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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 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.

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)

Teachers recommend solving Kerala Syllabus Plus Two Botany Previous Year Question Papers and Answers Pdf Board Model Paper 2023 to improve time management during exams.

Kerala Plus Two Botany Board Model Paper 2023 with Answers

I. Answer any 3 questions from 1 to 5. Each carries 1 score.

Question 1.
Choose the correct answer. The innermost wall layer of microsporangium that give nutrition to pollen grain is called _____.
(a) Epidermis
(b) Endothecium
(c) Tapetum
(d) Middle layer
Answer:
Tetum

Question 2.
Name the first transgenic cow that produces human protein enriched milk.
Answer:
Rosie

Question 3.
Fill in the blank:
The natural interconnection of different food chains is called ______ .
Answer:
Food web

Question 4.
Choose the correct answer.
Which of these enzymes removes nucleotides from the ends of a DNA?
(a) Exonuclease
(b) Endonuclease
(c) DNA Ligase
(d) DNA Polymerase
Answer:
Exonuclease

Question 5.
Fill in the blank:
The population interaction in which one species is harmed whereas the other is unaffected is known as
Answer:
Amensalism

II. Answer any 9 questions from 6 to 16. Each carries 2 scores.

Question 6.
Write two roles of Predators in ecosystem.
Answer:
1. They plays an important role in biological control and reduce the size ofprey populations.
2. Predators maintain species diversity in nature.

Question 7.
(a) What is emasculation?
Answer:
(a) Removal of anthers from the flower bud before dehiscence.
(b) It frevert self pollination

Question 8.
Bt-toxin does not kill Bacillus, but it kill insects. Write the reason.
Answer:
The Bt toxin is producéd by the bacteria as inactive protoxin. Due to alkaline PH of insect gut, insecticidal protein crystals are Solubilised and convert inactive protoxin into active toxin. Active Bt toxin binds to the midgut epithelial cells and make pores, then swelling, it causes lysis and leads to the death of insect.

Question 9.
(a) Write the name of the ecological pyramid given in the figure.
(b) This pyramid is always upright. Why?

Answer:
(a) Pyramid of energy.
(b) when energy flows from one trop hic level to the next some energy is lost as heat at each step
hence energy deceases in uccessive trophic levels.

Question 10.
How can we visualize DNA fragments in G el electrophoresis?
Answer:
Separated DNA fragmentcan bevisualised as orange coloured band under UV light exposure after staining with Ethydium Bromide.

Question 11.
(a) Expand ELISA.
(b) What is the principle of this process?
Answer:
(a) Enzyme linked immuno-sorbent assay
(b)Antigen-antibody interaction

Question 12.
Observe the figure of Monocot seed and label the parts marked as A, B, C & D

Answer:
A—Endosperm
B — Scutellum
C — Plumule
D — Radicle

Question 13.
Write the differnce between Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).
Answer:
Gross primary productivity : The rate of production of total organic matter during photosynthesis.
Net primary productivity : Gross primary productivity minus respiration losses. GPP — R = NPP

Question 14.
N_t+1 = N_t + [(B + 1) – (D + E)] is the equation to find the population density at a given time. What does B, I, D and E stands for?
Answer:
B — Natality I Birth rate
I — Immigration
D — Mortality I Death rate
E — Emigration

Question 15.
(a) Which’enzymes are known as ‘molecular scissors’?
(b) What is the use of thse enzymes in Genetic engineering?
Answer:
(a) Restriction enzymes
(b) They are used to cut at specific nucleotide sequence of DNA

Question 16.
Match the following:

A	B
Fragmentation	Water soluble inorganic nutrients go down into the soil
Leaching	Breakdown of detritus into smaller particles
Catabolism	Formation of inorganic nutrients from humus
Mineralisation	Bacterial and fungal enzymes degrade detritus into simpler inorganic substances

Answer:

A	B
Fragmentation	Breakdown of detritus into smaller particles
Leaching	Water soluble inorganic nutrients go down into the soil
Catabolism	Bacterial and fungal enzymes degrade detritus into simpler inorganic substances.
Mineralisation	Formation of inorganic nutrients from humus

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

Question 17.
(a) Observe the figure and write the name of two population growth curves marked as A & B.
(b) What does K and r stands for?

Answer:
(a) A— Exponential growth
B — Logistic growth.

