Category: Class 12

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 4 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 2.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is  ………….
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) Zero
(c)  \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}} \)
(d) infinite
Answer:
(b) Zero
Hint: The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 3.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is……………………..
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 4.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\overrightarrow{\mathrm{B}}\) is ……….


Answer:
(d)

Question 5.
In a series resonant RLC circuit, the voltage across 100Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 pF, then the voltage across L is……………….
(a) 600 V
(b) 4000 V
(c) 400 V
(d) IV
Answer:
(c) 400 V

Question 6.
During the propagation of electromagnetic waves in a medium:
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density

Question 7.
First diffraction minimum due to a single slit of width 1.0 x 10^-5 cm is at 30°. Then wavelength of light used is, …………………
(a) 400 Å
(b) 500 Å
(c) 600 Å
(d) 700 Å
Answer:
(b) 500 Å
Hint. For diffraction minima, d sin θ = nλ
\(\lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7}\)
λ = 500 Å

Question 8.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 9.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint: Kinetic energy of emitted electron depends on the frequency of incident radiation.

Question 10.
The ratio between the first three orbits of hydrogen atom is…………….
(a) 1:2:3
(b) 2:4:6
(c) 1:4:9
(d) 1:3:5
Answer:
(c) 1:4:9
Hint :
E_n = \(\frac{-13.6 \times z^{2}}{n^{2}}\)
n = 1; E₁ =- 13.6 eV/ atom
n = 2; E₂ = – 3.4 eV/ atom
n = 3; E₃ = -1.51 eV/atom                             ’
The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

Question 11.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint: Bohr theory could not explain the five structure of hydrogen spectrum.

Question 12.
If a half-wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
(a) 0° – 90°
(b) 90° – 180°
(c) 0° – 180°
(d) 0° – 360°
Answer:
(c) 0° – 180°

Question 13.
Diamond is very hard because ……………..
(a) it is covalent solid
(b) it has large cohesive energy
(c) high melting point
(d) insoluble in all solvents
Answer:
(b) it has large cohesive energy

Question 14.
The internationally accepted frequency deviation for the purpose of FM broadcasts.
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz

Question 15.
The blue print for making ultra durable synthetic material is mimicked from
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather
Answer:
(c) Parrot fish

Part – II

Answer any six questions in which Q. No 17 is compulsory.   [6 x 2 = 12]

Question 16.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines  is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 17.
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
Answer:
I = 32A, t= 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n =?

Question 18.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux φ_B
\(\phi_{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot\overrightarrow{\mathrm{A}}=\mathrm{B} \mathrm{A} \cos \theta=\mathrm{B} \perp \mathrm{A}\)

Question 19.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit.
It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 20.
State the laws of reflection.
Answer:

• The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
• The angle of incidence i is equal to the angle of reflection r. i = r

Question 21.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 22.
The radius of the 5th orbit of hydrogen atom is 13.25 A. Calculate the wavelength of the electron in the 5th orbit.
Answer:
2πr = nλ
2 x 3.14 x 13.25 Å = 5 x λ
.’. λ = 16.64 Å

Question 23.
Draw the output waveform of a full wave rectifier.
Answer:

Question 24.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Part – III

Answer any six questions in which Q.No. 28 is compulsory. [6 x 3 = 18]

Question 25.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.
This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Question 26.
What is electric power and electric energy?
Answer:
Electric power: It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
\({ P }=\frac { { W } }{ t } ={ VI }={ I }^{ 2 }{ R }=\frac { { V }^{ 2 } }{ { R } } \)
Electric energy: It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Vt = VIr joule = I²R? joule.

Question 27.
A bar magnet having a magnetic moment \(\overrightarrow{\mathrm{M}}\) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
Consider a bar magnet of magnetic moment \(\overrightarrow{\mathrm{M}}\). When a bar magnet first cut in two pieces along the axis, their magnetic moment is \(\frac{\overrightarrow{\mathrm{M}}}{2}\)

Question 28.
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
Answer:
L = 40 x 10^-3H; i = 0.1 sin (200 t – 40°); X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90° Therefore,
υ = V_m sin (ωt + 90°)
υ = 2.4 sin (200f – 40°+ 90°)
υ = 2.4 sin (200f +50°)volt

Question 29.
Writedown the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.

Question 30.
Two light sources have intensity of light as I_o. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I_o
The resultant intensity is, I = 4 I_o cos² (φ/2)
Resultant intensity when, ϕ = π/ 3, is I = 4I₀ cos² (π / 6) I
= 4I₀(√3 / 2)² =3I₀

Question 31.
Write the properties of cathode rays.
Answer:

• Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1
• It can be deflected by application of electric and magnetic fields.
• When the cathode rays are allowed to fall on matter, they produce heat.
• They affect the photographic plates and also produce fluorescence when they fall on   certain crystals and minerals.
• When the cathode rays fall on a material of high atomic weight, x-rays are produced.
• Cathode rays ionize the gas through which they pass.
• The speed of cathode rays is up \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the speed of light.

Question 32.
Distinguish between wireline and wireless communication.
Answer:

Wireline communication	Wireless communication
It is a point-to-point communication.	It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres.	It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected.	These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV.	Ex. mobile, radio or TV broadcasting and satellite communication.

Question 33.
What is the difference between Nano materials and Bulk materials?
Answer:

• The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
• When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
• For example, ZnO can be both in bulk and nano form.
• Though chemical composition is the; same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Part – IV

Answer all the questions.  [5 x 5 = 25]

Question 34.
(a) Calculate the electric field due to a dipole on its equatorial plane.
Answer:
Electric field due to an electric dipole at a point on the equatorial plane. Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure.

Since the point C is equi­distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \(\overrightarrow{\mathrm{E}}_{+}\) is along BC \(\overrightarrow{\mathrm{E}}_{-} \) and the direction of E is along CA. \(\overrightarrow{\mathrm{E}}_{+} \) and \(\overrightarrow{\mathrm{E}}_{-} \) and E are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right| \sin \theta \) and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| \sin \theta \) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of E₊ and E and its direction is \(\overrightarrow { { E } } _{ + }\quad and\quad \overrightarrow { { E } } _{ – }\) and its direction is along \(-\hat{p}\)

(b) How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bf and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ and to be compared are connected to the terminals M₁, N₁ and M₂, N₂, of the DPDT switch. The positive terminals of Bf ξ₁ and ξ₂ should be connected to the same end C.

The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁, is included in the secondary circuit and the balancing length l₁, is found by adjusting the jockey for zero deflection. Then the second cells ξ₂, is included in the circuit and the balancing length l₂, is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have
ξ₁ = Irl₁ ……………….. (1)
ξ₂ = Irl₂ ……………….. (2)
By dividing equation (1) by (2)
\(\frac{\xi_{1}}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Question 35.
(a) Discuss the working of cyclotron in detail.
Answer:
Cyclotron : Cyclotron is a device used accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle : When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction : The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { mv^{ 2 } }{ r } =qv{ B }\Rightarrow r=\frac { m }{ q{ B } } v\Rightarrow r\propto v\)
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the Very important condition in cyclotron operation is the resonance condition. It happens when the frequency { at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
\(f_{\text {osc }}=\frac{q \mathrm{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\text {osc }}}\)
The time Period of oscillation is \(\mathrm{T}=\frac{2 \pi m}{q \mathrm{B}}\)
The kinetic energy of the charged particle is \(\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{q^{2} \mathrm{B}^{2} r^{2}}{2 m}\)

Limitations of cyclotron

• the speed of the ion is limited
• electron cannot be accelerated
• uncharged particles cannot be accelerated

[OR]

(b) Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

(1) Induction stove : Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

(2) Eddy current brake : This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

(3) Eddy current testing : It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field. When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

(4) Electro magnetic damping : The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:

• Electromagnetic waves are produced by any accelerated charge.
• Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
• Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
• Electromagnetic waves travel with speed which is equal to the speed of light in vacuum
  or free space, \(c=\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =3\times 10^{ 8 }{ ms }^{ -1 }\)
• The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

[OR]

(b) Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
Experimental setup
Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

S₁ and S₂ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

Equation for path difference
Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the midpoint of the screen O is equidistant from S₁ and S₂. P is any point at a distance y  from O.

The waves from S₁ and S₂ meet at P either inphase or out-of-phase depending upon the path difference between the two waves. The path difference S between the light waves from S₁ and S₂ to the point P is, δ = S₂P and S₁P
A perpendicular is dropped from the point S₁, to the line S₂P at M to find the path difference more precisely.
δ = S₂P – MP = S₂M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S₂S₁ M are equal.
∠OCP = ∠S₂S₁ M = θ
In right angle triangle ΔS₁S₂M, the path difference, S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ ≈ tan θ ≈ θ From the right angle triangle ΔOCP, tan θ = y/D. The path difference, δ =dy/D

Question  37.
(a) Give the construction and working of photo emissive cell.
Answer:
Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

• It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
• The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
• A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

• When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer
• For a given cathode, the magnitude of
  the current depends on the intensity to incident radiation and
  the potential difference between anode and cathode.

[OR]

Question 37.
(b) Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O. This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then eE = eBv
\(v=\frac{E}{B}\) …………….. (1)

(ii) Determination of specific charge
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
\(e \mathrm{V}=\frac{1}{2} m v^{2} \Rightarrow \frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1) , we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ………..(2)
Substituting the value of E ,B and V, the specific charge vam be determined as
\(\frac{e}{m}=1.7 \times 10^{11} \mathrm{Ckg}^{-1}\)

(iii) Deflection of charge only due to uniform electric field
When the magnetic field is turned off, the deflection is only due to electric field. The
deflection in vertical direction is due to the electric force.
F_e = eE ………………….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion,
acceleration of the electron is
\(a_{e}=\frac{1}{m} \mathrm{F}_{e}\) ………………… (4)
Substituting equation (4) in equation (3),
\(a_{e}=\frac{1}{m} e E=\frac{e}{m} E\)
Let y be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be μ = 0 before entering the parallel electric plates. Let l be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
\(t=\frac{l}{v}\) …………(5)

Hence, the deflection y’ of cathod rays is( note : u = 0 and \(a_{e}=\frac{e}{m} \mathrm{E}\))

Therefore, the deflection y on the screen is
y α y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get

Substituting the values on RHS, the value of specific charge is calculated as
e/m = 1.7 x 10¹¹Ckg^-1

Question 38.
(a) What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED): LED is a p-n junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). It consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode.

When the p-n junction is forward biased, the conduction band electrons on n-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the. electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hv is emitted. For non- radiative recombination, energy is liberated in the form of heat.

The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.

[OR]

(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the Challenges and risk factors.

• ICT is widely used in increasing food productivity and farm management.
• It helps to optimize the use of water, seeds and fertilizers etc.
• Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
• Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining

• ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
• Information and communication technology provides audio-visual warning to the trapped underground miners.
• It helps to connect remote sites.

Students can Download Tamil Nadu 12th Maths Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Maths Model Question Paper 3 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If A is a 3 × 3 non-singular matrix such that AA^T = A^T A and B = A^-1 A^T, then BB^T = ________.
(a) A
(b) B
(c) I₃
(d) B^T
Answer:
(c) I₃

Question 2.
The rank of the matrix \(\left[\begin{array}{cc}
7 & -1 \\
2 & 1
\end{array}\right]\) is ________.
(a) 9
(b) 2
(c) 1
(d) 5
Answer:
(b) 2

Question 3.
The value of \(\sum_{i=1}^{13}\left(i^{n}+i^{n-1}\right)\) is ________.
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer:
(a) 1 + i

Question 4.
Which of the following is incorrect?
(d) Re(z) ≤ |z|
(b) Im (z) ≤ |z|
(c) \(z \bar{z}=|z|^{2}\)
(d) Re(z) ≥ |z|
Answer:
(d) Re(z) ≥ |z|

Question 5.
According to the rational root theorem, which number is not possible rational zero of 4x⁷ + 2x⁴ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d)5
Answer:
(c) \(\frac{4}{5}\)

Question 6.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to ______.
(a) tan² α
(b) o
(c) -1
(d) tan 2α
Answer:
(c) -1

Question 7.
The domain of the function defined by f(x) = sin^-1 \(\sqrt{x-1}\) is _______.
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Answer:
(a) [1, 2]

Question 8.
The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) is ________.
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac{1}{2}\) (a² + b²)
Answer:
(b) 2(a² + b²)

Question 9.
The directrix of the parabola x² = -4y is ________.
(a) x = 1
(b) x = 0
(c) y = 1
(d) y = 0
Answer:
(c) y = 1

Question 10.
If the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane x + 3y – αz + b = β, then (α, β) is ______
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Answer:
(b) (-6, 7)

Question 11.
If \(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}\) for non-coplanar vectors \(\vec{a}, \vec{b}, \vec{c}\) then ________.
(a) \(\vec{a}\) parallel to \(\vec{b}\)
(b) \(\vec{b}\) parallel to \(\vec{c}\)
(c) \(\vec{c}\) parallel to \(\vec{a}\)
(d) \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
Answer:
(c) \(\vec{c}\) parallel to \(\vec{a}\)

Question 12.
The maximum product of two positive numbers, when their sum of the squares is 200, is ________.
(a) 100
(b) \(25 \sqrt{7}\)
(c) 28
(d) \(24 \sqrt{14}\)
Answer:
(a) 100

Question 13.
If w(x,y, z) = x² (y – z) + y² (z – x) + z² (x – y), then \(\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z}\) is ________.
(a) xy + yz + zx
(b) x (y + z)
(c) y (z + x)
(d) 0
Answer:
(d) 0

Question 14.
If u (x, y) = x² + 3xy + y – 2019, then \(\left.\frac{\partial u}{\partial x}\right|_{(4,-5)}\) is equal to ______.
(a) -4
(b) -3
(c) -7
(d) 13
Answer:
(c) -7

Question 15.
The volume of solid of revolution of the region bounded by y² = x (a – x) about x-axis is ________.

Answer:
(d) \(\frac{\pi a^{3}}{6}\)

Question 16.
\(\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\) = _________.

Answer:
(c) \(\int_{0}^{2 a} f(x) d x\)

Question 17.
The differential equation of the family of curves y = Ae^x + Be^-x, where A and B are arbitrary constants is _______.

Answer:
(b) \(\frac{d^{2} y}{d x^{2}}-y=0\)

Question 18.
The differential equation corresponding to xy = c² where c is an arbitrary constant, is ______.
(a) xy”+ x = 0
(b) y” = 0
(c) xy’ + y = 0
(d) xy”- x = 0
Answer:
(c) xy’ + y = 0

Question 19.
If \(f(x)=\left\{\begin{array}{ll}
2 x, & 0 \leq x \leq a \\
0 & , \text { otherwise }
\end{array}\right.\)
is a probability density function of a random variable, then the value of a is _________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 20.
The proposition \(p \wedge(\neg p \vee q)\) is
(a) a tautology
(b) a contradiction
(c) logically equivalent to p ∧ q
(d) logically equivalent to p ∨ q
Answer:
(c) logically equivalent to p ∧ q

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Answer:
Let the two parts be x and (12 – x)
Given that x = \(\sqrt[3]{12-x}\)
Cubing on both side, x³ = 12 – x
x³ + x – 12 = 0

Question 22.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)\)
Answer:

Question 23.
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.
Answer:

Given radius = 5 cm and the circle is touching x axis
So centre will be (0, ± 5) and radius = 5
The equation of the circle with centre (0, ± 5) and radius 5 units is
(x – 0)² + (y ± 5)² = 5²
(i.e) x² + y² ± 10y + 25 – 25 = 0
(i.e) x² + y² ± 10y = 0

Question 24.
Find the length of the perpendicular from the origin to the plane.
\(\bar{r} \cdot(3 \vec{i}+4 \bar{j}+12 \vec{k})=26\).
Answer:
Taking the equation of the plane in cartesian form we get,
\((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\)
i.e. 3x + 4y + 12z – 26 = 0
The length of the perpendicular from (0, 0, 0) to the above plane is

Question 25.
Evaluate \(\lim _{x \rightarrow \pi / 2} \frac{\log (\sin x)}{(\pi-2 x)^{2}}\)
Answer:

Note that here l’ Hospitals rule, applied twice yields the result.

Question 26.
Evaluate \(\int_{-1}^{1} e^{-\lambda x}\left(1-x^{2}\right) d x\)
Answer:
Taking u = 1 – x² and v= e^-λx, and applying the Bernoulli’s formula, we get

Question 27.
Solve: \(\frac{d y}{d x}+2 y=e^{-x}\)
Answer:
Given that \(\frac{d y}{d x}+2 y=e^{-x}\)
This is a linear differential equation.
Here P = 2; Q = e
∫P dx = ∫2 dx = 2x
Thus, I.F = e^∫Pdx = e^2x
Hence the solution of (1) is \(y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c\)
That is, ye^2x = ∫e^-x e^2x dx + c (or) ye^2x = e^x + c (or) y = e^-x + ce^-2x is the required solution.

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for \((p \vee q) \vee \neg q\)
Answer:
truth table for \((p \vee q) \vee \neg q\)

Question 30.
Show that f(x, y) = \(\frac{x^{2}-y^{2}}{y^{2}+1}\) is continuous at every (x, y) ∈ R².
Answer:

Here, f satisfies all the three conditions of continuity at (a, b). Hence, f is continuous at every point of R² as (a, b) ∈ R².

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Form a polynomial equation with integer coefficients with \(\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}\) as a root.

Question 32.
Find the equation of the tangents from the point (2, -3) to the parabola y² = 4x

Question 33.
Find the vector and cartesian equations of the straight line passing through the points (-5, 2, 3) and (4,-3, 6).

Question 34.
Find the points on the curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y =1729.

Question 35.
Assuming log₁₀ e = 0.4343, find an approximate value of log₁₀ 1003.

Question 36.
Evaluate: \(\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cos ^{4} x d x\)

Question 37.
Solve the differential equation: \(\frac{d y}{d x}\) = e^x+y + x³ e^y

Question 38.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times, and X denote the number of heads.
(ii) A fair die is tossed 240 times, and X denote the number of times that four appeared.

