Plus Two

Reviewing Kerala Syllabus Plus Two Chemistry Previous Year Question Papers and Answers Pdf March 2020 helps in understanding answer patterns.

Kerala Plus Two Chemistry Previous Year Question Paper March 2020

Time: 2 Hours
Total Scores: 60

Answer any 7 questions from 1-9. Each carries 1 score. (7 × 1 = 7)

Question 1.
Which of the following lattices has the highest packing efficiency (assuming that atoms are touching each other)?
(a) Simple cubic
(b) Body centred cubic
(c) Face centred cubic
Answer:
(c) Face centered cubic

Question 2.
The limiting molar conductivity of weak electrolytes can be calculated by using the law
(a) Faraday’s law
(b) Kohlrausch law
(c) Henry’s law
(d) Raoult’s law
Answer:
(b) Kohlrausch law

Question 3.
The Bredig’s arc method is used to prepare which of the following solutions?
(a) Silver sol
(b) Gelatine sol
(c) CdS sol
(d) As₂S₃ sol
Answer:
(a) Silver sol

Question 4.
The product obtained by the reaction of calcium phosphide with water is
(a) Phosphoric acid
(b) Phosphine
(c) Phosphorous acid
(d) Phosphorus trichloride
Answer:
(b) Phosphipe

Question 5.
Among the following, which is more acidic?
(a) HCOOH
(b) CH₃CH₂COOH
(c) CH₃COOH
(d) CH₃CH₂CH₂COOH
Answer:
(a) HCOOH

Question 6.
In the presence of light, chloroform is slowly oxidised by air to an extremely poisonous gas called ___________
Answer:
Phosgene

Question 7.
Benzene diazonium chloride when treated with Cu₂Cl₂ and HCl, the product formed is chlorobenzene. This reaction is known as ___________
Answer:
Sand Meyer’s reaction

Question 8.
The monomer unit of natural rubber is ___________
Answer:
Isoprene or 2-methyl 1,3 butadiene

Question 9.
Name a substance which can be used as an antiseptic and disinfectant at different concentrations.
Answer:
Phenol, Chlorine or Sulfur dioxide

Answer any 10 questions from 10-22. Each carries 2 scores. (10 × 2 = 20)

Question 10.
Classify each of the following as being either a p-type or n-type semiconductor: (2 × 1 = 2)
(a) Ge doped with B
(b) Si doped with As
Answer:
(a) P type semiconductor
(b) N-type semiconductor

Question 11.
Schottky and Frenkel defects are two types of stoichiometric point defects shown by ionic solids. Give two points of difference between Schottky defect and Frenkel defect.
Answer:

Schottky Defect	Frenkel Defect
1. Due to the lack of an equal number of cations & anions from the lattice.	1. Due to the missing cation from the lattice, it is seen in the interstitial space.
2. Density decreases.	2. No change in density.

Question 12.
Complete the table by giving the value of Van’t Hoff factor ‘i’ for complete dissociation of solute. (4 × ½ = 2)

Salt	Vant Holf factor ‘i’ for complete dissociation of solute
NaCl	__________________
Al(NO3)3	__________________
K2SO4	__________________
Al2(SO4)3	__________________

Answer:
(a) NaCl (i = 2)
(b) Al(NO₃)₃ (i = 4)
(c) K₂SO₄ (i = 3)
(d) Al₂(SO₄)₃ (i = 5)

Question 13.
For a reaction A + B → C + D, the rate equation is, Rate = K [A]^3/2 [B]^1/2. Give the overall order and molecularity of reaction.
Answer:
(a) Order = 2 (\(\frac{3}{2}\) + \(\frac{1}{2}\) = \(\frac{4}{2}\) = 2)
Molecularity = 2 (one molecule A + one molecule B)

Question 14.
Give the general method used for the concentration of the following ores: (2 × 1 = 2)
(a) Bauxite ore
(b) Zinc sulphide ore
Answer:
(a) Bauxite – Leaching
(b) Zine Sulphide – Froth Floatation

Question 15.
Semiconductors of very high purity can be obtained by zone refining. Explain the principle behind zone refining. (2)
Answer:
The impurities are more soluble in the melt rather than the solid state of a metal.

Question 16.
The composition of bleaching powder is Ca(OCl)₂.CaCl₂.Ca(OH)₂.2H₂O. Give one method for the preparation of bleaching powder. (2)
Answer:
Bleaching powder is prepared by passing dry chlorine through slaked lime.
2Ca(OH)₂ + 2Cl₂ → [Ca(OCl)₂] + CaCl₂ + 2H₂O

Question 17.
(a) In d-block elements the radii of elements of third transition series are similar to those of the elements of second transition series. Give reason.
(b) Outer electronic configuration of Cu²⁺ ion is 3d⁹. Calculate its spin only magnetic moment. (2 × 1 = 2)
Answer:
(a) Lanthanoid contraction.
The overall decrease in atomic and ionic radii from lanthanum to lutetium is known as lanthanoid contraction.
(b) µ = \(\sqrt{n(n+2)}\)
n = number of unpaired electrons
= \(\sqrt{1(3)}\)
= \(\sqrt{3}\)
= 1.73 BM

Question 18.
Assign the primary valence and secondary valence of the central metal in [Ni(CO)₄].
Answer:
Primary valency = 0
Secondary valency = 4

Question 19.
Aryl halides are less reactive towards nucleophilic substitution reactions. Write any two reasons for the less reactivity of aryl halides.
Answer:
Resonance effect
Delocalisation of π electron in the benzene ring.

Question 20.
Ethanol and methoxymethane are functional isomers. But ethanol has a higher boiling point than methoxy-methane. Give reason.
Answer:
Due to the presence of hydrogen bonds.

Question 21.
Give a chemical test to distinguish between propanal and propanone.
Answer:
Propanal – Aldehyde
It gives a silver mirror with Tollens reagent.
or
With Fehling’s solution, it gives a red ppt.
or
With Schiffs reagent it gives a purple colour.
Propanone – Ketone does not give any of the above test.

Question 22.
Analgesics and antibiotics are drugs having different therapeutic actions. Define each class of drugs.
Answer:
Analgesics: Pain Killer
eg. Aspirin, Heroine, Paracetamol, Ibuprofen
Antibiotics: Drugs which kill or inhibit the growth of bacteria
eg. Penicillin, Amoxycillin, Azithromycin

Answer any 7 questions from 23-31. Each carries 3 scores. (7 × 3 = 21)

Question 23.
For ethanol-acetone mixture solute-solvent interaction is weaker than solute-solute and solvent-solvent interaction.
(a) Does this solution obey Raoult’s law?
(b) Give the vapour pressure-mole fraction graph for this solution.
Answer:
(a) No
(b) Positive deviation

Question 24.
The temperature dependence of the rate of a chemical reaction can be explained by the Arrhenius equation.
(a) Give Arrhenius equation.
(b) The rate of a chemical reaction doubles for an increase of 10 K in absolute temperature from 300 K. Calculate the activation energy (Ea)? [R: 8.314 J K^-1 mol^-1, log 2 = 0.3010]
Answer:
(a) Arrhenius equation is k = \(A \cdot e^{-E a / R T}\)

Question 25.
The existence of charge on colloidal particles is confirmed by the electrophoresis experiment.
(a) What is meant by electrophoresis?
(b) In the coagulation of a negative sol, the coagulating power is in the order Al³⁺ > Ba²⁺ > Na⁺. Name and state the rule behind this.
Answer:
(a) Electrophoresis: Movement of colloidal particles under an applied electric field is called electrophoresis.
(b) Hardy: Schulze rule
The greater the valency of the flocculating ion added, the greater is its power to cause precipitation.

Question 26.
Give the steps involved in the preparation of potassium dichromate (K₂Cr₂O₇) from chromite ore.
Answer:
1. Chromite ore is fused with sodium carbonate in free access of air to form sodium chromate.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2. Sodium chromate is acidified with sulpuric acid to give sodium dichromate.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
3. Sodium dichromate is treated with KCl to give potassium dichromate.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Question 27.
C is an isomer of [Pt(NH₃)₂Cl₂] is used to inhibit the growth of tumours.
(a) Give the IUPAC name of [Pt(NH₃)₂Cl₂].
(b) Give the structure of cis and trans isomers of [Pt(NH)₃Cl₂].
Answer:
(a) [Pt(NH₃)₂Cl₂] diammine dichlorido Plainum (II)
(b)

Question 28.
(a) Which is the major product obtained when 2-bromopentane is heated with alcoholic solution of potassium hydroxide?
(b) Name and state the rule that governs the formation of major product.
Answer:
(a) But-2-ene
(b) Saytzeff Rule: More symmetrical products are favoured.

Question 29.
Complete the following table: (3 × 1 = 3)

Answer:
1. CH₃CH₂NC or Ethyl Isocyanide
2. Hoffman Bromamide Reaction
3. C₆H₅NH₂/Aniline

Question 30.
(a) Vulcanisation is carried out to improve the physical properties of rubber. Explain the process of vulcanisation of rubber.
(b) Classify the following into addition and condensation polymers: (4 × ½ = 2)
PVC, nylon 66, teflon, terylene
Answer:
(a) Heating natural rubber in the presence of sulphur is known as vulcanisation.
(b) Addition polymers: PVC, Teflon
Condensation polymers: Nylon 66, Terylene

Question 31.
(a) Differentiate between globular and fibrous proteins. (2)
(b) The deficiency of which vitamin causes night blindness. (1)
Answer:
(a)

Globular Proteins	Fibrous Protein
1. Spherical.	1. Thread like shape.
2. Water soluble eg. Haemoglobin, Albumin	2. Insoluble in water eg. Keratin, Myosin

(b) Vitamin A

Answer any 3 questions from 32-35. Each carries 4 scores. (3 × 4 = 12)

Question 32.
Daniell cell converts the chemical energy liberated during the redox reaction to electrical energy.
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}^{2+}{ }_{(\mathrm{aq})} \rightarrow \mathrm{Zn}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})} ; \mathrm{E}_{\text {cell }}^0=1.1 \mathrm{~V}\)
(a) Identify the anode and cathode in Daniell cell. (1)
(b) Calculate the standard Gibbs energy (∆_rG°) for the reaction. (2)
(c) Give the Nernst equation of the above cell reaction. (1)
Answer:
(a) Anode: Zinc or Zn
Cathode: Cu or Copper
(b) ∆_rG° = -nFE°cell
= -2 × 96500 × 1.1
= -212300 J
(c) \(E_{\text {cell }}=E^{\circ} \text { cell }-\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}\right]\)

Question 33.
Account for the following:
(a) N₂ is less reactive at room temperature. (1)
(b) PCl₃ fumes in moisture. (1)
(c) Cl₂, is a powerful bleaching agent. (1)
(d) H₃PO₃ is dibasic
Answer:
(a) In N₂ molecule one N atom is triply bonded to another N (N ≡ N). To break this bond, a high bond dissociation energy is needed. So N₂ is less reactive at room temp.
(b) PCl₃ + 3H₂O → H₃PO₃ + 3HCl ↑
It is due to the presence of HCl formed.
(c) Cl₂ + H₂O → HClO
HClO → HCl + [O]
The nascent oxygen causes bleaching action.
(d)

(H₃PO₃) – Presence of two OH groups indicates two replaceable H⁺. So its basicity is two.

Question 34.
(a) A mixture of anhydrous ZnCl₂ and conc. HCl is an important reagent used to distinguish primary, secondary, and tertiary alcohols. How the above reagent is used to distinguish the three types of alcohols? (3)
(b) Predict the product formed in the reaction:

Answer:

Question 35.
Explain the following reactions:
(a) Rosenmund reduction (2)
(b) Cannizzaro reaction (2)
Answer:
(a) Acid chloride reacts with H₂ in the presence of Pd/BaSO₄ to give aldehyde, which is known as Rosenmund Reduction.

(b) Cannizaro Reaction
Aldehydes with no hydrogen atom undergo oxidation and reduction in the presence of strong alkali to produce a primary alcohol and a salt of carboxylic acid.

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf Board Model Paper 2020 helps in understanding answer patterns.

Kerala Plus Two Maths Board Model Paper 2020 with Answers

Time: 2 Hours
Total Score: 60 Marks

Question 1 to 8 carry 3 scores each. Answer any 7 questions.

Question 1.
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-3 & 1 & 0 \\
1 & 3 & -2
\end{array}\right]\)
(i) Which of the following is the order of the matrix A+B?
(a) 2 × 2
(b) 3 × 2
(c) 2 × 3
(d) 3 × 3
(ii) Find 3A
(iii) Evaluate 3A – B
Answer:
(i) (c) 2 × 3

(ii) 3A = \(\left[\begin{array}{lll}
3 & 6 & 9 \\
9 & 3 & 6
\end{array}\right]\)

(iii) 3A – B
= \(\left[\begin{array}{lll}
3 & 6 & 9 \\
9 & 3 & 6
\end{array}\right]\) – \(\left[\begin{array}{ccc}
-3 & 1 & 0 \\
1 & 3 & -2
\end{array}\right]\)
= \(\left[\begin{array}{lll}
6 & 5 & 9 \\
8 & 0 & 8
\end{array}\right]\)

Question 2.
(i) If y = sin^-1 x, Find \(\frac{d y}{d x}\)
(ii) Hence show that (1 – x²)\(\frac{d^2 y}{d x}\) – x\(\frac{d y}{d x}\) = 0
Answer:
(i) Given; y = sin^-1 x
\(\frac{d y}{d x}\) = \(\frac{1}{\sqrt{1-x^2}}\) …………….. (1)

Question 3.
(i) Which of the following is the solution of the differential equation \(\frac{d y}{d x}\) + sin x = 0
(a) y = C cos x
(b) y = cos x + C
(c) y = sin x + C
(d) y = C sin x
(ii) Form the differential equation representing the family of curves y = a sin(x + 6), where a and b are arbitrary constants.
Answer:
(i) (b) y = cos x + C

(ii) Given; y = a sin(x + b) …………… (1)
Differentiating (1) w.r.to x; we get
\(\frac{d y}{d x}\) = a cos (x + b)
Differentiating (2) w.r.to x; we get
\(\) = -a sin(x + b)
⇒ \(\frac{d^2 y}{d x^2}\) = -y

Question 4.
Using properties of determinants prove that
\(\left|\begin{array}{ccc}
b+c & a & a \\
b & c+a & b \\
c & c & a+b
\end{array}\right|\) = 4abc
Answer:
LHS = \(\left|\begin{array}{ccc}
b+c & a & a \\
b & c+a & b \\
c & c & a+b
\end{array}\right|\)
Apply R₁ → R₁ – R₂ – R₃
= \(\left|\begin{array}{ccc}
0 & -2 c & -2 b \\
b & c+a & b \\
c & c & a+b
\end{array}\right|\)
= 2c(ab + b² – bc) – 2 b(bc – c² – ac)
= 2abc + 2cb² – 2bc² – 2b²c + 2bc² + 2abc
= 4abc

Question 5.
(i) Find the maximum and minimum values of f, if any of the function f(x) = |x| + 3, x ∈ R
(ii) What is the absolute maximum value of the function f(x) – |x| + 3, x ∈ [-3, 2]?
Answer:

(i) The function is not differentiable at x = 0. But from the graph it is clear that it has only minimum at x = 0 and it is 3.

