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TN State Board 12th Physics Model Question Paper 2 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.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between points A and C is……………………………..
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) 1/4 μF

Answer:
(b) 2 μF

Question 2.
If the electric field in a region is given by \(\overrightarrow{\mathrm{E}}=5 \hat{i}+4 \hat{j}+9 \hat{k}\), then the electric flux through a surface of area 20 units lying in the
y – z plane will be …………………
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hint. The area vector
\(\overrightarrow{\mathrm{A}}=20 \hat{i} ; \overrightarrow{\mathrm{E}}=(5 \hat{i}+4 \hat{j}+9 \hat{k})\)
Flux ( φ)\(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\) = 5 x 20 = 100 units

Question 3.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

(a) π Ω
(b) \(\frac{\pi}{2} \Omega\)
(c) 2π Ω
(d) \(\frac{\pi}{4} \Omega\)
Answer:
(b) \(\frac{\pi}{2} \Omega\)

Question 4.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed co. Find the ratio of its magnetic moment with angular momentum is …………..
(a) M
(b) \(\frac{3}{\pi} \mathrm{M}\)
(c) \(\frac{2}{\pi} \mathrm{M}\)
(d) \(\frac{1}{2} \mathrm{M}\)

Answer:
(b) \(\frac{3}{\pi} \mathrm{M}\)

Question 5.
A proton enters a magnetic field of flux density 1.5 Wb/m² with a speed of 2 x 10⁷ m/s at angle of 30° with the field. The force on the proton will be ……………….
(a) 0.24 x 10^-12 N
(b) 2.4 x 10 ^-12 N
(c) 24 x 10^-12 N
(d) 0.024 x 10^-12 N
Answer:
(b) 2.4 x 10 ^-12 N
Hint: F = Bqv sin θ = 1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30°= 2.4 x 10^-12 N

Question 6.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\) Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) . The power factor of the of the circuit is ……………
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(c) 1

Question 7.
The inductance of a coil is proportional to…………………………………….
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 8.
The electric and magnetic fields of an electromagnetic wave are……………….
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other

Question 9.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will ……………
(a) get shifted downwards
(b) get shifted upward
(c) will remain the same
(d) data insufficient to conclude

Answer:
(b) get shifted upward

Question 10.
The wavelength λ_e of an electron and λ_p of a photon of same energy E are related ………..

Answer:
(d) \(\lambda_{p} \propto \lambda_{e}^{2}\)

Question 11.
A system consists of N_o nucleus at t = 0. The number of nuclei remaining after half of a half-life (that  is, at time \(t=\frac{1}{2} \mathrm{T}_{\frac{1}{2}}\)
Answer:

Hint:

Question 12.
In a pure semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called……………………………….
(a) acceptor
(b) donor
(c) intrinsic semiconductor
(d) extrinsic semiconductor
Answer:
(c) intrinsic semiconductor
Hint: Pure semiconductors are called intrinsic semiconductors.

Question 13.
The light emitted in an LED is due to……………………………
(a) Recombination of charge carriers
(b) Reflection of light due to lens action
(c) Amplification of light falling at the junction
Answer:
(a) Recombination of charge carriers

Question 14.
The frequency range of 3 MHz to 30 MHz is used for………………………………..
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
The materials used in Robotics are……………………..
(a) Aluminium and silver
(b) Silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Part – III

Answer any six questions. Question No. 20 is compulsory.   [6×2 = 12]

Question 16.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 17.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
\(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\)
The S.I. unit of current density. \(\frac{\mathrm{A}}{\mathrm{m}^{2}}(\text { or }) \mathrm{Am}^{-2}\)

Question 18.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation \((\overrightarrow{\mathrm{M}})\) induced in the material due to magnetising field \((\overrightarrow{\mathrm{H}})\)
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Question 19.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 20.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Answer:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, φ = 4 mWb = 4 x 10^-3 Wb
Induction of the coil , L
\(\mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{I}}=\frac{200 \times 4 \times 10^{-3}}{0.4}=\frac{800 \times 10^{-3}}{0.4}=2 \mathrm{H}\)

Question 21.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 22.
How many photons of frequency 10¹⁴ Hz will make up 19.86 J of energy?
Answer:
Total energy emitted per second = Power x time
19.863 = Power x is
∴ Power 19.86 W
Number of photons =

Question 23.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 10¹⁰ decays/s

Question 24.
A transistor having α =0.99 and V_BE = 0.7V, is given in the circuit. Find the value of the collector current.
Answer:

Part – III

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

Question 25.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 26.
A conductor of linear mass density 0.2 g m^-1 suspended ,by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the x page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s^-2 = 111.87.
Answer:

Question 27.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 28.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = (u)c

Question 29.
If the focal length is 150 cm for a glass lens, what is the power of the lens?
Answer:
Given: focal length,f = 150 cm (or) f= 1.5 m
Equation for power of lens is, P = 1/f
Substituting the values,
\(P = \frac{1}{1.5}\)= 0.067 diopter
As the power is positive, it is a converging lens.

Question 30.
A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.
Answer:
de-Broglic wavelength of the particle is \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\) \(\text { i.e. } \lambda \propto \frac{1}{\sqrt{m}}\)
As m_e <<m_p ,so λ_e >>λ_P
Hence protons have greater de- broglic wavelength

Question 31.
Distinguish between avalanche and zener breakdown.
Answer:

Avalanche Breakdown	Zener Breakdown
It occurs injunctions which are lightly and have wide depletion widths.	It occurs in junctions which are heavily doped and have narrow depletion widths.
It occurs at higher reverse voltages when thermally generated electrons get enough kinetic energy to produce more electrons by collision.	It occurs due to rupture of covalent bonds by strong electric fields set up in depletion region by the reverse voltage.
At reverse voltage above 6V breakdown is due to avalanche effect.	At reverse voltage below 6V breakdown is due to zener effect.
Electric field produced is weak in nature.	A strong electric field is produced
Charge carriers obtain energy from the applied potential.	Zener current is independent of applied voltage.