Question 39.
Let

be any three Boolean matrices of the same type. Find (A ∨ B) ∧C

Question 40.
Sketch the graph of y = sin\(\left(\frac{1}{3} x\right)\) for 0 ≤ x < 6π.

Part – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself ? (Use Cramer’s rule to solve the problem).
[OR]
(b) Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)

Question 42.
42. (a) Let z₁, z₂, and z₃ be complex numbers such that |z₁| = |z₂| = |z₃| = r > 0 and z₁ +z₂ + z₃ ≠ 0.
Prove that \(\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=r\)
[OR]
(b) Find all cube roots of \(\sqrt{3}\) + i.

Question 43.
(a) Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and \(\sqrt{3}\) are two of its zeros.
[OR]
(b) Let W(x, y, z) = x² – xy + 3 sin z, x, y, z∈R. Find the linear approximation at (2, -1, 0)

Question 44.
(a) Prove p → (q → r) ≡ (p ∧ q) → r without using truth table.
[OR]
(b) Evaluate \(\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin x+\cos x} d x\)

Question 45.
(a) Prove that \(\text { (i) } \tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2} \text { (ii) } \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}\)
[OR]
(b) Find two positive numbers whose product is 100 and whose sum is minimum.

Question 46.
(a) If X is the random variable with distribution function F (x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
\frac{1}{2}\left(x^{2}+x\right) & 0 \leq x<1 \\
1, & x \geq 1
\end{array}\right.\)
then find (i) the probability density function f(x) (ii) P(0.3 ≤ X ≤ 0.6)
[OR]
(b) Find the vertex, focus, equation of directrix and length of the latus rectum of the following: y² – 4y – 8x + 12 = 0

Question 47.
(a) The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what population may be anticipated in 2020? [log_e \(\left(\frac{16}{3}\right)\) = 0.2070; e^-0.42 = 1.52]
[OR]
(b) Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point (3, 6, -2), (-1, -2, 6) and (6, 4, -2).

Students can Download Tamil Nadu 12th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 3 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. the correct plot for electrostatic potential due to this spherical shell is ………

Answer:
(b)

Question 2.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1Ω. The resistance of the wire will be 2Ω at …………….
(a) 1154 K
(b) 1100 K
(c) 1400K
(d) 1127 K
Answer:

Question 3.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to ………..
(a) n²R
(b) \(\frac{R}{n^{2}}\)
(c) \(\frac{R}{n}\)
(d) nR
Answer:
(a) n²R
Hint :
Resistance in parallel combination, R= r/n = r = Rn
Resistance in series combination , R = nr = n²R

Question 4.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is……………………….
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
Answer:
(b) 7 μT

Question 5.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when the energy is stored equally between the electric and magnetic fields, is……………………….
(a) \(\frac{Q}{2} \)
(b) \(\frac{Q}{\sqrt{3}}\)
(c) \(\frac{Q}{\sqrt{2}}\)
(d) Q
Answer:
(c) \(\frac{Q}{\sqrt{2}}\)

Question 6.
The direction of induced current during electro magnetic induction is given by …………………………………..
(a) Faraday’s law
(b) Lenz’s law
(c) Maxwell’s law
(d)  Ampere’s law
Answer:
(b) Lenz’s law

Question 7.
In an electromagnetic wave in free space the rms value of the electric field is 3 V m . The peak value of the magnetic field is………………………..
(a) 1.414 x 10^-8 T
(b) 1.0 x 10^-8 T
(c) 2.828 x 10^-8 T
(d) 2.0 x 10^-8 T
Answer:
(a) 1.414 x 10^-8 T

Question 8.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1m
(b) 5m
(c) 3m
(d) 6m
Answer:
(b) 5m
Hint:
Resolution limit sin θ =

Question 9.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is ……………..
(a) six
(b) four
(c) five
(d) three
Answer:
(a) six
Hint:
Number of images formed \(n=\frac{360}{\theta}-1=\frac{360}{60}-1=5\)

Question 10.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is
{a) 4125 Å
(b) 3750  Å
(c) 6000 Å
(d) 2062.5 Å
Answer:
(b) 3750  Å

Question 11.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is…..

Answer:
(c)

Question 12.
The speed of the particle, that can take discrete values is proportional to ……………
(a) n^-3/2
(b) n^-3
(c) n^1/2
(d) n
Answer:
(d) n
Hint:
\(P=m v=\frac{n h}{2 a} ; V \propto n\)

Question 13.
Doping semiconductor results in ………………………………
(a) The decrease in mobile charge carriers
(b) The change in chemical properties
(c) The change in the crystal structure
(d) The breaking of the covalent bond
Answer:
(c) The change in the crystal structure

Question 14.
The signal is affected by noise in a communication system …………………….
(a) At the transmitter
(b) At the modulator
(c) In the channel
(d) At the receiver
Answer:
(c) In the channel

Question 15.
The technology used for stopping the brain from processing pain is…………………………..
(a) Precision medicine
(b) Wireless brain sensor
(c) Virtual reality
(d) Radiology
Answer:
(c) Virtual reality

Part – II

Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
Define ‘Electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
\(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
The electric field is a vector quantity and its Sl unit is Newton per Coulomb (NC^-1).

Question 17.
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 18.
Compute the intensity of magnetisation of the bar magnet whose mass, magnetic moment and density are 200 g, 2 A m² and 8 g cm^-3, respectively.
Answer:

Question 19.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in ‘ which the electric field and the electric flux are changing with time.

Question 20.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 21.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to. 81.9 x 10¹⁵ (Given: mass of proton is 1836 times that of electron).
Answer:

Question 22.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Answer:
Alpha decay: The alpha decay process symbolically in the following way
Beta decay: β decay is represented by \(_{ z }^{ A }X\rightarrow _{ z,-1 }^{ A }Y+e^{ + }+v\)
Gamma decay: The gamma decay is given \(\mathrm{z} \mathrm{X}^{*} \rightarrow_{Z}^{\mathrm{A}} \mathrm{X} \)+ gamma(y)rays

Question 23.
Explain the need for a feedback circuit in a transistor oscillator.
Answer:
The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.

Question 24.
Give the factors that are responsible for transmission impairments.
Answer:

• Attenuation
• Distortion (Harmonic)
• Noise

Part – III

Answer any six questions. Question No. 32 is compulsory. [6 x 3 = 18]

Question 25.
A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 10^s NC^-1. How much work is required to rotate all the water molecules from θ = 0° to 90°?
Answer:
When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) – U(0°)
As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule
from θ = 0° to 90°,
For one water molecule, W = – pE cos 90° + pE cos 0° = pE
W = 6.3 x10^-30 x 3 x 10⁵ = 18.9 x 10^-25
For 10²² water molecules, the total work done is W_tot = 18.9x 10 ²⁵ x 10²² = 18.9 x 10 ³ J

Question 26.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (V_B – V_A ). If the points A and B lie on the same equipotential surface, work done is zero because V_B – V_A

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to an equipotential surface.

Question 27.
An electronics hobbyist is building a radio which requires 150 ft in her circuit, but she has only 220 Ω 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Answer:
Required effective resistance = 150 ft
Given resistors of resistance, R = 220 ft, R = 79 ft, R = 92 ft
Parallel combination of R₁ and R₂

Question 28.
What is magnetic permeability?
Answer:
Magnetic permeability: The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.

Question 29.
A circular metal of area 0.03 m² rotates in a uniform magnetic field of 0.4 T. The axis of rotation passes through the centre and perpendicular to its plane and is also parallel to the field. If the disc completes 20 revolutions in one second and the resistance of the disc is 4 Ω calculate the induced emf between the axis and the rim and induced current flowing in the disc.
Answer:
A = 0.03 m²; B = 0.4T; f = 20rps; R = 4Ω
Area covered in 1 sec = Area of the disc x frequency
= 0.03 x 20 = 0.6 m²

Question 30.
Give the characteristics of image formed by a plane mirror.
Answer:

• The image formed by a plane mirror is virtual, erect, and laterally inverted.
• The size of the image is equal to the size of the object.
• The image distance far behind the mirror is equal to the object distance in front of it.
• If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as,\(n=\left(\frac{360}{\theta}-1\right)\)

Question 31.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?
Answer:

Question 32.
Assuming V_CEsat = 0.2 V and β = 50, find the minimum base current (I_B) required to drive the transistor given in the figure to saturation.
Answer:

V_CEsat = 0.2 V and β = 50
V_CE = V_CC -I_C R_C
0.2= 3 – I_c(1 k)

Question 33.
Mention any two advantages and disadvantages of Robotics.
Answer:
(i) Advantages of Robotics:

• The robots are much cheaper than humans.
• Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
• Robots are more precise and error free in performing the task.

(ii) Disadvantages of Robotics:

• Robots have no sense of emotions or conscience.
• They lack empathy and hence create an emotionless workplace.
• If ultimately robots would do all the work, and the humans will just sit and monitor them* health hazards will increase rapidly.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Answer:
Expression for electrostatic potential energy of the dipole in a uniform electric field:
Considera dipole placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\) placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\) . A dipole experiences a torque when kept in an uniform electric field E.

This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole. The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

This work done is equal to the potential energy difference between the angular positions 0 and 0′. U(θ) – (Uθ’) = ΔU = -pE cos θ + pE cos θ’.

If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos 90° = 0. The potential energy stored in the system of dipole kept in the uniform electric field is given by.
U = -pE cos θ = -p-E  ……………. (3)

In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.

The potential energy is maximum when the dipole is aligned anti-parallel (θ = Jt) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

[OR]

(b) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O.
Consider an element of length dl of the wire at a distance l from point O and \(\vec{r} \) be the vector joining the element dl with the point P. Let θ be the angle between \(d \vec{l} \text { and } \vec{r}\). Then, the magnetic field at P due to the element is
\(d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} l}{4 \pi r^{2}} \sin \theta\)(unit vector perpendicular to \(d \vec{l} \text { and } \vec{r}\)

The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors \(d \vec{l} \text { and } \vec{r}\)
(let it be \(\hat{n}\)). The net magnetic field can be determined by integrating equation with proper limits. \(\overrightarrow{\mathrm{B}}=\int d \overrightarrow{\mathrm{B}}\) From the figure, in a right angle triangle PAO,

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating \(d \overrightarrow{\mathrm{B}} \) by varying the angle from θ = φ₁ to θ =φ₂ is

For a an infinitely long straight wire, 1 =0 and 2 =, the magnetic field is \(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n} \) ………… (3)
Note that here \(\hat{n} \) represents the unit vector from the point O to P.

Question 35.
(a) Obtain an expression for motional emf from Lorentz force.
Answer:
Motional emf from Lorentz force: Consider a straight conducting rod AB of length l in a uniform magnetic field \(\vec{B}\) which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec{v}\) towards right side. When the rod moves, the free electrons present in it also move with same velocity \(\vec{v}\) in B. As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation  \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\)=-
\( e(\vec{v} \times \overrightarrow{\mathrm{B}}) \) …………. (1)

The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field \(\overrightarrow{\mathrm{E}}\), the coulomb force starts acting on the free electrons along AB and is given by
\(\overrightarrow{\mathrm{F}}_{\mathrm{E}}=-e \overrightarrow{\mathrm{E}}\) ……….. (2)

The magnitude of the electric field E keeps on increasing as long as accumulation of electrons at the end A continues. The force \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) and the coulomb force F balance each other and no further accumulation of free electrons at the end A takes place,
i.e

The potential difference between two ends of the rod is
V= El
V= vBl
Thus the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf

ε = B lv    ………….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.

[OR]

(b) Explain the Maxwell’s modification of Ampere’s circuital law.
Answer:
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.

If ‘A’ be the area of the capacitor plates and ‘cf be the charge on the plates at any instant ‘f during the charging process, then the electric field in the gap will be

But \(\frac{d q}{d t}\) is the rate of change of charge on the capacitor plates. It is called displacement current and given by \(\mathrm{I}_{d}=\frac{d q}{d t}=\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\)

This is the missing term in Ampere’s Circuital Law. The total current must be the sum of the
conduction current I_c Hence, the modified form of the ampere’s law.
\(\oint_{c} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0}\left(\mathrm{I}_{c}+\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\right)\)

Question 36.
(a) Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope: The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object objective but also in resolving two points on the object separated by a small distance d_min  Smaller the value of d_min better will be the resolving power of the microscope.

The radius of central maxima is, \(r_{0}=\frac{1.22 \lambda v}{a}\) . In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is d_min, then the magnification m is, \(m=\frac{r_{o}}{d_{\min }}\)


To further reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.

Such an objective is called oil immersed objective. The term n sin β is called numerical aperture NA.

[OR]

(b) Derive an expression for de Broglie wavelength of electrons.
Answer:
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\frac{1}{2} m v^{2}=e \mathrm{V}\)
Therefore, the speed v of the electron is \(v=\sqrt{\frac{2 e \mathrm{V}}{m}}\)
Hence, the de Broglie wavelength of the electron is \(\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}\)
Substituting the known values in the above equation, we get

For example, if an electron is accelerated through a potential difference of 100V, then its de Broglie wavelength is 1.227 Å. Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
\(\lambda=\frac{h}{\sqrt{2 m \mathrm{K}}}\)

Question 37.
(a) Explain the variation of average binding energy with the mass number by graph and discuss its features.
Answer:
We can find the average binding energy per nucleon \(\overline{\mathrm{BE}}\).
It is given by

The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. \(\overline{\mathrm{BE}}\) is plotted against A of all known nuclei.

 

Important inferences from of the average binding energy curve:

• The value of BE rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A= 56 (iron) and then it slowly decreases.
• The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
• For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive.  If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than the initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
• If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

[OR]

(b) State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Answer:
Laws of Boolean Algebra: The NOT, OR and AND operations are \(\bar{A}\), A + B, A.B are the Boolean operations. The results of these operations can be summarised as:

Complement law:

The complement law can be realised as \(\overline{\mathrm{A}}\) = A

Commutative laws
A + B = B +A
A . B = B . A

Associative laws
A + (B + C) = (A + B) + C
A . (B . C) = (A .B) . C

Distributive laws
A( B + C) = AB + AC
A + BC = (A + B) (A + C)
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.

Question 38.
(a) Modulation helps to reduce the antenna size in wireless communication – Explain. Antenna size:
Answer:
Antenna is used at both transmitter and receiver end. Antenna height is an important parameter to be discussed. The height of the antenna must be a multiple of \(\frac{\lambda}{4}\)
\(h=\frac{\lambda}{4}\) ……………….. (1)
where X is wavelength \(\left(\lambda=\frac{c}{v}\right), c\) c is the velocity of light and v is the frequency of the signal v to be transmitted.

An example
Let us consider two baseband signals. One signal is modulated and the other is not modulated. The frequency of the original baseband signal is taken as v = 10 kHz while the modulated signal is v = 1 MHz. The height of the antenna required to transmit the original baseband signal of frequency
v = 10kHz is

The height of the antenna required to transmit the modulated signal of frequency v = 1 MHz is

Comparing equations (2) and (3), we can infer that it is practically feasible to construct an antenna of height 75 m while the one with 7.5 km is not possible. It clearly manifests that modulated signals reduce the antenna height and are required for long distance transmission.

[OR]

(b) What are the possible harmful effects of usage of Nanoparticles? Why?
Answer:
Possible harmful effects of usage of Nanoparticles:

The research on the harmful impact of application of nanotechnology is also equally important and fast developing. The major concern here is that the nanoparticles have the dimensions same as that of the biological molecules such as proteins. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

The adsorbing nature depends on the surface of the nanoparticle. Indeed, it is possible to deliver a drug directly to a specific cell in the body by designing the surface of a nanoparticle so that it adsorbs specifically onto the surface of the target cell.

The interaction with living systems is also affected by the dimensions of the nanoparticles. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles. Nanoparticles can also cross cell membranes.

It is also possible for the inhaled nanoparticles to reach the blood, to reach other sites such as the liver, heart or blood cells. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition and surface characteristics.

Students can Download Tamil Nadu 12th Maths Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Maths Model Question Paper 2 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If A = \(\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]\) be such that λA^-1 = A, then λ is _______.
(a) 17
(b) 14
(c) 19
(d) 21
Answer:
(c) 19

Question 2.
If ω ≠ 1 is a cubic root of unity and (1+ ω)⁷ = A+ B ω, then (A, B) equals to _______.
(a) (1,0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer:
(d) (1, 1)

Question 3.
The value of z – \(\bar{Z}\) is ______.
(a) 2 Im (z)
(b) 2 i Im (z)
(c) Im (z)
(d) i Im (z)
Answer:
(b) 2 i Im (z)

Question 4.
If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if ________.
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a < 0
Answer:
(c) a < 0

Question 5.
sin(tan^-1 x), |x| < 1 is equal to _______.

Answer:
(d) \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is _____.
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)
Answer:
(a) (4, 7)

Question 7.
The axis of the parabola x² = – 4y is ______.
(a) y= 1
(b) x = 0
(c) y = 0
(d) x = 1
Answer:
(b) x = 0

Question 8.
The coordinates of the point where the line \(\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})\) meets the plane \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=3\) are _______.
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Answer:
(d) (5, -1, 1)

Question 9.
If the vectors \(\vec{a}=3 \vec{i}+2 \vec{j}+9 \vec{k}\) and \(\vec{b}=\vec{i}+m \vec{j}+3 \vec{k}\) are parallel then m is _________.

Answer:
(b) \(\frac{2}{3}\)

Question 10.
The minimum value of the function |3 – x | + 9 is ________.
(a) 0
(b) 3
(c) 6
(d) 9
Answer:
(d) 9

Question 11.
The curve y² = x² (1 – x²) has ______.
(a) an asymptote x = -1
(b) an asymptote x = 1
(c) two asymptotes x = 1 and x = -1
(d) no asymptote
Answer:
(d) no asymptote

Question 12.
If /(x, y, z) = xy + yz + zx, then f_x – f_z is equal to _______.
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 13.
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is _______.
(a) 0.2%
(b) 0.4%
(c) 0.04%
(d) 0.08%
Answer:
(b) 0.4%

Question 14.
The value of \(\int_{0}^{\pi} \sin ^{4} x d x\) is _______.