(ii) When the domain is restricted as [-3, 2], then f(x) has absolute maximum at x = -3 which is 6 and absolute minimum at x = 2 which is 5.

Question 6.
(i) Write a function which is not continuous at x = 0 and justify your answer
(ii) Check the continuity of the function

Answer:

Left limit and right limit are not equal. So not continuous.

(ii) For x < 0, f(x) – x + 2 and for x > 0 f(x) = – x + 2, which are polynomials so continuous. But the function is not defined at x = 0. Hence f(x) is continuous on its domain.

Question 7.
Consider the arrow diagram of the functions f and g.

(i) Check whether the functions f and g are bijective function? Justify.
(ii) Write the function go f.
(iii) Is go f a bijective function function? Justify.
Answer:
(i) f is not onto since d has no preimage, g is not one-one since c and d has same image z. Hence both f and g are not bijective.

(ii) go f = (1, x), (2, z), (3, y)

(iii) The domain of gof is {1, 2, 3} and range is {x, y, z}. So go f is a bijective function.

Question 8.
(i) If α, β, γ are the direction angles of a vector, then which the following can be α + β?
(a) 80°
(b) 60°
(c) 120°
(d) can’t be determined
(ii) Find a direction cosines of the line passing through the points (2, 8, 3) and (4, 5, 9).
Answer:
(i) (c) 120°. Since the sum any two direction angle should be greater than or equal to 90°.

(ii) The direction ratios of the line is 4 – 2, 5 – 8, 9 – 3.
2, -3, 6
Then direction cosines is
\(\frac{2}{\sqrt{4+9+36}}\), \(\frac{-3}{\sqrt{4+9+36}}\), \(\frac{6}{\sqrt{4+9+36}}\)
⇒ \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)

Questions 9 to 18 carry 4 scores each. Answer any 8.

Question 9.
(i) If tan^-1 x = \(\frac{\pi}{10}\), then the value of cot^-1 x is
(a) \(\frac{\pi}{5}\)
(b) \(\frac{2\pi}{5}\)
(c) \(\frac{3\pi}{5}\)
(d) \(\frac{4\pi}{5}\)
(ii) Find the value of
sin (tan^-1\(\frac{2}{3}\)) + cos (tan^-1√3)
Answer:
(i) (b) \(\frac{2\pi}{5}\), since cot^-1 x = \(\frac{\pi}{2}\) – tan^-1 x

(ii) sin (2 tan^-1\(\frac{2}{3}\)) + cos (tan^-1√3)
= sin (tan^-1 \(\frac{2 \times \frac{2}{3}}{1-\frac{2}{3} \times \frac{2}{3}}\)) + cos (tan^-1√3)
= sin (tan^-1\(\frac{12}{5}\)) + cos(tan^-1√3)
Draw a triangle to find tan^-1\(\frac{12}{5}\) = sin^-1\(\frac{12}{13}\)
= sin (sin^-1\(\frac{12}{13}\)) + cos \(\frac{\pi}{3}\)
= \(\frac{12}{13}\) + \(\frac{1}{2}\) = \(\frac{37}{26}\)

Question 10.
Let A = {-1, 0, 1}
(i) Give reason why the operation defined by a ⊗ b = \(\frac{a}{b}\) is not a binary operation on A.
(ii) (a) Write a binary operation * on A
(b) Find (-1 * -1) * -1.
(iii) How many binary operations are possible on A?
Answer:
(i) Since the division by zero is not defined.

(ii) (a) a * b = a × b
(b) (-1 * -1) *-1 = (1) * -1 = -1

(iii) Binary operation is function A × A to A .
Therefore number of function is 39.

Question 11.
(i) Find the equation of a line L passing through the points (-1, 0, 2) and (2, 1, 3).
(ii) If \(\vec{c}\) = î + ĵ + λk̂ be a vector perpendicular to the above line, then find λ.
(iii) Find the equation of a plane on which the line L lies.
Answer:
(i) The equation of the line is
\(\frac{x + 1}{2 + 1}\) = \(\frac{y – 0}{1 – 0}\) = \(\frac{z – 2}{3 – 2}\)
⇒ \(\frac{x + 1}{3}\) = \(\frac{y}{1}\) = \(\frac{z – 2}{1}\)

(ii) Since vector is perpendicular, we have
3 × 1 + 1 × 1 + 1 × λ = 0
3 + 1 + A = 0 ⇒ λ = -4

(iii) Equation of the plane passing through (-1, 0, 2) and perpendicular vector = \(\vec{c}\) = î + ĵ – 4k̂ is
1(x + 1) + 1(y – 0) – 4(z – 2) = 0
x + y – 4z + 1 + 8 = 0
x + y – 4z + 9 = 0

Question 12.
Find \(\frac{d y}{d x}\) of the following:
(i) y = sec (tan x)
(ii) x^y = y^x
Answer:
(i) \(\frac{d y}{d x}\) = sec (tan x) tan (tan x) sec² x

(ii) Given x^y = y^x
Take log on both sides:
y log x = x log y
Differentiating w r to x

Question 13.
(i) Find \(\int \tan ^{-1} x d x\)
(ii) Hence find the area bounded by the curve
y = tan^-1 x with X – axis from x = 0 and x = 1
Answer:

Question 14.
Solve the differential equation \(\frac{d y}{d x}\) = \(\frac{x^2+y^2}{2 x y}\)
Answer:
Given ; \(\frac{d y}{d x}\) = \(\frac{x^2+y^2}{2 x y}\)
Put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\)
v + x\(\frac{d v}{d x}\) = \(\frac{x^2+(v x)^2}{2 x(v x)}\)
v + x\(\frac{d v}{d x}\) = \(\frac{1+v^2}{2 v}\)

Question 15.
Consider a plane which is equidistant from the points (1, 2, 1) and (3, 4, 7).
(i) Which of the following is a point on the plane?
(a) (1, 3, 1)
(b) (4, 2, 3)
(c) (2, 3, 4)
(d) (1, 3, 6)
(ii) Find the equation of the above plane.
Answer:
(i) (c) (2, 3, 4)

(ii) The plane will pass through the point (2, 3, 4) and
will a direction ratio 3 – 1, 4 – 2, 7 – 1 ⇒ 2, 2, 6
then dr’s can be taken as 1, 1, 3
Hence the equation is
1(x – 2) + 1(y – 3) + 3(z – 4) = 0
x + y + 3z – 17 = 0

Question 16.
Let \(\vec{a}\) = 2î + ĵ – 3k̂ and \(\vec{b}\) = 4î + ĵ + k̂ be two vectors
(i) Find \(\vec{a}\) vector \(\vec{c}\) perpendicular to \(\vec{a}\) and \(\vec{b}\).
(ii) Find the volume of the parallelepiped with coinitial vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
(iii) If \(\vec{c}\) is rotated in such a way that it makes 60° with its initial direction, then what is the volume of the new parallelepiped formed?
Answer:
(i) \(\vec{c}\) = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 1 & -3 \\
4 & 1 & 1
\end{array}\right|\) = 4î – 14ĵ – 2k̂

(ii) volume = [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
= (\(\vec{a}\) × \(\vec{b}\) ×) . \(\vec{c}\) = \(\vec{c}\) . \(\vec{c}\) = |\(\vec{c}\)|²
= 16 + 196 + 4 = 216
When \(\vec{c}\) is rotated 60° the height of the parallelepiped will be reduced by half of the original volume.
Since Volume = \(\vec{c}\) × \(\vec{c}\)
= |\(\vec{c}\)| × |\(\vec{c}\)| cos(60°)
= |\(\vec{c}\)| × |\(\vec{c}\)| × \(\frac{1}{2}\)
Hence the new volume is 108.

Question 17.
(i) Evaluate \(\int_0^\pi x \sin x d x\)
(ii) Hence evaluate the area bounded by the curve y = x sin x between x = -π and x = π.
Answer:

Question 18.
In a ladies hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(i) Find the probability that she reads neither Hindi nor English newspapers.
(ii) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(iii) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer:
i) Let the events H- Hindi, E = English
P(H) = \(\frac{60}{100}\) = 0.6; P(E) = \(\frac{40}{100}\) = 0.4
P(H∩E) = \(\frac{20}{100}\) = 0.2
P(H∪E) = P(H) + P(E) – P(H∩E)
= 0.6 + 0.4 – 0.2 = 0.8
P(she reads neither Hindi nor English newspapers.)
= P(H’∩E’) = 1 – P(H∪E) = 1 – 0.8 = 0.2

(ii) P(E/H) = \(\frac{P(E \cap H)}{P(H)}\) = \(\frac{0.2}{0.6}\) = \(\frac{1}{3}\)

(iii) P(E/H) = \(\frac{P(E \cap H)}{P(E)}\) = \(\frac{0.2}{0.4}\) = \(\frac{1}{2}\)

Questions from 19 to 25 carry 6 scores each. Answer any 5.

Question 19.
(i) Construct a 3 × 3 matrix A, where elements are given by a_ij = 2i – j
(ii) Verify that C = A – A’ is a skew symmetric matrix.
(iii) Verify that C² is a symmetric matrix.
Answer:

Clearly (C²) = C² therefore symmetric.

Question 20.
Consider the matrix A = \(\left[\begin{array}{ccc}
1 & 3 & 6 \\
-1 & -1 & 2 \\
1 & 1 & 5
\end{array}\right]\)
(i) Find |A|
(ii) Verify that A × adj A = |A| I
(iii) Evaluate | A^-1|
Answer:
(i) |A| = \(\left[\begin{array}{ccc}
1 & 3 & 6 \\
-1 & -1 & 2 \\
1 & 1 & 5
\end{array}\right]\) = 14

(ii) C₁₁ = – 5 – 2 = -7; C₁₂ = -(5 – 2) = 7
C₁₃ = – 1 + 1 = 0; C₂₁ = -(15 – 6) = -9
C₂₂ = 5 – 6 = -1; C₂₃ = -(1 – 3) = 2
C₃₁ = – 6 + 6 = 12; C₃₂ = -(2 + 6) = -8
C₃₃ = – 1 + 3 = 2

(iii) A × A^-1 = I
|A × A^-1| = |I|
|A| × |A^-1| = 1
= \(\frac{1}{14}\)

Question 21.
Integrate the following with respect to x:
(i) \(\frac{1}{x^2-6 x+13}\)
(ii) \(\frac{\cos x}{(\sin x-1)(\sin x-2)}\)
Answer:

Question 22.
Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice.
(i) Write the probability distribution of X.
(ii) Find variance of X.
Answer:
(i) Sample space; n(S) = 36
X: Sum of the numbers obtained.
X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Question 23.
(i) Consider the function f(x) = 2x³ – 6x² + 1
(a) Find the equation of tangents parallel to X- axis.
(b) Find the intervals in which the function f is decreasing.
(ii) The length x of a rectangle is decreasing at the rate of 5 cm/s and the width y is increasing at the rate of 4 cm/s. When x = 8 cm and y = 6 cm, find the rale of change of the area of the rectangle.
Answer:
(i) Given; f(x) = 2x³ – 6x² + 1
(a) f'(x) = 6x² – 12x
For point at which tangents parallel to x axis is when f'(x) – 0 ⇒ 6x² – 12x = 0 ⇒ x = 0, 2
Then the tangents lines are y = f(0) = 1 ⇒ y = 1 and y = f(2) = -7 ⇒ y = -7

(b) The turning point x = 0, 2 divides the domain into the following intervals (-∞, 0), (0, 2), (2, ∞) ‘ Clearly, f'(x) < 0, x ∈ (0, 2). Hence f(x) is decreasing in the interval (0, 2)

(ii) Given, x and y be the length and breadth of the rectangle at an instant t.
\(\frac{d y}{d t}\) = -5, \(\frac{d x}{d t}\) = 4
Given; x = 8 and y = 6
A = xy ⇒ \(\frac{d A}{d t}\) = x\(\frac{d y}{d t}\) + y\(\frac{d x}{t}\)
⇒ \(\frac{d A}{d t}\) = (8) (4) + (6) (-5) – 2 cm²/s

Question 24.
Let \(\vec{a}\) = 2î – 4ĵ + 5k̂ and \(\vec{b}\) = î – 2ĵ – 8k̂ be two vectors
(i) Find \(\vec{a}\) vector \(\vec{c}\) representing a diagonal of the parallelogram with \(\vec{a}\) and \(\vec{b}\) as the adjacent sides.
(ii) Find the projection of \(\vec{b}\) on \(\vec{c}\).
(iii) Find the angle between the vectors \(\vec{c}\) and \(\vec{a}\)
Answer:
(i) Here \(\vec{c}\) can be \(\vec{a}\) + \(\vec{b}\)
\(\vec{c}\) = 3î – 6ĵ – 3k̂
(ii) the projection of \(\vec{b}\) on \(\vec{c}\) = \(\frac{\vec{b} \cdot \vec{c}}{|\vec{c}|}\)

Question 25.
Maximise: Z = 600x + 400y
Subject to the constraints:
x + 2y ≤ 12 ;2x + y ≤ 12
4x + 5y ≥ 20 ; x ≥ 0, y ≥ 0
(i) Draw the feasible region.
(ii) Solve the LPP?
Answer:

(ii) The corner points are A(5, 0), B(6, 0), C(4, 4) D(0, 6), E(0, 4)

Corner Point	Value of Z z = 600x + 400y
A(5, 0)	z = 600x + 400y = 3000 + 0 = 3000
B(6, 0)	z = 600x + 400y = 3600 + 0 = 36 00
C(4, 4)	z = 600x + 400y = 2400 + 1600 = 4000
0(0, 6)	z = 600x + 400y = 0 + 2400 = 2400
E(0, 4)	z = 600x + 400y = 0 + 1600 = 1600

The maximum value of z is 4000 at C(4, 4).

Teachers recommend solving Kerala Syllabus Plus Two Botany Previous Year Question Papers and Answers Pdf March 2024 to improve time management during exams.

Kerala Plus Two Botany Previous Year Question Paper March 2024

I. Answer any 3 questions from 1-4. Each carries 1 score. (3 × 1 = 3)

Question 1.
Choose the correct answer and fill up the blanks. In grasses the monocot embryo contains single cotyledons named. (Plumule, Radicle, Scutellum, Coleorhiza)
Answer:
Scutellum.

Question 2.
In Recombinant DNAtechnology, precipitated DNA can be separated by ______.
Answer:
Spooling

Question 3.
A single stranded DNA or RNA tagged with radioactive molecule is called
(a) Plasmid
(b) Probe
(c) Clone
(d) Vector
Answer:
b) Probe.