Question 32.
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.

Question 33.
What are black holes?
Answer:
Black holes are end stage of stars Which are highly dense massive object. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun. It has very strong gravitational force such that no particle or even light can escape from it. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other starts. Every galaxy has black hole at its center. Sagittarius A* is the black hole at the center of the Milky Way galaxy.

Part – IV

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

Question 34.
(a) How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies.

Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements
Δq₁, Δq₂, Δq₃ ….. Δq_n ……… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such change elements

Here Δ qi is the i^th charge element, r is the distance of the point P from the z^th charge element and \(\hat{r}_{i \mathrm{P}} \) is the unit vector from ith charge element to the point P.

However, the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit Δq → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form
\(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r^{2}} \hat{r}\)

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r}\) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\overrightarrow{\mathrm{F}}=q \overrightarrow{\mathrm{E}}\)

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = Q/L. Its unit is coulomb per meter (Cm^-1). The charge present in the infinitesimal length dl is dq = λ dl
The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on. a surface of area A, then surface charge density (charge per unit area) is \(\lambda=\frac{Q}{L}\). Its unit is coulomb per square meter (Cm^-2 ). The charge present in the infinitesimal area dA is dq = adA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by \(\rho=\frac{Q}{V}\) . Its unit is coulomb per cubic meter (Cm^-3 ). The charge present in the infinitesimal volume element dV is dq = ρdV.
The electric field due to a volume of total charge Q is given by

[OR]

(b) Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as
V = El
As we know, the magnitude of current density
\(\mathrm{J}=\sigma \mathrm{E}=\sigma \frac{\mathrm{V}}{l}\)
But \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\),so we write the equation as
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
By rearranging the above equations, we get
\(\mathrm{V}=\mathrm{I}\left(\frac{l}{\sigma \mathrm{A}}\right)\)

The quantity \(\frac{l}{\sigma \mathrm{A}}\) is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR …………….. (3)

Question 35.
(a) Calculate the magnetic held inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current carrying solenoid: Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\) = μ₀  x (total current enclosed by Amperian loop)

Since the elemental lengths along be and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals.

Since the magnetic field outside the   solenoid is zero, the integral
\(\int_{c}^{d} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=0\)
For the path along ab, the integral is

where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We can take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is

Let NI be the current passing through the solenoid of N turns, then

The number of turns per unit length is given by \(\frac{\mathrm{NI}}{\mathrm{L}}=n\) then
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
Since n is a constant for a given solenoid and μ_o is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

[OR]

(b) Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle: The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction: In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working: If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil. This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.

As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil ε_p is almost equal and opposite to the applied voltage υ_p and is given by
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by
\(\varepsilon_{\mathrm{s}}=-N_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ………………… (1)
where N_p and N_s are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then  ε_s = υ_s where υ_s is the voltage across secondary coil.
\(v_{s}=\varepsilon_{s}=-\mathrm{N}_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ……….. (2)

From equations (1) and (2),
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=K\) …………………… (3)

This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υ_p i_p= Output power υ_si_s
where i_pand i_s are the currents in the primary and secondary coil respectively. Therefore,
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=\frac{i_{p}}{i_{s}}\) ………………….. (4)

Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}=K\)

(i) If N_s> N_p ( or K > 1), ∴ V_s > V_p and I_s < I_p. This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.

(ii) If N_s < N_p (or K < 1) , ∴ V_s < V_p and I_s > I_p . This is step-down transformer where voltage is decreased and the current is increased.

Question 36.
(a) Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves: Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows. If the charged -particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves. Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means
E_y =E₀ sin (kz-ωt)
B_r = B₀ sin(kz – ωt) where, E_o and B_o are amplitude of oscillating electric and magnetic field,\(\hat{k} \) is a wave number, ω is the angular frequency of the wave and k (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.

Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between E_o and B_o is equal to the speed of electromagnetic wave, which is equal to speed of light c.
\(c=\frac{E_{0}}{B_{0}}\)

In any medium, the ratio of E_o and B_o is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
\(v=\frac{E_{0}}{B_{0}}<c\)
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.

[OR]

(b) Explain about compound microscope and obtain the equation for magnification.
Answer:
Compound microscope:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Magnification of compound microscope
From the ray diagram, the linear magnification due to the objective is,
\(m_{0}=\frac{h^{\prime}}{h}\)
from the figure ,\(\tan \beta=\frac{h}{f_{0}}=\frac{h^{\prime}}{L} \), then
\(\frac{h^{\prime}}{h}=\frac{L}{f_{0}} ; m_{o}=\frac{L}{f_{o}}\)

Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as f_o and f_e are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification m_e of the eyepiece is,
\(m_{e}=1+\frac{D}{f_{e}}\)

The total magnification m in near point focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)

If the final image is formed at infinity (normal focusing), the magnification m_e of the eyepiece is
\(m_{e}=\frac{D}{f_{e}}\)

The total magnification m in normal focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 37.
(a) Briefly explain the principle and working of electron microscope.
Answer:
Electron Microscope:
Principle:

• This is the direct application of wave nature of particles. The wave nature of the electron is used in the construction of microscope called electron microscope.
• The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
• De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes.
• As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
• Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:

• The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
• The electrons emitted from the source are accelerated by high potentials. The beam is made parallel by magnetic condenser lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
• With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science.

[OR]

(b) Discuss the process of nuclear fission and its properties.
Answer:

• When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
• The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
• The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.
• Uranium undergoes fission reaction in 90 different ways. The most common fission reactions of
  
• Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state. \(_{ 92 }^{ 236 }{ U }\) . But this excited state does not last longer than 10^-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Question 38.
(a) Transistor functions as a switch. Explain.
Answer:
The transistor in saturation and cut-off regions functions like an electronic switch that helps to. turn ON or OFF a given circuit by a small control signal.

