Explore topic-wise InterviewSolutions in Current Affairs.

1. (c) conflicts

2. (a) a society in an advanced state of social development

3. (a) motivated

4. (b) loose

5. (d) responsible

Correct option is (C) Public holidays

Fastest growing occupations: It is true that most of us were not aware of the current demands in the job market 10 years ago. As the emerging jobs and the required skill sets change all the professional qualifications become irrelevant. In addition to the basic degree, it is necessary to foresee the emerging trends in the job market. This paragraph shows that the topmost opportunity to grow is in the field of wind and solar energy. Technicians are needed in this sector. There is 100% chances of placement for those who qualify themselves in operating wind mills, designing, installing and maintenance of solar panels. Paramedical profession is likely to grow exponentially.

It means that those who pursue nursing, lab technician courses would get placed soon. Strikingly, Mathematicians and Statisticians wall be next in huge demand as 70% growth is for them only. As there is a boom in media services, T.V., radio, newspapers and periodicals, there is a rising demand for media persons and journalists. Against the common beleif, only 50% growth opportunity is there for professionals qualified in software development. It is heartening to note that cycle repairers and two wheeler mechanics will get opportunities next to software engineers. Though lakhs of graduates do B.Ed. course the growth opportunity in teaching jobs in colleges and schools is only 30%. The dismal picture for sportsmen and coaches is discouraging. Only 20% growth is predicted in placements for coaches and sportsmen. Finally only 10% growth is possible for those who specialise in food analysis and working in hotels.

(a) At the end of plus two exams students start thinking about their career path.

(b) To build a strong career, students must understand the prospective field and their primary interest. They need to qualify suitably by pursuing a degree course in their chosen area of interest.

(c) Personality development, communicative competence, required skill sets, keeping abreast of technological advancements in addition to qualification in the respective field are the integral aspects of a successful career.

(d) Cooking is called culinary art.

(e) Media is a popular carrer option because print and entertainment media have become a force to reckon with. Radio, TV channels, internet companies and advertisement companies are increasingly recruiting qualified professionals.

(f) Optometry, pathology, nursing, physiotherapy and dentistry are some popular paramedical courses.

(g) The number of international travellers visiting India is rapidly growing. This has resulted in the exponential growth of hospitality and service industry. A lot of career options such as house keeping, front-desk executives, tourism management etc. are now open.

(h) No. Sports journalists, dieticians, technical trainers and commentators need not be sports persons.

(i) Traditional

(j) International travellers

1. (b) disputable

2. (c) positive

3. (b) admirable qualities

4. (c) hated

5. (c) boiler

6. (a) rinsing

7. (d) restricting supply

8. (b) hanging freely

9. (d) detain

10. (a) mix thoroughly

1. (b) liked

2. (d) pessimistic

3. (c) deep

4. (b) departure

5. (c) inadequate

1. (b) almost the same

2. (d) making a solution

3. (d) of little depth

4. (c) unpleasant

5. (d) incompressible

6. (d) socially accepted behaviour

7. (d) supplementary

8. (c) luck

9. (c) foretelling

10. (b) squeezing out

The doctor injected an antivenom. The antivenom contains preformed antibodies which when injected act on the pathogen immediately provide protection by providing passive immunity

a. Adolescents are easily affected by (or are vulnerable to) peer pressure. Curiosity, need for adventure and excitement, and experimentation constitute common causes for motivation. A child’s natural curiosity motivates him/her to experiment. Television, movies, newspapers, internet also promote drug use. 

b. Addiction is the psychological attachment to certain effects such as euphoria or temporary feeling of well-being. Dependence is the tendency of the body to show withdrawal syndrome or symptoms if regular doses of drug/alcohol is abruptly discontinued.

1. They show variable oxidation state 

2. They form coloured ions 

3. They are used as catalyst.

4. They form alloys.

General characteristics of the long form of Periodic table :- 

(i) There are in all 18 vertical columns i.e. 18 groups in the long form periodic table. 

(ii) There are groups numbered from 1 to 18 from the left. 

(iii) There are seven horizontal rows called periods. 

(iv) The elements of groups 1, 2 and 13 to 17 are called main group elements. 

(v) The elements of group 3 to 12 are called transition elements.

S – block elements :- The elements in which the last electron enters the s – orbital of their outer most energy level are called s – block elements. It has elements of groups 1 and 2. The general electronic configuration of s – block elements is ns^1-2.

Correct option is a. 89° , 117°

28 g of N₂ molecule contain 2 × 6.023 × 10²³ atoms.

Hence, 1.4 g of N₂ molecules contain 

\(\frac{2\times6.023\times10^{23}}{28}\) = 1.04 = 6.023 x 10²² atoms.