Question 4.
Sun is the only source of energy for all ecosystem on earth. But one exception. What is that?
Answer:
Deep sea hydro-thermal ecosystem I Deep Sea.

II. Answer any 9 questions from 5-15. Each carries 2 scores. (9 × 2 = 18)

Question 5.
Observe the diagram of Pollen grain given below and answer the questions.

(a) Identify the cell ‘A’ and ‘B’.
(b) Write the two features of ‘A’.
Answer:
(a) A – Vegetative cell
B – Generative cell

(b) Vegetative cell is bigger / has abundant food reserve I has large irregularly shaped nucleus. (Any two features)

Question 6.
Name the special cellular thickening present in the synergid at micropylar end. Write its function.
Answer:
Filiform apparatus. It plays an important role in guiding the entry of pollen túbe into the synergids.

Question 7.
What are the two core techniques that enabled the birth of modern biotechnology?
Answer:
Genetic engineering. Bioprocess engineenng I Chemical engineering processes.

Question 8.
Explain any two methods by which recombinant DNA can be directly introduced into the host cell.
Answer:
Microinjection : Direct injection of recombinant DNA (rDNA) into the nucleus of an animal cell is called microinjection  It is the rDNA transfer method for animal cell.

Biolistics : Bombardment of plant cell with high velocity micro particle of gold or tungsten coated with DNA is called biolistics I It is the rDNA transfer method for plant cell.

Question 9.
The first trarisgenic cow p roduced human protein enriched milk. Name the cow and the
protein found in milk.
Answer:
The first transgenic cow is Rosie. The protein is aipha-lactalbumin I Human protein.

Question 10.
Match the following

A	B
1. Biopiracy	A. ADA deficiency
2. Gene Therapy	B. Basmathi Rice
3. RNA Interference	C. Cry gene
4. Bacillus thuringiensis	D. Melodegynae incognita

Answer:

A	B
1.Biopiracy	Basmati Rice
2. Gene Therapy	ADA deficiency
3. RNA Interference	Meloidogyne incognitia
4. Bacillus thurengiensis	Cry gene

Question 11.
the following graph shows two types of population growth curves.

(1) Name the growth curve ‘a’ and ‘b’.
(2) What does ‘K’ stands for?
Answer:
(1) a) Exponential growth I J shaped curve I Geometric growth model.
b) Logistic growth / Verhulst-Pearf Logistic Growth I Sigmoid Growth I S-shaped curve.

(2) K – Carrying capacity.

Question 12.
Parasites evolved special adaptations to live on host. What are they?
Answer:

• The loss of unnecessary sense organs
• presence of adhesive organs or suckers to cling on to the host
• loss of digestive system
• high reproductive capacity.

Question 13.

The figure depicts pyramid of energy.
(a) Pyramid of energy is always upright, can never be inverted. Justify
(b) Which are the other two ecological pyramids?
Answer:
(a) When energy flow from a particular trophic level to the next level some energy is lost as heat at each step. Only 10% of the energy is transferred to each trophic level from the lower trophic level according to law of 10%.

(b) Pyramid of numbers, Pyramid of biomass.

Question 14.
A list of different organism in an ecosystem are given below. Arrange them in 1^st, 2^nd and 3^rd and 4^th trophic level. (Phytoplankton, Man, Fish, Zooplankton)
Answer:
Phytoplankton → Zooplankton → Fish → Man

Question 15.
Rate of Biomass production in an ecosystem is called productivity. Productivity are divided into two, GPP and NPP.
(a) Expand GPP & NPP.
(b) Write the equation relating GPP with NPP.
Answer:
(a) GPP — Gross Primary ProductMty
NPP – Net Primary Productivity
b) GPP-R=NPR

III. Answer any 3 questions from 16 to 19. Each carries 3 scores. (3 × 3 = 9)

Question 16.
Observe the diagram and answer the following questions:

(1) Label‘a’,‘b’,‘c’and”d’.
(2) Name the diploid cell found in embryo sac.
Answer:
(1) a — Antipodals
b — polar nuclei
c — Synergids
d — Egg I Female gamete

(2) Central cell.

Question 17.
The genes of organism can be altered by Manipulation. Such organisms are called Genetically Modified Organism (GMO). List any 3 merits of G.M. Plants.
Answer:
Merits of G.M. Plants includes:

• Made crops tolerant to abiotic stress (cold, drought, salt & temperature)
• Develop pest resistance
• Helped to reduce post-harvest losses

Question 18.
Observe the figure given below:
(1) Fill in the blanks a, b, and c in the figure.
(2) Write the name of this process.
(3) What is the name of thermostable DNA polymerase enzyme used in this techniques?

Answer:
(1) a. Denaturation
b. Annealing
c. Extension.

(2) Polymerase Chain Reaction

(3) Taq Polym erase (from the bacteria Therm us aquaticus).

Question 19.
Given below are examples of some population interactions. Identify and define the interaction.
(a) Cattle egret and grazing cattle
(b) Fig tree and wasp
(c) Cuscuta growing on hedge plants.
Answer:
(a) Commensalism: It is +, O interaction In this interaction one species is benefitted and the other is neither behefitted nor harmed.

(b) Mutualism : It is +, + interaction In this interaction both the species are benefrited.

(c) Parasitism: It is +, – interaction I in this interaction only one species is benefited and the interaction is detrimental to the other species Interaction between host and parasite.

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2021 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2021

Answer the following questions from 1 to 45 upto a maximum score of 60.

I. Questions 1 to 8 carry 1 score each. (8 × 1 = 8)

Question 1.
Fill in the blanks:
” The force between two point charges is directly proportional to the product ……… of and inversely proportional to the ………. of the distance between them”.
Answer:
charges, square

Question 2.
The expression ΣB.AS = 0 is
(i) Gauss Law in Electrostatics
(ii) Gauss Law in Magnetism
(iii) Ampere’s circuital law ‘
(iv) Lenz’s law
Answer:
(ii) Gauss Law in Magnetism

Question 3.
The electromagnetic waves used in LASIK eye surgery is
(i) microwaves
(ii) ultraviolet rays
(iii) infra-red waves
(iv) gamma rays
Answer:
(ii) ultraviolet rays

Question 4.
Write Lens maker’s formula:
Answer:
\(\frac{1}{f}\) = (n-1) [\(\frac{1}{R_1}\) – \(\frac{1}{R_2}\)]

Question 5.
Name the property of light that proves its transverse nature.
Answer:
Polarisation

Question 6.
Write the equation for the wavelength of de Broglie wave associated with a moving particle.
Answer:
Wave length λ = \(\frac{h}{mV}\)

Question 7.
Energy of electron in the n^th orbit of hydrogen atom is E_n =\(\frac{13.6}{n^2}\) eV. What is the energy required to make electron free from first orbit of hydrogen atom?
Answer:
+ 13.6 eV

Question 8.
If radius of first electron orbit of hydrogen is a₀, radius of second electron orbit of hydrogen is ……… .
Answer:
a = 4a₀ [a = n²a₀; n = 2]

Questions from 9 to 22 carry 2 scores eqch. (14 × 2 = 28)

Question 9.
Calculate the electric potential at a point 9.0 cm away from a point charge of 4 × 10^-7 C.
Answer:
V = \(\frac{1}{4 \pi \varepsilon_0}\) \(\frac{q}{r}\)
= 9 × 10⁺⁹ × \(\frac{4 \times 10^{-7}}{9 \times 10^{-2}}\) = 4 × 10⁴ V

Question 10.
State Biot – Savart law and express it mathematically.
Answer:
dB = \(\frac{\mu_0}{4 \pi}\) \(\frac{\mathrm{Id} \ell \sin \theta}{\mathrm{r}^2}\)

Question 11.
Draw Wheatstone’s bridge and write its balancing condition.
Answer:

When bridge is balanced, \(\frac{P}{Q}\) = \(\frac{R}{S}\)

Question 12.
Determine the value of resistance R in the figure, assuming that the current through the galvanometer (G) is zero.

Answer:
When galvanometer current is zero,
We can write \(\frac{P}{Q}\) = \(\frac{R}{S}\)
\(\frac{R}{3}\) = \(\frac{40}{60}\)
R = 3 × \(\frac{40}{60}\) = 2 Ω

Question 13.
Write any two properties of nuclear force.
Answer:
(i) N uclear force is indepentant of charge
(ii) It is a short range force

Question 14.
Define half life of a radioactive sample. Write the equation that connects half life with disintegration constant.
Answer:
Half life is the time taken to reduce half of initial value of sample.
T_1/2 = \(\frac{0.693}{λ}\)

Question 15.
An air cored solenoid has 1000 turns per metre and carries a current of 2A. Calculate the magnetic intensity (H).
Answer:
n = 1000, I = 2 A
H = 1000 × 2 = 2000 Alm³

Question 16.
The behaviour of magnetic filed lines near two magnetic substances P and Q are shown below.

(a) From the figure identify paramagnetic substance.
Answer:
Substance Q

(b) Susceptibility of substance P is ………….. (positive /negative).
Answer:
Negative

Question 17.
Current in a curcuit fall from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced. Calculate the self inductance of the curcuit.
Answer:
ε = L \(\frac{dI}{dt}\) dI = 5 – 0
200 = L \(\frac{5 – 0}{0.1}\) dt = 0.1 ; ε = 200
L = \(\frac{200 \times 0.1}{5}\) = 4 H

Question 18.
Using a suitable ray diagram prove that the radius of curvature of a spherical mirror is twice its focal length.
Answer:

Consider a ray AB parallel to principal axis incident on a cpncave mirror at point B and is reflected along BF. The line CB is normal to the mirror as shown in the figure.
Let θ be angle of incidence and reflection.
Draw BD ⊥ CP,
In right angled ∆ BCD,
Tanθ = \(\frac{BD}{CD}\) ……….(1)
In right angled ∆ BFD,
Tan2θ = \(\frac{BD}{FD}\) ……….(2)
Dividing (1) and(2)
\(\frac{Tanθ}{Tan2θ}\) = \(\frac{CD}{FD}\) ………(3)
If θ is very small, then tanθ ≈ θ and tan2θ ≈ 2θ
The pointB lies very close to P. Hence CD ≈ CP and FD ≈ FP From (3) we get
\(\frac{2θ}{θ}\) = \(\frac{PC}{PF}\) = \(\frac{R}{f}\)
R = 2f

Question 19.
A light bulb of resistance 484 Ω is connected with 200 V ac supply. Find peak value of current through the bulb.
Answer:
R = 484 Ω
V = 220 V
I_rms = \(\frac{V}{R}\) = \(\frac{220}{484}\) = 0.45 A
I_rms = \(\frac{I_0}{\sqrt{2}}\)
I₀ = \(\sqrt{2} \times I_{m s}\)
= √2 × 0.45 = 0.64 A

Question 20.
Write any two postulates of Bohr model of hydrogen atom.
Answer:
1. Electrons revolve round the positively charged nucleus in circular orbits.
2. The electron which remains in a privileged path cannot radiate its energy.
3. The orbital angular momentum of the electron is an integral multiple of h/2π.
4. Emission or Absorption of energy takes place when an electron jumps from one orbit to another.

Question 21.
The symbol of a logic gate is given below. Identify the gate and write its truth table.

Answer:
OR gate
Tablee

Question 22.
When bulk pieces of conductors are subjected to changing magnetic flux, currents are induced in them.
(a) Write the name of this induced current.
Answer:
Eddy current

(b) Write any two practical applications of this current.
Answer:
(i) Magnetic breaking
(ii) Induction furnance

Questions from 23 to 34 carry 3 scores each. (12 × 3 = 36)

Question 23.
(a) Define electric dipole moment. (1)
Answer:
Electric dipole moment is product of magnitude of charge of dipole and its length.

(b) A system has two charges 2.5 × 10^-7 C and 2.5 × 10^-7 C located at points (0, 0, -15 cm) and (0,0, +15 cm), respectively.
Determine the magnitude and direction of electric dipole moment of the system. (2)
Answer:
q = 2.5 × 10^-7 C
2a = 2 × 15 = 30 cm
P = q × 2a = 2.5 × 10^-7 × 30 × 10^-2
= 75 × 10^-9 Cm
direction → -q to +q

Question 24.
(a) Write any two properties of electric field lines. (2)
Answer:
(i) Electric field lines never intercept each other.
(ii) Electric field lines are originated from positive charge and terminated at negative charge.

(b) Observe the figure and write the signs of the charges q₁ and q₂.

Answer:
q₁ is positive and q₂ is negative.

Question 25.
Derive an expression for the energy stored in a ca-pacitor in terms of capacitance and potential differ-ence across the capacitor.
Answer:
Energy of a capacitor is the work done in charging it. Consider a capacitor of capacitance ‘C’. Let ‘q’ be the charge at any instant and ‘V’ be the potential. If we supply a charge ‘dq’ to the capacitor, then work done can be written as,
dw = Vdq
dw = \(\frac{q}{C}\)dq (Since V = \(\frac{q}{C}\))
∴ total work done to change the capacitor (from 0 to Q) is
W = \(\int_0^Q \frac{q}{C} d q\)
W = \(\frac{1\left[q^2\right]_0^Q}{\mathrm{C} 2}\)
W = \(\frac{1}{C} \frac{Q^2}{2}\)
but Q = CV
W = \(\frac{1}{2}\) CV²
This work done is stored in the capacitor as electric potential energy.
Energy stored in the capacitor is,
U = \(\frac{1}{2}\) CV²

Question 26.
Write any one difference between polar and non-polar molecule: Give one example each for polar and non-polar molecule.
Answer:
In polar molecule, positive centre and negative centre does not coincide each other. But in non polar molecule, positive centre and negative centre coin-cide each other.
Example:
Polar molecule – H₂O, HCl
Non polar molecule – H₂, O₂

Question 27.
(a) Define angle of dip. (1)
Answer:
The angle between earths magnetic field with its horizontal is called dip.