Presence of dc source at the input (saturation region):
When a high input voltage (V. = +5V) is applied, the base current (I ) increases and in turn increases the collector current. The transistor will move into the saturation region (turned ON). The increase in collector current (I_c) increases the voltage drop across R_c .thereby lowering the output voltage, close to zero. The transistor acts like a closed switch and is equivalent to ON condition.

Absence of dc source at the input (cut-off region):
A low input voltage (V_in = OV), decreases the base current (I_B) and in turn decreases the collector current (I_c ). The transistor will move into the cut-off region (turned OFF). The decrease in collector current (I_c) decreases the drop across, thereby increasing the output voltage, dose to +5 V. The transistor acts as an open switch which is considered as the OFF condition.

It is manifested that, a high input gives a low output and a low input gives a high output. In addition, we can say that the output voltage is opposite to the applied input voltage. Therefore, a transistor can be used as an inverter in computer logic circuitry

[OR]

(b) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified.
They are

1. Amplitude modulation,
2. Frequency modulation, and
3. Phase modulation.

1. Amplitude Modulation (AM): If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure
(a) is the message signal or baseband signal that carries information, figure
(b) shows the high-frequency carrier signal and figure
(c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.

(ii) Frequency Modulation (FM):
The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. An increase in the amplitude of the ‘ baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave.

Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A, C). The increase in amplitude in the negative half cycle (B, D) reduces the frequency of the modulated wave (Figure (c)).

(iii) Phase Modulation (PM)
The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency.

The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

Students can download 10th Science Chapter 8 Periodic Classification of Elements Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 8 Periodic Classification of Elements

Samacheer Kalvi 10th Science Periodic Classification of Elements Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
The number of periods and groups in the periodic table are:
(a) 6, 16
(b) 7, 17
(c) 8, 18
(d) 7, 18
Answer:
(d) 7, 18

Question 2.
The basis of modem periodic law is ____.
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons.
Answer:
(a) atomic number

Question 3.
……….. group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th
Answer:
(a) 17th

Question 4.
_______ is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity.
Answer:
(b) ionic radii

Question 5.
Chemical formula of rust is:
(a) Fe0.xH₂O
(b) Fe04.xH₂O
(c) Fe₂O₃. xH₂O
(d) FeO
Answer:
(c) Fe₂O₃. xH₂O

Question 6.
In the aluminothermic process the role of Al is:
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent
Answer:
(b) reducing agent

Question 7.
The process of coating the surface of the metal with a thin layer of zinc is called ____.
(a) painting
(b) thinning
(c) galvanization
(d) electroplating.
Answer:
(c) galvanization

Question 8.
Which of the following inert gas has electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer:
(a) He

Question 9.
Neon shows zero electron affinity due to ____.
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density.
Answer:
(b) stable configuration of electrons

Question 10.
……….. is an important metal to form amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al
Answer:
(b) Hg

II. Fill in the blanks:

1. If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ………..
2. …………. is the longest period in the periodical table.
3. ………… forms the basis of modern periodic table.
4. If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of Cl atom is ………..
5. Among the given species A^– A⁺, and A, the smallest one in size is ……….
6. The scientist who propounded the modern periodic law is …………
7. Across the period, ionic radii ………… (increases,decreases).
8. ……….. and ………… are called inner transition elements.
9. The chief ore of Aluminium is ………….
10. The chemical name of rust is ………….
Answer:
1. ionic
2. 6th (sixth) period
3. Atomic number
4. 0.99 Å
5. A⁺
6. Dimitri Mendeleev
7. decreases
8. Lanthanides, Actinides
9. bauxite
10. hydrated ferric oxide

III. Match the following:

Answer:
A. (ii)
B. (v)
C. (iv)
D. (iii)
E. (i)

IV. True or False: (If false give the correct statement)

1. Moseley’s periodic table is based on atomic mass.
2. Ionic radius increases across the period from left to right.
3. All ores are minerals; but all minerals cannot be called as ores.
4. Aluminium wires are used as electric cables due to their silvery white colour.
5. An alloy is a heterogenous mixture of metals.
Answer:
1. False – Moseley’s periodic table is based on atomic number.
2. True
3. True
4. False – Aluminium wires are used as electric cables because it is a good conductor of heat and electricity.
5. False – An alloy is an homogeneous mixture of metals.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: The nature of bond in HF molecule is ionic.
Reason: The electronegativity difference between H and F is 1.9.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 3.
Assertion: An uncleaned copper vessel is covered with greenish layer. Reason: copper is not attacked by alkali.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

VI. Short answer questions:

Question 1.
A is a reddish brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Answer:
(A) is a reddish brown metal – Copper

A – Copper; B – Cupric oxide; C – Cuprous oxide.

Question 2.
A is a silvery white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer:
A – Silvery white metal – Aluminium

The alloys of aluminium, Duralumin and Magnalium are used in making the aircraft.
A – Aluminium; B – Aluminium oxide.

Question 3.
What is rust? Give the equation for the formation of rust.
Answer:
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface. This compound is known as rust and the phenomenon of formation of rust is known as rusting.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

Question 4.
State two conditions necessary for rusting of iron.
Answer:
(i) The presence of water and oxygen is essential for the rusting of iron.
(ii) Impurities in the iron, the presence of water vapour, acids, salts and CO₂ speeds up rusting.

VII. Long answer questions:

Question 1.
(a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.
Answer:
Caustic alkali is added to bauxite, to dissolve bauxite ore and obtain a solution of sodium aluminate.

(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Answer:
CaF₂ (Fluorspar) is added along with cryolite and alumina. It is added to reduce the high melting point of the electrolyte.