(b) K — Carrng capacity
r — Intrinsic rate of natural increase

Question 18.
Write three adaptations of insect pollinated fldvers.
Answer:
Pollen grains are sticky. Flowers are large, brightly colou red. They have scent and rich in nectar. Small flowers present in cluster to make them conspicuous.

Question 19.
(a) Writethe name of the process given inthefigure.
(b) What is the purpose of this process in rDNA technology?
(c) What is the name of thermostabte DNA polymerase enzyme used in this technique?

Answer:
(a) Polymerase Chain Reaction
(b) Used for Amplification of DNA
(c) Taq Polymerase

Question 20.
Write three uses of Trangenic animals.
Answer:

1. Used to study the normal physiology and development.
2. Used to understand the role of a gene in the development of a disease.
3. Transgenic animals (mice) are used in testing the safety of vaccines and chemicals.

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

Kerala Plus Two Botany Previous Year Question Paper March 2020

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

Question 1.
Which one of the following has the largest population in a food chain?
a) Products
b) Primary consumers
c) Secondary consumers
d) Tertiary consumers
Answer:
Producers or Primary consumers

Question 2.
Which among the fallowing is a selectable marker in pBR-322?
a) “Ori”
b) Hind III
c) amp^R
d) rop
Answer:
amp^R

Question 3.
____ is a better yielding semi dwarf variety of rice developed in India.
a) Sonalika
b) Kalyan Sona
c) IR-8
d) Jaya/iBco)
3. Jaya

Question 4.
What is a Clone?
Answer:
Morphologically and genetically similar offsprings are produced through asexual reproduction called as Clone

Question 5.
Fill up the blank suitably.
Mortality : No. of deaths in the population during a given period.
_________ : No. of births in the population during a given period.
Answer:
Natality or birth rate

II. Answer any9 questions from 6-16. Each carries 2 scores. (9 × 2 = 18)

Question 6.
What are the main consequences of Global warming?
Answer:
1. Melting of polar ice caps and Himalayan snow caps.
2. Rise in sea level and submerge many coastal areas.

Question 7.
What is the difference between hydrarch succession and xerarch succession.
Answer:
Xerarch succession : Succession that occurs in desert or xeric condition. The poineer community formed is Lichen.
Hydrarch : It is the ecological succession occur in the newly formed pond or lake . The poineer community formed is phytoplankton

Question 8.
Suggest any two methods for the disposal of solid wastes.
8. Two methods for the disposal of solid wastes
1. Sanitary landfills
2. Incineration

Question 9.
Match the columns A and B.

a) Rhizome	i) Agave
b) Bulbil	ii) Water hyacinth
c) Offset	iii) Ginger
d) Leaf buds	iv) BrvoDhvllum

Answer:

a) Rhizome	Ginger
b) Bulbil	Agave
c) Offset	Water hyacinth
d) Leaf buds	Bulbil

Question 10.
Identify the types of population growth noted in the graph as ‘A’ and ‘B’.

Answer:
a) J-shaped curve or Exponential growth
b) Sigmoid curve or Logistic growth

Question 11.
How does the inactive protoxin o£ Bacillus thurinaiensis gets converted into active toxin when an insect ingest it?
Answer:
Bt toxin protein exist as inactive protoxins but it is converted into an active form in the presence of the alkaline pH of insect gut.

Question 12.
Given below is a data showing number of individuals and dry weight of different trophic levels inva grassland ecosystem. Construct,
a) Pyramid of number
b) Pyramid of biomass
Answer:

Trophic Level	Number of individuals	Dry weight(Kg nr<sup>-2</sup>)
Primary Producer	5,842,000	809
Primary Consumer	7,08,000	37
Secondary Consumer	3,54,000	11
Tertiary Consumer	3	1.5

Answer:

Question 13.
How can you differentiate true fruits from false fruits?
Answer:
True fruits : Fruits develops only from the ovary eg. Mango
Fuse fruits : In this thalamus develops into fruit. eg. apple, strawberry, cashew, etc

Question 14.
What is biofortification? Write any two objectives of biofortification.
Answer:
biofortification
It is the breeding of crops with higher levels of vitamins minerals,proteins and healthierfats Objectives of improving nutritional quality
(i) Protein content and quality;
(ii) Oil content and quality

Question 15.
How can we make a host cell competent to receive a foreign gene or DNA?
Answer:
This is done by treating them with calcium ions and incubating the cells and recombinant DNA on ice, followed by placing them at 42°C Then putting them back on ice. This helps the bacteria to take up the recombinant DNA.