Answer:
(b) \(\frac{3 \pi}{8}\)

Question 15.
\(\int_{a}^{b} f(x) d x\) is _______.

Answer:
(d) \(\int_{a}^{b} f(a+b-x) d x\)

Question 16.
The degree of the differential equation \(y(x)=1+\frac{d y}{d x}+\frac{1}{1.2}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{1.2 .3}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is ________.
(a) 2
(b) 3
(c) 1
(d) 4
Answer:
(c) 1

Question 17.
In finding the differential equation corresponding toy = e^mx where m is the arbitrary constant, then m is ____.
(a) \(\frac{y}{y^{\prime}}\)
(b) \(\frac{y^{\prime}}{y}\)
(c) y’
(d) y
Answer:
(b) \(\frac{y^{\prime}}{y}\)

Question 18.
Let X be random variable with probability density function f(x) = \(\left\{\begin{array}{ll}
2 / x^{3} & x \geq 1 \\
0 & x<1
\end{array}\right.\)
Which of the following statement is correct
(a) both mean and variance exist
(b) mean exists but variance does not exist
(c) both mean and variance do not exist
(d) variance exists but mean does not exist
Answer:
(b) mean exists but variance does not exist

Question 19.
The random variable X has the probability density function f(x) = \(\left\{\begin{array}{cc}
a x+b, & 0<x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
and E(X) = \(\frac{7}{12}\), then a and b are respectively _______.
(a) 1 and \(\frac{1}{2}\)
(b) \(\frac{1}{2}\) and 1
(c) 2 and 1
(d) 1 and 2
Answer:
(a) 1 and \(\frac{1}{2}\)

Question 20.
A binary operation on a set S is a function from ________.
(a) S → S
(b)(S x S) → S
(c) S → (S x S)
(d) (S x S) → (S x S)
Answer:
(b)( S x S) → S

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Solve the following system of homogeneous equations.
3x + 2y + 7z = 0, 4x – 3y – 2z = 0, 5x + 9y + 23z = 0
Answer:
The matrix form of the above equation is

The augmented matrix [A, B] is

The above matrix is in echelon form. Here ρ(A, B) = ρ( A) < number of unknowns.
⇒ The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0 ……..(1)
-17y – 34z = 0 …….(2)
Taking z = t in (2) we get -17y – 34t = 0
⇒ -17y = 34t
⇒ y= \(\frac{34 t}{-17}\) = -2t
Taking z = t; y = -2t in (1) we get
3x + 2 (-2t) + 7t = 0
3x – 4t + 7t = 0
⇒ 3x = -3t ⇒ x = -t
So the solution is x = -t; y = -2t; and z = t, t∈R

Question 22.
Show that |3z – 5 + i| = 4 represents a circle, and, find its centre and radius.
Answer:
The given equation |3z – 5 + i| = 4 can be written as

It is of the form |z – z| = r and so it represents a circle, whose center and radius are \(\left(\frac{5}{3},-\frac{1}{3}\right)\) and 4/3 respectively.

Question 23.
Find the equation of the circle whose centre is (2, -3) and passing through the intersection of the line 3x – 2y = 1 and 4x + y = 27.
Answer:
Solving 3x – 2y = 1 and 4x + y = 27
Simultaneously, we get x = 5 and y = 7
∴ The point of intersection of the lines is (5, 7)
Now we have to find the equation of a circle whose centre is
(2, -3) and which passes through (5, 7)

∴ Required equation of the circle is
(x – 2)² + (y + 3)² = \((\sqrt{109})^{2}\)
⇒ x² + y² – 4x + 6y – 96 = 0

Question 24.
Find the intercepts cut off by the plane \(\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})=12\) on the coordinate axes.
Answer:
\(\vec{r} \cdot(6 \vec{i}+4 \vec{j}-3 \vec{k})=12\)
Compare the above equations into \(\vec{r} \cdot \vec{n}=q\) so q = 12
Let a, b, c are intercepts of x-axis, y-axis and z-axis respectively.
Clearly

x – intercept = 2; y – intercept = 3; z – intercept = -4

Question 25.
Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x² for 1 ≤ x ≤ 2.
Answer:
f(1) = 0 and f(2) = -2. Clearly f(x) is defined and differentiable in 1 < x < 2. Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that
f'(c) = \(\frac{f(2)-f(1)}{2-1}\) = 1 – 2c
That is, 1 – 2c = -2 ⇒ c = \(\frac{3}{2}\)

Question 26.
Show that the percentage error in the n^th root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.
Answer:

Question 27.
Solve the differential equation: tany \(\frac{d y}{d x}\) = cos (x + y) + cos (x -y)
Answer:
tan y \(\frac{d y}{d x}\) = cos (x + y) + cos(x – y)
tan y \(\frac{d y}{d x}\) = cos x cos y – sin x sin y + cos x cos y + sin x sin y
tan y \(\frac{d y}{d x}\) = 2 cos x cos y
seperating the variables
\(\int \frac{\tan y}{\cos y}\) dy = 2∫cos x dx ⇒ ∫sec y tan y dy = 2∫cos x dx
sec y = 2 sin x + c

Question 28.
The probability density function of X is given by \(f(x)=\left\{\begin{array}{cc}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.\)
Find the value of k.
Answer:

Question 29.
Construct the truth table for the following statement. \(\neg(p \wedge \neg q)\).
Answer:

Question 30.
Find an approximate value of \(\int_{1}^{1.5} x^{2} d x\) by applying the right-hand rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.
Answer:
Here a = 1; b = 1.5; n = 5; f(x) = x²
So, the width of each subinterval is

x₀ = 1; x₁ = 1.1; x₂ = 1.2; x₃ = 1.3; x₄ = 1.4; x₅ = 1.5
The Right hand rule for Riemann sum,
S = [f(x₁) + f(x₂) + f(x₃) + f(x₄) + f(x₅)] Δx
= [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)] (0.1)
= [1.21 + 1.44 + 1.69 + 1.96 + 2.25] (0.1)
= [8.55] (0.1)
= 0.855.

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Find a matrix A if adj (A) = \(\left[\begin{array}{ccc}
7 & 7 & -7 \\
-1 & 11 & 7 \\
11 & 5 & 7
\end{array}\right]\)

Question 32.
Obtain the Cartesian form of the locus of z = x + iy in the following case Im[(1 – i)z +1] = 0

Question 33.
If \(\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}, \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}\), show that \([\vec{a} \vec{b} \vec{c}]\) depends on neither x nor y.

Question 34.
The Taylor’s series expansion of f(x) = sin x about x = \(\frac{\pi}{2}\) is obtained by the following way.

Question 35.
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error in computing (i) the volume of the cube and (ii) the surface area of cube.

Question 36.
Evaluate \(\int_{0}^{1} \frac{\sin \left(3 \tan ^{-1} x\right) \tan ^{-1} x}{1+x^{2}} d x\)

Question 37.
Find the particular solution of (1 + x³) dy – x² ydx = 0 satisfying the condition y(1) = 2.

Question 38.
If X is the random variable with distribution function F(x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
x, & 0 \leq x<1 \\
1, & 1 \leq x
\end{array}\right.\)
then find (z) the Probability density function f(x)

Question 39.
Show that \(((\neg q) \wedge p) \wedge q\) is a contradiction.

Question 40.
Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.

Part – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) By using Gaussian elimination method, balance the chemical reaction equation:
C₂H₆ + O₂ → H₂O + CO₂.
[OR]
(b) \(\frac{d y}{d x}+\frac{3 y}{x}=\frac{1}{x^{2}}\), given that y = 2 when x = 1

Question 42.
(a) Find the real values of x and y for the equation \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i\)
[OR]
(b) Find the area between the line y = x + 1 and the curve y = x² – 1.

Question 43.
(a) Determine k and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
[OR]
(b) Evaluate: \(\int_{0}^{\frac{\pi}{2}} \frac{d x}{5+4 \sin ^{2} x}\)

Question 44.
(a) A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m, and the height at the edge of the road must be sufficient for a truck 4 m high to clear if the highest point of the opening is to be 5 m approximately. How wide must the opening be?
[OR]
(b) Using truth table check whether the statements \(\neg(p \vee q) \vee(\neg p \wedge q)\) and \(\neg p\) are logically equivalent.

Question 45.
(a) Find the value of cot^-1 x – cot^-1 (x + 2) = \(\frac{\pi}{12}\), x > 0
[OR]
(b) Verify Euler’s theorem for f(x, y) = \(\frac{1}{\sqrt{x^{2}+y^{2}}}\)

Question 46.
(a) Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
[OR]
(b) If X is the random variable with probability density function f(x) given by,
\(f(x)=\left\{\begin{array}{rc}
x+1, & -1 \leq x<0 \\
-x+1, & 0 \leq x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
then find (z) the distribution function f(x) (ii) P (-0.5 ≤ X ≤ 0.5)

Question 47.
(a) Sketch the graph of the function: y = \(x \sqrt{4-x}\)
(b) The velocity v, of a parachute falling vertically satisfies the equation, \(v \frac{d v}{d x}=g\left(1-\frac{v^{2}}{k^{2}}\right)\)
where g and k are constants. If v and x are both initially zero, find v in terms of x.

Students can Download Tamil Nadu 12th Biology Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 4 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
A Plant called ‘X’ possesses small flower with reduced perianth and versatile anther. The probable agent for pollination would be _________.
(a) water
(b) air
(c) butterflies
(d) beetles
Answer:
(b) air

Question 2.
“Gametes are never hybrid”. This is a statement of _______.
(a) Law of dominance
(b) Law of independent assortment
(c) Law of segregation
(d) Law of random fertilization
Answer:
(c) Law of segregation

Question 3.
In which techniques Ethidium Bromide is used?
(a) Southern Blotting techniques
(b) Western Blotting techniques
(c) Polymerase Chain Reaction
(d) Agarose Gel Electrophoresis
Answer:
(d) Agarose Gel Electrophoresis

Question 4.
Which of the following statement is correct?
(a) Agar is not extracted from marine algae such as seaweeds
(b) Callus undergoes differentiation and produces somatic embryoids
(c) Surface sterilization of explants is done by using mercuric bromide
(d) pH of the culture medium is 5.0 to 6.0
Answer:
(d) pH of the culture medium is 5.0 to 6.0

Question 5.
The term pedogenesis is related to _______.
(a) Fossils
(b) Water
(c) Population
(d) Soil
Answer:
(d) Soil

Question 6.
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancer?
(a) Ammonia
(b) Methane
(c) Nitrous oxide
(d) Ozone
Answer:
(d) Ozone

Question 7.
A wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat is rich in _______.
(a) iron
(b) carbohydrates
(c) proteins
(d) vitamins
Answer:
(c) proteins

Question 8.
Statement A: Coffee contains caffeine.
Statement B: Drinking coffee enhances cancer.
(a) A is correct, B is wrong
(b) A and B – Both are correct
(c) A is wrong, B is correct
(d) A and B – Both are wrong
Answer:
(b) A and B – Both are correct

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Draw and label the structure of a typical pollen grain.
Answer:

Question 10.
What is test cross? Why it is done?
Answer:
Test cross is crossing an individual of unknown genotype with a homozygous recessive. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character.

Question 11.
You are working in a biotechnology lab with a bacterium namely E.coli. How will you cut the nucleotide sequence? Explain it.
Answer:
The DNA nucleotide sequence can be cut using Restriction endonucleases (RE). Restriction

Question 12.
What is ecological hierarchy? Name the levels of ecological hierarchy.
Answer:
The interaction of organisms with their environment results in the establishment of grouping of organisms which is called ecological hierarchy.

Question 13.
Mutagens are the substances that induces mutation. Name any two physical and chemical mutagens.
Answer:
UV short waves, X-rays – Physical mutagens.
Nitromethyl, Urea – Chemical mutagens.

Question 14.
What is pseudo cereal? Give an example.
Answer:
The term pseudo-cereal is used to describe foods that are prepared and eaten as a whole grain, but are botanical outliers from grasses. Example: quinoa. It is actually a seed from the Chenopodium quinoa plant, belongs to the family Amaranthaceae.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
“Endothecium is associated with dehiscence of anther” Justify the statement.
Answer:
The inner tangential wall develops bands (sometimes radial walls also) of α cellulose (sometimes also slightly lignified). The cells are hygroscopic. The cells along the junction of the two sporangia of an anther lobe lack these thickenings. This region is called stomium. This region along with the hygroscopic nature of endothecium helps in the dehiscence of anther at maturity.

Question 16.
What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.

Uses of genetic mapping:

• It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
• They are useful in predicting results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 17.
How synthetic seeds are developed?
Answer:
Artificial seeds or synthetic seeds (synseeds) are produced by using embryoids (somatic embryos) obtained through in vitro culture. They may even be derived from single cells from any part of the plant that later divide to form cell mass containing dense cytoplasm, large nucleus, starch grains, proteins, and oils, etc. To prepare the artificial seeds different inert materials are used for coating the somatic embryoids like agrose and sodium alginate.

Question 18.
Discuss the three zones of a lentic ecosystem.
Answer:
There are three zones, littoral, limnetic and profundal. The littoral zone, which is closest to the shore with shallow water region, allows easy penetration of light. It is warm and occupied by rooted plant species. The limnetic zone refers the open water of the pond with an effective penetration of light and domination of planktons.

The deeper region of a pond below the limnetic zone is called profundal zone with no effective light penetration and predominance of heterotrophs. The bottom zone of a pond is termed benthic and is occupied by a community of organisms called benthos (usually decomposers).

Question 19.
Write a short note on clean development mechanism.
Answer:
Clean Development Mechanism (CDM) is defined in the Kyoto protocol (2007) which provides project based mechanisms with two objectives to prevent dangerous climate change and to reduce green house gas emissions. CDM projects helps the countries to reduce or limit emission and stimulate sustainable development.

An example for CDM project activity, is replacement of conventional electrification projects with solar panels or other energy efficient boilers. Such projects can earn Certified Emission Reduction (CER) with credits / scores, each equivalent to one tonne of CO₂, which can be counted towards meeting Kyoto targets.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Describe dominant epistasis with an example.
Answer:

Dominant Epistasis – It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different pair of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of a gene at another locus is known as epistatic. The gene whose expression is interfered by non-allelic genes and prevents from exhibiting its character is known as hypostatic. When both the genes are present together, the phenotype is determined by the epistatic gene and not by the hypostatic gene.

In the summer squash the fruit colour locus has a dominant allele ‘W’ for white colour and a recessive allele ‘w’ for coloured fruit. ‘W’ allele is dominant that masks the expression of any colour. In another locus hypostatic allele ‘G’ is for yellow fruit and its recessive allele ‘g’ for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotype WWgg is crossed with yellow fruit with genotype wwGG, the F₁ plants have white fruit and are heterozygous (WwGg). When F₁ heterozygous plants are crossed they give rise to F₂ with the phenotypic ratio of 12 white : 3 yellow : 1 green.

Since W is epistatic to the alleles ‘G’ and ‘g’ the white which is dominant, masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16). Double recessive ‘wwgg’ will give green fruit (1/16). The Plants having only ‘G’ in its genotype (wwGg or wwGG) will give the yellow fruit(3/16).

[OR]

(b) Point out the significance of plant succession.
Answer:
Significance of Plant Succession:

• Succession is a dynamic process. Hence an ecologist can access and study the seral stages of a plant community found in a particular area.
• The knowledge of ecological succession helps to understand the controlled growth of one or more species in a forest.
• Utilizing the knowledge of succession, even dams can be protected by preventing siltation.
• It gives information about the techniques to be used during reforestation and afforestation.
• It helps in the maintenance of pastures.
• Plant succession helps to maintain species diversity in an ecosystem.
• Patterns of diversity during succession are influenced by resource availability and disturbance by various factors.
• Primary succession involves the colonization of habitat of an area devoid of life.
• Secondary succession involves the re-establishment of a plant community in disturbed area or habitat.
• Forests and vegetation that we come across all over the world are the result of plant succession.

Question 21.
(a) Compare the various types of Blotting techniques.
Answer:

[OR]

(b) Explain different types of hybridization.
Answer:
Types of Hybridization:
According to the relationship between plants, the hybridization is divided into.
1. Intravarietal hybridization – The cross between the plants of same variety. Such crosses are Useful only in the self-pollinated crops.

2.. Intervarietal hybridization – The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization. This technique has been the basis of improving self-pollinated as well as cross pollinated crops.

3. Interspecific hybridization – The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization. It is commonly used for transferring the genes of disease, insect, pest and drought resistance from one species to another.
Example: Gossypium hirsutum x Gossypium arboreum – Deviraj.

4. Intergeneric hybridization – The crosses are made between the plants belonging to two different genera. The disadvantages are hybrid sterility, time consuming and expensive procedure. Example: Raphanobrassica and Triticale.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Select the correct production site and action site of Relaxin.
(a) Hypothalamus and pituitary gland
(b) Pituitary gland and Pelvic joints and cervix
(c) Placenta and pelvic joint and cervix
(d) Hypothalamus and placenta
Answer:
(c) Placenta and pelvic joint and cervix

Question 2.
Fusion of young individuals produced immediately after the mitotic division of adult parent cell is called _______.
(a) Merogamy
(b) Anisogamy
(c) Hologamy
(d) Paedogamy
Answer:
(d) Paedogamy

Question 3.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number ______.
(a) 20
(b) 21
(c) 23
(d) 19
Answer:
(b) 21

Question 4.
DNA finger printing techniques was developed by _______.
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
(d) Hershey and Chase
Answer:
(b) Alec Jeffreys

Question 5.
Identify the correct sequence of periods from oldest to youngest
(a) Cambrian → Permian → Devonian → Silurian → Ordovician
(b) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian → Devonian → Permian

Question 6.
Spread of cancerous cells to distant sites is termed as _________.
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 7.
Assertion (A): Streptomycin is an antibiotic.
Reason (R): Antibiotic are microbial chemicals inhibits the growth of pathogenic microbe.
(a) A is right R is wrong
(b) R explains A
(c) A and R are wrong
(d) A is wrong but R is right
Answer:
(b) R explains A

Question 8.
World Ozone Day was observed on _________.
(a) September 16^th
(b) October 12^th
(c) December 1^th
(d) August 18^th
Answer:
(a) September 16^th

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 10.
State any two unique features of ELISA test.
Answer:
ELISA is highly sensitive and can detect antigen even in nanograms.
ELISA test does not require radioisotopes or radiation counting apparatus.