1 molecule of CH₄ = 6 + 4 = 10 electrons, 16 g of CH₄ contains 10 × 6.022 × 10²³ = 6.022 × 10²⁴

Since 22400 cm³ of hydrogen at STP weight = 2g

Therefore, 112 cm³ of hydrogen at STP weight

= \(\frac{2}{22400}\) x 112 = 0.01 g

Average atomic mass = \(\frac{11\times81+10\times19}{100}\)

= 10.81

(a) Soft water

(a) Towards centre

SBR stand for styrene(S)butadiene (B) rubber (R).

(1) 

(i) Natural rubber is cis-polyisoprene. It is obtained by polymerization of isoprene units at 1, 4 positions. In rubber molecule, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chain to come closer. Therefore, only weak vander Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

(ii) Cis-1, 4 polyisoprene (Natural rubber), due to this cis configuration about the double bonds, has the adjacent chain that do not fit together well (there is no close packing of adjacent chains). The only force that interact is the weak or low strength of van der Waals’ forces.

(iii) Cis-polyisoprene has a coiled structure in which the various polymer chains are held together by weak van der Waals’ forces.

(2) 

(i) Vulcanization of rubber is a process of improvement of the rubber elasticity and strength by heating it in the presence of sulphur, which results in three dimensional cross-linking of the chain rubber molecules (polyisoprene) bonded to each other by sulphur atoms.

(ii) Vulcanisation makes rubber more elastic and more stiffer. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus rubber get stiffened.

(iii) The improved properties of vulcanised rubber are 

(i) high elasticity 

(ii) low water absorption tendency,

(iii) resistance to oxidation.

The German Chemist, Johann Dobereiner in early 1829 was the first to consider the idea of trends among properties of element.

The middle element of each of the triads had an atomic weight about half way between the atomic weights of the other two. Also the properties of the middle element were in between those of the other two members. Dobereiner’s relationship is known as the haw of triads.

Bonds formed between electropositive atoms are metallic, whereas those formed between electronegative atoms are covalent. If there is a difference in electronegativity values of two bonded atoms a polar covalent bond is formed where percent ionic character increases with the increase in the difference in electronegativity. When ∆EN > 1.7, the bond is said to have an ionic character greater than or equal to 50% and is referred to as an ionic bond (a bond between an electropositive and an electronegative atom).

• Polarisation of the anion by the cation in an ionic compound leads to the development of a covalent character in ionic bonds. Fajan's rules for estimating covalent character states that small cations, large anions, large charge on either ion and cations with non-noble gas configuration give rise to the covalent character of ionic bonds.

Li and Mg; Be and Al.

Answer is Np and Pu.

i. Transuranic elements are the man-made elements with atomic number higher than 92 (the atomic number of uranium). These elements are prepared by nuclear reactions which involve transformation of naturally occuring elements.

ii. The elements from atomic numbers 104 (Rf) to 112 (Uub) are identified. Elements upto atomic number 118 (Uuo) are synthesized. 

They are listed in the following table.

The series of elements from Thorium (Z = 90) to Lawrencium (Z = 103) in which 5f orbitals are progressively filled are called actinoids.

The elements are called actinoid because many physical and chemical properties are similar to actinium which is the prototype of the actinoids.

An alloy is a solid or solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to posses different physical properties than those of the component elements.

An important alloy of lanthanoids is Mischmetal It contains lanthanoids (4 - 95%), iron (5%) and traces of S, C, Si, Ca and Al.

Uses:

1. Mischmetal is used in cigarettes and gas lighters
2. It is used in flame throwing tanks
3. It is used in flame bullets and shills.

The approximate percentage of iron in misch metal is 5

Misch metal is an alloy of lanthanum metal and iron and traces of S, Ca, C or Al. It is used in making bullets and lighter flint.

CH₃-CH=CH-COOH

But-2-en-1-oic acid or 2-butenoic acid

1 alcohols with molecular formula C₄H₁₀O are chiral in nature.

CH₃CH₂ CH₂ COOH 

Butanoic acid

Given cost of 1 dozen roses = Rs. 2

Number of roses bought by the florist = 100 dozens

Thus, cost price of 100 dozen roses = 2 x 100 = Rs. 200

Roses left after discarding the mutilated ones = 80 dozens

Calculating the price at which the florist should sell the 80 dozen roses in order to make a profit of 20%, we have

Profit % = ((SP-CP)/CP) x100 = ((SP-200)/200) x 100

40 = SP – 200

SP = Rs. 240

Therefore, the SP of the roses should be Rs. 240/80 = Rs. 3 per dozen.

We know that SP is more than CP so it’s a gain.As the formula states Gain = SP – CP

713 – 620 = 93

Gain% = (Gain × 100) / CP

= (93 × 100) / 620

= 15%

Cost price of 10 article = selling price of 9 article

Let CP of 1 article = Rs. x

Selling price of 9 article = 10x

Selling price of 1 article = \(\frac{10x}{9}\)

Hence,

Profit = \(\frac{10x}{9}\) - x = \(\frac{x}{9}\)

= Gain% = \(\frac{gain}{cost\,price}\)x 100 = \(\frac{\cfrac{x}{9}}{x}\)x 100 = \(11\frac{1}{9}\)%

As the formula states SP = ((100 – Loss %) /100) × CP= ((100-(40/3))/100) × 645

= ((260/3)/100) × 645

= (260/300) × 645

=559

∴ The selling price is 559

Let CP = Rs. x SP = Rs. 240

Let profit be Rs. P.