Question 2.
The electronic configuration of metal A is 2, 8, 18, 1.
The metal A when exposed to air and moisture forms B a green layered compound. A with con. H₂SO₄ forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Answer:
Metal (A) with electronic configuration- 2, 8, 18, 1 is copper.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂ (B)
Copper carbonate (Green layer)

(A) – Copper (Cu)
(B) – Copper Carbonate (CuCO₃. Cu(OH)₂)
(C) – Copper Sulphate (CuSO₄)
(D) – Sulphur dioxide (SO₂)

Question 3.
Explain the smelting process.
Answer:
The roasted ore of copper is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.
2 FeS + 3O₂ → 2 FeO + 2 SO₂
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂
Cu₂O + FeS → Cu₂S + FeO
FeO + SiO₂ → FeSiO₂ (slag)

VIII. HOT questions:

Question 1.
Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions.
Answer:
Metal A belongs to Period 3 and Group 13. So metal ‘A’ is aluminium.
(A) in red hot condition reacts with steam to form ‘B’.

‘A’ with strong alkali forms ‘C’

(A) – Aluminium
(B) – Aluminium oxide
(C) – Sodium meta aluminate

Question 2.
Name the acid that renders aluminium passive. Why?
Answer:
Dilute or concentrated nitric acid (HNO₃) renders aluminium passive. Because nitric acid does not attack aluminium but it renders aluminium passive due to the formation of an oxide film on its surface.

Question 3.
(a) Identify the bond between H and F in HF molecule.
Answer:
Ionic, because the electronegativity difference is more than 1.7.

(b) What property forms the basis of identification?
Answer:
Electronegativity.

(c) How does the property vary in periods and in groups?
Answer:
In a period, from left to right the electronegativity increases because of the increase in the nuclear charge.
In a Group, from top to bottom, the electronegativity decreases because of the increase in size of the elements.

Samacheer Kalvi 10th Science Periodic Classification of Elements Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
The shortest period in the periodic table contains elements.
(a) 18
(b) 8
(c) 2
(d) 32
Answer:
(c) 2

Question 2.
Group number of carbon family is _____.
(a) 13
(b) 15
(c) 17
(d) 14.
Answer:
(d) 14.

Question 3.
The ore forming elements, chalcogens are present in ……… group of the modern periodic table.
(a) 18th
(b) 1st
(c) 2nd
(d) 16th
Answer:
(d) 16th

Question 4.
Valency of all the alkali metals is _____.
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 5.
The largest atom in the 2nd period of the periodic table is:
(a) Li
(b) Be
(c) F
(d) Ne
Answer:
(a) Li

Question 6.
The covalent radii of Hydrogen, if the distance between the Hydrogen nuclei of the molecule is 0.74 Å is:
(a) 1.58 Å
(b) \(\frac{0.74}{4}\) Å
(c) 0.37 Å
(d) 0.74 Å
Answer:
(c) 0.37 Å

Question 7.
Pick out the correct ionic radii in increasing order for the following species – Na, Cl, Na⁺, Cl^– _____.
(a) Na > Cl > Na⁺ > Cl^–
(b) Cl > Na > Na⁺ > Cl^–
(c) Cl^– > Na > Na⁺ > Cl
(d) Cl < Na⁺ < Cl^– < Na.
Answer:
(d) Cl < Na⁺ < Cl^– < Na.
Hint:
Na = 186 pm,
Cl = 91 pm,
Na⁺ = 116 pm,
Cl^– = 167 pm.

Question 8.
In the third period, the first ionization potential is of the order:
(a) Na > Al > Mg > Si > P
(b) Mg > Na > Si > P > Al
(c) Na < Al < Mg < Si < P
(d) Na < Al < Mg < Si < P
Answer:
(c) Na < Al < Mg < Si < P

Question 9.
Which one of the following is the least electronegative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen

Question 10.
Which is a widely used a scale to determine the electronegativity?
(a) Pauling scale
(b) Moseley scale
(c) Mendeleev scale
(d) none of these.
Answer:
(a) Pauling scale

Question 11.
Which one of the following orders of ionic radii is correct?
(a) H^– > H⁺ > H
(b) Na⁺ > F^– > O^2-
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Question 12.
The percentage of carbon in Pig iron is:
(a) < 0.25%
(b) 0.25 – 2%
(c) 2 – 4.5%
(d) > 5%
Answer:
(c) 2 – 4.5%

Question 13.
The chemical formula of clay is _____.
(a) Al₂O₃
(b) Al₂O₃.2H₂O
(c) Al₂O₃. 2SiO₂.2H₂O
(d) Al₂O₃. 2SiO₂.H₂O.
Answer:
(c) Al₂O₃. 2SiO₂.2H₂O

Question 14.
The temperature in the combustion zone is maintained at:
(a) 1500°C
(b) 400°C
(c) 1000°C
(d) 1380°C
Answer:
(a) 1500°C

Question 15.
Oil used in Froth floatation method is _____.
(a) pine oil
(b) natural oil
(c) crude oil
(d) Synthetic oil.
Answer:
(a) pine oil

Question 16.
The first most abundant metal present in the Earth crust is:
(a) Iron
(b) Aluminium
(c) Zinc
(d) Copper
Answer:
(b) Aluminium

Question 17.
……….. metal is used for making calorimeters.
(a) Copper
(b) Tin
(c) Mercury
(d) Iron
Answer:
(a) Copper

Question 18.
More reactive metal is _____.
(a) Zn
(b) Fe
(c) Ag
(d) Na.
Answer:
(d) Na.