(b) At a particular place the horizontal and vertical components of earth’s magnetic field are found to be equal. What is the value of dip at this place? (2)
Answer:
B_H = B_V
B cosθ = B sinθ
cosθ = sinθ
∴ θ = 45°

Question 28.
In the figure shown below
(a) Which are the resistors Connected in parallel? (1)
Answer:
6Ω and 3Ω

(b) Calculate the current drawn from the cell. (2)

Answer:
6Ω and 3Ω resistors are’connected in parallel. Hence effecting resistance
R = \(\frac{6 \times 3}{6+3}\) = \(\frac{18}{9}\) = 2Ω
This 2Ω connected 8Ω in series.
.-. Total resistance = 8 + 2 = 10Ω
Current I = \(\frac{\text { Total voltage }}{\text { Total resistance }}\)
I = \(\frac{24}{10}\) = 2.4 A

Question 29.
Using Ampere’s circuital law show that the intensity of magnetic field at an axial point near the centre of a current carrying solenoid is B = μ₀n I.
Answer:

Consider a solenoid having radius ‘r’. Let ‘n’ be the number of turns per unit length and I be the current flowing through it.
In order to find the magnetic field (inside the solenoid ) consider an Amperian loop PQRS. Let ‘e ‘ be the length and ‘b’ the breadth.
Applying Amperes law, we can write

(since RS is completely out side the solenoid, for which B = 0)
Substituting the above values in eq (1 ),we get
Bl = μ₀ I_ene …….(2)
But I_ene = n l I
where ‘nl’ is the total number of turns that carries current I (inside the loop PQRS)
∴ eq (2) can be written as
Bl = μ₀ nI l
B = μ₀ nI
If core of solenoid is filled with a medium of relative permittivity nr then
B = μ₀μ_rnl

Question 30.
Write a circuit diagram explain how a moving coil galvanometer can be converted to an ammeter.
Answer:
A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it.
Theory: Let G be the resistance of the galvanometer, giving full deflection for a current I_g.
To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In thisoafirange- ment Ig current flows through Galvanometer and remaining (I-Ig) current flows through shunt resistance.
Since G and S are parallel
P.d Across G = p.d across S
Ig × G = (I-Ig)S

Question 31.
Prove that when an alternating voltage is applied to an inductor, the current thrcjugh it lags behind voltage by an angle \(\frac{\phi}{2}\).
Answer:

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage.

Let the applied voltage be
V = V₀ sin ωt ………(1)
Due to the flow of alternating current through coil, an emf, L\(\frac{dI}{dt}\) is produced in the coil. This induced emf is equaland opposite to the applied emf (in the case of ideal inductor)
ie. L\(\frac{dI}{dt}\) = V₀ sin ωt
dI = \(\frac{dI}{dt}\) sin ωt dt
Integrating, we get
I = \(\frac{V_0}{L \omega}\) cos ωt
I = \(\frac{V_0}{L \omega}\) Sin (ωt – \(\frac{\phi}{2}\))
I = I₀ Sin (ωt – \(\frac{\phi}{2}\)) ………(2)
Where I₀ = \(\frac{V_0}{L \omega}\)
The term Lω is called inductive reactance. Comparing eq(1) and eq(2), we can understand that, the current lags behind the voltage by an angle 90°.

Question 32.
(a) The current due to time varying electric field is called ……… (1)
Answer:
displacement current

(b) An electromagneticjwave travels in free space with a velocity of 3 × 10⁸ m/s. At a particular point in space and timte, magnitude of intensity of electric field is 6.3 V/m. What is magnitude of magnetic field at this point? (2)
Answer:
C = \(\frac{E}{B}\)
3 × 10⁸ = \(\frac{63}{B}\)
Magnetic field, B = \(\frac{63}{3 × 10^8}\) = 2.1 × 10^-8 T

Question 33.
Using Huygens wave theory prove that angle of incidence is equal tojangle of reflection.
Answer:
Reflection of plane wave by a plane surface.

AB is the incident wavefront and CD is the reflected wavefront, ‘i’ is the angle of incidence and ‘r’ is the angle of reflection. Let c₁ be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
The time taken for the ray to travel from P to Q is
t = \(\frac{\mathrm{PO}}{\mathrm{C}_1}+\frac{\mathrm{OQ}}{\mathrm{C}_1}\) …….(1)
t = \(\frac{A O \sin i}{C_1}+\frac{O D \sin r}{C_1}\)

O is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero.

\(\frac{A O}{C_1}\) (sin i – sin r) = 0
sin i – sin r = 0
sin i = sinr i = r

Question 34.
(a) Write Einstein’s photoelectric equation.
Answer:
hυ = hυ₀ + \(\frac{1}{2}\)mv²

(b) Using this equation show that, “photoelectric emission is not possible if the frequency of incident radiation is less than threshold – frequency”. (2)
Answer:
hυ = hυ₀ + \(\frac{1}{2}\)mv²
If incident frequency is less than threshold frequence (υ < υ₀), kinetic energy of photoelectron becomes negative, which means that no photo emission takes place.

Reviewing Kerala Syllabus Plus Two Chemistry Previous Year Question Papers and Answers Pdf March 2021 helps in understanding answer patterns.

Kerala Plus Two Chemistry Previous Year Question Paper March 2021

Time: 2 Hours
Total Scores: 60

Answer the following questions from 1 to 40 upto a maximum score of 60.
Answer questions from 1 to 11. Each carries 2 scores. (11 × 2 = 22)

Question 1.
(i) Which of the following is an anisotropic solid? (1)
(A) NaCl
(B) Glass
(C) Rubber
(D) Plastic
(ii) Glass is called pseudo solid. Give reason. (1)
Answer:
(i) NaCl
(ii) Because it is a supercooled liquid.

Question 2.
An element has a cubic close-packed structure.
(i) What is the coordination number of each atom? (1)
(ii) Give the total number of voids in N mol of it. (1)
Answer:
(i) Total no. of surrounding atoms = coordination number in ccp = 12
(ii) 3N voids or 3 × 6.022 × 10²³

Question 3.
A mixture of two liquids A and B form an ideal solution. Draw the vapour pressure composition curve for this solution. (2)
Answer:

Question 4.
(i) The electrolyte used in Leas-storage battery ______________ (1)
(ii) Give one example of a primary cell. (1)
Answer:
(i) H₂SO₄ (38%)
(ii) Drycell, Mercury cell

Question 5.
What is a zero-order reaction? Give the unit of rate constant for zero order reaction. (2)
Answer:
If the sum of powers of concentration terms in the rate equation is zero it is called zero order reaction.
Unit of K for zero order reaction is ie., mol L^-1 s^-1.

Question 6.
Classify the following as homogeneous and heterogeneous catalysis.

Answer:
(A) In homogenous catalysis reactants and catalysts are in the same phase.

(B) In heterogenous catalysis reactants and catalysts are in different phases.

Question 7.
(i) Which of the following ore can be concentrated by the froth floatation method? (1)
(A) Bauxite
(B) Siderite
(C) Cuprite
(D) Zinc blende
(ii) Zinc and Mercury are low-boiling liquids. Name the technique used to refine these metals. (1)
Answer:
(i) D. Zinc blend
(ii) Distillation

Question 8.
(i) Name the important Oxo acid of Nitrogen. (1)
(ii) Name the method used for the manufacture of this acid. (1)
Answer:
(i) Nitric Acid (HNO₃)
(ii) Ostwald’s Process

Question 9.
Give a reason for the following:
(i) PCl₃ fumes in moist air. (1)
(ii) PCl₅ is highly reactive. (1)
Answer:
(i) PCl₃ reacts with moisture to give HCl. Which is fuming nature.
PCl₃ + 3H₂O → H₃PO₃ + HCl
(ii) Due to the presence of two axial Cl atoms in trigonal bipyramidal structure. Axial bonds are longer than equatorial bonds.

Question 10.
(i) Write the IUPAC name of K₂[Zn(OH)₄] (1)
(ii) Metal present in chlorophyll is _____________ (1)
Answer:
(i) Potassium tetra hydroxo zincate (II) or Potassium tetra hydroxido zincate (II).
(ii) Mg (Magnesium)

Question 11.
Identify the main product in the following reactions:

Answer:

Questions from 12-29 carry 3 scores each. (18 × 3 = 54)

Question 12.
Define unit cell. Calculate the number of particles per unit cell in the Body-centered cube and Face-centered cube. (3)
Answer:
Unit cell: Smallest repeating unit of a crystal.
B.C.C
No. of particles at the corner = 8
Corner particles contribute \(\frac{1}{8}\)
i.e. \(\frac{1}{8}\) × 8 = 1
In addition to this one particle is present in the centre without sharing. So total particles present is \(\frac{1}{8}\) × 8 + 1 = 2

F.C.C
Corner particles contribute \(\frac{1}{8}\)
∴ No. of particles due to corner atoms \(\frac{1}{8}\) × 8 = 1
Face centred particles contribute \(\frac{1}{2}\)
No. of particles due to face centred particles 6 × \(\frac{1}{2}\) = 3
Total no. of particles = 1 + 3 = 4

Question 13.
(i) What type of magnetic substances are used to make permanent magnets? (1)
(ii) Draw the schematic alignment of magnetic moments in ferromagnetic and ferrimagnetic substances. (2)
Answer:
(i) Ferromagnetic substances
(ii) Ferromagnetic – ↑ ↑ ↑ ↑
Magnetic moments are in the same direction.
Ferrimagnetic – ↑↓ ↑↓ ↑↑
Magnetic moments are unequal and in opposite directions.

Question 14.
State Henry’s Law. Give two applications of it. (3)
Answer:
The amount of gas dissolved in a liquid is directly proportional to the pressure applied p = k_H.x
p – pressure, k_H – Henry’s constant, x – mole fraction.
Applications:
(i) Soda water.preparation or soft drinks.
(ii) Bends in Scuba divers

Question 15.
(i) Daniel’s cell is represented as
\(\mathrm{Zn}(\mathrm{~s}) / \mathrm{Zn}_{(\mathrm{aq})}^{2+} \| \mathrm{Cu}_{(\mathrm{aq})}^{2+} \mid \mathrm{Cu}(\mathrm{~s})\)
Write the Nernst equation for Daniel’s cell. (1)
(ii) The conductivity of 0.2 M solution of KCl at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity. (2)
Answer:

= 124 S cm² mol^-1

Question 16.
Variations of molar conductivity (λ_m) versus concentration (√c) for strong and weak electrolytes are given below:

(i) Identify I and II as strong and weak electrolytes. (1)
(ii) What does \(\lambda_{\mathrm{m}}^0\) indicate? (1)
(iii) Suggest a method to determine \(\lambda_{\mathrm{m}}^0\) for the electrolyte II. (1)
Answer:
(i) I for strong electrolytes
II for weak electrolyte
(ii) Molar conductivity at infinite dilution – \(\lambda_{\mathrm{m}}^0\)
(iii) Kohlrausch’s law

Question 17.
(i) The vapour pressure of pure liquids A and B is 400 mm and 600 mm of Hg respectively, calculate the vapour pressure of the solution in which the mole fraction of B is 0.4. (2)
(ii) Which of the following is true for an ideal solution? (1)
(A) ΔH_mix > 0
(B) ΔH_mix = 0
(C) ΔV_mix > 0
(D) ΔH_mix < 0
Answer:

Question 18.
The integrated rate equation for a first-order reaction is
K = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
(i) What is half life period? (1)
(ii) Derive an expression for the half-life period of a first-order reaction. (2)
Answer:
(i) Period required for a radioactive substance to become exactly half of its initial concentration.

Question 19.
(i) Write any two characteristics of Chemisorption. (2)
(ii) Why are finely powdered substances more effective adsorbents than their crystalline form? (1)
Answer:
(i) Irreversible, specific
(ii) Surface area increases more adsorption occurs.

Question 20.
Differentiate between the following:
(i) Calcination and Roasting
(ii) Pig iron and Wrought iron
(iii) Mineral and Ore
Answer:
(i) Calcination: Heating in the limited supply of air.
Roasting: Heating in the presence of plenty of air.
(ii) Wrought iron: The purest form of iron.
Pig iron: Impure iron obtained from the blast furnace.
(iii) Ore: A mineral from which metals can be extracted profitably.
Minerals: Form of metals present in the earth’s crust.

Question 21.
Potassium dichromate is a very useful oxidizing agent.
(i) Name the ore of Potassium dichromate. (1)
(ii) Explain the preparation of Potassium dichromate from Sodium chromate. (2)
Answer:
(i) Chromite Ore/FeCr2O4

1. Iron chromate is treated with Na₂CO₃ in the presence of air to give Sod. chromate.
2. Sod. chromate is acidified with dil. HCl to give sod. dichromate.
3. Sod. dichromate is treated with KCl to get potassium dichromate.

Question 22.
(i) Account for the following:
A. Zr and Hf have identical radii. (1)
B. Transition metals are very good catalysts. (1)
(ii) Calculate the spin-only magnetic moment of \(M_{(a q)}^{2+}\) iron (Z = 27). (1)
Answer:
(i) A. Due to lanthanide contraction.
B. Ability to show variable oxidation state.

Question 23.
Explain the following reactions:
(i) Rienrier-Tieman reaction.
(ii) Williamson’s synthesis.
Answer:
Reimer-Tiemann Reaction:

Williamson’s Synthesis
Alkyl halide + Sod. alkoxide → ether R-X + R’ONa → R – OR’

Question 24.
(i) How are the following conversions carried out?
A. Propene to propane-2-ol.
B. Ethanal to Ethanol. (2)
(ii) Name the enzyme which converts glucose to ethanol. (1)
Answer:
(i) A. Propene to propan-2-ol.

B. Ethanal to Ethanol

(ii) Zymase

Question 25.
(i) The test to distinguish Propanal and Propanone is (1)
(A) Tollens’ test
(B) Lucas test
(C) Hinsbergtest
(D) Bromine-Water test
(ii) Which is more reactive towards nucleophilic addition, CH₃CHO or C₆H₅-CHO? Give reason. (2)
Answer:
(i) (A) Tollens’ test
(ii) CH₃CHO
C₆H₅CHO is less reactive due to the presence of phenyl group.

Question 26.
Identify the products and name the reactions.

Answer:
(i) Cannizzaro reaction.

(ii) Aldol condensation

Question 27.
(i) Classify the following into monosaccharides and disaccharides. (1)
Ribose, Fructose, Maltose, Sucrose.
(ii) How is starch different from glycogen? (1)
(iii) Name the two hormones which work together to regulate glucose levels in the body. (1)
Answer:
(i) Monosaccharides: Ribose, fructose
Disaccharides: Maltose, Sucrose
(ii) Starch storage of carbohydrate plants.
Glycogen – Animal starch. Carbohydrates are stored in animals.
(iii) Insulin and glucagon

Question 28.
Write the monomeric units and one uses each of the following polymers:
(i) PVC
(ii) Teflon
(iii) Nylon-6,6
Answer:
(i) PVC – Vinyl chloride, for making pipes
(ii) Teflon – Tetra fluoro ethene for making nonstick utensils
(iii) Nylon 6,6 – Adipic acid + hexamethylene diammine for making net.