Question 11.
Define Anaphylaxis.
Answer:
Anaphylaxis is the classical immediate hypersensitivity reaction. It is a sudden, systematic, severe and immediate hypersensitivity reaction occurring as a result of rapid generalized mast-cell degranulation

Question 12.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

Question 13.
What is S – D sequence?
Answer:
The 5′ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 14.
Expand (a) GIFT (b) ICSI
Answer:
GIFT – Gamete Intra – Fallopian Transfer
ICSI – Intra-cytoplasmic sperm injection

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Write a short note on encystment in amoeba.
Answer:
During unfavorable conditions (increase or decrease in temperature and scarcity of food) Amoeba withdraws its pseudopodia and secretes a three-layered, protective, chitinous cyst wall around it and becomes inactive. This phenomenon is called encystment. When conditions become favourable, the encysted Amoeba divides by multiple fission and produces many minute amoebae called pseudopodiospore or amoebulae.

The cyst wall absorbs water and breaks off liberating the young pseudopodiospores, each with a fine pseudopodia. They feed and grow rapidly to lead an independent life.

Question 16.
Draw a schematic representation of human oogenesis.
Answer:

Question 17.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

• Sex-education in school and public forums.
• Promoting the uses of contraception.
• Compulsory sterilization for mentally retarded and criminals.
• Egg donation.
• Artificial insemination by donors.
• Prenatal diagnosis of genetic disorders and performing MTP.
• Gene therapy.
• Cloning.
• Egg/sperm donation of healthy individuals.

Question 18.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.
1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.

2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 19.
What is Q₁₀ value? How it is calculated?
Answer:
The effect of temperature on the rate of reaction is expressed in terms of temperature coefficient or Q₁₀ value. The Q₁₀ values are estimated taking the ratio between the rate of reaction at X°C and rate of reaction at (X-10°C).

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Differentiate between r-selected species and k-selected species.
Answer:
r – selected species:

• Smaller sized organisms
• Produce many offspring
• Mature early
• Short life expectancy
• Each individual reproduces only once or few times in their life time
• Only few reach adulthood
• Unstable environment, density independent

k – seleced species:

• Larger sized organisms
• Produce few offspring
• Late maturity with extended parental care
• Long life expectancy
• Can reproduce more than once in lifetime
• Most individuals reach maximum life span
• Stable environment, density dependent

[OR]

(b) Give a detailed account on ethanol production by microbes and the uses of ethanol.
Answer:
Ethanol production:
Saccharomyces cerevisiae (Yeast) is the major product of ethanol.

Since ethanol is used for industrial, laboratory and fuel proposes, it is called as industrial alcohol.

Organism used: Saccharomyces cerevisiae, bacteria like Zymomonas mobilis and Sarcina ventriculi.
Substances used: Molasses, Com, Potatoes, wood waste.

Process of ethanol production:
Step-1: Milling of fees stock.
Step-2: Adding fungal (Aspergillus) amylase to break down starch into sugar.
Step-3: Yeast is added to convert sugar into ethanol.
Step-4: Distillation yield 96% concentrated ethanol.

Uses of Ethanol:
Ethanol and bio-diesel are the two commonly used first generation bio-fuels.
Ethanol is used as fuel, mainly as bio-fuel additive for gasoline.

Question 21.
(a) How DNA is packed in an eukaryotic cell?
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (H1) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an H1 molecule. Chromatin lacking H1 has a beads-on-a-string appearance in which DNA enters and leaves the nucleosomes at random places. H1 of one nucleosome can interact with H1 of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chromatin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucleosome, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different H1 molecules. DNA is a solenoid and packed about 40 folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of proteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In a typical nucleus, some regions of chromatin are loosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

[OR]

(b) Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more complex molecules. He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment.

The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy. The atmosphere was oxygen free and the combination of CO₂, NH₃ and UV radiations gave rise to organic compounds. The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers.

They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive ‘ atmosphere was reducing and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Students can Download Tamil Nadu 12th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 1 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is …………….
(a) 80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 80 x 10^-17 J

Question 2.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Question 3.
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P which is at exactly at  R/2 distance between two coils is …………..

Answer:

Question 4.
An electron moves on a straight line path XY as shown in the figure. The coil abed is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil?

(a) The current will reverse its direction as the electron goes past the coil
(b) No current will be induced
(c) abcd
(d) adcb
Answer:
(a) The current will reverse its direction as the electron goes past the coil

Question 5.
The resistance of an ideal ammeter is ……………
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 6.
In a step-down transformer the input voltage is 22 kV and the output voltage is 550 V. The ratio of the number of turns in the secondary to that in the primary is ……………
(a) 1 : 20
(b) 20 : 1
(c) 1 : 40
(d) 40 : 1
Answer:
(c) 1 : 40

Question 7.
If the magnetic monopole exists, then which of the Maxwell’s equation to be modified?

Answer
\(\oint \overrightarrow { { E } } \cdot d\overrightarrow { { A } } =0\)

Question 8.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint. Let d₁ = 5 cm and d₂ = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d₁n + d₂n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 9.
When a ray of light enters a glass slab from air
(a) its wavelength decreases
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint: \(Wavelength,\lambda =\frac { { Velocity } }{ { Frequency } } =\frac { u }{ v } \)
When light travels from air to glass, frequency ν remains unchanged, velocity u decreases and hence wave length λ also decreses

Question 10.
A particle of mass 3 x 10^-6 g has the same wavelength as an electron moving with a velocity
6 x 10⁶ m s^-1. The velocity of the particle is…………………………………………
(a) 1.82 x 10^-18 ms^-1
(b) 9 x 10^-2 ms^-1
(c) 3 x 10^-31 ms^-1
(d) 1.82 x 10¹⁵ ms^-1
Answer:
(d) 1.82 x 10¹⁵ ms^-1
Hint: de – Broglie wavelength of electron

Question 11.
If the nuclear radius of ²⁷A1 is 3.6 fermi, the approximate unclear radius of ⁶⁴Cu is
(a) 2.4
(b) 1.2
(c) 4.8
(d) 3.6
Answer:
(c) 4.8
Hint :

Question 12.
Energy of characteristic X-ray is a consequence of ………….
(a) energy of projectile electron
(b) thermal energy of target
(c) transition in target atoms
(d) none of the above
Answer:
(c) transition in target atoms

Question 13.
The specific characteristic of a common emitter amplifier is ……………
(a) High input resistance
(b) Low power gain
(c) Signal phase reversal
(d) Low current gain
Answer:
(c) Signal phase reversal

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called ………………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
The particle size of ZnO material is 30 nm. Based on the dimension it is classified as ……………..
(a) Bulk material
(b) Nanomaterial
(c) Soft material
(d) Magnetic material
Answer:
(b) Nanomaterial

Part – II

Answer any six questions in which Q. No 17 is compulsory. [6 x 2 = 12]

Question 16.
Write a short note on the superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\(\overrightarrow{\mathrm{F}}_{1}^{\mathrm{tot}}=\overrightarrow{\mathrm{F}}_{12}+\overrightarrow{\mathrm{F}}_{13}+\overrightarrow{\mathrm{F}}_{14}+\cdots+\overrightarrow{\mathrm{F}}_{1 n}\)

Question 17.
If an electric field of magnitude 570 N C^-1, is applied in the copper wire, find the acceleration experienced by the electron.
Answer:

Question 18.
State Ampere’s circuital law.
Answer:
The line integral of magnetic field over a closed loop is p0 times net current enclosed by the loop.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\)

Question 19.
Write down the equation for a sinusoidal voltage of §0 Hz and its peak value is 20 V. Draw the corresponding voltage versus time graph.
Answer:

Question 20.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time.

Question 21.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 22.
How many photons per second emanate from a 50 mW laser of 640 nm?
Answer:

Question 23.
An ideal diode and a 5 Ω resistor are connected in series with a 15 V power supply as shown in figure below. Calculate the current that flows through the diode.
Answer:
The diode is forward biased and it is an ideal one. Hence, it acts like a closed switch with no barrier voltage.Therefore, current that flows through the diode can be calculated using Ohm’s law.

V = IR
\(I=\frac{V}{R}=\frac{15}{5}=3 \mathrm{A}\)

Question 24.
What do you mean by Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

Part – III

Answer any six questions in which Q.No. 26 is compulsory. [6 x 3 = 18]

Question 25.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 26.
If the resistance of coil is 3 Ω at 20°C and a = 0.004/°C then determine its resistance at 100°C.
Answer:
R₀ =3Ω = 100°C, T₀ = 20°C
α = 0.004/°C, R_T = ?
Rr= R₀(1+∝ (T-T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
⇒ R₁₀₀ = 3(1 + 0.32)
R₁₀₀= 3(1.32)
⇒ R₁₀₀ = 3.96 Ω

Question 27.
State ‘Tangent Law’.
Answer:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Question 28.
What are step-up and step-down transformers?
Answer:
If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.

Question 29.
One type of transparent glass has refractive index 1.5. What is the speed of light through this glass?
Answer:

Question 30.
What is a photo cell? Mention the different types of photocells.
Answer:
Photocells: Photo electric cell or photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect.
Types:

• Photo emissive cell
• Photo voltaic cell
• Photo conductive cell

Question 31.
Half lives of two radioactive elements A and B are 20 minutes and 40 minutes respectively. Initially, the samples have equal number of nuclei. Calculate the ratio of decayed numbers of A and B nuclei after 80 minutes.
Answer:
80 minutes = 4 half lives of A = 2 half live of B
Let the initial number of nuclei in each sample be N.

Question 32.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

Intrinsic	Extrinsic
These are pure semiconducting tetravalent crystals.	These are semiconducting tetravalent crystals doped with impurity atoms group III (or) V
Their electrical conductivity is low.	Their electrical conductivity is high.
There is no permitted energy state between valence and conduction band.	There is no permitted energy state of the impurity atom between valence and conduction band.
Their electrical conductivity depends on temperature.	Their electrical conductivity depends on temperature as well as dopant concentration.

Question 33.
Distinguish between Nanoscience and Nanotechnology.
Answer:

Nanoscience	Nanotechnology
Nanoscience is the science of objects with typical sizes of 1-100 nm. Nano means one- billionth of a metre that is 10-9 m. If matter is divided into such small objects the mechanical, electrical, optical, magnetic and other properties change.	Nanotechnology is a technology involv­ing the design, production, characteriza­tion, and applications of nano structured materials.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Obtain the expression for electric field due to an uniformly charged spherical shell. Electric field due to a uniformly charged
Answer:
spherical shell: Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a) At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially
outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,

The electric field \(\overrightarrow { { E } } { and }\quad d\overrightarrow { { A } } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\overrightarrow { { E } } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But = total area of Gaussian surface  = 4πr² Substituting this value in equation (2)

 

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b): At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\overrightarrow{\mathrm{E}}=\frac{Q}{4 \pi \varepsilon_{0} R^{2}} \hat{r}\) ………. (4)

Case (c) At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law.

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes
E = 0    (r < R)  ………….. (6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

[OR]

(b) Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:

An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.
Applying Kirchhoff’s current rule to junction B.
I₁ – I_G – I₃ = 0   ………….. (1)
Applying Kirchhoff’s current rule to junction D.
I₂ + I_G – I₄ = 0   ………….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁P + I_GG – I₂R = 0  ………….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁P + I₃Q – I₄S – I₂R = 0  ………….. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B through galvanometer (I_Q = 0). Substituting I_G = 0 in equation, (1), (2) and (3), we get
I₁ = I₃  ………………….. (5)
I₁ = I₄ ……………………. (6)
I₁P =I₂R ……………….(7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₁S – I₁R = 0
I₁( P + Q) = I₂( P + S) …………………(8)
Dividing equation (7)

This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance  (fourth one) can be determined.

Question 35.
(a) Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Answer:
Consider a bar magnet NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength q_m and separated by a distance of 21. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole (q_mC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb’s law of magnetism as follows:

The force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=-\mathrm{F}_{\mathrm{N}} \cos \theta \hat{i}+\mathrm{F}_{\mathrm{N}} \sin \theta \hat{j}\) ……………. (1)
Where \(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{r^{\prime 2}}\) . The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is

From equation (1) and equation (2), the net force at point C is \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\mathrm{F}_{\mathrm{S}}\). This net force is equal to the magnetic field at the point C.

If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r>> l, then,

Therefore , using equation (7) in equation (6) ,we get
\(\overrightarrow{\mathrm{B}}_{\text {equatorial }}=-\frac{\mu_{0}}{4 \pi} \frac{p_{m}}{r^{3}} \hat{i}\)
Since in genaral , the magnetic filed at equatorial points is given by

Note that magnitude of B_axial is twice that of magnitude of B_equatoral and the direction of B_axial and B_equatoral are opposite.

[OR]

Question 35.
(b) Give the advantage of AC in long distance power transmission with an example.
Answer:
Advantages of AC in long distance power transmission: Electric power is produced in a large scale at electric power stations with the help of AC generators. These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places. Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission.

But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (i²R) in the transmission lines which are hundreds of kilometer long. This power loss can be tackled either by reducing current I or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases die cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.

Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent. At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer.

Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.

Illustration: An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.

Case (i) : P = 2 MW; R = 40 Ω; V = 10 kV
Power, P= VI
Power loss = Heat produced = I²R = (200)² x 40 = 1.6 x 10⁶ W
% of power loss = \(\frac{1.6 \times 10^{6}}{2 \times 10^{6}} \times 100 \%\) = 0.8 x 100% = 80%

Case (ii): P = 2 MW; R = 40 Ω; V = 100 kV
∴ Current, \(I=\frac{P}{V}=\frac{2 \times 10^{6}}{100 \times 10^{3}}=20 \mathrm{A}\)
Power loss = Heat produced = I²R = (20)² x 40 = 0.016 x 10⁶ W
% of power loss = \(\frac{0.016 \times 10^{6}}{2 \times 10^{6}} \times 100 \%=\) = 0.008 x 100% = 0.8%

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

[OR]

(b) Obtain the equation for radius of illumination (or) Snell’s window.
Answer:
The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation n₂ sin i = n₂ sin r for the refraction happening at the point B on the boundary between the two media is,
n₁ sin i_c = n₂ sin 90° ………………. (1)
n₁ sin i_c = n₂
∵ sin 90° = 1
sin i_c = \(\frac{n_{2}}{n_{1}}\) ………………….(2)
From the right angle triangle ΔABC,

Question 37.
(a) Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Failures of classical wave theory:
Answer:
From Maxwell’s theory, light is an electromagnetic wave consisting of coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behaviour. Let us try to explain the experimental observations of photoelectric effect using wave picture of light.

When light is incident on the target, there is a continuous supply of energy to the electrons. According to wave theory, light of greater intensity should impart greater kinetic energy to the liberated electrons (Here, Intensity of light is the energy delivered per unit area per unit time). But this does not happen. The experiments show that maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.

According to wave theory, if a sufficiently intense beam of light is incident on the surface, electrons will be liberated from the surface of the target, however low the frequency of the radiation is. From the experiments, we know that photoelectric emission is not possible below a certain minimum frequency. Therefore, the wave theory fails to explain the existence of threshold frequency.

Since the energy of light is spread across the wavefront, the electrons which receive energy from it are large in number. Each electron needs considerable amount of time (a few hours) to get energy sufficient to overcome the work function and to get liberated from the surface. But experiments show that photoelectric emission is almost instantaneous process (the time lag is less than 10^-9 s after the surface is illuminated) which could not be explained by wave theory.

[or]

(b) Obtain the law of radioactivity.
Answer:
Law of radioactive decay:
At any instant t, the number of decays per unit time, called rate of decay \(\left(\frac{d \mathrm{N}}{d t}\right) \) to the number of nuclei at the same instant.
\(\frac{d \mathrm{N}}{d t} \propto \mathrm{N}\)

By introducing a proportionality constant, the relation can be written as
\(\frac{d \mathrm{N}}{d t}=-\lambda \mathrm{N}\) ……………………. (1)

Here proportionality constant X is called decay constant which is different for different radioactive sample and the negative sign in the equation implies that the N is decreasing with time. By rewriting the equation (1), we get
dN = -XNdt …………… (2)

Here dN represents the number of nuclei decaying in the time interval dt. Let us assume that at time t = 0 s, the number of nuclei present in the radioactive sample is N_o. By integrating the equation (2), we can calculate the number of undecayed nuclei N at any time t.
From equation (2), we get

N = N_o e^-λt …………… (4)
[Note: e^lnx = e^y ⇒ x = e^y ]
Equation (4) is called the law of radioactive decay. Here N denotes the number of undecayed nuelei present at any time t and N₀ denotes the number of nuclei at initial time t = 0. Note that the number of atoms is decreasing exponentially over the time. This implies that the time taken for all the radioactive nuclei to decay will be infinite. Equation (4) is plotted.

We can also define another useful quantity called activity (R) or decay rate which is the number of nuclei decayed per second and it is denoted as R = \(R=\left| \frac { dN }{ dt } \right| \)

Note : That activity R is a positive quantity. From equation (4), we get.

The equation (6) is also equivalent to radioactive law of decay. Here R₀ is the activity of the sample at t = 0 and R is the activity of the sample at any time t. From equation (6), activity also shows exponential decay behavior. The activity R also can be expressed in terms of number of undecayed atoms present at any time t. From equation (6), since N = N₀ e ^-λt we write
R = λN    ………………… (7)
Equation (4) implies that the activity at any time t is equal to the product of decay constant and number of undecayed nuclei at the same time t. Since N decreases over time, R also decreases.