Question 19.
The chief ore of Iron is:
(a) Magnesite
(b) Galena
(c) Cinnabar
(d) Haematite
Answer:
(d) Haematite

Question 20.
The metal which melts at room temperature is:
(a) Zinc
(b) Lead
(c) Gallium
(d) Tin
Answer:
(c) Gallium

Question 21.
Conversion of bauxite into alumina is _____.
(a) Hall’s process
(b) Alumino thermic process
(c) Baeyer’s process
(d) Bessemerisation process.
Answer:
(c) Baeyer’s process

Question 22.
………. metal can be cut with knife.
(a) Potassium
(b) Gallium
(c) Mercury
(d) Gold
Answer:
(a) Potassium

Question 23.
………. is not a good conductor of heat and electricity.
(a) Silver
(b) Tungsten
(c) Copper
(d) Aluminium
Answer:
(b) Tungsten

Question 24.
Electrolyte used in Hall’s process ______.
(a) Pure alumina + molten cryolite + fluorspar
(b) Pure alumina + molten bauxite + fluorspar
(c) Pure bauxite + molten cryolite + fluorspar
(d) Pure bauxite + molten Haematite + fluorspar.
Answer:
(a) Pure alumina + molten cryolite + fluorspar

Question 25.
The foaming agent used for froth floatation process is:
(a) Coconut oil
(b) Pine oil
(c) Sodium chloride
(d) Groundnut oil
Answer:
(b) Pine oil

Question 26.
Three elements A, B and C are having the electronic configuration Is² 2s¹, Is² 2s² and Is² 2s² 2p¹ respectively. Which element will have the lowest ionization energy?
(a) A
(b) B
(c) C
(d) B and C
Answer:
(a) A

Question 27.
Metal used in household utensils is ______.
(a) Al
(b) Co
(c) Fe
(d) Na.
Answer:
(a) Al

Question 28.
Which one of the following pair is a metalloid?
(a) Na and K
(b) F and Cl
(c) Cu and Hg
(d) Si and Ge
Answer:
(d) Si and Ge

Question 29.
The highly metallic element will have the configuration of:
(a) 2, 8, 7
(b) 2, 8, 8, 5
(c) 2, 8, 8, 1
(d) 2, 8, 2
Answer:
(c) 2, 8, 8, 1

Question 30.
The metal used in electroplating is ______.
(a) Cu
(b) Al
(c) Fe
(d) Co.
Answer:
(a) Cu

Question 31.
The flux which is used when the gangue present in the ore is acidic:
(a) Silica
(b) Calcium oxide
(c) Calcium silicate
(d) Cuprous sulphide
Answer:
(b) Calcium oxide

Question 32.
Matte is a mixture of:
(a) Cu₂O + Cu₂S
(b) Cu₂O + FeS
(c) Cu₂S + FeS
(d) Cu₂O + PbS
Answer:
(c) Cu₂S + FeS

Question 33.
Fe reacts with dilute nitric acid in cold condition to give ______.
(a) Ferrous nitride
(b) Ferrous nitrate
(c) Ferric nitride
(d) Ferric nitrate.
Answer:
(b) Ferrous nitrate

Question 34.
In the brass alloy, which is solvent?
(a) Zn
(b) Co
(c) Ag
(d) Cu.
Answer:
(d) Cu.

II. Fill in the blanks:

1. The major component of the matte is ………….
2. The modern periodic table is based on ………..
3. The valency of alkali metals is …………
4. The unreactive elements are present in group ………..
5. In the 2nd period, the smallest atom is ……….
6. The size of a cation is ………… than the neutral atom.
7. ……….. is the unit of ionization energy.
8. The ionization energy ……… down the group in the periodic table.
9. The electron affinities of noble gases are …………
10. ………. is the process of extracting the ore from the Earth’s crust.
11. Galena is the ore of …………..
12. The silvery white metal which is a good conductor of heat and electricity is …………
13. The slag formed during Bessemerisation process is ………….
14. Blister copper contains ………. percentage of copper.
15. Haematite ore is concentrated by ……….. washing.
Answer:
1. Cu₂S
2. atomic number
3. one
4. 18
5. Neon
6. smaller
7. KJ/mol
8. decreases
9. zero
10. Mining
11. lead
12. aluminium
13. Iron silicate or FeSiO₃
14. 98%
15. hydraulic

III. Match the following:

Question 1.
Match the column I with column II.

Answer:
A. (iv)
B. (v)
C. (i)
D. (ii)
E. (iii)

Question 2.
Match the column I with column II.

Answer:
A. (ii)
B. (iv)
C. (v)
D. (iii)
E. (i)

Question 3.
Match the column I with column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (i)
E. (ii)

Question 4.
Match the column I with column II.

Answer:
A. (v)
B. (iv)
C. (ii)
D. (iii)
E. (i)

IV. True or false. (If false give the correct statement)

1. Alkali metals are generally extracted by the electrolysis of their ores in fused state.
2. Every mineral is an ore but every ore is not a mineral.
3. Slag is a product formed during smelting by combination of flux and impurities.
4. Reactive metals occur in native state.
5. Malachite is a sulphide ore of copper.
6. Lanthanides are present in the 6th group of the periodic table.
7. Atomic radius increases as we go across the period due to increase in size.
8. As the positive charge increases, the size of the cation decreases.
9. If the difference in electronegativity is greater than 1.7, the bond is considered to be covalent.
10. Siderite is the carbonate ore of calcium.
Answer:
1.True
2. True
3. True
4. False – Reactive metals always occur in the combined state.
5. False – Malachite is the carbonate ore of copper.
6. False – Lanthanides are present in the 6th period of the periodic table.
7. False – Atomic radius increases as we go across the period due to decrease in size.
8. True
9. False – If the difference in electronegativity is greater than 1.7, the bond is considered to be ionic.
10. False – Siderite is the carbonate ore of Iron.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: Zinc blende is concentrated by Froth floatation process.
Reason: Zinc blende is a sulphide ore.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: In thermite welding, aluminium powder and Fe₂O₃ are used.
Reason: Aluminium powder is a strong reducing agent.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 3.
Assertion: To design the body of an aircraft, aluminium alloys are used.
Reason: Aluminium becomes passive when it is treated with dil or con.HNO₃
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

Question 4.
Assertion: Tinstone and the impurity wolframite are seperated by magnetic separation.
Reason: Tinstone is magnetic and wolframite is non-magnetic in nature.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 5.
Assertion: Bauxite is purified by leaching.
Reason: Bauxite undergoes thermal decomposition.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(b) Assertion is correct, Reason is wrong.