Question 29.
(i) Explain the role of the following as food additives: (1)
(A) BHT
(B) Saccharin
(ii) A low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Give one example. (2)
Answer:
(i) A. BHT (Butylated hydroxytoluene) – It is used as an antioxidant in food.
B. Saccharin – It is used as an artificial sweetener in food.
(ii) Tranquilizer

Questions from 30-40 carry 4 scores each. (11 × 4 = 44)

Question 30.
(i) Explain the following terms: (3)
A. Schottky defect
B. Frenkel defect
C. F-centre
(ii) Which of the following shows both Schottky and Frenkel defects? (1)
(A) KCl
(B) AgCl
(C) AgBr
(D) NaBr
Answer:
(i) A. Schottky defect: Equal no. of cations and anions are missed from the lattice and density decreases.
B. Frenkel defect: Cations are missed from the lattice and are seen in the interstitial site. No change in densities.
C. f-centres: Electron-trapped anion vacancies are known as F-centres.
(ii) AgBr

Question 31.
(i) What are colligative properties? (2)
(ii) 400 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such solution at 300 K is found to be 2.57 × 10^-4 atm. Calculate the molar mass of the protein. (R = 0.0821 L atm K^-1 mol^-1) (2)
Answer:
(i) Colligative properties are the properties that depend only on the number of solute particles and not on their nature.
(ii) Molar mass (M₂) = \(\frac{w_2 R T}{\pi V}\)
Here w₂ = 1.26 g, V = 400 cm³ = 0.4 L, T = 300 K, π = 2.57 × 10^-4 atm and R = 0.0821 L atm K^-1 mol^-1
So, M₂ = \(\frac{1.26 \times 0.0821 \times 300}{2.57 \times 10^{-4} \times 0.4}\) = 301885

Question 32.
(i) Explain the construction and working of the H₂-O₂ fuel cell. (3)
(ii) Write two methods to prevent corrosion of metals. (1)
Answer:
(i) Two graphite (electrodes) are placed in conc.NaOH solution (electrolyte). The electrodes are connected to the battery. H₂ gas is passed through the anode and O₂ gas is bubbled through the cathode.
Reactions at anode 2H₂ + 4OH^– → 4H₂O + e^–
and at Cathode O₂ + 2H₂O + 4e^– → 4OH^–
The overall reaction is 2H₂ (g) + O₂ (g) → 2H₂O (l)
(ii) The methods to prevent corrosion of metals are:

• By giving a non-metallic coating on the metal surface with paint, varnish, etc.
• By coating the metal surface with electro-positive metals like zinc, magnesium, etc.
• By coating it with an anti-rust solution. (any two)

Question 33.
(i) Write the Arrhenius equation. (1)
(ii) The rate of a reaction doubles when the temperature is increased from 298 K to 308 K. Calculate the activation energy. (2)
(iii) Give two differences between order and molecularity. (1)
Answer:
(i) The Arrhenius equation is k = \(\mathrm{A} \cdot \mathrm{e}^{-E a / R T}\)
(ii) We know that, \(\log \frac{k_2}{k_1}=\frac{E a}{2.303 R} \frac{\left[T_2-T_1\right]}{T_1 \cdot T_2}\)
Here T₁ = 298 K, T₂ = 308 K, and R = 8.314 J K^-1 mol^-1
Suppose k₁ = x, then k₂ = 2x
Then \(\frac{\log 2 x}{x}=\frac{E a}{2.303 \times 8.314} \frac{[308-298]}{298 \times 308}\)
E_a = \(\frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}\) = 52897.78 J mol^-1.
(iii)

Order	Molecularity
The sum of powers to which the concentration terms are raised in the rate-low expression.	Total number of molecules taking part in the reaction.
It can be zero or fractional.	It cannot be zero or fractional.

Question 34.
(i) What are (yophilic and lyophobic sols? Give one example for each type.
(ii) Explain the different types of emulsions. (2)
Answer:
(i) In lyophilic sols, the force of attraction between the dispersed phase and the dispersion medium is strong.
Eg. Starch solution, gum, gelatin, starch, rubber, etc. in the suitable dispersion medium.
But in lyophobic sols, the force of attraction between the dispersed phase and the dispersion medium is weak. eg. Arsenic sulfide (As₂S₃) sol, Sulpher soi, and metal sols like gold sol, silver sol, etc.
(ii) Emulsions are of two types: Oil in water (O/W) type and Water in oil (W/O) type
In oil in water type emulsion, oil is dispersed. phase and water is the dispersion medium, eg. milk
Inthe water in a type emulsion, water is the dispersed phase and oil is the dispersion medium, eg. butter and cream.

Question 35.
(i) Explain the steps involved in the leaching of Bauxite ore. (3)
(ii) What is the role of cryolite in the extraction of Aluminium? (1)
Answer:
(i) Bauxite ore is powdered and treated with NaOH to get sodium Aluminate. To this CO₂ is added and hydrated Al₂O₃ is precipitated it is filtered and dried.

(ii) Cryolite is added to lower the melting point of alumina and to increase the conductivity.

Question 36.
(i) Give the preparation and structure of XeF₂. (2)
(ii) Which of the following does not exist: (1)
(A) XeOF₄
(B) XeF₄
(C) XeO₄
(D) NeF₂
(iii) Why ICl is more reactive than I₂? (1)
Answer:
(i) Excess amount of Xe reacts with F₂ at about 673 K and 1 bar pressure to produce XeF₂.
Or, The equation:

The structure is linear as follows:

(ii) (D) NeF₂
(iii) The I-Cl bond is weaker than the I-I bond

Question 37.
(i) List the various structural isomerisms possible for coordination compounds. (2)
(ii) [Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3-, is weakly paramagnetic. Explain. (2)
Answer:
(i) The different types of structural isomerism shown by co-ordination compounds are:
1. Ionisation isomerism
2. Linkage isomerism
3. Solvate or hydrate isomerism
4. Co-ordination isomerism
(ii) This is because [Fe(H₂O)₆]³⁺ is an outer orbital complex while [Fe(CN)₆]^3- is an inner orbital complex/H₂O is a weak field ligand and hence electron pairing does not occur while CN^– is a strong field ligand and hence electron pairing occurs due to greater number of unpaired electrons in [Fe(H₂O)₆]³⁺ than that in [Fe(CN)₆]^3-.

Question 38.
(i) Give two differences between S_N1 and S_N2 reactions. (1)
(ii) Arrange 1-chloropropane, 2-chloropropane, and 1-chlorobutane in the increasing order of their boiling points. (1)
(iii) Give one use of chloroform. (1)
Answer:
(i)

SN1	SN2
1. Takes place in two steps.	1. Takes place in one step.
2. Through carbocation formation.	2. Through intermediate.
3. Retention occurs.	3. Inversion occurs.

(ii) 2-chloropropane < 1-chloropropane < 1-chloro-butane
(iii) Chloroform is used as a solvent for the production of freon refrigerant, as an anesthetic.

Question 39.
(i) How will you prepare Benzaldehyde from the following: (3)
A. Toluence
B. Benzene
C. Benzoyl Chloride
(ii) Identify the product obtained when Acetic acid is heated with P₂O₅. (1)
Answer:
(i) Etards reaction

Gatterman-Koch reaction

Rosenmund’s reaction

Question 40.
(i) Which of the following amines cannot be prepared by Gabriel Phthalimide synthesis? (1)

(ii) Explain the method to distinguish primary, secondary, and tertiary amines. Also, write the chemical equations involved. (3)
Answer:

(ii) Hinsberg’s test
Benzenesulphonyl chloride – Hinsberg reagent
1° amine + C₆H₅SO₂Cl → A precipitate soluble in alkali
2° amine + C₆H₅SO₂Cl → A precipitate insoluble in alkali
3° amine + C₆H₅SO₂Cl → No reaction

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2021 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2020

Time : 2 1/2 Hours
Maximum : 80 scores

I. Answer all questions from 1 to 6. Each carries 1 Score. (6 × 1 = 6)

Question 1.
How capacitance changes if the distance between the plates of a parallel plate capacitor is halved?
a) Does not change
b) Becomes half
c) Doubled
d) Becomes one fourth
Answer:
c) Doubled

Initial capaciatatance C₀ = \(\frac{A \varepsilon_0}{d}\)
New capaciatatance C₁ = \(=\frac{A \varepsilon_0}{d / 2}\) = \(\frac{2 A \varepsilon_0}{d}\)
C₁ = 2 C₀

Question 2.
The path of a charged particle entering parallel to uniform magnetic field will be
a) Circular
b) helical
c) Straight line
d) None of these
Answer
c) Straight line

Force acting on the charge F = q(\(\vec{υ} \times \vec{B}\))
= qυB
in this case θ = 0, Hence F = 0
No force acting on the partical, hence partical continues its path.

Question 3.
Coefficient of mutual inductance of two coils is 1H. Current in one of the coils is increased from 4 to 5 A in 1ms. What average emf will bw induced in the other coil?
a) 100V
b) 2000V
c) 100V
d) 200V
Answer:
a) 100V

Induced emf ε = M \(\frac{\mathrm{dI}}{\mathrm{dt}}\)
M = 1 H,dI = 5 – 4, dt = 1 × 10^-3
ε = \(=1 \times \frac{(5-4)}{1 \times 10^{-3}}\) = \(\frac{1}{10^-3}\)
= 10³
= 1000 V

Question 4.
Total internal reflection may be observed if
a) Light ray is travelling from denser medium to rarer medim
b) Light ray is travelling from rarer medium to denser medium
c) light ray is travelling from any medium to another medium
Answer:
a) Light ray is travelling from denser medium to rarer medim

Question 5.
Optical fibres make use of the phenomenon of ……….
a) refraction
b) total internal reflection
c) interference
d) diffraction
Answer:
b) total internal reflection

Question 6.
The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model.
a) much greater than
b) not different from
c) much less than
Answer:
a) much greater than

Question 7.
A Permanent electric dipole of dipole moment P is placed in a uniform external electric field E, as shown in Figure.

(a) Redraw the figure and show the magnitude and direction of force actng on the charges.
Answer:

(b) Write an expresson of the torque acting on this dipole in vector form.
Answer:
Torque acting on the dipole τ = \(\overrightarrow{\mathbf{P}} \times \overrightarrow{\mathbf{E}}\)

Question 8.
Ampere’s theorem helps to find the magnetic field in a region around a current carrying conductor.
a) Write the expression of Ampere’s theorem.
Answer:
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} \ell}=\gamma_0 \mathrm{I}\)

b) Draw a graph showing the variation of intensity of magnetic field with the distance from the axis of a current carrying conductor.
Answer:

Question 9.
A magnetised needle in uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however experiences a force of attraction in addition to a torque. Why?
Answer:
A bar magnet produces non – uniform field. Hence iron nail experiences a force of attraction in addition to a torque.

Question 10.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5.0 cm at a distance of 50cm from its mid-point? The magnetic moment of the bar magnet is 0.40 Am².
Answer:
If magnet is short,
Magnetic field along the equatorial line
B_E = \(\frac{\mu_0}{4 \pi} \frac{M}{r^3}\)
\(\frac{\mu_0}{4 K} \) = 10^-7, M = 0.40 Am², r = 50 cm = 0.5 cm
∴ B_E = 10^-7 × \(\frac{0.40}{(0.5)^3}\)
B_E = 3.2 × 10^-7 T
Magnetic field along the axial line
B_axial = \(\frac{\mu_0}{4 \pi} \frac{2 M}{2 r^3}\)
B_axial = 2 B_E
= 2 × 3.2 × 10^-7 T
= 6.4 × 10^-7 T

Question 11.
A magician during a show makes a glass lens with n = 1.47 disappear in a liquid.
a) What is the refractive index of the liquid?
Answer:
1.47

b) Could the liquid be water?
Answer:
No, the reflactive index of the water is 1.33

Question 12.
Explain why the bluish colour predominates in a clear sky.
Answer:
Blue colour undergoes for more scattering.

Question 13.
Match the following:
Tablee 1
Answer:
Tablee 2

Question 14.
Diodes are cone of the building elements of electronic circuits. Some type of diodes are shown in the figure.

a) Identify rectifier diode from the figure.
Answer:
(iii)

b) Draw the circuit diagram of a forward biased rectifier diodes are shown in the figure.
Answer:

Question 15.
The given figure shows the various propagation modes of e.m. waves in communication.

a) Write the names of propagation modes in A, B, C.
Answer:
A – ground wave
B – sky wave
C – space wave

b) Why transmission of TV signals via sky wave is not possible?
Answer:
The frequencies below 40 MHz is reflected by ionosphere. Tv signals have frequences above 54 MHz. Hence sky wave propagation;is not possible in Tv transmission.

Answer any 6 questions from 16 to 23. Each carries 3 scores. (6 × 3 = 18)

Question 16.
An infinitely long straight wire with uniform linear charge density is shown in figure.

a) Draw a Gaussian surface in order to calculate the electric field at P and mark direction of electric field at this point.
Answer:

b) Derive an expression to calculate electric field at this point P.
Answer:

Consider a thin infinitely long straight rod
conductor having charge density λ. [ λ = \(\frac{q}{l}\)]
To find the electric field at P ,we imagine a Gaussian surface passing through P.
Then according to Gauss’s law we can write,
\(\int \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{~s}}\) = \(\frac{1}{\varepsilon_0}\) q
\(\int E d s \cos \theta\)= \(\frac{1}{\varepsilon_0}\) q (θ = 0°)
\(E \int d s\)= \(\frac{\lambda I}{\varepsilon_0}\) q (since q = λ.I)
Integrating over the Gaussian surface, we get (we need not integrate the upper and lower surface because, electric lines do not pass through these surfaces.)

Question 17.
Three resistors R₁, R₂, R₃ are to be combined as shown in the figures.

a) Identify the series and parallel combinations,
Answer:
Fig 1 – parallel combination
Fig 2 – series combination

b) Which combination has lowest effective resistance?
Answer:
Parallel combination

c) Arrive at the expression for the effective resistance of parallel combination.
Answer:
Consider three resistors R₁, R₂ and R₃ connected in parallel across a pd of V volt. Since all the resistors are connected across same terminals, pd across all the resistors are equal.
Img 18
As the value of resistors are different current will be different in each resistor and is given by Ohm’s law
Current through the first resistor
I₁ = \(\frac{V}{R_1}\)
Current through the second resistor
I₂ = \(\frac{V}{R_2}\)
Current through the third resistor
I₃ = \(\frac{V}{R_3}\)

Total current through the combination is I = \(\frac{V}{R}\), where R is the effective resistance of parallel combination.
Total current through the combination = the sum of current through each resistor
I = I₁ + I₂ + I₃
Substituting the values of current we get
\(\frac{V}{R}\) = \(\frac{V}{R_1}\) + \(\frac{V}{R_2}\) + \(\frac{V}{R_3}\)
Eliminating V from all terms on both sides of the equations, we get
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

Question 18.
a) State Faraday’s law of electomagnetic induction.
Answer:
Faraday’s Law of induction
Faraday’s law of electromagnetic induction states that the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by
\(\epsilon=\frac{d \phi}{d t}\)
If the coil contain N turns, the total induced emf is given by
\(\epsilon=N \frac{d \phi}{d t}\)

b) How does the magnetic energy stored in an inductor and electrostatic energy stored in a capacitor related to their respective field strengths?
Answer:
In capacitor energy is stored in electronic field
E = \(\frac{1}{2}\) cv²
In inductor energy is stored in magnetic field
E = \(\frac{1}{2}\)LI²

Question 19.
A typical plane electromagnetic wave propagating along the Z direction is shown in figure.

a) Write the equation for electric and magnetic fields.
Answer:
This em wave is travelling in z direction and direction of oscillation of electric field is along x. Hence equation for electric field
E_x = E₀ Sin(Kz-wt)
The direction of oscillation of magnetic field is along the y direction, hence
B_y = B₀ Sin (Kz-wt)

b) Write the methods of production of radio waves and microwaves. Write any one use of these waves. (1+2)
Answer:
Radio wave: Rapid accelartion and deacceleration of electrons in aerials. These are used in radio and televisiion communication systems.