Now, profit % = 20%

Since Profit % = (Profit/CP) x 100

20 = (P/x) x 100

P = 20x/100 = x/5

Profit = SP – CP = 240 – x

P = 240 – x

x /5 = 240 – x

240 = x + x/5

240 = 6×15

x = 1200/6

x = 200

So, CP = Rs. 200

New SP = Rs. 275 and CP = Rs. 200

Profit % = {(SP – CP)/CP} x 100

{(275 – 200)1200} x 100 = (75/200) x 100

= 7500/200

= 37.5%

We have,

27x² – 12y²

= 3 [(3x)² – (2y)²]  (By using the formula (a² – b²) = (a-b) (a+b))

= 3 (3x + 2y) (3x – 2y)

We have,

16x² – 25y²

= (4x)² – (5y)²   (By using the formula (a² – b²) = (a + b) (a – b) we get,)

= (4x + 5y) (4x – 5y) 

We have,

144a² – 289b²

= (12a)² – (17b)²  (By using the formula (a² – b²) = (a-b) (a+b))

= (12a + 17b) (12a – 17b)

We have,

(x + 2y)² – 4 (2x – y)²

= (x + 2y)² – [2 (2x – y)]²

= [(x + 2y) + 2 (2x – y)] [x + 2y – 2 (2x – y)] 

= (x + 4x + 2y – 2y) (x – 4x + 2y + 2y)

= (5x) (4y – 3x)

We have,

(2a – b)² – 16c²

= (2a – b)² – (4c)²  (By using the formula (a² – b²) = (a-b) (a+b))

= (2a – b + 4c) (2a – b – 4c)

We have,

125x² – 45y²

= 5 (25x² – 9y²)

= 5 [(5x)² – (3y)²] (By using the formula (a² – b²) = (a-b) (a+b))

= 5 (5x + 3y) (5x – 3y)

The Arithmetic Mean between two numbers is the number which when placed between them forms an arithmetic progression with them. Thus if x is the arithmetic mean of two given numbers a and b, then a, x, b form an A.P.

∴ x – a = b – x ⇒ x = \(\frac{a+b}{2}\)

Ex. 

The arithmetic mean between – 4 and 6 is \(\frac{6+(-4)}{2}\) = 1.

Let the A.P. be a, a + d, a + 2d, ... Let l be the last term and S the required sum. 

Then,

S = \(\frac{n}{2}\) (a + l ) = \(\frac{ Number\, of\, terms}{2}\) (First term + Last term)

= \(\frac{n}{2}\) (a + a + (n – 1) d) = \(\frac{n}{2}\) (2a + (n-1)d),

where n is the number of terms, a is first term and d is common difference. 

Also, nth term = Sum of n terms – Sum of (n – 1) terms 

i.e., T_n = S_n – S_n–1. 

Ex. 

The sum of 20 terms of the the A.P. 1, 3, 5, 7, 9.... is 

S20 = \(\frac{20}{2}\) (2 ×1 + (20 - 1)× 2)  (∵ a = 1, d = 2, n = 20) 

= 10 × 40 = 400

nth term of an A.P.: 

The nth term of an A.P. a, a + d, a + 2d, a + 3d, ... is 

T_n = a + (n – 1)d 

where T_n denotes nth term, a the first term, n the number of terms and d the common difference. 

Also, common difference d = T_n – T_n–1.

Ex. 

The 9th term of the A.P.: 2, 5, 8.... is

T₉ = 2 + (9 – 1) × 3 = 2 + 24 = 26.

Note : Here a = 2, d = 3.

Let three numbers in G.P. be \(\frac{a}{r}\), a, ar … (i)

Given, \(\frac{a}{r}\). a. ar = 216

⇒ a³ = 216 = 6³

⇒ a = 6

Therefore sum of their product in pair are:

\(\frac{a}{r}\) × a + a × ar × \(\frac{a}{r}\) = 156

⇒ \(\frac{36}{r}\) + 36r + 36 = 156                 (∴ a = 6)

⇒ \(\frac{36}{r}\) + 36r = 120

⇒ \(\frac{3}{r}\) + 3r = 10

⇒ 3r² − 10r + 3 = 3

⇒ (3r − 1)(r − 3) = 0

⇒ r = \(\frac{1}{3}\), 3

Substituting value of a and r in eq (i), we get

18, 6, 2 when r = \(\frac{1}{3}\)

and 2, 6,18 when when r = 3