Question 16.
Early diagnosis is essential for the effective treatment of a disease. Write any two molecular diagnostic methods.
Answer:
Early diagnosis of disease is possible by:

1. Polymerase Chain Reaction (PCR) and
2.  Enzyme Linked Immuno-sorbent Assay (ELISA)

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

Question 17.
Define the following terms:
a) Autogamy
b) Geitonogamy
c) Xenogamy
Answer:
(a) Auto gamy: It is the transfer of pollen grains from the anther to the stigma of the same flower.

(b) Geitonocjamy: Transfer of palIen grains from the anther to the stigma of anotherfiower of the same plant.

(C) Xen oaamy : Transfer of pollen grains from anther to the stigma of a different plant. This type of pollination occurs betwecn genetically different species.

Question 18.
Ozone depletion in stratosphere is a serious ; environmental issue.
a) What is good ozone?
b) How do CFCs degrade ozone molecule is stratosphere?
Answer:
a) Good’ ozone- This ozone is found in the upper part of the atmosphere (stratosphere), and it acts as a shield and absorbing ultraviolet radiation from the sun.

b) The CFC produced in the upper part of the atmosphere is degraded by the action of UV. If splits and release active dorme. This active dorme destroys ozone and makes holes.

Question 19.
The following are the. diagrams of two important processes used in rDNA technology.
a) Identify A and B
b) Write the uses of A and B in rDNA technology.

Answer:
a) A – Polymerase chain reaction
B – Gel electrophoresis

b) A-Polymerase Chain Reaction is helpful to produce multiple copies (1 billion conies) òf the
gene of interest.

B- Gel electrophoresis
The cut fragments of DNA are separated by this technique.

Question 20.
Given below are examplesfor some ecological! population interactions. Place them under sUitable columns provided below.
a) Abingdon tortoise and goat.
b) Cuscuta and host tree.
c) Fig tree and wasp.
d) Algae and fungi in Lichens
e) Belanus and Chathabalus bermacles
f) Lice on humans.

Competition	Parasitism	Mutualism

Answer:

Competition	Parasitism	Mutualism
Abingdon tortoise and goat.	Cuscuta and host tree.	Fig tree and wasp.
Belanus and  Chathamalus bernacles.	Lice on humans.	Algae and fungi in Lichens.

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

Kerala Plus Two Botany Previous Year Question Paper March 2021

Answer the following questions from 1 to 31 upto a maximum score of 30.
I. Each questions from 1 to 7 carries 1 score.

Question 1.
Choose the correct answer
The plant which flowers only once in their life time:
(a) Mango
(c) Coconut
(b) Bamboo
(d) Jackfruit
Answer:
Bamboo

Fill in the blank:
Question 2.
During sexual reproduction, the process of syngamy results in the formation of a diploid cell called _____
Answer:
Zygote

Question 3.
Name the process of breeding crops for improved food quality.
Answer:
Biofortification

Question 4.
Choose the correctly mathced pair from the following:
(a) Gel electrophoresis – Used to produce biological products in large quantity.
(b) Bioreactors – Separation of DNA fragments.
(C) PCR – Amplification of gene
Answer:
PCR-Amplification of gene

Question 5.
Observe the relationship between the first two terms and fill in the blank.
Mycorrhizae : Mutualism
Lice on humans : _______
Answer:
Parasitism

Question 6.
Fill in the blank.
The portion of embryonal axis above the level of cotyledons in dicot embryo is known as ______ .
Answer:
Epicotyl

Question 7.
Which of the following is a curative method of ADA deficiency?
(a) ELISA
(b) Autoradiography
(c) PCR
(d) Bonemarrow transplantation
Answer:
Bone marrow transplantation

II. Each question from 8 to 26 carries 2 scores each. (19 × 2 = 38)

Question 8.
Differentiate monecious and dioecious plants with an example for both.
Answer:

Monecious plants	Dioecious plants
Male and female	Male and female
flowers present on the same plant.	flowers present on different plants.
Example – Coconut	Example – Papaya

Question 9.
EcoRI is an example of Restriction enzyme. List out the criteria for naming a Restriction enzyme.
Answer:
First capital letter represents the genus name of the bacteria(Ecoli). Second two letters represent the species name of the bacteria ‘R’ represents the strain of the bacteria. ‘I’ the roman number represents the order of isolation of enzyme from the bacteria.