Question 38.
(a) Explain the construction and working of a full wave rectifier.
Answer:
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two p-n junction diodes, a center tapped transformer, and a load resistor (R_L). The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G is at zero potential and N is at negative potential. This forward biases diode D₁ and reverse biases diode D₂. Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁ AGC As a result, positive half cycle of the voltage appears across R_L in the direction G to C.

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D, and reverse biases diode D_r Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC. As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature. The efficiency (r|) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

[OR]

(b) Explain the three modes of propagation of electromagnetic waves through space. Propagation of electromagnetic waves:
Answer:
The electromagnetic wave transmitted by the transmitter travels in three different modes to reach the receiver-according to its frequency range:
(i) Ground wave propagation (or) surface wave propagation (nearly 2 kHz to 2 MHz)
(ii) Sky wave propagation (or) ionospheric propagation (nearly 3 MHz to 30 MHz)
(iii) Space wave propagation (nearly 30 MHz to 400 GHz)

(i) Ground wave propagation
If the electromagnetic waves transmitted by the transmitter glide over the surface of the earth to reach the receiver, then the propagation is called ground wave propagation. The corresponding waves are called ground waves or surface waves.

 

Increasing distance: The attenuation of the signal depends on

• power of the transmitter
• frequency of the transmitter, and
• condition of the earth surface.

Absorption of energy by the Earth: When the transmitted signal in the form of EM wave is in contact with the Earth, it induces charges in the Earth and constitutes a current. Due to this, the earth behaves like a leaky capacitor which leads to the attenuation of the wave.

Tilting of the wave: As the wave progresses, the wavefront starts gradually tilting according to the curvature of the Earth. This increase in the tilt decreases the electric field strength of the wave. Finally, at some distance, the surface wave dies out due to energy loss.

The frequency of the ground waves is mostly less than 2 MHz as high frequency waves undergo more absorption of energy at the earth’s atmosphere. The medium wave signals received during the day time use surface wave propagation.

It is mainly used in local broadcasting, radio navigation, for ship-to-ship, ship-to-shore communication and mobile communication.

(ii) Sky Wave Propagation:
The mode of propagation in which the electromagnetic waves radiated from an antenna, directed upwards at large angles gets reflected by the ionosphere back to earth is called sky wave propagation or ionospheric propagation. The corresponding waves are called sky waves.

The frequency range of EM waves in this mode of propagation is 3 to 30 MHz. EM waves of frequency more than 30 MHz can easily penetrate through the ionosphere and does not undergo reflection. It is used for short wave broadcast services. Medium and high frequencies are for long-distance radio communication. Extremely long distance communication, is also possible as the radio waves can undergo multiple reflections between the earth and the ionosphere. A single reflection helps the radio waves to travel a distance of approximately 4000 km.

Ionosphere acts as a reflecting surface. It is at a distance of approximately 50 km and spreads up to 400 km above the Earth surface. Due to the absorption of ultraviolet rays, cosmic ray, and other high energy radiations like a, (3 rays from sun, the air molecules in the ionosphere get ionized. This produces charged ions and these ions provide a reflecting medium for the reflection of radio waves or communication waves back to earth within the permitted frequency range. The phenomenon of bending the radio waves back to earth is nothing but the total internal reflection.

(iii) Space wave propagation:
The process of seeding and receiving information signal through space is called space wave communication. The electromagnetic waves of very high frequencies above 30 MHz are called as space waves. These waves travel in a straight line from the transmitter to the receiver. Hence, it is used for a line of sight communication (LOS).

For high frequencies, the transmission towers must be high enough so that the transmitted and received signals (direct waves) will not encounter the curvature of the earth and hence travel with less attenuation and loss of signal strength. Certain waves reach the receiver after getting reflected from the ground.

Students can Download Tamil Nadu 12th Biology Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 3 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identity the mismatched pair regarding the anther walls.
(a) Epidermal layer – Protective in function
(b) Endothecium layer – Helps in dehiscence of anther
(c) Middle layer – Persistent layer
(d) Tapetum – Nutritive in function
Answer:
(c) Middle layer – Persistent layer

Question 2.
Extra nuclear inheritance is a consequence of presence of genes in _______.
(a) Mitochondria and chloroplasts
(b) Endoplasmic reticulum and mitochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(b) Endoplasmic reticulum and mitochondria

Question 3.
How many map units separate two alleles A and B, if the recombination frequency is 0.09?
(a) 900 cM
(b) 90 cM
(c) 9 cM
(d) 0.9 cM
Answer:
(c) 9 cM

Question 4.
Plasmids are _______.
(a) circular protein molecules
(b) required by bacteria
(c) tiny bacteria
(d) confer resistance to antibiotics
Answer:
(d) confer resistance to antibiotics

Question 5.
Solar energy used by green plants for photosynthesis is only ________.
(a) 2 – 8%
(b) 2 – 10%
(c) 3 – 10%
(d) 2- 9%
Answer:
(b) 2 – 10%

Question 6.
One of the chief reasons among the following for the depletion in the number of species making endangered is _____.
(a) over hunting and poaching
(b) green house effect
(c) competition and predation
(d) habitat destruction
Answer:
(d) habitat destruction

Question 7.
Assertion (A): Genetic variation provides the raw material for selection.
Reason (R): Genetic variations are differences in genotypes of the individuals.
(a) (A) is right and (R) is wrong
(b) (A) is wrong and (R) is right
(c) Both (A) and (R) are right
(d) Both (A) and (R) are wrong
Answer:
(c) Both (A) and (R) are right

Question 8.
Groundnut is native of ______.
(a) Philippines
(b) India
(c) North America
(d) Brazil
Answer:
(d) Brazil

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1 ,
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 10.
What are the materials used to grow microorganism like Spirulina.
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Question 11.
Compare Redifferentiation with Dedifferentiation.
Answer:
Redifferentiation:
A process by which an already differentiated cell undergo further differentiation to form another type of cell.

Dedifferentiation:
A process of reversion of cells (differentiated cells) to meristematic cells leading to formation of callus.

Question 12.
Loamy soil is ideal for crop cultivation – Justify.
Answer:
Loamy soil is ideal soil for cultivation, since it consists of 70% sand and 30% clay or silt or both. It ensures good retention and proper drainage of water. The porosity of soil provides adequate aeration and allows the penetration of roots.

Question 13.
Mention any four environmental benefits of Rain Water Harvesting.
Answer:

• Promotes adequacy of underground water and water conservation.
• Mitigates the effect of drought.
• Improves groundwater quality and water table / decreases salinity.
• Reduces soil erosion as surface run-off water is reduced.

Question 14.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Yes, drinking coffee in moderation enhances the health of a person. Caffeine enhances release of acetylcholine in brain, which in turn enhances efficiency. It can lower the incidence of fatty liver diseases, cirrhosis and cancer. It may reduce the risk of type 2 diabetes.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
How the flowers of salvia are adopted for mellitophily?
Answer:
Pollination in Salvia (Lever mechanism): The flower of Salvia is adapted for Bee pollination. The flower is protandrous and the corolla is bilabiate with 2 stamens. A lever mechanism helps in pollination. Each anther has an upper fertile lobe and lower sterile lobe which is separated by a long connective which helps the anthers to swing freely. When a bee visits a flower, it sits on the lower lip which acts as a platform.

It enters the flower to suck the nectar by pushing its head into the corolla. During the entry of the bee into the flower the body strikes against the sterile end of the connective. This makes the fertile part of the stamen to descend and strike at the back of the bee. The pollen gets deposited on the back of the bee. When it visits another flower, the pollen gets rubbed against the stigma and completes the act of pollination in Salvia.

Question 16.
What is the difference between mis-sense mutation and non-sense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.
Give an account on germplasm conservation. .
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank and DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 18.
How are microbial innoculants used to increase the soil fertility?
Answer:
Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil. They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose.

They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers. .

Question 19.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Find out the molecular explanation for the wrinkled pea seeds used by Mendel.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in r allele. In the homozygous mutant form of the gene (rr) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas. The wrinkled seed accumulates more sucrose and high water content.

Hence the osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.

[OR]

(b) Write in detail about Remote sensing and its uses.
Answer:
Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life for classification of land use patterns and studies, identification of biodiversity rich or less areas for futuristic works on conservation and maintenance of various species including commercial crop, medicinal plants and threatened plants. .

Specific uses:

• Helps predicting favourable climate, for the study of spreading of disease and controlling it.
• Mapping of forest fire and species distribution.
• Tracking the patterns of urban area development and the changes in Farmland or forests over several years.
• Mapping ocean bottom and its resources.

Question 21.
(a) Explain various edaphic factors that affect vegetation.
Answer:
The important edaphic factors which affect vegetation are as follows:

1. Soil moisture: Plants absorbs rain water and moisture directly from the air.
2. Soil water: Soil water is more important than any other ecological factors affecting the distribution of plants. Rain is the main source of soil water. Capillary water held between pore spaces of soil particles and angles between them is the most important form of water available to the plants.
3. Soil reactions: Soil may be acidic or alkaline or neutral in their reaction. pH value of the soil solution determines the availability of plant nutrients. The best pH range of the soil for cultivation of crop plants is 5.5 to 6.8.
4. Soil nutrients: Soil fertility and productivity is the ability of soil to provide all essential plant nutrients such as minerals and organic nutrients in the form of ions.
5. Soil temperature: Soil temperature of an area plays an important role in determining the geographical distribution of plants. Low temperature reduces use of water and solute absorption by roots.
6. Soil atmosphere: The spaces left between soil particles are called pore spaces which contains oxygen and carbon-di-oxide.
7. Soil organisms: Many organisms existing in the soil like bacteria, fungi, algae, protozoans, nematodes, insects and earthworms, etc., are called soil organisms.

[OR]

(b) Describe the procedure involved in Blue-White colony selection methods.
Answer:
Blue- White Colony Selection Method is a powerful method used for screening of recombinant plasmid. In this method, a reporter gene lacZ is inserted in the vector. The lacZ encodes the enzyme β-galactosidase and contains several recognition sites for restriction enzyme.

β-galactosidase breaks a synthetic substrates called X-gal (5-bromo-4-chloroindolyl- β-D- galacto-pyranoside) into an insoluble blue coloured product. If a foreign gene is inserted into lacZ, this gene will be inactivated. Therefore, no-blue colour will develop (white) because p-galactosidase is not synthesized due to inactivation of lacZ.

Therefore, the host cell containing r-DNA form white coloured colonies on the medium contain X-gal, whereas the other cells containing non-recombinant DNA will develop the blue coloured colonies. On the basis of colony colour, the recombinants can be selected.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify the proper sequence.
(a) juvenile phase, senescent phase, vegetative phase
(b) juvenile phase, maturity phase, senescent phase
(c) vegetative phase, maturity phase, juvenile phase
(d) senescent phase, juvenile phase, vegetative phase
Answer:
(b) juvenile phase, maturity phase, senescent phase

Question 2.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?

Answer:

Question 3.
Darwin’s finches are an excellent example of _______.
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) geographical isolation
Answer:
(c) adaptive radiation

Question 4.
A 30 year old woman has bleedy diarrhoea for the past 14 hours, which one of the following organisms is likely to cause this illness?
(a) Streptococcus pyogenes
(b) Clostridium difficile
(c) Shigella dysenteriae
(d) Salmonella enteritidis
Answer:
(c) Shigella dysenteriae

Question 5.
Which of the following microorganism is used for the production of citric acid in industries?
(a) Lactobacillus bulgaris
(b) Penicillium citrinum
(c) Aspergillus niger
(d) Rhizopus nigricans
Answer:
(c) Aspergillus niger

Question 6.
The interaction in nature, where one gets benefit on the expense of other is __________.
(a) Predation
(b) Mutualism
(c) Amensalism
(d) Commensalism
Answer:
(d) Commensalism

Question 7.
Which of the following region has maximum bio-diversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 8.
The thickness of stratospheric ozone layer is measured in ____________.
(a) Sieverts units
(b) Melson units
(c) Dobson units
(d) Beaufort Scale
Answer:
(c) Dobson units

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
What is parthenogenesis? Give two examples from animals.
Answer:
Development of an egg into a complete individual without fertilization is known as parthenogenesis. It was first discovered by Charles Bonnet in 1745.
E.g. Honey bees, Aphis.

Question 10.
Mention the production site and action site of following hormones.
(a) GnRH (b) Relaxin
Answer:

Hormone	Production Site	Action Site
GnRH	Hypothalamus	Pituitary gland
Relaxin	Placenta	Pelvic joints and cervix

Question 11.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 12.
State Van’t Hoff’s rule.
Answer:
Van’t Hoff’s rule states that with the increase of every 10°C, the rate of metabolic activity is doubled or the reaction rate is halved with the decrease of 10°C.

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 14.
List any four adverse effect of noise.
Answer:

1. High blood pressure
2. Stress related ailments
3. Sleep disruption
4. Hearing impairment

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Ban- body.

Question 16.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

• The structural gene codes for proteins, rRNA and tRNA required by the cell.
• Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
• The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 17.
Explain the principles of Lamarckian theory.
Answer:

• The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
• The theory of inheritance of acquired characters – Characters that are developed during the life time of an organism are called acquired characters and these are then inherited.

Question 18.
List the causative agent, mode of transmission and symptoms for Diphtheria and Typhoid.
Answer:

Question 19.
ELISA is a technique based on the principles of antigen-antibody reactions. Can this technique be used in the molecular diagnosis of a genetic disorder such as Phenylketonuria?
Answer:
Yes, ELISA test can be done to diagnose phenylketonuria. The affected person does not produce the enzyme phenylalanine hydroxylase. If specific antibodies are developed against the enzyme and ELISA is performed, the unaffected person will show positive result due to antigen and antibody reaction, whereas the affected individual produces negative result.

[Note: phenylketonuria is an inherited metabolic disorder that causes the accumulation of Phenylalanine (an amino acid) in body cells due to defect in the synthesizing of an enzyme phenylalanine hydroxylase]

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) The following is the illustration of the sequence of ovarian events (a-i) in a human female.

(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) A- Primordial follicle; B- Primary follicle; C- Secondary follicle; D-Tertiary follicle; E- Mature graafian follicle; F- Ovulation (release of egg); G- Empty Graafian follicle; H- Corpus luteum; I – Corpus albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle
H – Corpus luteum
During development of ovum, the primary follicle gets surrounded by many layers of granular cells and forms a new layer called secondary follicle.

Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

[OR]

(b) Write short notes on the following.
(i) Brewer’s yeast
(ii) Ideonella sakaiensis
(iii) Microbial fuel cells
Answer:
(i) Brewer’s yeast – Saccharomyces cerevisiae is a widely used fungal species in preparation and softening of bakery products like dough.

(ii) Ideonella sakaiensis is a bacterium is used to recycle PET plastics. The enzyme PETase and MHETase in the bacterium breakdown the PET plastics into terephthalic acid and ethylene glycol.

(iii) A microbial fuel cell is a bio-electrochemical system that drives an electric current by using bacteria and mimicking bacterial interaction found in nature. Microbial fuel cells work by allowing bacteria to oxidize and reduce organic molecules. Bacterial respiration is basically one big redox reaction in which electrons are being moved around.

A MFC consists of an anode and a cathode separated by a proton exchange membrane. Microbes at the anode oxidize the organic fuel generating protons which pass through the membrane to the cathode and the electrons pass through the anode to the external circuit to generate current.

Question 21.
(a) Write the salient features of Human Genome Project.
Answer:

• Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
• An average gene consists of3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
• Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
• The chromosomal organization of human genes shows diversity. .
• There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes, whereas chromosome ‘Y’ has 231 genes.
• Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

[OR]

(b) Tropical regions are rich in biodiversity. Why?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

• Warm tropical regions between the tropic of Cancer and Capricorn on either side of equator possess congenial habitats for living organisms.
• Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
• The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
• The average rainfall is often more than 200 mm per year.
• Climate, seasons, temperature, humidity, photo periods are more or less stable and
  encourage both variety and numbers. .
• Rich resource and nutrient availability.

Students can Download Tamil Nadu 12th Biology Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 2 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 2.
pBR 322, BR stands for ________.
(a) Plasmid Bacterial Recombination
(b) Plasmid Bacterial Replication
(c) Plasmid Boliver and Rodriguez
(d) Plasmid Baltimore and Rodriguez
Answer:
(c) Plasmid Boliver and Rodriguez

Question 3.
Which of the given plant produces cardiac glycosides?
(a) Calotropis
(b) Acacia
(c) Nepenthes
(d) Utricularia
Answer:
(a) Calotropis

Question 4.
Methane is _______ times as effective as CO₂ at trapping heat.
(a) 5
(b) 10
(c) 20
(d) 100
Answer:
(c) 20

Question 5.
Statement 1: Arachis hypogea belongs to Fabaceae
Statement 2: It is a native of Brazil.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the Statements are correct
(d) Both the Statements are incorrect
Answer:
(c) Both the Statements are correct

Question 6.
Which of the following scientist developed world’s first cotton hybrid?
(a) Dr. B.P. Pal
(b) C.T. Patel
(c) Dr. K. Ramiah
(d) N.G.P. Rao
Answer:
(b) C.T. Patel

Question 7.
Ecosystem is the structural and functional unit of ecology. This statement was given by _______.
(a) Tansley
(b) Odum
(c) Charles Elton
(d) Edwin
Answer:
(b) Odum

Question 8.
Name the scientist(s) who rediscovered the Mendelian work? .
(i) Hugo de Vries (ii) Carl Correns (iii) Tschermak (iv) T.H. Morgan
(a) (i), (ii)and (iv) (b) (i), (ii)and (iii) (iii) ii, iii, iv (d) i, iii and iv
Answer:
(b) (i), (ii)and (iii)

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Detached leaf of Bryophyllum produces new plants. How?
Answer:
In Bryophyllum, the leaf is succulent and notched on its margin. Adventious buds develop at these notches and are called epiphyllous buds. They develop into new plants forming a root system and become independent plants when the leaf gets decayed.

Question 10.
Crossing over occurs only in germinal cells. Yes or No? Support your answer.
Answer:
No. Though crossing over is a common process in germinal cells rarely it also occurs in somatic cells during mitosis. Such crossing over is called mitotic crossing over or somatic crossing over.

