Explore topic-wise InterviewSolutions in Current Affairs.

• Trypnosoma cruzi – Blood – Chagas disease 
• Plasmodium vivax – RBC and Liver cells – Benign tertian malaria

1. Class – Calcarea (Calcispongiae) 

• The skeleton is of spiculues made up of CaCO3. 

• Examples – Leucosolenia, Syncon, Grantia, Leucilla etc. 

2. Class – Hexactinillida (Hyalospongiae) 

• The skeleton is of spicules made up of silica. 

• Examples – Hyalomena, Euplectella, Pheronema etc. 

3. Class Demospongiae 

• The skeleton is of sponging fibres. 

• Examples – Euspongia, Spongilla, Chalina, Cliona Spheceospongia etc.

Interstitial cells

Sporozoa class of protozoa, locomotory orgenelles are absent.

Coelenterata

Filariaworm elephantiasis in human beings.

Animals of trilobita subphylum of phylum arthropoda are extinct (fossilized).

A linear equation in two variables is represented algebraically as ax + by + c = 0, where a \(\ne\) 0, b \(\ne\) 0.Graphically it represents a straight line.

The given equations are : 

2x + 3y –1 = 0 and (p + q)x + (2p – q)y – 21 = 0

Here, a₁ = 2, b₁ = 3, c₁ = –1 and a₂ = p + q, b₂ = 2p – q, c₂ = – 21 

For infinite solutions,

\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\)= \(\frac{c_1}{c_2}\) \(\Rightarrow\) \(\frac{2}{p+q}\) = \(\frac{3}{2p-q}\) = \(\frac{-1}{-21}\) \(\Rightarrow\) \(\frac{2}{p+q}\) = \(\frac{1}{21}\) and \(\frac{3}{2p-q}\) = \(\frac{1}{21}\)

\(\Rightarrow\) p + q = 42

Adding (1) and (2), we get

3p = 105 \(\Rightarrow\) p = 35

Putting p = 35 in (1), we get

35 + q = 42 \(\Rightarrow\) q = 42 - 35 = 7

\(\therefore\) p = 35, q = 7

(b) (4, 6)

\(\frac{X}{2}\) + \(\frac{y}{3}\) = 4

\(\Rightarrow\) 6 x \(\frac{X}{2}\) + 6 x \(\frac{y}{3}\) =  6 x 4

\(\Rightarrow\) 3x + 2y = 24 ..........(i)

Given, x + y = 10  \(\Rightarrow\) y = 10 – x .....(ii)

Substituting the value of y in (i), we get

3x + 2 (10 – x) = 24

\(\Rightarrow\)3x + 20 – 2x = 24

\(\Rightarrow\) x = 24 – 20 = 4

\(\therefore\) From eqn (ii), y = 10 – 4 = 6.

If a₁x + b₁ y + c₁ = 0 

a₂x + b₂y + c₂ = 0 

is a pair of linear equations in two variables x and y such that :

(i) \(\frac{a_1}{a_2}\) \(\ne\) \(\frac{b_1}{b_2}\),then the pair of linear equations is consistent with a unique solution, i.e., they intersect at a point.

(ii) \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) \(\ne\) \(\frac{c_1}{c_2}\),then the pair of linear equations inconsistent with no solution, i.e., they represent a pair of parallel lines.

(iii) \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\), then the pair of linear equations are consistent with infinitely many solutions, i.e., they represent coincident lines.

To solve the simultaneous linear equations algebraically. We use the following methods : 

(i) Substitution method 

(ii) Elimination method

Simultaneous linear equations in two variables can be solved by : 

(i) Algebraic method 

(ii) Graphical method

(b) x + y = 7 and x + 2y = 8

Total number of notes = 7 

\(\Rightarrow\)x + y = 7 

Total value of notes = Rs 40 

\(\Rightarrow\)5x + 10y = 40 

\(\Rightarrow\)x + 2y = 8

(d) several solutions

Check for \(\frac{a_1}{a_2}\)\(\ne\)\(\frac{b_1}{b_2}\), \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)\(\ne\)\(\frac{c_1}{c_2}\)

and \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\)

Here a₁ =2, b₁ =3,c₁= -5

a₂ = 4, b₂ = 6, c₂ = –10.

