}{1+\sqrt{3x+2}}dx\)

Let 3x + 2 = t²

∴ 3dx = 2+dt

dx = \(\frac {2t}{3}dt\)

∴ I = \(\int \frac {1-t}{1+t}. \frac {2t}{3}dt\)

= \(\frac 23 \int \frac {t-t^2}{1+t}dt\)

= \(\frac 23 \int {t-t^2} - \frac {2}{1+t}dt\)

= \(\frac 23[\frac {-t^2}{2} + 2t - 2log[1+t]]+c\)

= \(\frac 13[t^2+4t-4log[1+t]] + c\)

= \(\frac 13 [-(3x+2) + 4\sqrt{3x+2} - 4log [1t\sqrt{3x+2}]]+c\)

(by putting t = \(\sqrt{3x+2}\))

∴ \(\int \frac {1-\sqrt{3x+2}}{1+\sqrt{3x+2}}dx \)

\(= \frac 13 [-(3x+2) +4\sqrt{3x+2}-4log [1+\sqrt3x+2]]+c\)

\(\frac{y^2+1}{y^2}=1+\frac1{y^2}\neq1\)

\((\because\frac1{y^2}\,never\,be\,0)\)

Hence, \(\frac{y^2+1}{y^2}\) is not equal to 1. 

\(\int\frac{dx}{x^2+x+1}=\int\cfrac{dx}{((x + \frac12)^2+\frac34)}\)

\(=\cfrac{1}{\frac{\sqrt3}2}tan^{-1}\left(\cfrac{x+\frac12}{\frac{\sqrt3}2}\right)+c\)

\(=\frac2{\sqrt3}tan^{-1}(\frac{2x+1}{\sqrt3})+c\)

\(\int\frac{x^2+1}{(x+1)^2}dx\) 

= \(\int \frac{x^2+1+2x-2x}{(x+1)^2}dx\) 

= \(\int \frac{(x+1)^2}{(x+1)^2}dx\) - \(\int \frac{2x+2-2}{(x+1)^2}dx\)

= \(\int dx\) - \(\int \frac{2(x+1)}{(x+1)^2}dx\) + \(2\int \frac{1}{(x+1)^2}dx\) 

= \(\int dx\) - \(2\int \frac{1}{(x+1)}dx + 2\) + \(2\int \frac{1}{(x+1)^2}dx\)

= x - 2 log (x + 1) - \(\frac{2}{x+1}\) + C

\(\int \frac{Cos2x-1}{cos2x+1}dx\)

= \(\int \frac{1-2\,sin^2x-1}{2cos^2-1+1}dx\)

= \(\int \frac{-2\,sin^2x}{2\,cos^2x}dx\)

= \(-\int tan^2x\,dx\)

= \(-\int (sec^2x-1)\,dx\)

= \(-\int sec^2xdx+\int dx\)

= \(x-tanx+c\)

\(\int_0^1\frac{dx}{3x-4}\) 

= \([\frac{log|3x-4|}{3}]^1_0\)

= \(\frac{1}{3}\)(log 1 - log 4)

= \(\frac{-log\,4}{3}\)(∵ log 1 = 0)

Mosquitoes have six legs like other insects.

Calculation:

Given: f = {(1, 1), (2, 3), (0, - 1), (- 1, - 3)} is a function from Z to Z where Z is the set of integers such that f(x) = ax + b for some integers a and b.

As (1, 1) ∈ f ⇒ x = 1 and y = 1 will satisfy the equation y = f(x) = ax + b ⇒ a + b = 1

Similarly (2, 3) ∈ f ⇒ x = 2 and y = 3 will satisfy the equation y = f(x) = ax + b ⇒ 2a + b = 3

Now by solving the equations a + b = 1 and 2a + b = 3, we get a = 2 and b = - 1

Hence, option 4 is the correct answer.

g(x) = e^x . The domain is the set of all real numbers. Now taking log on both sides, we get lny=x. Now logarithm function is only valid for all real numbers greater than 0. Thus the values of 'y' include all real numbers greater than 0.
Hence range of g(x) is R⁺ or (0,∞).

Answer: {(a,1),(d,3),(b,2),(b,3)}

(b) Rann of Kutch, Gujarat

We have,

bcosθ = a

⇒ cosθ = \(\frac{a}{b}\)

∴ sinθ = \(\sqrt{1-cos^2θ}\)

= \(\sqrt{1-\frac{a^2}{b^2}}\) 

= \(\sqrt{\frac{b^2-a^2}{b}}\).