Question 11.
Why do we use cryoprotectants in conservation process? Name any two cryoprotectant.
Answer:
Cryoprotectants are the protective agents that are used to protect the cells or tissues from the stress of freezing temperature. E.g: Sucrose, dimethyl sulphoxide.

Question 12.
What is Eltonian pyramid?
Answer:
Eltonian pyramid or Ecological pyramid is a graphic representation of the trophic structure and function at successive trophic levels of an ecosystem.

Question 13.
What do you understand by the term good ozone and bad ozone?
Answer:
The ozone layer of troposphere is called bad ozone.
The ozone layer of stratosphere is called good ozone because the layer acts as an shield for absorbing UV rays coming from sun which is harmful for living organisms causing ONA damage.

Question 14.
A person got irritation while applying chemical dye. What would be your suggestion for alternative?
Answer:
If a grey haired person is allergic on using chemical dyes then he can go for natural dyes like Henna. Henna is an organic dye obtained from leaves and young shoots of Lawsonia inermis. The principal colouring matter is Tacosone’ which is harmless and causes no irritation on skin.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Differentiate between incomplete dominance and codominance.
Answer:
Incomplete Dominance:

• In incomplete dominance, neither of the allele is not completely dominant to another allele rather combine and produce new trait.
• New phenotype is formed due to character blending (not alleles)
• Example : Pink flowers of Mirabilis Jalapa

Co-dominance:

• In co-dominance, both the alleles in heterozygote are dominant and the traits are equally expressed (joint expression)
• No formation of new phenotype rather both dominant traits are expressed, conjointly.
• Example : Red and white flowers of camellia.

Question 16.
What are the features that a vector must possess to facilitate cloning?
Answer:
The following are the features that are required to facilitate cloning into a vector.

• Origin of replication (ori): This is a sequence from where replication starts and piece of DNA when linked to this sequence can be made to replicate within the host cells.
• Selectable marker: In addition to ori the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
• Cloning sites: In order to link the alien DNA, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.

Question 17.
Define somatic embryogenesis? Give any two of its applications.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids.

Applications:

• Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
• Somatic embryoids can be used for the production of synthetic seeds.

Question 18.
What is TSM? How it is classified and what does it focuses on?
Answer:
TSM stands for Traditional Systems of Medicines India has a rich medicinal heritage. A number of Traditional Systems of Medicine (TSM) are practiced in India some of which come from outside India. TSM in India can be broadly classified into institutionalized or documented and non-institutionalized or oral traditions. Institutionalized Indian systems include Siddha and Ayurveda which are practiced for about two thousand years.

These systems have prescribed texts in which the symptoms, disease diagnosis, drugs to cure, preparation of drugs, dosage and diet regimes, daily and seasonal regimens. Non-institutional systems, whereas, do not have such records and or practiced by rural and tribal peoples across India. The knowledge is mostly held in oral form. The TSM focus on healthy lifestyle and healthy diet for maintaining good health and disease reversal

Question 19.
How do sacred groves help in the conservation of biodiversity?
Answer:
These are the patches or grove of cultivated trees which are community protected and are based on strong religious belief systems which usually have a significant religious connotation for protecting community. Each grove is an abode of a deity mostly village God Or Goddesses like Aiyanar or Amman. 448 grooves were documented throughout Tamil Nadu, of which 6 groves (Banagudi shola, Thirukurungudi and Udaiyankudikadu, Sittannnavasal, Puthupet and Devadanam) were taken up for detailed floristic and faunistic studies.

These groves provide a number of ecosystem services to the neighbourhood like protecting watershed, fodder, medicinal plants and micro climate control.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Draw a flow chart depicting the various types of ploidy.
Answer:

[OR]

(b) Mention the application of biotechnology.
Answer:

• Biotechnology is one of the most important applied interdisciplinary sciences of the 21st century. It is the trusted area that enables us to find the beneficial way of life.
• Biotechnology has wide applications in various sectors like agriculture, medicine, environment and commercial industries.
• This science has an invaluable outcome like transgenic varieties of plants e.g. transgenic cotton (Bt-cotton), rice, tomato, tobacco, cauliflower, potato and banana.
• The development of transgenics as pesticide resistant, stress resistant and disease resistant varieties of agricultural crops is the immense outcome of biotechnology.
• The synthesis of human insulin and blood protein in E.coli and utilized for insulin deficiency disorder in human is a breakthrough in biotech industries in medicine.
• The synthesis of vaccines, enzymes, antibiotics, dairy products and beverages are the products of biotech industries.
• Biochip based biological computer is one of the successes of biotechnology.
• Genetic engineering involves genetic manipulation, tissue culture involves aseptic cultivation of totipotent plant cell into plant clones under controlled atmospheric conditions.
• Single cell protein from Spirulina is utilized in food industries.
• Production of secondary metabolites, biofertilizers, biopesticides and enzymes.
• Biomass energy, biofuel, bioremediation and phytoremediation for environmental biotechnology.

Question 21.
(a) Differentiate Primary succession and Secondary succession.
Answer:
Primary succession:

• Developing in an barren area.
• Initiated due to a biological or any other external factors.
• No soil, while primary succession starts
• Pioneer species come from outside environment.
• It takes more time to complete.

Secondary succession:

• Developing in disturbed area.
• Starts due to external factors only.
• It starts where soil covers is already present.
• Pioneer species develop from existing environment.
• It takes comparatively less time to complete.

(b) Enumerate the characters of anemophilous flowers
Answer:
Anemophilous plants have the following characteristic features:

• The flowers are produced in pendulous, catkin-like or spike inflorescence.
• The axis of inflorescence elongates so that the flowers are brought well above the leaves.
• The perianth is absent or highly reduced.
• The flowers are small, inconspicuous, colourless, not scented, do not secrete nectar.
• The stamens are numerous, filaments are long, exerted and versatile.
• Anthers produce enormous quantities of pollen grains compared to number of ovules available for pollination. They are minute, light and dry so that they can be carried to long distances by wind.
• In some plants anthers burst violently and release the pollen into the air. Example: Urtica.
• Stigmas are comparatively large, protruding, sometimes branched and feathery, adapted to catch the pollen grains. Generally single ovule is present.
• Plant produces flowers before the new leaves appear, so the pollen can be carried without hindrance of leaves.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Which one of the following groups includes sexually transmitted diseases caused by bacteria only?
(a) Syphilis, gonorrhoea and candidiasis
(b) Syphilis, chlamydiasis and gonorrhoea
(c) Syphilis, gonorrhoea and trichomoniasis
(d) Syphilis, trichomoniasis and pediculosis
Answer:
(b) Syphilis, chlamydiasis and gonorrhoea

Question 2.
ABO blood group in man is controlled by __________.
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Hershey and Chase experiment with bacteriophage showed that ____________.
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 4.
Modern man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 5.
Allergy involves _______.
(a) IgE
(b) IgG
(c) IgA
(d) IgM
Answer:
(a) IgE

Question 6.
Recombinant Factor VIII is produced in the ______ of the Chinese Hamster.
(a) Liver cells
(b) blood cells
(c) ovarian cells
(d) brain cells
Answer:
(c) ovarian cells

Question 7.
Match with correct pair

(a) A – ii, B – iv, C – i, D – iii
(b) A – ii, B – iii, C – i, D – iv
(d) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – ii, C – i, D – iii
Answer:
(b) A – ii, B – iii, C – i, D – iv

Question 8.
Assertion (A): Snake is a stenotherm.
Reason (R): Organism can tolerate narrow temperature fluctuations.
(a) Both A and R are correct R explain A
(b) A is correct R is incorrect
(c) A and R are correct, R doesn’t explains A
(d) Both A and R are incorrect
Answer:
(a) Both A and R are correct R explain A

PART – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Mention the differences between spermiogenesis and spermatogenesis.
Answer:
Spermiogenesis:
Transformation of spermatids into mature sperm.

Spermatogenesis:
Spermatogenesis is the sequence of events in the seminiferous tubules of testes that produces male gametes, the sperms.

Question 10.
Draw and label a gemmule of sponge.
Answer:

Question 11.
State Wiener’s Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^Z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R², R⁰, R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 12.
Why t-RNA is called an adapter molecule?
Answer:
The transfer RNA (t-RNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule. This term was postulated by Francis Crick.

Question 13.
Why Red list is prepared periodically?
Answer:
The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity .
• Preparing conservation priorities and help in conservation of action
• Information on international agreements on conservation of biological diversity

Question 14.
Write notes on (a) Eutrophication (b) Algal Bloom
Answer:
Eutrophication refers to the nutrient enrichment in water bodies leading to lack of oxygen and will end up in the death of aquatic organisms.
Algal Bloom is an excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Explain Foetal-ejection reflex.
Answer:
As the pregnancy progresses, increase in the oestrogen concentration promotes uterine contractions. These uterine contractions facilitate moulding of the foetus and downward movement of the foetus. The descent of the foetus causes dilation of cervix of the uterus and vaginal canal resulting in a neurohumoral reflex called Foetal ejection reflex or Ferguson reflex.

This initiates the secretion of oxytocin from the neurohypophysis which in turn brings about the powerful contraction of the uterine muscles and leads to the expulsion of the baby through the birth canal.

Question 16.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 17.
State immunological surveillance theory.
Answer:
The concept of immunological surveillance postulates that the primary function of the immune system is to “seek and destroy” malignant cells that arise by somatic mutation. The efficiency of the surveillance mechanism reduces either as a result of ageing or due to congenital or acquired immuno deficiencies, leads to increased incidence of cancer. Thus, if immunological surveillance is effective, cancer should not occur. The development of tumour represents a lapse in surveillance.

Question 18.
Write a short note on Pathaneer.
Answer:
In some parts of South India, a traditional drink called pathaneer is obtained from fermenting sap of palms and coconut trees. A common source is tapping of unopened spadices of coconut. It is a refreshing drink, which on boiling produces jaggery or palm sugar. When pathaneer is left undisturbed for few hours it gets fermented to form toddy with the help of naturally occurring yeast, to form a beverage that contains 4 percent alcohol. After 24 hours toddy becomes unpalatable and is used for the production of vinegar.

Question 19.
Elucidate the methodology of ELISA test.
Answer:
During diagnosis the sample suspected to contain the antigen is immobilized on the surface of an ELISA plate. The antibody specific to this antigen is added and allowed to react with the immobilized antigen. The anti-antibody is linked to an appropriate enzyme like peroxidase.

The unreacted anti-antibody is washed away and the substrate of the enzyme (hydrogen peroxidase) is added with certain reagents such as 4-chloronaphthol. The activity of the enzyme yields a coloured product indicating the presence of the antigen.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) What are IUD’s? Explain its way of functioning. Also describe their types.
Answer:
Intrauterine Devices (IUDs) are inserted by medical experts in the uterus through the vagina. These devices are available as copper releasing IUDs, hormone releasing IUDs and nonmedicated IUDs. IUDs increase phagocytosis of sperm within the uterus. IUDs are the ideal contraceptives for females who want to delay pregnancy. It is one of the popular methods of contraception in India and has a success rate of 95 to 99%.

Copper releasing IUDs differ from each other by the amount of copper. Copper IUDs such as Cu T-380 A, Nova T, Cu 7, Cu T 380 Ag, Multiload 375, etc. release free copper and copper salts into the uterus and suppress sperm motility. They can remain in the uterus for five to ten years.

Hormone-releasing IUDs such as Progestasert and LNG – 20 are often called as intrauterine systems (IUS). They increase the viscosity of the cervical mucus and thereby prevent sperms from entering the cervix.

Non-medicated IUDs are made of plastic or stainless steel. Lippes loop is a double S-shaped plastic device.

[OR]

(b) Explain how Urey – Miller’s experiment supports the origin of life?
Answer:

Urey and Miller (1953), paved way for understanding the possible synthesis of organic compounds that led to the appearance of living organisms is depicted in the Figure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode.

A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of gases (ammonia, methane and hydrogen) in the large chamber that was connected to the boiling water. The steam condensed to form water which ran down the ‘U’ Diagrammatic representation of Urey-Miller’s experiment tube. Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified.

Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 21.
(a) Discuss any two Allosomal abnormalities in human beings.
Answer:
Allosomal abnormalities in human beings:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities.
Several sex chromosomal abnormalities have been detected. E.g. Klinefelter’s syndrome and
Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males):
This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47, XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females):
This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45 ,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome:

[OR]

(b) Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
1. Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than the large population due to inbreeding depression (less adaptivity and variation)

2. Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow water marine invertebrates disappeared.

3. Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, over exploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction.

The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

Students can Download Tamil Nadu 12th Commerce Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Commerce Model Question Paper 4 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part -I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
Drucker stresses jobs of management.
(a) three
(b) two
(c) four
(d) Six
Answer:
(a) three

Question 2.
Choose the odd one out?
(a) Directing
(b) Motivating
(c) Innovation
(d) Controlling
Answer:
(c) Innovation

Question 3.
…………… harmonises the goal of an individual with the organisation’s goal.
(a) MBO
(b) MBE
(c) MBA
(d) MBM
Answer:
(a) MBO

Question 4.
Indian organised financial system consist of sectors.
(a) Two
(b) Four
(c) Three
(d) Six
Answer:
(b) Four

Question 5.
Capital market otherwise called as …………..
(a) Securities market
(b) Share market
(c) Debenture market
(d) Money market
Answer:
(a) Securities market

Question 6.
……………………. deals with the financial assets and securities whose maturity period does not exceed one year.
(a) Money market
(b) Capital market
(c) Stock exchange
(d) Government bonds
Answer:
(a) Money market

Question 7.
The consumer is to be protected against any ……………..
(a) unfair practices of trade
(b) family functions
(c) profit making firm
(d) loss in business
Answer:
(a) unfair practices of trade

Question 8.
The National Commission should have members.
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(d) 6

Question 9.
………………. is the latest outcome of liberalization.
(a) Privatisation
(b) Globalisation
(c) Foreign collaboration
(d) None of these
Answer:
(c) Foreign collaboration

Question 10.
A bill of exchange drawn on a specified banker is ……………
(a) promissory note
(b) cheque
(c) hundi
(d) share
Answer:
(b) cheque

Question 11.
The entrepreneurs classified on the basis of types of business are …………….
(i) Industrial entrepreneur
(ii) Technical entrepreneur
(iii) Professional entrepreneur
(iv) Business entrepreneur
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b) (i) and (iv)

Question 12.
Board meetings to be conducted minimum times in a year.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 13.
A public company shall have a minimum of ……………. directors.
(a) Twelve
(b) Seven
(c) Three
(d) Two
Answer:
(c) Three

Question 14.
Which of the following types of shares are issued by a company to rains capital from the existing shareholders?
(a) Equity shares
(b) Rights shares
(c) Preference shares
(d) Bonus shares
Answer:
(b) Rights shares

Question 15
should cover aspects like sources of finance, technical know-how, source of labours and raw material, market potential and profitability.
(a) Technical Report
(b) Finance Report
(c) Project Report
(d) Progress Report
Answer:
(c) Project Report

Question 16.
Find the odd one out is context of Trading Entrepreneur.
(a) Selling
(b) Commission
(c) Buying
(d) Manufacturing
Answer:
(d) Manufacturing

Question 17.
Which of the below is not classified into managerial function.
(a) Planning
(b) Marketing
(c) Organising
(d) Controlling
Answer:
(c) Organising

Question 18.
Section 6 of Negotiable Instruments Act 1881 deals with ……………
(a) Promissory Note
(b) Bills of exchange
(c) Cheque
(d) None of the above
Answer:
(c) Cheque

Question 19.
……………… results from the removal of barriers between national economics to encourage the flow of goods, services, capital and labour.
(a) privatisation
(b) Liberalisation
(c) Globalisation
(d) Foreign Trade
Answer:
(c) Globalisation

Question 20.
The state commission can entertain complaints where the value of the goods or services and the compensation, if any claimed exceed.
(a) ₹ 2 lakhs but does not exceed ₹ 5 lakhs
(b) ₹ 20 lakhs but does not exceed ₹ 1 crore
(c) ₹3 lakhs but does not exceed ₹ 5 lakhs
(d) ₹ 4 lakhs but does not exceed ₹ 20 lakhs
Answer:
(d) ₹ 4 lakhs but does not exceed ₹ 20 lakhs

PART – II

Answer any seven in which Question No.30 is compulsory. [7 x 2 = 14]

Question 21.
State the meaning of Authority.
Answer:
Authority means the right of a superior to give the order to his subordinates. This is the issue of commands followed responsibility for their consequences.

Question 22.
How is price decided in a secondary market?
Answer:
Financial markets allow for the determination of the price of the traded financial asset through the interaction of buyers and sellers. They provide a signal for the allocation of funds in the economy, based on the demand and supply, through the mechanism called price discovery processes.

Question 23.
What is meant by Remiser?
Answer:
Remiser is an agent of a member of a stock exchange. He obtains business for his principal i.e. the member.

Question 24.
What is Human Resource Management ?
Answer:
Human Resource Management is a function of management concerned with hiring, motivating and maintaining people in an organisation. It focuses on people in an organisation.

Question 25.
Define Marketer.
Answer:
“A person whose duties include the identification of the goods and services desired by a set of consumers, as well as the marketing of those goods and services on behalf of a company”.

Question 26.
What is Caveat Emptor?
Answer:
‘Caveat emptor’ is a Latin term that means “let the buyer beware.” Similar to the phrase “sold as is,” this term means that the buyer assumes the risk that a product fails to meet expectations or have defects.

Question 27.
Expand VUCA.
Answer:
VUCA – volatility, uncertainty, complexity and ambiguity.

Question 28.
What is meant by goods?
Answer:
The term goods mean every kind of movable property other than actionable claim and money.

Question 29.
List down the managerial functions of entrepreneurs.
Answer:

• Planning
• Organising
• Directing
• Controlling
• Coordination

Question 30.
What do you mean by Equity Share?
Answer:
The share of a company which do not have any preferential rights with regard to dividend and repayment of share capital at the time of liquidation of a company, is called as equity share or ordinary share.