VI. Short answer questions:

Question 1.
State the modern periodic law.
Answer:
The physical and chemical properties of the elements are the periodic function of their atomic number.

Question 2.
‘X’ is a silvery white metal. X reacts with O₂ to form Y. The same compound is obtained from the metal on reaction with steam with the liberation of hydrogen gas. Identify X and Y.
Answer:
(i) Silvery white metal ‘X’ is Aluminium.
(ii) It reacts with O₂ to form ‘Y’

(iii) Y can also be obtained on reaction with steam with the liberation of H₂.

Question 3.
Write any four characteristics of periods.
Answer:

• In a period, the electrons are filled in the same valence shell of all elements.
• As the electronic configuration changes along the period, the chemical properties of the elements also change.
• The atomic size of the elements in a period decreases from left to right
• In a period, die metallic character of the element decreases, while their non-metallic character increases.

Question 4.
Write the Principle of Hydraulic washing.
Answer:
The difference in the densities or specific gravities of the ore and the gangue is the main principle behind this method.

Question 5.
What are coinage metals?
Answer:
Copper, silver and gold are called coinage metals, as they are used in making coins and jewellery.

Question 6.
How will you separate tinstone from wolframite?
Answer:
Magnetic separation method. Tinstone is magnetic in nature.
Method: The crushed ore is placed over a conveyer belt which rotates around two metal wheels, one of which is magnetic. The magnetic particles are attracted to the magnetic wheel and fall separately apart from the non¬magnetic particles.

Question 7.
What are ores?
Answer:
The mineral from which a metal can be readily and economically extracted on a large scale is said to be ore.
eg. Bauxite Al₂O₃.2H₂O is the ore of Aluminium

Question 8.
Define electronegativity.
Answer:
It is the tendency of an element in a covalent bond to attract the shared pair of electrons towards itself. It is a relative property.

Question 9.
In what period and group will an element with z = 118 will be present?
Answer:
Elements with z = 118 will be present in Period number ‘7’ and Group number 18.

Question 10.
Why flux is added during metallurgy?
Answer:
Flux is the substance added to the ore to reduce the fusion temperature and to remove impurities.
e.g. CaO, SiO₂

Question 11.
State the trends in the electronegativity in a Group and period.
Answer:
In a Group: Electronegativity decreases in a group because of the increased number of energy levels.
In a Period: The electronegativity increases because the increase in the nuclear charge.

Question 12.
Write a note about smelting.
Answer:
Smelting is a process of reducing the roasted metallic oxide to metal in a molten condition. In this process, impurities are removed by the addition of flux as slag.

Question 13.
Write the formula of the ores of Aluminium.
Answer:

Question 14.
Explain the action of Aluminium with air.
Answer:
On heating at 800°C, aluminium bums in the air very brightly forming its oxide and nitride.
4Al + 3O₂ → 2Al₂O₃ (Aluminium oxide)
2Al + N₂ → 2AlN (Aluminium nitride).

Question 15.
What happen when Aluminium reacts with steam?
Answer:
When steam is passed over red hot aluminium, H₂ gas is evolved.

Question 16.
Write the reaction of Aluminium with Sodium hydroxide?
Answer:

Question 17.
Explain the electrolytic refining of copper.
Answer:
Cathode: A thin plate of pure copper metal.
Anode: A block of impure copper metal.
Electrolyte: Copper sulphate solution + dilute H₂SO₄
When an electric current is passed through the electrolytic solution, pure copper gets deposited at the cathode and the impurities are settled at the bottom of the anode as anode mud.

Question 18.
Mention the uses of Aluminium.
Answer:
Aluminium is used in

1. household utensils
2. electrical cable industry
3. making aeroplanes and other industrial machine parts.

Question 19.
What are the methods employed to make an alloy?
Answer:

1. By fusing the metals together. Eg: Brass is made by melting zinc and copper.
2. By compressing finely divided metals. Eg: wood mexai.

Question 20.
Write the components of wood metal.
Answer:
Wood metal is an alloy of Lead, Tin, Bismuth and Cadmium.

Question 21.
What are the uses of copper?
Answer:

• Copper is used in manufacturing electric cables and other electric appliances.
• Copper is used for making utensils, containers, calorimeters and coins.
• Copper is used in electroplating.
• Copper is alloyed with gold and silver for making coins and jewels.

Question 22.
Give example for non-ferrous copper and aluminium alloys. Non-ferrous copper alloys: Brass (Cu, Zn), Bronze (Cu, Sn)
Answer:
Non-ferrous aluminium alloys: Duralumin (Al, Mg, Cu, Mn), Magnalium (Al, Mg)

Question 23.
How is rust formed?
Answer:
When iron is exposed to moist air, it forms a layer of brown hydrated Ferric oxide on its surface. This compound is known as rust.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

Question 24.
Why are the alloys prepared?
Answer:

1. To modify appearance and colour.
2. To modify chemical activity.
3. To lower the melting point.
4. To increase hardness and tensile strength.
5. To increase resistance to electricity.

Question 25.
Define corrosion.
Answer:
Corrosion is the gradual destruction of metals by chemical or electrochemical reaction with the environment.

Question 26.
What are alloys? How are they prepared?
Answer:

• An alloy is a homogeneous mixture of a metal with other metals or with non-metals that are fused together. e.g. Brass is an alloy of zinc (solute) in copper (solvent)
• Alloys are prepared by fusing the metals together.
• Alloys are prepared by compressing finely divided metals one over the other.

Question 27.
Which is known as Wet corrosion or Electrochemical corrosion?
Answer:
The corrosive action in the presence of moisture is called wet corrosion. It occurs as a result of electrochemical reaction of metal with water or aqueous solution of salt or acids or bases.

Question 28.
Write a note on Cathodic protection.
Answer:
It is the method of controlling corrosion of a metal surface protected is coated with the metal which is easily corrodible. The easily corrodible metal is called Sacrificial metal to act as anode ensuring cathodic protection.