Question 10.
Expand MOET. Write its significance.
Answer:
MOET – Multiple Ovulation Embryo Transfer Technology.
Significance
To improve successful production of hybrids.
Overcoming the problems of normal mating.

Question 11.
Different stages of development in a dicot embryo are given below. Arrange them in the correct sequential order. Heart shaped embryo, globular embryo, mature embryo, pro-embryo
Answer:
Pro-embryo → Globular embryo → Heart shaped embryo → Mature embryo

Question 12.
Observe the diagrams (a) and (b) given below. Identify the type of gametes. Explain the feature of gametes seen in diagram ‘b’.
Answer:

(a) Isogàmetes or Homogametes
(b) Heterogametes
Features of heterogametes
Morphologically distinct gametes i.e., male gamete is the antherozoid or sperm and Female gamete is egg or ovum.

Question 13.
The females of placental mammals exhibit cyclical changes in the activities of ovaries and accessory ducts during the reproductive phase. Specify two types of cyclical changes present in these organisms with appropriate examples.
Answer:

• Oestrus cycle
  1.  It is shown by Non-primate mammals eg. Cows, Tiger etc.
• Menstrual cycle
  1. It is shown by Primates eg. Man Monkey etc.

Question 14.
Write any two critical research areas of bio technology.
Answer:
(i) Providing the best catalyst in the form of improved organism usually a microbe or pure enzyme.
(ii) Creating optimal conditions through engineering for a catalyst to act.

Question 15.
What is biomagnification? Name any one chemical that causes biomagnification.
Answer:
Biomagnification It refers to increase in concentration of the toxicant at successive trophic level. Eg. DDT, Mercury etc.

Question 16.
List any four reasons for deforestation.
വനനശീകരണത്തിനുള്ള നാലു കാരണങ്ങൾ എഴുതുക.
Answer:

1. Conversion of forest to agricultural land so as to feed the growing human population.
2. Trees are axed for timber.
3. Trees are axed for firewood.
4. Slash and burn agriculture or Jhum cultivation.

Question 17.
Fill in the blanks with appropriate symbols (+, -, 0) or name of interaction.

Species A	Species B	Name of population interaction
(a)  _____	(b)  ____	Mutualism
+	(c) ____	Parasitism
+	0	(d)  _____

Answer:
(a) +
(b) +
(c) –
(d) Commensalism

Question 18.
What is the structural difference between proinsulin and mature functional insulin?
Answer:
Proinsulin

• Consists of three polypeptide chains: chain A and chain B and chain C.
• Non functional Mature Insulin
• Consists of two polypeptide chains: chain A and chain B linked together by disulphide bridges.

Question 19.
The diagram given below is E. coli cloning vector pBR322.

Identify the selectable markers present in it. Explain the significance of ‘ori’.
Answer:
amp^R and tet^R ori – Origin of replication or DNA sequence from where replication starts.

Question 20.
State the important steps in decomposition.
Answer:
Fragmentation, leaching, catabolism, Humification and mineralization.

Question 21.
Write four peculiarities of insect pollianted flowers.
Answer:

• Flowers are Large Colourful
• Fragrant With nectar
• Pollen grains sticky
• Small flowers are clustered into an inflorescence

Question 22.
Banana is a parthenocarpic fruit. What are partheno carpic fruits? How can be parthenocarpy induced?
Answer:
Parthenocarpic fruit is fruit formed without fertilization, they are Seedless fruits. Parthenocarpy can be induced by application of growth hormones.

Question 23.
Self incompatibility is an outbreeding mecahnism seen in plants. Explain self incompatibility.
Answer:
Self incompatibility means the inhibition of pollen germination on stigmatic surface of the same flower. It is the Genetic mechanism to prevent self pollination.

Question 24.
Based on the nature of the habitat, succession of plants may be hydrarch or xerarch.
(a) Which is the pioneer community in xerarch succession?
(b) Explain hydrarch succession.
Answer:
(a) Lichen
(b) Hydrarch succession is the transformation of hydric area into a mesic (medium water condition) area.