Microwaves: Microwaves are produced by special vaccum tubes. Microwaves ovens are domestic applica¬tion of those waves.

Question 20.
The figure shows the image formation of an object in simple microscope.

a) Find out the object distance and image distance from the figure.
Answer:
OQ is the object distance.
OQ₁ is the image distance.

b) Derive an equation for magnifying power of the simple microscope.
Answer:
A simple microscope is a converging lens of small focal length.

Working: The object to be magnified is placed very close to the lens and the eye is positioned close to the lens on the other side. Depending upon the position of object, the position of image is changed.
If the image is formed at ‘D’, (D = 25 cm) we can take u = -D. Hence the lens formula can be written as
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)
1 – \(\frac{v}{u}\) = \(\frac{v}{f}\)
\(\frac{v}{u}\) = 1 – \(\frac{v}{f}\)
m = 1 – \(\frac{v}{f}\)
The image is formed at D, ie. v = -D
m = 1 – \(\frac{-D}{f}\)
m = 1 + \(\frac{D}{f}\)

Question 21.
The atomic line spectra of hydrogen atom is shown in figure.

Answer:
A – Lyman series
B – Balmor series
C – Paschen series

Question 22.
Spontaneous and continuous disinflation of a nucleus of a heavy element with the emission of certain types of radiation is known as radioactivity,
a) The radioactive isotope ‘D’ decays according to the sequence.

If the mass number and atomic number of D₂ are 172 and 71 respectively, what are the (i) Mass number, (ii) atomic number of D.
Answer:

When an α (₂He⁴) emitted, mass number 4 and atomic number 2 is reduced from the parent nucleus. Hence pamet nucleus,
D₁ = ₇₃Y¹⁷⁶
When is emitted, atomic number of daughter nuclei increases by 1 unit and mass numbers constant
Hence D = ₇₂Z¹⁷⁶

atomic number of D is 72, and mass number is 176.

b) State radioactive decay law.
Answer:
The number of nuclei undergoing decay per unit time is proportional to number of nuclei in the sample at that time.

c) Write he relation connecting half-life and mean life of radioactive element.
Answer:
T_1/2 = 0.693 τ
T_1/2 is the half life
τ is the mean life

Question 23.
In the broadcast of communication, modulation is necesary.
a) What do you mean by modulation?
Answer:
Modulation is the process of superprosing a low ferquency signal on a high frequency carrier wave.

b) Explain any two reason why modulation is necessary?
Answer:
i) Size of the antenna can be reduced.
ii) Effectiven power radiated from the source can be increased.

Answer any 2 questions from 24 to 26, Each carries 4 scores. (2 × 4 = 8)

Question 24.
Three capacitors of capacitances 2 pF, 3pF are connected in parallel.
a) Write the SI unit of capacitance.
Answer:
Farad

b) Calculate the effective capacitance of the combination.
Answer:
C = C₁ + C₂ + C₃
C = (2+3+4) PF
C = 9 PF

c) Determine the charge on each capacitor if the combination is connected to a 100V supply
Q = CV
Q₁ = C₁V
= 2 PF × 100 = 2 × 10^-10 C
Q₂ = C₂V = 3 PF × 100 = 3 × 10^-10 C
Q₃ = C₃V = 4 PF × 100 = 4 × 10^-10 C

Question 25.
A rectangular loop o ara A and carrying a steady current I is placed in a uniform magnetic field.
a) Derive the expression of torque, acting on the loop.
Answer:
Torque on a rectangular current loop in uniform magnetic filed

Consider a rectangular coil PQRS of N turns which is suspended in a magnetic field, so that it can rotate (about yy1). Let l be the length (PQ) and ‘b’ be the breadth (QR).
When a current ‘l’ flows in the coil, each side produces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque.
Which can be written as
τ = Force × ⊥ distance …….(1)
But, force = BIl ……..(2) [since θ = 90°]
And from AQTR, we get
⊥ distance (QT) = b sin θ ………..(3)
Substituting the vales of eq (2) and eq (3) in eq(1)
we get
τ = BIl b sin θ
= BIA sin θ [since lb = A (area)]
τ = IAB sin θ
τ = m B sin θ [since m = IA]
If there are N turns in the coil, then
τ = NIAB sin θ

b) Increasing the current sensitivity may not necessarily increase the voltage sensitivity, of a galvanometer. Justify.
Answer:
Current sensitivtity can be increased by incraesing number of turns. When number of turns doubble, the resistance of the wire will also be doubbled. Hence the voltage sensitivity does not change.

Question 26.
The work function of caesium metal is 2.14eV. When light of frequency 6 × 10¹⁴ Hz is incident On the metal surface, photoemission of electrons occurs. (h = 6.6 × 10¹⁴)
a) Define work function
Answer:
Work function is the amount of energy required fora electron tojust escape from a metal surface.

b) Calculate the maximum kinetic energy of the emitted electrons.
Answer:
hυ = ϕ₀ + KE
6.6 × 10³⁴ × 6 × 10^-20 = 2.14 × 1.6 × 10^-19 + KE
39.6 × 10^-20 = 3.424 × 10^-19 + KE
KE = 5.36 × 10^-19 J
KE = 3.35 ev
KE = ev₀
3.35 ev = ev₀
v₀ = 3.35 v

Answer any 3 questions from 27 to 30, Each carries 5 scores. (3×5 = 15)

Question 27.
A Wheatstone bridge is shown in figure.

a) Derive a relation connecting the four resistors for the galvanometer to give zero or null deflection.
Answer:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I, I₁,I₂,I₃ and I₄ be the four currents passing through P,R,Q and S respectively.

Working :
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ……….(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₄S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄ ………(2)
Using KirchofFs second law in loop ABDA and BCDB, weget
V_AB = V_AD ………(3)
and V_BC = V_DC —-(4)
Substituting the values from eq(1) into (3) and (4), weget
I₁P = I₂R …….(5)
I₃Q = I₄S …….(6)
Dividing Eq(5)byEq(6)
\(\frac{\mathrm{I}_1 P}{\mathrm{I}_3 Q}\) = \(\frac{I_2 R}{I_4 S}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\)
This is called Wheatstone condition,

Teachers recommend solving Kerala Syllabus Plus Two Botany Previous Year Question Papers and Answers Pdf Board Model Paper 2020 to improve time management during exams.

Kerala Plus Two Botany Board Model Paper 2020 with Answers

I. Answer any 3 questions from 1 – 5. Each carries (3 × 1 = 3)

Question 1.
Choose the correct answer and fill in the blank.
_______ is the female gametophyte of angiosperms.
a) Embryo sac
b) Nuceluus
c) Integument
d) Pollen grain
Answer:
Embryosac

Question 2.
Choose the correct answer. Natural aging of a lake by nutrient enrichment of its water is called
a) Biomagnification
b) Algal bloom
c) Succession
d) Eutrophication
Answer:
Eutrophication

Question 3.
Fill in the blank :
The wheat variety having high protein content used as donor in bio-fortification is ______
Answer:
Atlas 66

Question 4.
Name the type interaction between an orchid plant and Mango tree.
Answer:
Commensalism

Question 5.
Find the odd one.
Hilsa, Sardine, Rohu, Mackerel
Answer:
Rohu

II. Answer any 9 questions from 6-16. Each carries 2 scores.( 9 × 2 = 18)

Question 6.
The innermost wall layer of Microsporangium is Tapetum.
a) What is it is function?
b) Features of Tapetal cells.
Answer:
a) It nourishes the pollen grain
b) It has dense cytoplasm, It posses prominent nucleus

Question 7.
Match the following:

A	B
Chlamydomonas	Gemmule
Conidia	Zoospore
Sponage	Bulbil
Hydra	Pencillium
	Buds

Answer:

A	B
Chlamydomonas	Zoospore
Conidia	Penicillium
Sponage	Gemmule
Hydra	Buds

Question 8.
Explain the terms:
a) Micro propagation
b) Totipotency
Answer:
a) It is the production of large number of plants within short time in tissue culture,
b) The ability of a cell to form whole plant.

Question 9.
Explain the separation of DNA fragments using gel electrophoresis.
Answer:
In Gel electrophoresis, negatively charged DNA fragments are separated by forcing them to move towards the anode under an electric field through a agarose matrix. The DNA fragments separate according to their sizd in agarose gel.

Question 10.
Write notes on:
a) Micro injection
b) Biolistics
Answer:
a) Micro-injection Recombinant DNA is directly injected into the nucleus of an animal cel.,

b) Biolistics – Cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA. It is suitable for plants.

Question 11.
Amplification of gene of interest is done using PCR.
a) Expand PCR
b) Mention the three steps of this process.
Answer:
a) Polymerase chain reaction

b) 1. Denaturation
2. Annealing
3. Extension

Question 12.
Observe the diagram and answer A and B.

Answer:
a) Pro-insulin
b) Mature insulin

Question 13.
a) Define transgenic animals.
b) Write any two uses of transgenic animals.
Answer:
a) Animals that possess manipulated gene (genetically modified) and express foreign gene are called as transgenic animals.
b) Transgenic animals are used in the field of

1. Biological products
2. Study of disease and
3. Normal physiology and development

Question 14.
Humification and Mineralisation occur during decomposition in the soil. Write the difference between these two processes.
Answer:
Humification
It is process by which simplified detritus is changed into dark coloured amorphous substance called humus. It is the reservoir of nutrients.

Minerilisation
It invoves the release of inorganic subtances (Water, CO₂, etc.) and other nutrients (NH₂⁺, Ca⁺⁺, Mg⁺⁺, K₊, etc.) in the soil.

Question 15.
a) Construct a grazing food chain using the following organisms: Bird, Grass, Grasshopper
b) Write the trophic level of grasshopper.
Answer:
a) Grass → GrasshopperBird
b) Second trophic level or primary consumer

Question 16.
Why is CNG better than diesel or petrol?
Answer:

1. CNG is better than diesel.
2. It burns most efficiently.
3. CNG is cheaper than petrol or diesel.
4. It cannot be adulterated like petrol or diesel.

III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 × 3 = 9)

Question 17.
Briefly describe the adaptations of desert plants.
Answer:

1. Thick cuticle on their leaf surfaces .
2. Stomata arranged in deep pits to minimise water loss.
3. They show CAM pathway (they open stomata during night and closed during daytime).
4. Leaves are reduced to spines and the flattened stems do photosynthesis.

Question 18.
Density of a population fluctuates due to changes in four basic processes.
a) Complete the diagram by adding the points ABC and D.
b) Explain A and D.

Answer:
a)

1. Natality
2. Mortality
3. Immigration
4. Emigration

b) A-lmmigration-Movement of individuals into the population.
D-Emigration- Movement of individuals out of the population.

Question 19.
Air pollution is a serious environmental issue. Give three harmful effects of air pollution.
Mention the out-breeding devices in plants to prevent self pollination, (any 3 points).
Answer:

1. Reduce the growth and yield of crops.
2. It cause premature death of plants.
3. Affect respiratory system of animals and humans!

Question 20.
Mention the out-breeding devices in plants to prevent self pollination. (any 3 points)
Answer:

1. Pollen release and stigma receptivity are not at the same time.
2. Anther and stigma are placed at different positions.
3. Self-incompatibility.
4. Production of unisexual flowers

Teachers recommend solving Kerala Syllabus Plus Two Botany Previous Year Question Papers and Answers Pdf Board Model Paper 2021 to improve time management during exams.

Kerala Plus Two Botany Board Model Paper 2021 with Answers

Answer the following questions from 1 to 31 upto a maximum score of 30.
1. Questions 1 to 7 carry 1 score each.

Question 1.
Which among the following is a vegetative propagule?
(a) Rhizome
(b) Gemmules
(c) Zoospores
(d) Conidia
Answer:
Rhizome

Fill in the blank :
Question 2.
The protective wall layer of fruit is known as ____.
Answer:
Pericarp

Question 3.
Fruit which develop from any part of the flower other than ovary is called ____.
Answer:
False fruit

Question 4.
Name any two products obtained through Beekeeping.
Answer:
Honey and bee wax

Question 5.
Fill in the blank:
The restriction enzyme EcoRI isolated from the bac-terium _____.
Answer:
Escherichia coli RY 13

Question 6.
In recombinant DNAtechnology, DNAfragments are joined by the help of the enzyme.
(a) DNA ligase
(b) DNA polymerase
(c) Restriction enzyme
(d) Taq polymerase
Answer:
DNA ligase

Question 7.
Select the non-parasitic organism from the list given below:
(a) Lice
(b) Cuscuta
(c) Epiphytic orchid.
(d) Ticks
Answer:
Epiphytic orchid

II. Questions 8 to 26 carry 2 scores each.

Question 8.
In some organisms, female gamete undergoes de-velopment to form new organisms without fertiliza-tion. Name the phenomenon. Give an example.
Answer:
Parthenogenesis
eg. Honeybee and rotifers

Question 9.
Chromosome number in meiocytes and gametes of some organisms are given in the table. Fill in the blanks.

Name of organism	Chromosome number in meiocytes	Chromosome number in qametes
a. Rice	–	12
b. Onion	–	8
c. Apple	34	–
d. Maize	20	–

Answer:

Name of organism	Chromosome number in meiocytes	Chromosome number in qametes
a. Rice	24	12
b. Onion	16	8
c. Apple	34	17
d. Maize	20	10

Question 10.
There are two types of cyclic events that happens in the reproductive phase of female organismsamong placental mammals. Which are the cyclic events and give one example for each.
Answer:
Menstrual cycle-Primates eg. monkeys, apes Oesrous cycle – Non primates eg. cow, sheep

Question 11.
The outer layer of pollengrains is made up of’ sporopollenin. What is the importance of sporopollenin?
Answer:
It is one of the most resistant organic material which can withstand high temperature strong acid and al-kali. So the pollen grains are well preserved as fos-sils.

Question 12.
Differentiate between Autogamy and Xenogamy.
Answer:
Atutogamy: It is the transfer of pollen grains from the anther to the stigma of the same flower.
Xenogamy: Transfer of pollen grains from anther to the stigma of a different plant. This type of pollina-tion occurs between genetically different species.

Question 13.
If the female parent produces bisexual flowers, emas-culation is necessary in artificial hybridization.
(a) What is emasculation?
(b) Write down the importance of emasculation.
Answer:
(a) Anthers are removed before the dehiscence of bisexual flowers.

(b) It is helpful to avoid self pollination and fertilisation.