Question 29.
What are the methods used to prevent corrosion?
Answer:
Corrosion of metals is prevented

• by coating with paints
• by coating with oil and grease
• by alloying with other metals
• by the process of galvanization
• by electroplating
• by sacrificial protection

Question 30.
A reddish brown metal ‘A’ reacts with dil.HCl in the presence of O₂ and forms the compound ‘B’. ‘B’ can also be prepared by heating the metal A with Cl₂. Identify A and B.
Answer:
Reddish brown metal ‘A’ is copper.
(A) reacts with dil.HCl in the presence of O₂ and forms CuCl₂ which is ‘B’.

(B) can also prepared by the action of Cl₂.

Question 31.
Write the uses of copper.
Answer:

1. It is extensively used in manufacturing electric cables and other electric appliances.
2. It is used for making utensils, containers, calorimeters and coins.
3. It is used in electroplating.
4. It is alloyed with gold and silver for making coins and jewels.

Question 32.
Write the name and formula of the ores of iron.
Answer:

Question 33.
Define periodicity.
Answer:
The electronic configurations of elements help us to explain the periodic recurrence of physical and chemical properties. Anything which repeats itself after a regular interval is called periodic and this behaviour is called periodicity.

Question 34.
What happens in the combustion zone during the extraction of iron.
Answer:
The temperature in the combustion zone is 150°C. In this region coke bums 02 to form CO₂, when the charge comes in contact with a hot blast of air.

Question 35.
Explain the reactions taking place in the reduction zone.
Answer:
In the upper region of reduction zone, the temperature is at 400°C. In this region CO reduces ferric oxide to form spongy iron.

Question 36.
Define Metallic radius.
Answer:
It is defined as half the distance between the nuclei of adjacent metal atoms.

Question 37.
Complete the following reactions.

1. 4Fe + 10HNO₃ → 4Fe(NO₃)₂ + ………. + 3H₂O
2. 2Fe + 6H₂SO₄ → Fe₂(SO₄)₃ + ………. + 6H₂O

Answer:

1. NH₄NO₃
2. 3SO₂

Question 38.
What happens when steam is passed over red hot iron?
Answer:
When steam is passed over red hot iron magnetic oxide is formed.
3Fe + 4H₂O (steam) → Fe₃O₄ + 4H₂↑

Question 39.
Define Electron affinity.
Answer:
Electron affinity is the amount of energy released when a gaseous atom gains an electron to form its anion. It is also measured in kJ / mol.

Question 40.
Complete the table.
Answer:

Question 41.
Define Metallurgy.
Answer:
Metallurgy is a science of extracting metals from their ores and modifying the metals into alloys for various uses, based on their physical and chemical properties and their structural arrangement of atoms.

Question 42.
Write a short note on leaching or chemical process.
Answer:
This method is employed when the ore is in a very pure form. The ore is treated with a suitable reagent such that the ore is soluble in it but the impurities are not. The impurities are removed by filtration. The solution of the ore, ie., the filtrate is treated with a suitable reagent which precipitates the ore.
E.g. Bauxite Al₂O₃.2H₂O, the ore of aluminium.

Question 43.
Relate all the four columns of the table with their unique properties.
Answer:

Question 44.
Guess Who I am?
(i) I am preserved in Kerosene.
Answer:
Sodium

(ii) My ore is leached with NaOH.
Answer:
Aluminium

(iii) I sacrifice myself to protect my friend Iron.
Answer:
Magnesium

(iv) I am being used in propellers
Answer:
Nickel steel

Question 45.
Explain the method of making alloys.
Answer:

• By fusing the metals together. E.g. Brass is made by melting zinc and copper.
• By compressing finely divided metals. E.g. Wood metal: an alloy of lead, tin, bismuth and cadmium powder is a fusible alloy.

Question 46.
Write the differences between a mineral and a ore.
Answer:

VII. Long answer questions:

Question 1.
Write the reactions taking place during Bessemerisation of copper.
Answer:
2 FeS + 3O₂ → 2FeO + 2SO₂↑

2Cu₂S + 3O₂ → 2 Cu₂O + 2 SO₂↑
2Cu₂O + Cu₂S → 6 Cu + SO₂↑

Question 2.
How do electronegativity values help to find out the nature of bonding between atoms?
Answer:

• If the difference in electronegativity between two elements is 1.7, the bond has 50% ionic character and 50 % covalent character.
• If the difference is less than 1.7, the bond is considered to be covalent.
• If the difference is greater than 1.7, the bond is considered to be ionic.

Question 3.
Explain Froth floatation with diagram.
Answer:
Principle: This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method. Eg: Zinc blende (ZnS).

Question 4.
Explain the Baeyer’s process of conversion of Bauxite into alumina.
Answer:
(i) Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150°C to obtain sodium meta aluminate.
(ii) On diluting sodium meta aluminate with water, a precipitate of aluminium hydroxide is formed.
(iii) The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.

Question 5.
Explain the Hall’s Process of electrolytic reduction of alumina with diagram.
Answer:
Hall’s Process:
Aluminium is produced by the electrolytic reduction of fused alumina (Al₂O₃) in the electrolytic cell.
Cathode : Iron tank linked with graphite
Anode : A bunch of graphite rods suspended in molten electrolyte.
Electrolyte : Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)

Temperature: 900 – 950 °C
Voltage used: 5 – 6 V
Overall reaction: 2 Al₂O₃ → 4 Al + 3O₂↑

Question 6.
Write the reaction involved in the middle region of blast furnace during the extraction of iron.
Answer:
The Middle Region (Fusion Zone): The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂.

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

Question 7.
What are the three different types of iron? Write their uses.
Answer:
(i) Pig iron (Iron with 2-4.5% of carbon): It is used in making pipes, stoves, radiators, railings, manhole covers and drain pipes.
(ii) Steel (Iron with < 0.25% of carbon): It is used in the construction of buildings, machinery, transmission cables and T. V towers and in making alloys.
(iii) Wrought iron (Iron with 0.25-2% of wraught carbon): It is used in making springs, anchors and electromagnets.