Question 25.
Organisms belonging to different trophic levels are listed below. Arrange the organisms under appropri- ate headings in the table provided. Man, phytoplankton, Fish, Zooplankton

First Trophic level	Second Trophic level	Third Trophic level	Fourth Trophic level

Answer:

1st trophic level	2nd trophic level	3rd trophic level	4th trophic level
Phytoplankton	Zooplankton	Fish	Man

Question 26.
Catalytic converters are fitted in automobiles for reducing emission of poisonous gases.
(a) Name the catalysts used in catalytic converters
(b) What happens to the automobile exhaust when it passes through catalytic converters?
Answer:
(a) Platinum, Palladium and Rhodium.
(b) When exhaust passes through the catalytic converter
1. Unburnt hydrocarbons are converted into carbon dioxide and water.
2. Carbon monoxide is converted into Carbon dioxide.
3. Nitric oxide is converted into Nitrogen gas.

III. Each questions 27 to 31 carries 3 scores. (5 × 3 = 15)

Question 27.
The four basic processes that affect the population density are given below. Natality, Mortality, Emigration, Immigration. Which of the two processes contribute to a decrease in population density and explain them.
Answer:
Mortality and Emigration

1. Mortality – It is the number of deaths in the population during a given period.
2. Emigration – It is is the number of individuals of the population who left the habitat.

Question 28.
Bt cotton is a transgenic plant with insecticidal protein named Bt toxin.
(a) Name the specific gene which produce Bt toxin.
(b) Name the organism from which this gene is isolated.
(c) Explain the mechanism behind the insect resistance in Bt cotton.
Answer:
(a) Cry gene

(b) Bacillus thuringiensis

(c) Cry gene produce insecticidal protein crystals that exists as inactive protoxins, Once an insect ingest the protoxin it is converted into an active toxin due to the alkaline pH of the gut, The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually cause death of the insect.

Question 29.
A typical microsporangium is surrounded by four wall layers. Name the wall layers and state the function. of the innermost wall layer.
Answer:
Epidermis, Endothecium, middle layers & Tapetum Tapetum-Nourishes the developing pollen grains.

Question 30.
Different steps in rDNA technology are given below. Arrange them in correct sequential order.
(a) Ligation of DNA frament into a vector
(b) Culturing of host cells in a medium at large scale
(c) Isolation of DNA
(d) Transferring the rDNA into the host
(e) Fragmentation of DNA
(f) Isolation of desired DNA fragment
Answer:
(c) Isolation of DNA
(e) Fragmentation of DNA
(f) Isolation of desired DNA fragments
(a) Ligation of DNA fragment into a vector
(d) Transferring the rDNA into the host.
(b) Culturing of host cells in a medium at large scale

Question 31.
Dairying is the management of animals for milk and its products for human consumption. List out various management measures that should be undertaken for a successful management of dairy farm.
Answer:

1. Selection of good breeds having high yielding potential and resistance to diseases.
2. Provide adequate water.
3. Maintained disease free condition.
4. Feeding should be in a scientific manner (in the quality and quantity of fodder).
5. Regular visit of a veterinary doctor.

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

Kerala Plus Two Botany Previous Year Question Paper March 2022

Part – I

A. Answer any 3 questions from 1 to 4. Each carries 1 score.
Choose the correct answer:

Question 1.
Innermost wall layer of Micro-sporangium is
(a) Middle layer
(b) Epidermis
(c) Tapetum
(d) Endothecium
Answer:
Tapetum

Question 2.
Fill in the blank.
The first restriction endonuclease enzyme isolated is ______ .
Answer:
Hindu

Question 3.
In polar seas, seals have a thick layer of fat. Name the fat layer seen below their skin.
Answer:
Blubber

Question 4.
Choose the correct answer:
The green plants in the ecosystem are the chief
(a) Decomposers
(c) Producers
(b) Consumers
(d) Herbivores
Answer:
Producers

B. Answer all questions from 5 to 6. Each carries 1 score.

Question 5.
Fill in the blank.
The time period from birth to the natural death of an organism is known as _____.
Answer:
Life span

Question 6.
Expand ELISA
Answer:
Enzyme Linked Immuno-sorbent Assay

Part-II

A. Answer any two questions from 7 to 9. Each carries 2 scores.

Question 7.
What are the four main objectives of Bio-fortification?
Answer:

1. Increased protein content and quality.
2. Increased oil content and quality
3. Increased vitamin content
4. Increased micronutrient and mineral content

Question 8.
Mention the functions of GEAC (Genetic Engineering Approval Committee)
Answer:
They take decisions regarding the validity of GM research and safety of introducing GM-organisms for public services.