(d) an infinite number of solutions

Here a₁ = a, b₁ = b, c₁ = – (2a² – 3b² ) = 3b² – 2a²

a₂ = 1, b₂ = 2, c₂ = – (2a – 6b) = 6b – 2a

\(\frac{a_1}{a_2}=\frac{a}{1}\), \(\frac{b_1}{b_2}=\frac{b}{2}\),\(\frac{c_1}{c_2}\)= \(\frac{3b^2-2a^2}{6b-2a}\)

2a = b \(\Rightarrow\) a = \(\frac{b}{2}\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b}{2}\), \(\frac{b_1}{b_2}=\frac{b}{2}\) and 

\(\frac{c_1}{c_2}\)= \(\frac{3b^2-2\times\frac{b^2}{4}}{6b-2\times\frac{b}{2}}\)= \(\frac{3b^2-\frac{b^2}{2}}{6b-b}\) = \(\frac{\frac{5b^2}{2}}{5b}\)=\(\frac{b}{2}\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

(a) \(\frac{1}{2} \) and \(-\frac{1}{2} \)

(4)^{x + y} = 1 \(\Rightarrow\) (4)^{x + y} = 4⁰ \(\Rightarrow\) x + y = 0  .........(i)

(4)^{(x – y)} = 4 \(\Rightarrow\) (4)^{x – y} = 4¹ \(\Rightarrow\) x – y = 1 .........(ii)

Adding eqn (i) and (ii), 2x = 1

\(\Rightarrow\) x = \(\frac{1}{2}\)

\(\therefore\) From eqn (i),\(\frac{1}{2}\) + y = 0

\(\Rightarrow\) y = -\(\frac{1}{2}\)

(c) x = 2, y = 4

2x + y = 8 ............…(i)

– 4x + 3y = 4 ..............(ii)

Multiplying eqn (i) by 2 and adding to eqn (ii), we get

4x + 2y – 4x + 3y = 16 + 4

\(\Rightarrow\) 5y = 20

\(\Rightarrow\) y = 4

Putting y = 4 in (i), we get 2x + 4 = 8

\(\Rightarrow\) 2x = 4

\(\Rightarrow\) x = 2.

(d) (7, –3)

x – y = 10

\(\Rightarrow\) x = y + 10

Substitute this value of x in eqn 2x + 3y = 5.

so, that we get the solution.

(b) 2 and 3

Let \(\frac{1}{X}\) = a, \(\frac{1}{y}\) = b, Then, the given equations reduce to:

2a + 3b = 2 ....…(i)

6a + 18b = 9 ….(ii)

Now solve eqn (i) and (ii) to find the values of a and b, and then x and y.

(b) k \(\ne\) – 4

In the given system of equations,

a₁ = 2, b₁ = 3, c₁ = –5 

a₂ = k, b₂ = –6, c₂ = – 8

For a unique solution.

\(\frac{a_1}{a_2}\)\(\ne\)\(\frac{b_1}{b_2}\) \(\Rightarrow\) \(\frac{2}{k}\) \(\ne\) \(\frac{3}{-6}\) \(\Rightarrow\) 3k \(\ne\) -12

\(\Rightarrow\) k\(\ne\) -4

People’s rights: The right entitled to the people by the constitution.

Equity: Th quality of being fair or Impartial, fairness.

Sink: An environment’s ability to absorb and render harmless waste and pollution.

Gross Domestic Product: The total money value of all the goods and services produced In the country during a year which exclude the money from abroad.

Per capita income: The National Income of a country d1vI(d by Its population gives per capita income.

Social indicators: The Indicators which show some effect on society either with change In Income or any other changes.

Human Development Index: A composite Index based on life exptary, general health level, literacy rate, and education sanitation facilities besides per capita income.

Environment’s source: The potential of an environment to provide the function natural resources like land, water, minerals, ores, products from trees and animals, etc.

Unsafe drinking water: The water that Is contaminated from chemical and industrial waste.