Part – III

Answer any seven in which question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
What are the principles of Taylor?
Answer:
Principles of scientific management propounded by Taylor are:

• Science, Not Rule of Thumb
• Harmony, Not Discord
• Mental Revolution
• Cooperation, Not individualism
• Development of each and every person to his or her greatest efficiency and prosperity.

Question 32.
Write a note on Secondary Market.
Answer:
Secondary Market is the market for securities that are already issued. Stock Exchange is an important institution in the secondary market.

Question 33.
Explain National Stock Market System. (NSMS)
Answer:
National stock market system was advocated by the High Powered Group. It is headed by Shri. Pherwani (popularly known as Pherwani Committee). At present the National Stock Market comprises the following:

• National Stock Exchange of India Limited (NSE)
• Stock Holding Corporation of India Limited (SHCIL)
• Securities Trading Corporation of India (STCI)

Question 34.
Marketer is an innovator? Do you agree?
Answer:
Marketer seeks to distinguish his products/services by adding additional features or functionalities to the existing product, modifying the pricing structure, introducing new delivery pattern, creating new business models, introducing change in production process and so on.

Question 35.
What is meant by artificial scarcity?
Answer:
There are certain situations where the shop-keepers put up the board ‘No Stock” in front of their shops, even though there is plenty of stock in the store. In such situations consumers who are desperate to buy such goods have to pay hefty price to buy those goods and thus earning more profit unconscientiously.

Question 36.
State the framework of Corporate Governance in India.
Answer:
The Indian Corporate Governance framework requires listed companies

• To have independent directors on the board; At least one third of the directors have to be independent directors.
• To have at least one independent woman director,
• To disclose all deals and payments to related parties.
• To disclose details of managerial compensation.

Question 37.
Discuss the implied conditions and warranties in sale of goods contract.
Answer:
In every contract of sale, there are certain expressed and implied conditions and warranties. The term implied conditions means conditions which can be inferred from or guessed from the context of the contract.Following are the implied conditions:

• Conditions as to Title
• Conditions as to Description
• Sale by Sample
• Conditions as to Quality or Fitness
• Conditions as to Merchantability
• Condition as to Wholesomeness
• Condition Implied by Trade Usage Following are the implied warranties:
• Quiet Possession
• Free from Any Encumbrances
• Warranty in the case of Dangerous Goods

Question 38.
List the commercial functions of Entrepreneur and explain them shortly.
Answer:
Commercial Functions of Entrepreneur:
Production or Manufacturing: Under production function, entrepreneur has to take decision relating to selection of factory site, design and layout, type of products to be manufactured, research and development, product design, etc.

Marketing: Entrepreneur has to carry out following functions pertaining to marketing aspect namely consumer research, product planning and development, standardisation, packaging, pricing, warehousing, distribution, promotion etc.

Accounting: Entrepreneur has to arrange to prepare trading and profit and loss account in order to know the profit or loss incurred out of operation of the business and prepare balance sheet to know the financial status of business at a particular day.

Finance: In the sphere of financial function, an entrepreneur has to take decisions like choosing the right type of financing, framipg the best dividend policy, acquiring of funds, efficiently managing fixed and current assets, maximising shareholders wealth and investing of funds efficiently and effectively.

Human Resource Management: Entrepreneur has to estimate the manpower needs of the enterprise and accordingly decide the size of manpower required for various slots of organisational structure.

Question 39.
What is issue of shares at discount? What conditions should be fulfilled?
Answer:
When the shares are issued at a price below the face value they are said to be issued at a discount. For example, a share having the face value of Rs 10 is issued at Rs 8. The companies act 2013, prohibits the issue of shares at discount (Section 53), except sweat Equity share.

Question 40.
What do you understand by ‘Poll’?
Answer:
Poll means tendering or offering vote by ballot to a specially appointed officer, called the polling officer. Under the Companies Act, poll means exercising voting right in proportion to shareholder’s contribution to the paid-up capital of a limited company having a share capital.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41.
(a) Discuss the implications of span of management.
Answer:
The Span of Management has two implications:
(i) Influences the complexities of the individual manager’s job.
(ii) Determine the shape or configuration of the Organisation.
The span of management is related to the horizontal levels of the organisation structure. There is a wide and a narrow span of management. With the wider, span, there will be less hierarchical levels, and thus, the organisational structure would be flatter. Whereas, with the narrow span, the hierarchical levels increases, hence the organisational structure would be tall.

• Both these organisational structures have their advantages and the disadvantages. But however the tall organisational structure imposes more challenges.
• Since the span is narrow, which means less number of subordinates under one superior, requires more managers to be employed in the organisation. Thus, it would be very expensive in terms of the salaries to be paid to each senior.
• With more levels in the hierarchy, the communication suffers drastically. It takes a lot of time to reach the appropriate points, and hence the actions get delayed.
• Lack of coordination and control because the operating staff is far away from the top management.

[OR]

(b) What are the functions of Financial Markets?
Answer:
Intermediary Functions
(i) Transfer of Resources: Financial markets facilitate the transfer of real economic resource from lenders to ultimate borrowers.

(ii) Enhancing Income: Financial markets allow lenders earn interest/dividend on their surplus
investible funds and thus contributing to the enhancement of the individual and the national income.

(iiI) Productive Usage: Financial markets allow for the productive use of the funds borrowed and thus enhancing the income and the gross national production.

(iv) Capital Formation: Financial markets provide a channel through which new savings flow to aid capital formation of a country.

(v) Price Determination: Financial markets allow for the determination of the price of the traded financial asset through the interaction of buyers and sellers.

(vi) Sale Mechanism: Financial market provides a mechanism for selling of a financial
asset by an investor so as to offer the benefits of marketability and liquidity of such assets.

(vii) Information: The activities of the participants in the financial market result in the generation and the consequent dissemination of information to the various segments of the markets, so as to reduce the cost of transaction of financial assets.

Financial Functions

• Providing the borrowers with funds so as to enable them to carry out their investment plans.
• Providing the lenders with earning assets so as to enable them to earn wealth by deploying the assets in productive ventures.
• Providing liquidity in the market so as to facilitate trading of funds

Question 42.
(a) Explain the various types of New Financial Institutions.
Answer:
(i) Venture Fund Institutions: Venture capital financing is a form of equity financing designed especially for funding new and innovative project ideas. Venture capital funds bring into force the hi-technology projects which are converted into commercial  production.

(ii) Mutual Funds: Financial institutions that provide facilities for channeling savings of small investors into avenues of productive investments are called ‘Mutual Funds’.

(iii) Factoring Institutions: “Factoring” is an arrangement whereby a financial institution provides financial accommodation on the basis of assignment/sale of account receivables.

(iv) Over the Counter Exchange of India (OTCEI): The OTCEI was set up by a premier
financial institution to allow the trading of securities across the electronic counters throughout the country.

(v) National Stock Exchange of India Limited (NSEI): NSEI was established in 1992 to function as a model stock exchange. The Exchange aims at providing the advantage of nation-wide electronic screen based “scripless” and “floorless” trading system in securities.

(vi) National Clearance and Depository System (NCDS): Under the scripless trading system, settlement of transactions relating to securities takes place through a book entry.

(vii) National Securities Depositories Limited: The NSDL was set up in the year 1996 for achieving a time bound dematerialization as well as rematerialization of shares.

(viii) Stock Holding Corporation of India Limited (SHCIL): Stock Holding Corporation of India Limited (SHCIL) aims at serving as a central securities depository in respect of transactions on stock exchanges. The Corporation also takes up the administration of clearing functions at a national level.

[OR]

(b) What are the features of Government Securities?
Answer:
(i) Agencies: Government securities are issued by agencies such as Central Government, State Governments, semi-government authorities like local Government authorities.
(ii) RBI Special Role: RBI takes a special and an active role in the purchase and sale of these securities as part of its monetary management exercise.
(iii) Nature of Securities: Securities offer a safe avenue of investment through guaranteed payment of interest and repayment of principal by the Government.
(iv) Liquidity Profile: The liquidity profile of gilt-edged securities varies. Accordingly liquidity profile of securities issued by Central Government is high.
(v) Tax Rebate: A striking feature of these securities is that they offer wide-range of tax incentives to investors.

(vi) Market: As each sale and purchase has to be negotiated separately, the Gilt-Edged Market is an Over-The-Counter Market.
(vii) Forms: The securities of Central and State Government take such forms as inscribed stock or stock certificate, promissory note and bearer bond.
(viii) Participants: The participants in Government securities market include the Government sector comprising Central and State Governments
(ix) Trading: Small and less active, banks and corporate holders who purchase and sell Government securities on the stock exchanges participate in trading.
(x) Issue Mechanism: The Public Debt Office (PDO) of the RBI undertakes to issue government securities.
(xi) Issue opening: A notification for the issue of the securities is made a few days before the public subscription is open.
(xii) Grooming Gradual: It is the acquisition of securities nearing maturity through the stock exchanges by the RBI.
(xiii) Switching: It is the purchase of one security against the sale of another security carried out by the RBI in the secondary market as part of its open market operations.
(xiv) Auctioning: A method of trading whereby merchants bid against one another and where the securities are sold to the highest bidder.

Question 43.
(a) Distinguish between Stock Exchange and Commodity Exchange.
Differences between Stock Exchange and Commodity Exchange.

	Stock Exchange	Commodity Exchange
(0	Meaning: Stock Exchange is an organised market for purchase and sale of industrial and financial securities.	A commodity exchange is an exchange where commodities are purchased and sold. E.g.: Gold, Silver, Rice, Wheat
(ii)	Function: Providing easy marketability.	Offering hedging or price insurance services.
(Hi)	Object: It is facilitating capital formation.	It will facilitate goods flow through risk reduction.
(iv)	Participants: Investors and Speculators are participating in stock exchange.	Producers, dealers, traders are involved in the commodity exchange.
(v)	Articles traded: Industrial securities such as stocks and bonds and government securities.	Only durable and graded goods are traded in this.

[OR]

(b) Differentiate HR from HRM.

S.No.	Human Resource (HR)	Human Resource Management (HRM)
1.	Meaning: Human resource, i.e., the employees create other sources.	Managing the human resource is known as human resource management.
2.	Human resource exhibits innovation and creativity.	Human resource management has universal relevance.
3.	Human resource alone can think, act and analyse.	The accomplishment of organisational goal is made possible through human resource in an organisation.
4.	Human resources are movable.	Human resource management is intangible function.
5.	Human resource can work as team.	Human resource management focuses on the development of manpower through training . and development programmes.

Question 44.
(a) Explain the important methods of interview.
Answer:
Interview means a face to face interaction between the interviewer and interviewee.
Interview may be of various types:-

• Preliminary Interview: It is conducted to know the general suitability of the candidates who have applied for the job.
• Structured Interview: In this method, a series of questions is to be asked by the interviewer. The questions may be pre-prepared.
• In depth Interview: This interview is conducted to test the level of knowledge of the interviewee in a particular field.
• Panel Interview: Where a group of people interview the candidate. The panel usually comprises the chair person, subject expert, psychological experts and so on.
• Stress Interview: This type of interview is conducted to test the temperament and emotional balance of the candidate.
• Online Interview: Due to tremendous growth in information and communication technology, interviews are conducted by means of internet via Skype, Google duo, Whatsapp.

[OR]

(b) What is your contribution to promote the market in the modern society?
Answer:
Market is a place where buyers and sellers gather for purchase and sale.
Market may be of Local market, national market and international or Global market.
To develop and promote the market the following are needed:

• Eligible and satisfied and customer is needed.
• Quality and durable goods are to be marketed.
• Recent trends like E-marketing, online marketing are to be encouraged.
• After sales service are to be provided to durable goods.
• Customers are to be financed for buying costly articles.
• New Innovations and marketing research are to be introduced to develop the market.

Question 45.
(a) Discuss any two new methods of marketing with its advantages.
Answer:
Rural Marketing: Rural marketing is a process of developing pricing, promoting and distributing rural specific goods and serv ices leading to exchange with rural customers.There is inflow of goods into rural markets for production and consumption and there is also outflow of products to urban areas.
The rural to urban flow consists of agricultural products like rice, wheat and sugar etc.

Service Marketing: Service marketing is a special branch of marketing. It denotes the processing of selling service goods like telecommunication, banking, insurance, healthcare, tourism and professional services. The service products are mostly intangible. The unique feature of services marketing warrant different strategies compared with the marketing of physical goods.

[OR]

(b) Write about five important consumer legislations.
Answer:
To protect the consumers from the unfair traders, the government passed various legislative Acts – They are follows:

• The Indian Contract Act, 1982 was passed to bind the people on the promise made in the contract.
•  The Essential Commodities Act, 1955 protects the consumers against artificial shortages created by the sellers by hoarding the goods.
• The Prevention of Food Adulteration Act, 1954 checks the adulteration of food articles and ensures purity of goods supplied.
•  Weights and Measures Act, 1958 protects the consumer against malpractices of underweight dr under measurement.

Question 46.
(a) Explain the term District Forum and explain the functions of District Forum.
Answer:
As per the Consumer Protection Act 1986, a district forum is established in each and every district to solve the problems in the concerned district. Present or Retired district judge is the president of District forum

Functions:

• The complaints relating to the district can be solved by district forum.
• Compensation that can be claimed is less than Rs. 20 lakhs.
• Sometimes the district forum also may pass orders against traders.

[OR]

(b) Distinguish between Condition and Warranty.

Basis for Difference	Condition	Warranty
1. Meaning	It is a stipulation which is essential to the main purpose of the contract of sale.	It is a stipulation which is collateral to the main purpose of contract.
2. Significance	Condition is necessary to the contract that the breaking of which cancels out the contract.	The violation of warranty will not revoke the contract.
3. Transfer of Ownership	Ownership on goods cannot be transferred without fulfilling the conditions.	Ownership on goods can be transferred on the buyer without fulfilling the warranty.
4. Remedy	In case of breach of contract, the affected party can cancel the contract and claim damages.	In the case of breach of warranty, the affected party cannot cancel the contract but can claim damages only.
5. Treatment	Breach of condition may be treated as breach of warranty.	Breach of warranty cannot be treated as breach of condition.

Question 47.
(a) Explain the different kinds of endorsements.
Answer:
When the person signs on the back of the instrument to transfer his interest, it is known as endorsement. The endorsement are of various types:

(i) Blank or general endorsement : When the endorser puts his mere signature on the back of an instrument without mentioning the name of the person to whom the endorsement is made, it is called Blank Endorsement

(ii) Endorsement in full or special endorsement: If the endorser, in addition to his signature, mentions the name of the person to whom it is endorsed, is known as endorsement in full or special endorsement.

(iii) Conditional endorsement: When the endorser of a negotiable instrument makes his liability dependent upon the happening of an event which may or may not happen, it is called conditional endorsement.

(iv) Restrictive endorsement : When an endorsement restricts or prohibits further negotiability of the instrument, it is, called Restrictive Endorsement.

(v) Partial Endorsement :Where the endorsement seeks to transfer only a part of the amount payable under the instrument, the endorsement is called Partial Endorsement.

[OR]

(b) Write the difference between Share Certificate and Share Warrant.
Answer:

S.No.	Share Certificate	Share Warrant
1.	A share certificate is an instrument in writing for the legal proof of the ownership.	A share warrant is a negotiable instrument, issued against fully paid up shares.
2.	Every company must issue share certificate to its shareholders.	There is no compulsion of the issue of share warrants by the company.
3.	Normally, the holder of the share certificate is to be the member of the company.	Generally, the holder of the share warrant is not the member of the company.
4.	The share certificate is issued by the company within three months of the allotment of shares.	Shares warrant can be issued only when the shares are fully paid up.
5.	No need to authorized in the Articles of Association to issue share certificate.	The issue of a share warrant must be authorized in the Articles of Association of the company.

 

Students can Download Tamil Nadu 12th Commerce Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Commerce Model Question Paper 3 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part -I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
How many times a security can be sold in a secondary market?
(a) Only one time
(b) Two times
(c) Three times
(d) Multiple times
Answer:
(d) Multiple times

Question 2.
Money market Institutions are ………………
(a) Investment Houses
(b) Mortgage Banks
(c) Reserve Bank of India
(d) Commercial Banks and Discount Houses
Answer:
(d) Commercial Banks and Discount Houses

Question 3.
Risk in the Money Market is ………………
(a) High
(b) Market Risk
(c) Low credit and Market Risk
(d) Medium Risk
Answer:
(c) Low credit and Market Risk

Question 4.
A pessimistic speculator is ………………
(a) Stag
(b) Bear
(c) Bull
(d) Lame Duck
Answer:
(b) Bear

Question 5.
An optimistic speculator is ……………
(a) Bull
(b) Bear
(c) stag
(d) Lame Duck
Answer:
(a) Bull

Question 6.
SEBI is empowered by the Finance Ministry nominate ……………. mumbers on the governing body of every stock exchange.
(a) 5
(b) 3
(c) 6
(d) 7
Answer:
(b) 3

Question 7.
The process of converting physical shares into electronic from is called from is called …………..
(a) Dematerialisation
(b) Delisting
(c) Materialisation
(d) Debarring
Answer:
(a) Dematerialisation

Question 8.
Labour turnover is the rate at which employees the organisation.
(a) enter
(b) leave
(c) salary
(d) none of the above
Answer:
(b) leave

Question 9.
Which of the following test is used to measure the various characteristics of the candidate?
(a) physical Test
(b) psychological Test
(c) attitude Test
(d) proficiency Tests
Answer:
(b) psychological Test

Question 10.
When trainees are trained by supervisor or by superior at the job is called …………..
(a) Vestibule training
(b) Refresher training
(c) Role Pay
(d) Apprenticeship training
Answer:
(d) Apprenticeship training

Question 11.
Stock Exchange Market is also called …………..
(a) Spot Market
(b) Local Market
(c) Security Market
(d) National Market
Answer:
(c) Security Market

Question 12.
Marketing mix means a marketing program that is offered by a firm to to its target ………….. to earn profits through satisfaction of their wants.
(a) Wholesaler
(b) Retailer
(c) Consumer
(d) Seller
Answer:
(c) Consumer

Question 13.
Effective use of Social media marketing increase conversion rates of ……………..
(a) Customer to buyers
(b) Retailer to customers
(c) One buyer to another buyer’s
(d) Direct contact of marketer
Answer:
(a) Customer to buyers

Question 14.
The Consumer protection Act came into force with effect from ……………
(a) 1.1. 1986
(b) 1.4.1986
(c) 15.4. 1987
(d) 15.4.1990
Answer:
(c) 15.4. 1987

Question 15.
Which of the following is not consumer right summed up by John F. Kennedy
(a) Right to safety
(b) Right to choose
(c) Right’to consume
(d) Right to he informed
Answer:
(a) Right to safety

Question 16.
The National Consumer Disputes Redressal Commission has jurisdiction to entertain complaints where the value of goods/services complained against and the compensation, if any, claimed is………………..
(a) Exceeding ₹ 1 crore
(b) Exceeding ₹ 10 lakhs
(c) Exceeding ₹ 5 lakhs
(d) Exceeding ₹ 12 lakhs
Answer:
(a) Exceeding ₹ 1 crore

Question 17.
New Economic Policy was introduced in the year
(a) 1980
(b) 1991
(c) 2013
(d) 2015
Answer:
(b) 1991

Question 18.
In which of the following types, the ownership is immediately transferred to buyer?
(a) When goods are ascertained
(b) When goods are appropriate
(c) Delivery to the carrier
(d) Sale or return basis
Answer:
(c) Delivery to the carrier

Question 19.
………………. cannot be a bearer instrument.
(a) Cheque
(b) Promissory Note
(c) Bills of exchange
(d) None of the above
Answer:
(a) Cheque

Question 20.
Which of the below is not classified into managerial functions?
(a) Planning
(b) Marketing
(c) Organising
(d) Controlling
Answer:
(c) Organising

Part – II

Answer any seven in which Question No. 30 is compulsory. [7 x 2 = 14]

Question 21.
What do you mean by Span of Management?
Answer
The Span of Management refers to the number of subordinates who can be managed efficiently by a superior. Simply, the manager having the group of subordinates who report him directly is called as the span of management.

