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According to The Theory of natural selection, the organisms are in competition for limited resources (struggle for existence).

Only organisms with favourable variations survive. And this leads to the evolution of new species with favourable variations. They become accustomed to the new environment.

This new species produce more number of offsprings (overproduction).

Here the pests have variations to overcome the DDT. Organisms with favourable variations survive and lead to the evolution of new species which can survive DDT.

(c) → (a) → (b) → (d) → (f) → (e)

For example, say a borrower has not paid on a loan for 100 days. That would be considered a classified loan, as would a loan where a borrower has only made small partial payments for more than 90 days. When a loan is classified, it is reported on the lender’s accounting records with its classification status.

Nitrogen dioxide (NO₂)

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

b) Assertion is wrong but Reason is correct. 

M = \(\frac{n}{v}\); M = 0.5,V = 250 ml = 0.25 L

0.5 = \(\frac{n}{0.25}\) ; n = 0.5 x 0.25 = 0.125

GMM of Na₂CO₃ = 23 x 2 + 12 x 1 + 16 x 3 = 106 g

∴ Mass of Na₂ CO₃ in the solution of 250 ML

with 0.5 Na₂CO₃ = 0.125 x 106 = 13.25g

Lead nitrate on heating decomposes to lead monoxide, nitrogen dioxide and oxygen. 

2Pb (NO₃)₂ +∆ → 2PbO + 4NO₂ + O₂ 

NO₂ is liberated, along with oxygen, as a reddish brown gas. The residue left is lead monoxide (brown). It is a decomposition reaction.

Correct answer is (a) Brown fumes of nitrogen dioxide

\(2 pb (NO_3)_2\) \(\overset{\text{heat}}{\longrightarrow}\) \(\underset{\text{Light yellow}}{2pbO↑} + \underset{\text{Reddish-Brown}}{4NO2↑}\) _{\(+ \underset{\text{colourless}}{O_2\uparrow}\)}

Hence, The brown fumes will be nitrogen dioxide.

Correct option is b) Assertion is wrong but Reason is correct.

(d) Li₂CO₃

Li₂CO₃ is least stabel.

Correct option is d) ₹ 2,000

Correct option is d) ₹ 2,000

(a) Ester 

(b) Alcohol

2AgBr_(s) \(\overset{sunlight}{\longrightarrow}\) 2Ag_(s) + Br_2(g)

Len's maker's formula 1/f = (n-1) [1/R₁ - 1/R₂]

R₁, R₂ – Radii of curvature 

n – Refractive index;

f – Focal length

(a) Ohm's law : The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature. 

(b) R = V/i = 5/1 = 10/2 = 5 Ohms.

X : Silver Bromide 

This reaction is "Photo chemical reaction."

Zinc is above hydrogen rvhereas copper is below hydrogen in the activity series of metals. That is why zinc displaces hydrogen from dilute hydrochloric acid, while copper does not.
              Zn + HCl ------->ZnCl₂ + H₂
              Cu + HCI ------> No reaction

Correct option is (a) 0 

Oxidation number of Cr in chromate ion (CrO₄²) is +6 

Oxidation number of Cr in dichromate ion (Cr₂O₇^2-) is also +6

HClO is strongest oxidising agent amount.

(a) One-half of the maximum velocity 

Correct answer is (i) Viscosity.

(B) Chloramphenicol

Answer is B. OKT-3

B. Blue colonies represent non-recombinant bacteria

A. Transfection

(A) 1-methylcyclohexene

(A) 1,2-dimethylcyclohexene

d. None of these

pH required for the growth of bacteria is 6.8 – 7.2.

d. Lysogenic factor 

(a) C–O is more polar because O is more electronegative than N. 

(b) C–O is more polar because O is more electronegative than S.

In elimination reactions, the most highly substituted alkene predominates.

The ion that is required in trace amounts for the growth of bacteria is Cobalt.

b. Concentration of growth factor have a linear relationship with the growth of the organism

Correct option is (3) Chitin and polysaccharides

Answer : Folic acid reverses its toxic effects

The general formula of Polysaccharides is (C₆H₁₀O₅)_n, 40≤n≤3000.

The most important vitamin for the growth of bacteria is B-complex.

a. Lag phase

Mycotoxins are formed during the end of Lag phase.