Natural resources: The resources that are naturally existing substances like land, water, minerals, ores, etc,

Carrying capacity of the environment: The capacity of the environment to support economic production and consumption In the future.

The environmental resources have been used up and damaged to an unprecedented extent in the course of economic growth.

(iii) maximum Principal quantum number of any element of the period.

(i) group 1& 2 elements belong to this block 

(ii) Having general formula =ns^1-2 

(iii) Group 1 element called alkali metals. Group 2 called alkaline earth metals 

(a) Soft and highly reactive.

(b) Low ionization enthalpy, so highly electropositive 

(c) Give colors to flame.

 (i) some of them are non-metals especially those heading all the groups. 

(ii) Moving down the group metallic character increases 

(iii) Many of them are highly electronegative and form a covalent compound

Groundwater overuse is particularly found in Punjab and Western U.P, hard rock plateau areas of central and south India, some coastal areas, and urban settlements.

The environment’s ability to absorb and render harmless waste and pollution is “Sink function”.

If the water drawn up is more than that is being recharged, then it is obvious that after some time no more groundwater is left.

(b) wind energy.

Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings.

(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (Δ_iH₁) and a positive electron gain enthalpy (Δ_egH). 

(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (Δ_iH₁) and a low negative electron gain enthalpy (Δ_egH). 

(c) Element III is likely to be the most reactive non–metal as it has a high first ionization enthalpy (Δ_iH₁) and the highest negative electron gain enthalpy (Δ_egH). 

(d) Element V is likely to be the least reactive non–metal since it has a very high first ionization enthalpy (Δ_iH₂) and a positive electron gain enthalpy (Δ_egH). 

(e) Element VI has a low negative electron gain enthalpy (Δ_egH). Thus, it is a metal. Further, it has the lowest second ionization enthalpy (Δ_iH₂). Hence, it can form a stable binary halide of the formula MX₂(X=halogen). 

(f) Element I has low first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen). 

(a) Element V has highest ionization enthalpy. Hence it is the least reactive element.

(b) The element having the lowest first ionization enthalpy will tend to lose electron easily and hence would be a reactive metal. Thus, II in the most reactive metal.

(c) The non-metals have high ionization enthalpies (but less than that of noble gases). The element III is thus most reactive non-metal.

(d) The element IV is least reactive non-metal.

(e) Metals have relatively lower volumes of ionization enthalpies. The first ionization enthalpy of elements in group II is higher than that of element in group I. Since the element M forms a stable binary halide of the formula MX₂, M should belong to the group II of the periodic table. The sum of IE₁ and IE₂ for a group 2 element is less than that of its neighbours. Keeping this in mind, VI appears to be metal capable of forming a stable halide of the formula MX₂.

(f) I element can form a predominantly stable covalent halide of the formula MX (X = halogen).

Given: Velocity of light in the first medium = v₁ = 1.5 × 10⁸ m/s,

velocity of light in the second medium = v₂ = 0.75 × 10⁸ m/s,

refractive index of the second medium with respect to the first medium = ₂n₁ = ?

₂n₁ = \(\frac{v_1}{v_2}\)

₂n₁ = \(\frac{1.5\times10^8}{0.75\times10^8}\) = 2

Hence, the refractive index of the second medium with respect to the first medium is 2.

[Note : The absolute refractive index of the second medium = \(\frac{3\times10^8m/s}{0.75\times10^8m/s}\) = 4

(greater than that of diamond, not likely).]

Laws of reflection of light:

1. The incident ray and the reflected ray of light are on the opposite sides of the normal to the reflecting surface at the point of incidence and all the three are in the same plane.

2. The angle of incidence j and the angle of reflection are equal in measure.

Data: v = 1.5 × 10⁸ m/s,

c = 3 × 10⁸ m/s, n = ?

n = \(\frac{c}{v}=\frac{3\times10^8m/s}{1.5\times10^8m/s}\) = 2

This is the absolute refractive index of the medium.

Reflection of light: When light is incident on the surface of an object, in general, it is deflected in different directions. 

This process is called reflection of light.