Question 14.
Fusion of polar nuclei with male gamete in double fertilisation result in formation of endosperm.
(a) Write down the function of endosperm.
(b) Write briefly about the endosperm development in coconut.
Answer:
(a) It gives nutrition to the developing embryo.

(b) PEN undergo successive nuclear divisions to give rise to free nuclei. This stage of endosperm v development is called free nuclear endosperm. Later cell wall formation occurs and the endosperm becomes cellular.

Question 15.
‘Hisardale’ is a new breed of sheep developed in Punjab.
(a) Identify the method by which the breed is devel-oped.
(b) Name the parental breeds of’Hisardale’.
Answer:
(a) Crossbreeding
(b) Bikaneriewes and Marino rams

Question 16.
In r-DNA technology, amplification of gene is done by a process called PCR.
(a) Expand PCR
(b) What are the three main steps involved in PCR?
Answer:
(a) Polymerase chain reaction
(b) Denaturation, Annealing and extension.

Question 17.
Observe the picture given below:
(a) Identify the instrument in the figure.
(b) Write any one use of this instrument.

Answer:
(a) Bioreactor

(b) It is the large culture vesel for large Scale production of Recombinant protein, enzyme etc.

Question 18.
List any two uses of GMO (Genétically Modified Or ganism) in agriculture.
Answer:
(i) Made crops more tolerant to abiotic stresses (cold, drought, salt, heat).
(ii) Reduced reliance on chemical pesticides (pest-resistant crops).

Question 19.
Expand GEAC. Mention its function.
Answer:
GEAC-Genetic Engineering Approval Committee. They take decisions regarding the validity of GM research and the safety of introducing GM organisms for public services.

Question 20.
Comment on Brood Parasitism with an example.
Answer:
In Brood parasitism parasitic bird lays its eggs in the nest of its host and the host incubate them. The eggs of the parasitic bird resemble the host’s egg in size and colour. So the host bird keep up it in their nest. Example of brood parasitism are cuckoo (koel) and the crow.

Question 21.
The figure shows simplified model of phosphorus cycle. Analyse the figure and fill up the blanks.

Answer:
a) Rock Minerals
(b) Producers

Question 22.
The figure given below indicates a pyramid of energy. Why pyramid of energy is always upright in
position?

Answer:
Pyramid of energy is always upright because when energy flows from a particular tropic level to the next tropic level. Some energy is always lost as heat at each step.

Question 23.
What is ecological succession? Differentiate Hydrarch succession and Xerarch succession?
Answer:
Ecological succession-It is the successive replace- – ment of plant communities in an area over a p eriod of time. Hydrarch – It involves the ecological succession in the newly formed pond or lake. Pioneer communi ty’- Phytoplanktons. Xerarch – It involves the ecological succession on bare rock surfaces. Pioneer community – Lichen.

Question 24.
‘Jhum’ cultivatjon/Slah and bum agriculture is an agricultural practice in north esatern states of India. How this practice enhances deforestation?
Answer:
In slash and burn agriculture, the farmers cut down the trees of the forest and burn the plant remains. The ash is used as a fertiliser and theland is then used for farming or cattle grazing. After cultivation, the area is left for several years so as to allow its recovery. The farmers then move on to other areas and repeat this process.

Question 25.
Write any four measures for controlling Global Warming.
Answer:

1. Cutting down use of fossil fuel.
2. Improving efficiency of energy usage.
3. Preventing deforestation and planting trees.
4. Slowing down the growth

Question 26.
In 1980 Government of India introduced a management programme, JFM.
(a) Expand JFM
(b) What is its significance?
Answer:
(a) Joint Forest Managment(JFM)

(b) This is helpful to local communities for getting
benefit of various forest products for the return of services and thus the forest can be conserved in a sustainable manner.

III. Questions 27 to 31 carey 3 scores each.

Question 27.
Pollination is the transfer of pollengrams from anther to stigma of pistil. Write any three adaptations seen in wind pollinated plants.
Answer:

1. Pollen grains are light and no-sticky.
2. Flowers have a single ovule in each ovary.
3. They possess well-exposed stamens and feathery stigma.

Question 28.
Different steps of plant breeding is given below;
(1) Testing, release and commercialization of new cultivais.
(2) Cross hybridization among the selected parents.
(3) Evaluation and selection of parents.
(4) Collection of variability
(5) Selection and testing of superior recombinants.
Create a flow chart of plant breeding from the above steps in sequential manner.

Answer:
(a) Evaluation and selection of parents.
(b) Cross hybridization among the selected parents.
(c) Selection and testing of superior recombinants.

Question 29.
Given diagram shows a technique used in r-DNA technology.

(a) Identify the process
(b) What is the purpose of this technique in r-DNA technokgy?
(c) Name the stain used in this technique for visualizing DNA under UV-light.
Answer:
(a) Gel electrophoresis
(b) In this cut fragments of DNA are separated.
(c) Ethidium bromide

Question 30.
Bt-cotton is a transgenic plant
(a) What does ‘Bt’ stands for?
(b) Explain the mechanism of insect resistance in Bt-cotton
Answer:
(a) Bacillus thuringiensis
(b) Bt toxin protein exist as inactive protoxins but it is converted into an active of form in the presence of the alkaline pH of insect gut. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis atid results in the death of insect.

Question 31.
Opuntia is a plant well adapted to desert conditions. Write down the various adaptation found in desert
plants.
Answer:
Desert plants have a thick cuticle on their leaf surfaces and stomata arranged in deep pits to minimise water loss through transpiration. They also have CAM pathway in which they open stomata during night and closed during time.

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf Board Model Paper 2021 helps in understanding answer patterns.

Kerala Plus Two Maths Board Model Paper 2021 with Answers

Time: 2 Hours
Total Score: 60 Marks

Answer the following questions from 1 to 29 up to a maximum score of 60.

Part – A

Questions from 1 to 10 carry 3 scores each. (10 × 3 = 30)

Question 1.
Find the value of x.
\(\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\) = \(\left|\begin{array}{cc}
2 & 4 \\
5 & 1
\end{array}\right|\) (3)
Answer:
Expanding on both sides
2x² – 24 = 2 – 20
2x² – 24 = -18
2x² = 24 – 18 = 6
x² = 3
x = ±√3

Question 2.
A = \(\left|\begin{array}{cc}
3 & 1 \\
-2 & 1
\end{array}\right|\)
(i) Find (adj.A) (1)
(ii) Also prove that A(adj A) = 5I (2)
Answer:

Question 3.
Determine the value of the constant k so that the function
is continuous at x = 2 (3)
Answer:
Given is continuous at x = 2
∴ At x = 2, LHL = RHL = f(2)
We have f(2) = k(2)² = 4 K _________ (1)
RHL = \(\lim _{x \rightarrow 2^{+}} f(x)\) = 3 _________ (2)
∴ From (1) and (2)
4K = 3
K = \(\frac{3}{4}\)

Question 4.
Verify Mean Value Theorem if f(x) = x² – 4x – 3, is the interval [1, 4]. (3)
Answer:
Since f(x) is a polynomial function, it is continuous in [1, 4]
f(x) = x² – 4x – 3
f'(x) = 2x – 4
∴ It is differentiable in (1, 4)
f(a) = f(1) = 1² – 4(1) – 3 = 1 – 4 – 3 = -6
f(b) = f(4) = 4² – 4(4) – 3 =16 – 16 – 3 = -3
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
2c – 4 = \(\frac{- 3 – 6}{ 4 – 1}\) = \(\frac{- 3 + 6}{3}\)
2c – 4 = \(\frac{3}{3}\) = 1
2c = 5
c = \(\frac{5}{2}\) ∈ (1, 4)
Hence Mean Value Theorem Verified.

Question 5.
The radius of a circle is increasing uniformly at the rate of 5 cm/s. Find the rate at which the area of the circle is increasing when the radius is 8 cm. (3)
Answer:
Let r be the radius and A be the area of the circle.
Then \(\frac{d r}{d t}\) = 5 cm/s
Let A = πr²
\(\frac{d A}{d t}\) = 2πr. \(\frac{d r}{d t}\) = 2πr × 5
= 10πr
\(\left.\frac{\mathrm{dA}}{\mathrm{dt}}\right]_{\mathrm{r}=8}\) = 10π × 8
= 80π cm²/s

Question 6.
Find the unit vector in the direction of vector \(\vec{\mathrm{PQ}}\), where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively. (3)
Answer:

Question 7.
Find the vector and cartesian equations of the planes that passes through the point (1, 4, 6) and the normal î – 2ĵ + k̂ (3)
Answer:
Given(x₁, y₁, z₁) = (1, 4, 6)
Direction ratios of normal <a, b, c> = <1, -2, 1>
∴ Equation is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
1(x – 1) – 2(y – 4) + 1(z – 6) = 0
x – 1 – 2y + 8 + z – 6 = 0
x – 2y + z + 1 = 0
Vector equation is \(\vec{r}\) . (î – 2ĵ + k̂) = -1

Question 8.
(i) Find the principal value of sin^-1(\(\frac{1}{\sqrt{2}}\)) (1)
(ii) Evaluate tan^-1(1) + cos^-1(\(\frac{-1}{2}\)) + sin^-1(\(\frac{-1}{2}\)) (2)
Answer:
(i) sin^-1(\(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{4}\)

(ii) We have sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)
∴ cos^-1(\(\frac{-1}{2}\)) + sin^-1(\(\frac{-1}{2}\)) = \(\frac{\pi}{2}\)
tan^-1(1) + cos^-1(\(\frac{-1}{2}\)) + sin^-1(\(\frac{-1}{2}\))
= \(\frac{\pi}{4}\) + \(\frac{\pi}{2}\) = \(\frac{3 \pi}{4}\)

Question 9.
Find the equation of line joining (1, 2) and (3, 6) using determinants. (3)
Answer:
Equation of line joining (x₁, y₁) and (x₂, y₂) is \(\left|\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x & y & 1
\end{array}\right|\) = 0
i.e., \(\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 6 & 1 \\
x & y & 1
\end{array}\right|\) = 0
Expanding along R₁,
1(6 – y) – 2(3 – x) + 1(3y – 6x) = 0
6 – y – 6 + 2x + 3y – 6x = 0
– 4x + 2y =0
ie., 2x – y = 0

Question 10.
Find the general solution of the differential equation.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x² (3)
Answer:
Given \(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
It is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q
Where P = \(\frac{1}{x}\), Q = x²
Integrating factor (I. F) = \(e^{\int P d x}\)
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x
Solution is y (I. F) = ∫Q(IF) dx
yx = ∫x².x dx
ie., xy = ∫x³ dx
xy = \(\frac{x^4}{4}\) + C

Part – B

Questions from 11 to 22 carry 4 scores each. (12 × 4 = 48)

Question 11.
(i) Construct a 2 × 2 matrix A = [a_ij], whose elements are given by a_ij = \(\frac{(i+j)^2}{2}\) (2)
(ii) If \(\left[\begin{array}{cc}
a-b & d \\
2 a-b & c
\end{array}\right]\) = \(\left[\begin{array}{cc}
-1 & 4 \\
0 & 5
\end{array}\right]\) then find a,b,c and d. (2)
Answer:

(ii) Given \(\left[\begin{array}{cc}
a-b & d \\
2 a-b & c
\end{array}\right]\) = \(\left[\begin{array}{cc}
-1 & 4 \\
0 & 5
\end{array}\right]\)
a – b = -1 ________ (1) d = 4
2a – b = 0 ________ (2) c = 5
Solving (1) and (2), a = 1 b = 2

Question 12.
A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 2
\end{array}\right]\) B = \(\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\) C = \(\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]\)
Find (i) 3A – C (2)
(ii) AB (2)
Answer:

Question 13.
(i) tan^-1 x + tan^-1 y = …………………… (1)
(ii) Prove that tan^-1\(\frac{2}{11}\) + tan^-1\(\frac{7}{24}\) = tan^-1\(\frac{1}{2}\) (3)
Answer:

Question 14.
Find \(\frac{d y}{d x}\).
(i) y = sin(cos(x²)) (2)
(ii) x² + xy + y² = 100 (2)
Answer:
(i) Given y = sin [cos(x²)]
\(\frac{d y}{d x}\) = cos[cos(x²)] . \(\frac{d}{d x}\)cos[cos(x²)]
= cos[cos(x²)] – sin(x²) \(\frac{d}{d x}\) .(x²)
= -cos[cos(x²)]. sin(x²).2x
= -2x sin(x²).cos[cos(x²)}

(ii) Given x² + xy – y² = 100
Differentiating with respect to x,
2x + \(\frac{x d y}{d x}\) + y + 2y.\(\frac{d y}{d x}\) = 0
(x + 2y)\(\frac{d y}{d x}\) = – 2x – 2y
\(\frac{d y}{d x}\) = \(\frac{-(2 x+y)}{x+2 y}\)

Question 15.
(i) Find the maximum value of the function f(x) = sin x + cos x in [0, π/2]. (2)
(ii) Find the intervals in which the function f(x) = x² + 2x – 5 is increasing or decreasing.
Answer:
(i) Given f(x) = sin x + cos x
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
= – (sin x + cos x)
f'(x) = 0 ⇒ cos x – sin x = 0
cos x = sin x
x = \(\frac{\pi}{4}\); (∵ x ∈[0, \(\frac{\pi}{2}\)]
f”(\(\frac{\pi}{4}\)) = -(sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\))
= – (\(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)) = -(\(\frac{2}{\sqrt{2}}\)) = -√2 < 0
∴ f(x) has a maximum value at x = \(\frac{\pi}{4}\)
Also f(0) = sin 0 + cos 0 = 1
f(\(\frac{\pi}{2}\)) = sin\(\frac{\pi}{2}\) + cos\(\frac{\pi}{2}\) = 1
f(\(\frac{\pi}{4}\)) = sin\(\frac{\pi}{4}\) + cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = √2
∴ Maximum value of f(x) = √2

(ii) Given f(x) = x² + 2x – 5
f'(x) = 2x + 2
f'(x) = 0 ⇒ 2x + 2 = 0
x = -1

∴ Intervals are (-∞, -1) and (-1, ∞)
At (-∞,-1), f'(-2) = -4 + 2 = -2 < 0 At (-1, ∞), f'(0) = 0 + 2 = 2 > 0
∴ f(x) is decreasing in (-∞, -1) and increasing in (-1, ∞)

Question 16.
(i) Write the order of the differential equation.
(\(\frac{d^2 s}{d t^2}\)) + 3(\(\frac{d s}{d t}\))³ + 4 = 0
(ii) Find the general solution of the differential equation.
sec² x . tan y dx + sec² y . tan x . dy = 0
Answer:
(i) Order = 2