Question 22.
What is Mutual Fund?
Answer
Financial institutions that provide facilities for channeling savings of small investors into avenues of productive investments are called ‘Mutual Funds’. A mutual fund company invests the funds pooled from shareholders and gives them the benefit of diversified investment portfolio and a reasonable return.

Question 23.
Who is called a Broker?
Answer
Broker is a commission agent. He acts as an intermediary between buyers and sellers of securities. He charges a commission for his services from both the parties.

Question 24.
What is Mentoring training method?
Answer
Mentoring is the process of sharing knowledge and experience of an employee. The focus in this training is on the development of attitude of trainees. It is mostly used for managerial employees.

Question 25.
Define E-Marketing.
Answer
“E-Marketing is achieving marketing objectives through use of digital technologies like Internet, world wide web, e-mail, wireless media, and management of digital customer data and electronic customer management systems.

Question 26.
What is caveat Venditor?
Answer
Today, most sales in the U.S. fall under the principle of caveat venditor, which means “let the seller beware,” by which goods are covered by an implied warranty of merchantability.

Question 27.
What is Privatisation.
Answer
Privatization is the incidence or process of transferring ownership of a business enterprise, agency or public service from the government to the private sector.

Question 28.
List three characteristics of a promissory Note.
Answer
Characteristics of a Promissory Note:

• A promissory note must be in writing.
• The promise to pay must be unconditional.
• It must be signed by the maker.

Question 29.
List down the promotional functions of entrepreneurs.
Answer

• Discovery of Idea
• Determining the business objectives
• Detailed Investigation
• Choice of form of enterprise
• Fulfilment of the formalities
• Preparation of Business Plan
• Mobilisation of funds
• Procurement of Machines and Materials

Question 30.
Name the companies required to appoint KMP?
Answer
Following Companies are required to appoint KMP:

Part – III

Answer any seven in which question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
What determines the span of management?
Answer:
The Span of Management has two implications:

• Influences the complexities of the individual manager’s job.
• Determine the shape or configuration of the organisation.
• There is a wide and a narrow span of management.

Question 32.
Bring out the scope of financial market in India.
Answer:
The financial market provides financial assistance to individuals, agricultural sectors, industrial sectors, service sectors, financial institutions like banks, insurance sectors, provident funds and the government as a whole. With the help of the financial market all the above stated individuals, institutions and the Government can get their required funds in time. Through the financial market the institutions get their short term as well as long term financial assistance. It leads to overall economic development.

Question 33.
Explain the types of Treasury Bills?
Answer:
Treasury Bills are issued to the public and other financial institutions for meeting the short-term financial requirements of the Central Government. Treasury Bills may be classified into three. They are:

• 91 days Treasury Bills
• 182 days Treasury Bills and
• 64 days Treasury Bills 34.

Question 34.
Explain National Stock Exchange. (NSE)
Answer:
National Stock Exchange was incorporated in November, 1992. It uses satellite link to spread trading throughout the country. Through computer network, members’ orders for buying and selling within prescribed price are matched by central computer.

Question 35.
Name the types of Selection Test.
Answer:
Selection tests are of two types:

• Ability Tests and Personality Tests. Ability tests can further be divided into: aptitude test, achievement test, intelligence test, and judgement test.
• Personality tests can further be divided into: interest test, personality inventory test, projective test or thematic appreciation test, and attitude test.

Question 36.
Write the importance of Consumerism.
Answer:
Importance of consumerism lies in

• Awakening and uniting consumers
• Discouraging unfair trade practices
• Protecting against exploitation
• Awakening the government.
• Effective implementation of consumer protection laws
• Providing complete and latest information
• Discouraging anti-social activities

Question 37.
What are the advantages of disinvestment?
Answer:
Disinvestment in PSUs: The Govt, has started the process of disinvestment in those PSUs which had been running into loss. It means that Govt, has been selling out these industries to private sector. So disinvestment is a system of privatizing government enterprises.

Question 38.
Explain the promotional functions of entrepreneur.
(i) Discovery of Idea: The first and foremost function of entrepreneur is idea generation. A person may conceive his own ideas or develop the ideas contributed by others. Ideas can be generated through several ways like own experience and exposure of entrepreneur, keen observation of environment, education, training, market survey, environmental scanning and so on.

(ii) Determining the business objectives: Entrepreneur has to develop business objectives in the backdrop of nature of business and type of business activity i.e. nature of business, manufacturing or trading, type of business organisation chosen so that he/she can organise the venture in accordance with the objectives determined by him/her.

(iii) Detailed Investigation: Entrepreneur should investigate commercial feasibility of the product proposed to be produced and conduct market study to ascertain the potential demand for the product.

(iv) Choice of form of enterprise: Entrepreneur has to choose the appropriate form of organisation suited to implement the venture. There are various forms of organisation namely sole proprietor, partnership, company and co-operatives etc. which are in existence.

(v) Fulfilment of the formalities: Having chosen the appropriate type of organisation, entrepreneur has to take necessary steps to establish the form of organisation chosen. As regards sole trader, the formalities are barest minimum. In the case of partnership firm, entrepreneur has to arrange for partnership deed and he has to get the deed registered.

(vi) Preparation of Business Plan: Entrepreneur has to prepare a business plan or project report
of the venture that he is proposing to take up.

(vii) Mobilisation of funds: Entrepreneur has to take steps to mobilise capital needed to implement the venture. Entrepreneur has to estimate the fixed capital and working capital required for running the project.

(viii) Procurement of Machines and Materials: Entrepreneur has to locate the various sources of supply of machineries, equipments and materials.

Question 39.
What is Causal Vacancy?
Answer:
If a director is absent from India, for a period which is not less than three months, then it is called casual vacancy. It may be filled by the appointment of alternate director. The appointment must be authorised by the articles of association by passing a resolution in the meeting.

Question 40.
Give any three cases in which an ordinary resolution need to be passed.
Answer:
An ordinary resolution is one which can be passed by a simple majority, i.e. if the members of votes cast by members, entitled to vote in favour of the resolution is more than the votes cast against the resolution.
Ordinary Resolution is required for the following matters:

• To change or rectify the name of the company
• To alter the share capital of the company
• To redeem the debentures

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41
(a) Explain Lombard Street and Wall Street.
Answer:
Lombard Street
Lombard Street is in London. It is a street notable for its connections with the merchants, banking and insurance industries of London. From bank junction, where nine streets converge by the bank of England, Lombard Street runs southeast for a short distance.

Wall Street
Wall Street is a street in New York. It is the original home of the New York Stock exchange.
Also it is the historic headquarters of the largest U.S. brokerages and investment banks.
Wall Street is also used as a collective name for the financial and investment community.

[OR]

(b) Discuss the Operating functions of HRM.
Answer:
Operating functions of HRM:

• Procurement: Acquisition deals with job analysis, human resource planning, recruitment, selection, placement and promotion.
• Development: It includes performance appraisal, training, executive development, and organisational development.
• Compensation: It deals with job evaluation, wage and salary administration, incentives, bonus schemes.
• Retention: This is made possible through health and safety, social security, job satisfaction and quality of work life.
• Maintenance: This encourages employees to work with job satisfaction, reducing labour turnover, for human resource.

Question 42.
(a) Differentiate Recruitment and Selection.
Answer:

Basis for comparison	Recruitment	Selection
1. Meaning	It means searching candidates for the right job.	It refers to the process of selecting the suitable candidates and offering them job.
2. Objective	Inviting large number of candidates to apply for the vacant post.	Picking up the most suitable candidates.
3. Method	It is an economical method.	It is an expensive method.
4. Contractual relation	It involves the communication of vacancies. There is no contractual relation.	It creates contractual relation between employer and employee.
5. Process	Recruitment process is very simple.	Selection process is very complex and complicated.
6. Time	Requires less time since it involves merely identifying vacancies and advertising them.	It is more time-consuming because each and every candidate has to be tested on various aspects before finally being selected.

[OR]

(b) Explain the principles of placement.
Answer:
The following are the principles of placement:

• Job First, Man Next: Man should be placed on the job according to the requirements
  of the job.
• Job Offer: The job should be offered to the man based on his qualification.
• Terms and conditions: The employee should be informed about the terms and conditions of the organisation.
• Aware about the Penalties: The employee should also be made aware of the penalties if he / she commits a mistake.
• Loyalty and Co-operation: When placing a person in a new job, an effort should be made to develop a sense of loyalty and co-operation in him.

Question 43.
(a) What is Marketing?
Answer:

• Marketing is the performance of buying activities that facilitate to more flow of goods and services. It is one of the oldest profession in the world.
• Objective: The traditional objective of marketing had been to make the goods available at places where they are needed.
• Later on this idea was changed from ‘exchange’ to “Satisfaction of human wants”. Marketing is linking the consumer and the producer. Also marketing helps to maintain economic stability and economic development.

[OR]

(b) What are the salient features of the Consumer Protection Act 1986?
Answer:
Salient features of The Indian Consumer Protection Act, 1986:

• Protecting consumers against products and services which are harmful to the health of the consumers.
• Ensuring consumers with supply of goods at fair quality.
• Ensuring the availability of goods in correct quantity and right size.
• Protecting the consumers against pollution of various kinds.
• Ensuring that consumers are charged fair price.
• Protecting the consumers against unfair trade practices of unscrupulous trader.

Question 44.
(a) What are the responsibilities of consumers?
Answer:
Rights and responsibilities are two sides of the same coin. The responsibilities of consumers are listed below:

• The consumer must pay the price of the goods according to the terms and conditions.
• The consumer has the responsibility to make the seller to deliver the goods in time.
• The consumer has to bear any loss, which may arise to the seller, due to delay in taking delivery.
• The consumer has to follow the instructions and precautions while using the product.
• The consumer must collect the cash receipt as a proof of goods purchased from the seller.
• The consumer must file a complaint with the seller about the defects in products or deficiency in service.

[OR]

(b) Discuss in detail the rights of an unpaid seller against the goods.
Answer
A seller is deemed to be an unpaid seller when :
(a) the whole of the price has not been paid or
(b) a bill of exchange or other negotiable instrument given to him has been dishonoured. Rights of an Unpaid Seller against the Goods: Where the Property in the Goods has Passed to the Buyer:
Right of Lien: An unpaid seller has a right to retain the goods till he receives the price. But to exercise this lien-
(i) He must be in possession of goods
(ii) The goods must have been sold without any condition.

• Right of Stoppage in Transit: Where the seller has delivered the goods to a carrier or other bailee for the purpose of transmission to the buyer, but the buyer has not acquired them, then the seller can stop the goods.
• Right of Resale: The unpaid seller can resell the goods

(iii) Where they are of a perishable nature or
(iv) Where the seller has expressly reserved the right of resale in the contract itself.

Question 45.
(a) Discuss the problems faced by Women Entrepreneurs?
Answer:
There is a tremendous growth in the women entrepreneurship in India. But there are certain problems met by women entrepreneurs. They are as follows
(i) Problem of Finance: The external sources of funds for the women is limited because they do not generally own properties in their own name. They are depending on their own savings and loan from friends and relatives.
(ii) Lack of Education: Illiterate and semi-literate women entrepreneurs face difficulties in respect of accounts, money matters, marketing and day-to-day operations

(iii) Lack of Network Support: The success of business depends on the support of family members, friends and relatives. But it is reported that the women entrepreneurs get very limited support in times of crisis.
(iv) Stiff Competition: They have to face acute competition for their goods from organised sector and from their male counterparts.
(v) Lack of Information: The lack of knowledge or limited knowledge about subsidies, concessions and incentives given by Government will affect the business.

[OR]

(b) Discuss the Preparation of a project report.
Answer:
An entrepreneur can get the report prepared by technical consultancy organisations or by auditors or by government agencies. The report should include the following:
(i) Technical feasibility: It should contain the description of product, raw materials, quality control measures, water, power and transport.
(it) Economic viability: It involves compilation of demand for domestic and export market.
(iii) Financial viability: It should cover the aspects like: Non-recurring expenses like cost of land, and building. Recurring expenses like wages and salaries.
(iv) Managerial competency: Entrepreneurs has to include the mechanism for managing the venture in the project report.

(v) Provisional registration certificate: He has to apply for provisional registration certificate. It will be issued after certain conditions for a period of one year.
(vi) Permanent registration certificate: If the venture has commenced production, or is ready to commence production, it is eligible to get permanent registration certificate.
(vii) Statutory licence: He should obtain Municipal Licence from the authority concerned.
(viii) Power connection: He has to make application to Assistant Divisional Engineer of Electricity Board for power connection.
(ix) Arrangement of finance: Entrepreneur requires two types of finance namely Long term and Short term.

Question 46.
(a) Explain how director of a company can be removed from the office?
Answer:
A Director of Company can be removed from his Office before the expiry of his term by

• the Shareholders
• the Central Government
• the Company Law Board

(i) Removal by shareholders (Sec- 169): A company may, by giving a special notice and passing an ordinary resolution, remove a director before the expiry of his period of office.
(ii) Removal by the Central Government: The Central Government has been empowered to remove managerial personnel from office on the recommendation of the Company Law Board under the following situations:

• Where a person concerned in the conduct and management of the affairs of a company has been guilty of fraud and negligence.
• If the business is managed by a person without sound business principles.
• Where the business of a company has been managed by such a person, who likes to cause injury or damage to the business.

(iii) Removal by the Company Law Board: If an application has been made to the Company Law Board against the oppression and mismanagement of the company’s affairs by a director, then the Company Law Board may order to terminate the director.

[OR]

(b) Describe the different types of resolutions which company may pass in the suitable matters required for each type of resolution.
Answer:
A motion, with or without the amendments which is put to vote at a meeting and passed with the required quorum becomes resolution. Resolution may be classified into three types. They are: Ordinary resolution, Special resolution and resolution requiring special notice.

(i) Ordinary Resolution: An ordinary resolution is one which can be passed by a simple majority.
Ordinary Resolution is required for the following matters:

• To change or rectify the name of the company
• To alter the share capital of the company
• To redeem the debentures
• To declare the dividends
• To appoint the directors

(ii) Special Resolution: A special resolution is the one which is passed by not less than 75% of majority.
Special Resolution is required for the following matters:

• To change the registered office of the company
• To alter the Articles of Association
• To commence any new business
• To appoint the auditor for the company

(iii) Resolution requiring Special Notice: There are certain matters specified in the Companies Act, 2013 which may be discussed at a general meeting only if a special notice is given at least 14 days before the meeting.

The following matters require special notice:

• To remove a director before the expiry of his period
• To appoint a director in the place of a director so removed

Question 47.
(a) Discuss the significance of understanding business environment and the internal factors affecting business.
Answer:
A business in order to remain successful and competitive has to adapt to the constantly changing environment. The significance of understanding the business environment is as follows:

• Helps in formulating strategy and future planning.
• The analysis of business environment helps a business to identify new opportunities.
• Environment scanning helps the firms to identify threats which affect the business.

Internal factors of environment:

• Value system: The success of an organisation depends upon the sharing of value system by all members.
• Vision and objectives: The vision and objectives of a business guides its operations and strategic decisions.
• Management structure: The structure of management/board and their style of functioning, the composition of the board.
•  Internal power relations: This refers to the internal power relations that exist in an organisation. The relations among board members, and the CEO and employees, shareholders are the factors affecting in taking decisions.

[OR]

(b) What is Voluntary Consumer Organisations? Explain its Functions.
Answer:
Voluntary consumer organisations refer to the organisation formed voluntarily by the consumers to protect their rights and interests.
Functions:

• Collecting Data on Different Products: These organizations collect samples of different products from time to time and test them.
• Filing Suit on Behalf of Consumers: If a consumer is not able to protest regarding his complaints, these organisations file case in the court, on behalf of a consumer.
• Protests against Adulteration: The consumer organisations play a significant role in eliminating the evil of adulteration, hoarding, black-marketing.
• Helping Educational Institutions: These organizations advice the educational institutions the way to prepare courses of study.
• Extending Support to Government: Consumer organisations keep informing the government agencies about adulteration, artificial scarcity, inferior quality products.