Question 9.
Fill in the blanks with suitable terms:

Asexual Reproduction
Asexual Reproductive Structure	Examples
Zoospores	(a)
(b)	Penicillium
(c)	Hydra
Gemmules	(d)

Answer:

Asexual Reproductive Structures	Examples
Zoospoores	Chiamydomonas
Conidia	Penicillium
Buds	Hydra
Gemmules	Sponges

B. Answer any two questions from 10 to 13. Each carries 2 scores.

Question 10.
Comment on Self-incompatibility.
Answer:
Pollen from the same flower does not germinate in the stigma. This genetic mechanism helps to prevent self polination. Pollen from the same flower does not germinate in the stigma. This genetic mechanism helps to prevent self polination.

Question 11.
Introduction of semi-dwarf varieties increased production of rice and wheat. Complete the table with the names of any two semi-dwarf varieties of wheat and rice.

Rice	Wheat
–	–
–	–

Answer:
(i) Wheat – Sonalika and Kalyan Sona.
(ii) Rice – Jaya and Ratna

Question 12.
Fill in the blanks.

Population Interaction
Species – A	Species – B	Name of Interaction
(a)    _____	(b)	Competition
(c) _____	_____	Predation
_____	0	(d) ______

Answer:

Question 13.
Name the type of succession that occurs in wet area. What is the climax community in that succession?
Answer:
Hydrarch
Climax community – Forest

Part -III

A. Answer any three questions from 14 to 17. Each carries 3 scores.

Question 14.
Explain the method of genetically engineered human insulin production by ‘Eli Lilly’ in 1983.
Answer:
Eli Lilly Company prepared DNA sequences corresponding to A and B chain of insulin. A and B Chain’DNA were introduced into plasmids of E. coil to produce the Apoly peptide and B poly peptide chains, Chain A and B were extracted and combined by creating disulphide bonds.

Question 15.
List out some of the adaptations seen in desert plants.
Answer:

• They have a thick cuticle
• Leaf reduced to spine
• Sunken stomata to minimize water loss through transpiration.

Question 16.
What are the consequences of deforestation?
Answer:

• Increased carbon dioxide concentration
• Loss of Biodiversity.
• Disturbs hydrologic Cycle.
• Soil erosion.

Question 17.
The figure shows the pyramid of energy. Pyramid of energy is always upright. Why?

Answer:
When energy flows from one trophic level to the next level, some energy is lost as heat at each step a 10%.

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

Question 18.
CNG is widely used in vehicles in Delhi.
(a) Expand CNG.
(b) What are the advantages of using CNG than Diesel?
Answer:
(a) Compressed Natural Gas
(b) It burns efficiently

• It is cheaper than petrol or diesel
• It cannot be siphoned off by thieves.
• It cannot be adultered like petrol or diesel.

Part – IV

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

Question 19.
Pollination is an important mechanism found in flowering plants.
(a) Define pollination
(b) Write short note on:
(1) Autogamy
(2) Geitonogamy
(3) Xenogamy
Answer:
(a) Transfer of pollen grains from the anther to the stigma of a pistil is termed as pollination.
(b)

1. Autogamy – Transfer of pollen grain from the anther to the stigma of the same flower.
2. Geitonogamy – The transfer of pollen grains from the anther to the stigma of different flowers of the same plant
3. Xenogamy – The transfer of pollen grains from the anther to the stigma of different plants of the same species.

Question 20.
Restriction enzymes are widely used in Biotechnolgoy.
(a) Name two kinds of restriction enzymes.
(b) Explain the naming procedure of restriction enzymes with one example.
Answer:
(a) Exonucleases & Endonucleases

(b) Eg:EcoRI – derivedfromEsche,ichiacoliRYl3
E – Genus (Escherichia)
co – Species (coli)
R – strain of bacteria
I – order in which enzyme is Isolated.

Teachers recommend solving Kerala Syllabus Plus Two Botany Previous Year Question Papers and Answers Pdf March 2023 to improve time management during exams.

Kerala Plus Two Botany Previous Year Question Paper March 2023

Answer any 3 questions from 1 to 5. Each carries 1 score.

Choose the correct answer:
Question 1.
The residual persistent nucellus is called _____. (Pericarp, Tapetum, Scutellum, Perisperm)
Answer:
Perisperm

Question 2.
Fill in the blank :
The process of breaking down of detritus into smaller particles is called _____.
Answer:
Fragmentation

Question 3.
Name the technique used to separate the fragments of DNA.
Answer:
Gel electrophoresis

Question 4.
Fill in the blanks:
Interaction between sea anemone and clown fish is an example for _____ interaction.
Answer:
Commensalism

Question 5.
Name the ‘cry’ gene that control corn borer insects.
Answer:
crylAb

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

Question 6.
Differentiate between grazing food chain and detritus food chain.
Answer:
Grazing Food Chain
It starts with producer belongs to first trophic level

Detritus Food Chain
It starts with dead organic matter belongs to first trophic level.