∴ cotθ = \(\frac{cosθ}{sinθ}\)

= \(\frac{\frac{a}{b}}{\frac{\sqrt{b^2-a^2}}{b}}\) 

= \(\frac{a}{\sqrt{b^2-a^2}}\)

And cosecθ = \(\frac{1}{sinθ}\)

= \(\frac{b}{\sqrt{b^2-a^2}}\)

∴ cosecθ + cotθ = \(\frac{b}{\sqrt{b^2-a^2}}\) + \(\frac{a}{\sqrt{b^2-a^2}}\)

= \(\frac{b+a}{\sqrt{b^2-a^2}}\).

a)0.5

b)0.25

c)0.75

d)0.3

this answer 

a)0.5

The correct option (2) 2⁵⁰(2⁵⁰ – 1)

Explanation:

Product of elements in A is even = (total subjects) – (Number of subject of all whose element are odd) 

= 2¹⁰⁰ – 2⁵⁰

= 2⁵⁰(2⁵⁰ – 1)

Given:

A : B = 2 : 3

B : C = 8 : 9

C : D = 3 : 4

C = Rs. 4347 

Calculation:

A sum of Rs. x is divided between A, B, C and D

⇒ A : B = 2 : 3 = (2 × 8 × 3)/(3 × 8 × 3)

⇒ 48/72

⇒ B : C = 8 : 9 = (3× 8 × 3)/(9 × 3 × 3)

⇒ 72/81

⇒ C : D = 3 : 4 = (9 × 3 × 3)/(4 × 9 × 3)

⇒ 81/108

⇒ A : B : C : D = 48 : 72 : 81 : 108

⇒ A : B : C : D = 16 : 24 : 27 : 36

According to question C =  Rs. 4347

⇒ 27 unit = Rs. 4347

⇒ 103 unit = Rs. x

⇒ x = (4347 × 103)/27

⇒ x = Rs. 16583

∴ The value of x is Rs. 16583

 Given:

The ratio of the sum, difference, and product of two numbers is 11: 1: 120.

 Calculation:

Let the two numbers be a and b, and the value of sum is 11x, the difference is 1x and the product is 120x. Now apply the above-given things.

⇒ a + b = 11x      ----(1)         and a × b = 120x      ----(3)

⇒ a - b = 1x       ----(2)     (Now add (1) & (2))

⇒ a = 6x and by putting in any equations we get b = 5x

Now put these values of a and b in equation (3)

⇒ 6x × 5x = 120x

⇒ x = 4

As x is 4, than the value of a and b will be

a = 6×4 = 24, b = 5×4 = 20

Now the squares of these two numbers a and b is

24² = 576, 20² = 400, Their sum is

a² + b² = 576 + 400 = 976

∴ The sum of the square of 2 numbers is 976

Given:

A and B are paid Rs. 4400

A = 120% of B

Concept used:

Total = xA + yB

x and y are ratio

Calculation:

A = 120% of B

⇒ A = 6/5 of B

⇒ 5A = 6B

⇒ A/B = 6/5

⇒ A : B = 6 : 5

A + B = Rs. 4400

⇒ 11 → 4400

⇒ 1 → 400

⇒ 5 → 2000

⇒ B = Rs. 2000

∴ B paid Rs. 2000

∵ The given relation between a, b and c is

a : b : c = 2 : 3 : 1

Let the numbers be a = 2x, b = 3x, and c = x

Now, the sum of the numbers is 120

So, a + b + c = 120

⇒ 2x + 3x + x = 120

⇒ 6x = 120

So, b = 3x = 60

∴ The required number b is 60

Calculation:

The series follows following pattern

7 × 4 + 5 = 33

33 × 4 + 5 = 137

137 × 4 + 5 = 553

553 × 4 + 5 = 2217

2217 × 4 + 5 = 8873

∴ The value of A is 2217

Calculation:

The series follows following pattern

7 × 4 + 5 = 33

33 × 4 + 5 = 137

137 × 4 + 5 = 553

553 × 4 + 5 = 2217

2217 × 4 + 5 = 8873

∴ The value of ? is 2217

Given:

The given number is 723 + 722 + 721 + 720

Concept used:

A given natural number can be expressed as 2^a.3^b.5^c.7^d...... and so on 

Where a, b, c, d ...... and so on are whole numbers

Total number of factors = (a + 1) × (b + 1) × (c + 1) × (d + 1) ...... and so on

Calculation:

Given number is 723 + 722 + 721 + 720

The number can be expressed as 7²⁰ × (7³ + 7² + 7¹ + 1)

The number can be expressed as 7²⁰ × (343 + 49 + 7 + 1) = 7²⁰ × 400

The number can be expressed as 7²⁰ × 2⁴ × 5² 

Total number of factors = (20 + 1) × (4 + 1) × (2 + 1)

Total number of factors = 21 × 5 × 3 = 315

∴ Total number of factors is 315

Formula used:

(a + b)² = a² + b² + 2ab

Concept used:

Numbers that follow each other continuously in the order from smallest to largest are called consecutive numbers.

For example: 1, 2, 3, 4, 5, and so on

Calculation:

Let the three consecutive numbers be x, x + 1, and x + 2

Now, A/Q

⇒ (x + 2)² – x(x + 1) = 25

⇒ x² + 4 + 4x – x² – x = 25

⇒ 3x + 4 = 25

⇒ 3x = 21

⇒ x = 7

∴ The largest number is 9

Given :

X and Y are natural numbers other than 1, and Y is greater than X

Calculations :

1 < X < Y 

Let X be 2 and Y be 3         (X and Y are natural numbers)

Now put value in each option to get the values 

Option(1) ⇒ XY = 2 × 3 = 6 

Option(2) ⇒ X/Y = 2/3 

Option(3) ⇒ Y/X = 3/2 

Option(4) ⇒ (X + Y)XY = 5/6 

We can see that option 1 represents the largest value among all 

∴ Option1 will be the correct choice.