(ii) Given sec² x tan y dx + sec² y tan x dy = 0
sec² x tan y dx = -sec² y tan x dy
\(\frac{\sec ^2 x}{\tan x}\)dx = \(\frac{\sec ^2 y}{\tan y}\)dy
Which is variable seperable
∴ Solution is \(\frac{\sec ^2 x}{\tan x}\)dx = \(\frac{\sec ^2 y}{\tan y}\)dy
ie., log |tan x| = – log |tan y| + log C
tan x = \(\frac{C}{\tan y}\)
tan x. tan y = C

Question 17.
Find a unit vector perpendicular to each of the vector \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\), where \(\vec{a}\) = 3î + 2ĵ + 2k̂ and \(\vec{b}\) = î + 2ĵ – 2k̂. (4)
Answer:
Let \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\) = 4î + 4ĵ
\(\vec{d}\) = \(\vec{a}\) – \(\vec{b}\) = 2î + 4k̂
Vector perpendicular to \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\) is \(\vec{c}\) × \(\vec{d}\)
ie., \(\vec{c}\) × \(\vec{d}\) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
4 & 4 & 0 \\
2 & 0 & 4
\end{array}\right|\) = 16î – 16ĵ – 8k̂
\(\vec{c}\) × \(\vec{d}\) = \(\sqrt{256+256+64}\)
= \(\sqrt{32}\)
= 24
∴ Required unit vector = \(\frac{\vec{c} \times \vec{d}}{|\vec{\mathrm{c}} \times \vec{\mathrm{d}}|}\)
= \(\frac{16 \hat{i}-16 \hat{j}-8 \hat{k}}{24}\)
= \(\frac{2}{3}\)î – \(\frac{2}{3}\)ĵ – \(\frac{1}{3}\)k̂

Question 18.
Find the shortest distance between the pair of lines
\(\vec{r}\) = (i + 2j + 3k) + λ(i – 3j + 2k)
\(\vec{r}\) = (4i + 5j + 6k) + µ(2i + 3j + k) (4)
Answer:
Let \(\vec{a}\)₁ = î + 2ĵ + 3k̂,
\(\vec{b}\)₁ = î – 3ĵ + 2k̂
\(\vec{a}\)₂ = 4î + 5ĵ + 6k̂,
\(\vec{b}\)₂ = 2î + 3ĵ + k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = 3î + 3ĵ + 3k̂

Question 19.
Let A and B be two independent events such that P(A) = \(\frac{1}{7}\) and P(B) = \(\frac{1}{2}\). Find
() P(A∩B) (1)
(b) P(A∪B) (2)
(c) P[A∩B’)∩(B∩A’)] (1)
Answer:
Given P(A) = \(\frac{1}{7}\) P(B) = \(\frac{1}{5}\)
∴ P(A’) = \(\frac{6}{7}\) P(B) = \(\frac{4}{5}\)
(i) P(A∩B) = P(A).P(B) = \(\frac{1}{7}\) × \(\frac{1}{5}\) = \(\frac{1}{35}\)

(ii) P(A∪B) = 1 – P(A’). P(B’)
= 1 – \(\frac{6}{7}\) × \(\frac{4}{5}\) = 1 – \(\frac{24}{35}\) = \(\frac{11}{35}\)

(iii) P((A∩B’)∩(B∩A’))
Since A∩B’ and B∩A’ are,
mutually exclusive events,
(A∩B’)∩(B∩A’) = Φ
∴ P[(A∩B’)∩(B∩A’)] = 0

Question 20.
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relatIon. (4)
Answer:
We have
R = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1),
(3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5)}
Since (a, a) ∈ R ∀ a ∈ A, R is reflexive.
(a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A
∴ R is symmetric.
(a, b) ∈ R, (b,c) ∈ R ⇒ (a, c) ∈ R ∀ a,b,c ∈ A
∴ R is transitive.
Hence R is an equivalence relation.

Question 21.
Find \(\frac{d y}{d x}\) of the function y^x = x^y. (4)
Answer:
Given y^x = x^y
taking ‘log’ on both sides
log y^x = log x^y
ie., x log y = y log x
Differentiating with respect to x,

Question 22.
Find
(i) ∫x log x dx (2)
(ii) ∫x² sin x dx (2)
Answer:

Part – C

Questions from 23 to 29 carry 6 scores each. (7 × 6 = 42)

Question 23.
Express the matrix A = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix. (6)
Answer:
Given A = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) A’ = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\)
Let P = \(\frac{1}{2}\)(A + A’) = \(\frac{1}{2}\)\(\left[\begin{array}{ccc}
12 & -4 & 4 \\
-4 & 6 & -2 \\
4 & -2 & 6
\end{array}\right]\) symmetric
Q = \(\frac{1}{2}\)(A – A’) = \(\frac{1}{2}\)\(\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)Skew symmetric
∴ A = P + Q

Question 24.
Solve the following system of equations by Matrix Method:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4 (6)
Answer:

Question 25.
(i) State whether the function f : R → R defined by f(x) = 3 – 4x is a bijective function or not. Justify your answer. (4)
(ii) Let f : R → R and. g : R → R be defined by
f(x) = 2x +1 and g(x) = x². Then find gof and fog.(2)
Answer:
(i) Given f(x) = 3 – 4x
f(x¹) = f(x²) ⇒ 3 – 4x¹ = 3 – 4x²
⇒ -4x¹ = -4x²
⇒ x¹ = x²
∴ f is one – one
Let y = 3 – 4x
4x = 3 – y
x = \(\frac{3 – y }{4}\)
f(x) = f(\(\frac{3 – y }{4}\)) = 3 – 4(\(\frac{3 – y }{4}\)) = y
∴ For every y ∈ R, there exists x ∈ R such that f(x) = y
∴ f is onto
Hence f is a bijective function

(ii) Given f(x) = 2x +1, g(x) = x²
(gof)(x) = g[f(x)] = g(2x + 1)
= (2x + 1 )² = 4x² + 4x + 1
(fog)(x)= f[g(x)] = f(x²)
= 2x² + 1

Question 26.
Find the equation of the tangent line to the curve
y = x² – 2x – 7 which is
(i) Parallel to the line 2x – y + 9 = 0 (3)
(ii) Perpendicular to the line 5y – 15x = 13 (3)
Answer:
Given y = x² – 2x + 7
\(\frac{d y}{d x}\) = 2x – 2
(i) Tangent is parallel to 2x – y + 9 = 0
∴ Slope of tangent = slope of given line
\(\frac{d y}{d x}\) = \(\frac{-2}{-1}\)
2x – 2 = 2
2x = 4
x = 2
When x = 2, y = (2)² – 2 (2) + 7 = 7
∴ Point is (2, 7)
Equation of tangent is dy
y – y₀ = \(\frac{d y}{d x}\)(x – x₀)
y – 7 = 2(x – 2)
y – 7 = 2x – 4
2x – y + 3 = 0

(ii) Tangent is perpendicular to 5y – 15x = 13
ie., – 15 x + 5y – 13 = 0
∴ Slope of tangent × slope of given line = -1
\(\frac{d y}{d x}\) × 3 = -1
\(\frac{d y}{d x}\) = \(\frac{-1}{3}\)
2x – 2 = \(\frac{-1}{3}\)
2x = 2 – \(\frac{1}{3}\) = \(\frac{5}{3}\), 5. x = \(\frac{5}{6}\)
When x = \(\frac{5}{6}\), y = \(\frac{25}{36}\) – \(\frac{10}{6}\) + 7 = \(\frac{217}{36}\)
Point is (\(\frac{5}{6}\), \(\frac{217}{36}\))
∴ Equation of tangent is
y – y₀ = \(\frac{d y}{d x}\)(x – x₀)
y – \(\frac{217}{36}\) = –\(\frac{1}{3}\) (x – \(\frac{5}{6}\))
36y – 217 = -12x + 10
12x + 36y – 227 = 0

Question 27.
Find the following :
(i) \(\int \frac{1}{1+x^2} d x\) (1)
(ii) \(\int \frac{d x}{x^2-6 x+13}\) (2)
(iii) \(\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x\) (3)
Answer:

Question 28.
(i) Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant. (3)
(ii) Find the area of the region bounded by the two parabolas y = x² and y2 = x. (3)
Answer:
(i) Required area

Given y² = x _________ (1)
y = x² __________ (2)
Point of intersection is given by solving (1) & (2)
ie., √x = x² ⇒ x = 0, 1
When x = 0, y = 0
x = 1, y = 1
∴ Points of intersections are (0, 0) and (1, 1)

Question 29.
Solve Linear Programming Problem (LPP) graphically.
Maximize : z = 3x + 2y
Subject to constraints: x + 2y ≤ 10 (6)
3x + y ≤ 15
x, y ≥ 0
Answer:
Corresponding equation are
x + 2y = 10 ________ (1)

x	0	10
y	5	0

3x + 2y = 15 _________ (2)

x	0	5
y	15	0

Solving (1) and (2)
The point of intersection is (4, 3)

∴ z has maximum when x = 4 and y = 3
Maximum value = 18

Teachers recommend solving Kerala Syllabus Plus Two Botany Previous Year Question Papers and Answers Pdf Board Model Paper 2022 to improve time management during exams.

Kerala Plus Two Botany Board Model Paper Model 2022 with Answers

Part – I

A. Answer any 3 questions from 1 to 4. Each carries 1 score. (3 × 1 = 3)

Question 1.
Observe the relationship of the first pair and fill in the blank.
Restriction enzyme: Cutting of DNA
_______ : Joining of DNA fragments.
Answer:
DNA Ligase

Question 2.
Choose the correct answer and fill up the blank. The single cotyledon of monocotyledons are called
(plumule, radicle, scutellum, coleorrhiza)
Answer:
Scutellum

Question 3.
Name the association between fungi and roots of higher plants.
Answer:
Mycorrhiza

Question 4.
Which among the following equation is related with net primary productivity?
(a) GPP + R = NPP
(b) NPP-R = GPP
(c) GPP + NPP = R
(d) GPP-R = NPP
Answer:
GPP – R = NPP

B. Answer all questions from 5 to 6. Each carries 1 score. (2 × 1 = 2)

Question 5.
Fill in the blank.
Animals that have their DNA altered and express a foreign gene are known as ____.
Answer:
Transgenic animals

Question 6.
Choose the correct answer.
The protective, outer thick wall of the fruits in angiosperms. (Perisperm, Endosperm, pericarp, Integument)
Answer:
Pericarp

Part – II

Answer any two questions from 7 to 9. Each carries 2 scores. (2 × 2 = 4)

Question 7.
Observe the given diagram showing the process of maturation of insulin.

(a) Explain the structure of Proinsulin.
(b) What are the change that occur in Proinsulin when it becomes mature insulin?
Answer:
(a) Proinsulin composed of polypeptide chains A and B are connected with an extra stretch-called the C peptide.
(b) C peptide is removed during maturation.

Question 8.
Match the following:

Asexual Reproductive Structures	Organisms
(a) Zoospores	Hydra
(b) Buds	Penicillium
(c) Conidia	Sponges
(d) Gemmules	Chlamydomonas

Answer:

Asexual Reproductive Structures	Organisms
(a) Zoospores	Chlamydomonas
(b) Buds	Hydra
(c) Conidia	Penicillium
(d) Gemmules	Sponges

Question 9.
MOET is an advanced animal breeding technique for herd improvement. Expand MOET. Which hormone is used for superovulation?
Answer:
Multiple Ovulation Embryo Transfer Technology. FSH (Follicle stimulating hormone)

B. Answer any two questions from 10 to 13. Each carries 2 scores. (2 × 2 = 4)

Question 10.
Distinguish euryhaline and stenohaline organisms.
Answer:
Euryhaline organisms :- They can tolerate wide range of salinity.
Stenohaline organisms :- They can tolerate narrow range of salinity.

Question 11.
In many plants more than one embryo is seen in one seed. Name the phenomenon with an example,
Answer:
Polyembryony : Example Seeds of orange or citrus

Question 12.
Write any two points that are important for successful poultry farm management.
Answer:

• Selection of good breeds
• Provide proper food & water

Question 13.
Given below are the different stages of hydrarch succession. Forest, marshmeadow, phytoplankton, Reed swamp, submerged free floating stage, scrub, submerged plant stage. Write down the pioneer species and climax community. Arrange the stages in correct sequential order.
Answer:
Pioneer species – Phytoplankton.
Climax community – Forest.
Phytoplankton → Submerged plant stage → Submerged freefloating plant stage → Reed → swamp → Marsh- Meadow → Scrub → Forest stage.

Part – III

Answer any three questions from 14to 17. Each carries 2 scores. (3 × 3 = 9)

Question 14.
The equation of population density is given below.N_t+1 = N_t + [(BB + I) – (D + E)]
(a) What does D and E stand for?
(b) Write two factors which increase the population density.
Answer:
(a) D – Mortality
E – Emigration

(b) Natality & Immigration

Question 15.
Increase in the level of green house gases lead to overheating of earths atmosphere.
(a) Name the phenomenon.
(b) State any two consequences of this phenomenon.
Answer:
(a) Global warming
(b) Affects weather & climate ( El Nino effect), Melting of polar ice caps & Himalayan snow caps that leads to rise in sea level.

Question 16.
Observe the given ecological pyramid in an aquatic ecosystem.

Identify the type of pyramid and reason out why the pyramid is inverted.
Answer:
Pyramid of biomass Here biomass of primary producers (phytoplanktons) is much less than primary consumers (zooplanktons)

Question 17.
Bt-cotton is an example of genetically engineered plant. Name the gene responsible for Bt-toxin production. Explain how Bt-toxin kill the insect.
Answer:
Cry gene
When an insect ingest the inactive insecticidal protein crystals, it is converted into an active toxin due to the alkaline pH of the gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually leads to death of the insect.

B. Answer the following question. Carries 3 scores. (1 × 3 = 3)

Question 18.
Vehicular emission is the major cause of air pollution.
(a) Write four steps taken by Delhi Government for reducing vehicular pollution.
(b) What is CNG?
Answer:
(a)

1. Use of low-sulphur petrol & diesel,
2. Use of catalytic converter
3. Use of CNG & Use of unleaded petrol etc.

(b) Compressed Natural Gas

Part – IV

Answer any one question from 19 to 20. Carries 5 scores. (1 × 5 = 5)

Question 19.
The given diagram depicts the separation of DNA fragments.

(a) Identify the technique.
(b) How can you visualize the separated DNA fragments in this technique?
(c) What is elution?
Answer:
(a) Gel electrophoresis
(b) Separated DNA fragments can be visualised after staining with Ethidium bromide followed by exposure to UV light.
(c) Separated bands of DNA are cut out from the agarose gel and extracted from gel piece.

Question 20.
Given below is the diagram of a mature embryosac in angiosperm.

Label the parts a,b, c and d.
What is double fertilization? Mention the ploidy of zygote and PEN.
Answer:
a – Synergids
b – Secondary nucleus or Polar nuclei
c – Antipodals
d – Filiform apparatus

Two types of fusion like syngamy and triple fusion are together known as Double fertilization Zygote – Diploid (2n), PEN – Triploid (3n)