Question 7.
List out any two devices in plants, which help to discourage self-pollination and to encourage cross pollination.
Answer:

1. Pollen release and stigma receptivity are not synchronized.
2. Anther and stigma are placed at different position.

Question 8.
How can we make bacterial cell competent for trans- formation with recombinant DNA?
Answer:
Bacterial cells are treated with divalent cation such as Ca²⁺ to increase cell permeability. Then these cells are treated with recombinant DNA on ice. The cells and rDNA in ice are allowed to heat at 42°C Then return back to ice bath. Repeat this process.

Question 9.
Observe the given figure of ovule and label the parts (a), (b), (c) and (d).

Answer:
(a) Funicle
(b) Micropyle
(c) Embryo sac / Female gametophyte
(d) Chalaza

Question 10.
(a) Which is the thermostable enzyme used in PCR?
Answer:
Taq polymerase

(b) Name the bacterium from which the enzyme isolated.
Answer:
Thermus aquaticus.

Question 11.
Define productivity. What are the factors affecting primary productivity?
Answer:
The rate of biomass production is called productivity. Factors affecting primary productivity.

1. Environmental factors.
2. Availability of nutrients.

Question 12.
Explain the mechanism of insect resistance in Bt. cotton.
Answer:
Due to cry gene expression, insecticidal protein crystals are produced. In the Alkaline pH of insect
gut convert inactive protoxin into active toxin. Active Bt toxin binds to the midgut epithelium and causes cell lysis leading to death of insect.

Question 13.
Different types of population interaction has been observed in population. Write the type of interaction observed among the following species:

Species A	Species B	Type of Interaction
Orchid ophrys	Bees	(a)  _______
Ticks	Dogs	(b) _______
Barnacles	Whale	(e) _______
Fungus	Photosynthetic Algae	(d) _______

Answer:
(a) Mutualism
(b) Parasitism
(c) Commensalism
(d) Mutualism

Question 14.
The figure given below shows pyramid of energy. Pyramid of energy is always upright, can never be inverted. Give reason.

Answer:
When energy flow from one trophic level to the next level some energy is lost as heat at each step as 10%.

Question 15.
What are Genetically Modified Organisms(GMO)? Mention any one merit of GM plants.
Answer:
An organism whose genetic material is altered is called Genetically Modified Organism.
Made crops tolerant to abiotic stress.

Question 16.
The following graph shows two types of population growth curves.
(a) Name the growth curves (a)& (b)
Answer:
(a) (a)- Exponential growth
(b) Logistic growth

(b) What does ‘K’ stand for?

Answer:
K – Carrying capacity

III. Answer any 3 questions from 17 to 20. Each carries 3 scores.

Question 17.
Explain how Eli Lilly company produced genetically engineered insulin.
Answer:
Eli Lilly Company isolated DNA sequences corresponding to A and B chain of insulin. A and B Chain DNA were introduced in plasmids of E.coli to produce the polypeptide A and B chains: Polypeptide chain A and B were extracted. and combined by creating disulphide bonds to form genetically engineered insulin.

Question 18.
Wind pollinated flowers have many peculiarities. Write down any three such peculiarities.
Answer:
Flowers with well exposed stamens. Large feathery stigma to trap air-borne pollen grains. Flowers contain single ovule in one ovary and numerous flowers packed into an inflorescence e.g. corn cob.

Question 19.
Given below is a schematic representation which shows the factors influencing population density:

(i) Label (a) and (b)
Answer:
(a) – Mortality
(b) – Emigration

(ii) Which are the factors that contribute to an in crease in population density?
Answer:
Natality and Immigration

(iii) Define Natality.
Answer:
Natality – The number of births during a given period.

Question 20.
EcoRI is isolated from Escheritia coli RY 13. Explain the naming procedure of EcoRI.
Answer:
(E) – First letter represent genus of the bacteria.
(co) – First two letters represent the species of the organism.
(R) – Represent strain of bacteria.
Roman number (I) – Order in which enzyme is isolated.