Question 8.
What is corrosion? Write the chemistry behind the formation of rust.
Answer:
(i) The slow and steady destruction of a metal by chemical or electro chemical reaction with the environment.
(ii) When the surface of iron is exposed to moisture and other gases present in the atmosphere, the following chemical reaction takes place.
Fe → Fe²⁺ + 2e^–
O₂ + 2H₂O + 4e^– → 4OH^–
O₂ + 4H⁺ + 4e^– → 2H₂O
The Fe²⁺ ions are oxidised to Fe³⁺ ions.
The Fe³⁺ ions combine OH^– ions to form Fe(OH)₃. This becomes rust which is hydrated ferric oxide with the formula Fe₂O₃.xH₂O. It is a reddish brown substance.

Question 9.
Explain the methods of preventing corrosion.
Answer:
(i) Alloying : The metals can be alloyed to prevent the process of corrosion. Eg: Stainless Steel

(ii) Surface Coating : It involves application of a protective coating over the metal. It is of the following types:
(a) Galvanization: It is the process of coating zinc on iron sheets by using electric current.
(b) Electroplating: It is a method of coating one metal over another metal by passing electric current.
(c) Anodizing: It is an electrochemical process that converts the metal surface into a decorative, durable and corrosion resistant. Aluminium is widely used for anodizing process.
(d) Cathodic Protection: It is the method of controlling corrosion of a metal surface protected is coated with the metal which is easily corrodible. The easily corrodible metal is called sacrificial metal to act as anode ensuring cathodic protection.

Question 10.
Discuss the main featured of Periods in the modern periodic table (or) long form of periodic table.
Answer:
The horizontal rows are called periods.There are seven periods in the periodic table.

1. First period (Atomic number 1 and 2): This is the shortest period. It contains only two elements (Hydrogen and Helium).
2. Second period (Atomic number 3 to 10): This is a short period. It contains eight elements (Lithium to Neon).
3. Third period (Atomic number 11 to 18): This is also a short period. It contains eight elements (Sodium to Argon).
4. Fourth period (Atomic number 19 to 36): This is a long period. It contains eighteen elements (Potassium to Krypton). This includes 8 normal elements and 10 transition elements.
5. Fifth period (Atomic number 37 to 54): This is also a long period. It contains 18 elements (Rubidium to Xenon). This includes 8 normal elements and 10 transition elements.
6. Sixth period (Atomic number 55 to 86): This is the longest period. It contains 32 elements (Caesium to Radon). This includes 8 normal elements, 10 transition elements and 14 inner transition elements (Lanthanides).
7. Seventh period (Atomic number 87 to 118): Like the sixth period, this period also accommodates 32 elements. Recently 4 elements have been included by IUPAC.

Question 11.
Discuss the main feature of Groups in the long form of periodic table.
Answer:
(i) The vertical columns in the periodic table starting from top to bottom are called groups. There are 18 groups in the periodic table.

(ii) Based on the common characteristics of elements in each group, they can be grouped as various families.

(iii) The Lanthanides and Actinides, which form part of Group 3 are called inner transition elements.

(iv) Except ‘group O’, all the elements present in each group have the same number of electrons in their valence shell and thus have the same valency. Eg: all the elements of group 1 have one electron in their valence shells (Is¹). So, the valency of all the alkali metals is ‘ 1’.

(v) As the elements present in a group have identical valence shell electronic configurations, they possess similar chemical properties.

(vi) The physical properties of the elements in a group such as melting point, boiling point and density vary gradually.

(vii) The atoms of the ‘group 0’ elements have stable electronic configuration in their valence shells and hence they are unreactive.

VIII. Hot Questions:

Question 1.
Why noble gases have zero electron affinity value?
Answer:
Noble gases show no tendency to accept electrons because the outers and p orbitals of noble gases are completely filled. No more electrons can be added to them and hence their electron affinities are zero.

Question 2.
Arrange the following ions in order of their increasing ionic radii.
Answer:
Li⁺, Mg²⁺, K⁺ Al³⁺
Al³⁺ < Li⁺ < Mg²⁺ < K⁺

Question 3.
Cationic radius is smaller than its corresponding neutral atom. Why?
Answer:
When a neutral atom lose one or more electrons it forms a cation.
Na → Na⁺ + e^–
The radius of this cation (r_Na+)is decreased than its parent atom (r_Na).
When an atom is charged to cation, the number of nuclear charges becomes greater than the number of orbital electrons. Florence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

Students can download Maths Chapter 1 Relations and Functions Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f = {(x, y) |x,y; ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Answer:
X = {1,2,3,….}
Y = {1,2,3,….}
f = {(1,2) (2, 4) (3, 6) (4, 8) ….}
Domain = {1, 2, 3, 4 ….}
Co – Domain = {1, 2, 3, 4 ….}
Range = {2, 4, 6, 8 }
Yes this relation is a function.

Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation
R = {(x,f(x)) |x ∈ X, f(x) = x² + 1}
is a function from X to N?
Answer:
f(x) = x² + 1
f(3) = 3² + 1 = 9 + 1 = 10
f(4) = 4² + 1 = 16 + 1 = 17
f(6) = 6² + 1 = 36 + 1 = 37
f(8) = 8² + 1 = 64 + 1 = 65
yes, R is a function from X to N

Question 3.
Given the function f: x → x² – 5x + 6, evaluate
(i) f(-1)
(ii) f(2a)
(iii) f(2)
(iv) f(x – 1)
Solution:
Give the function f: x → x² – 5x + 6.
(i) f(-1) = (-1)² – 5(1) + 6 = 1 + 5 + 6 = 12
(ii) f(2a) = (2a)² – 5(2a) + 6 = 4a² – 10a + 6
(iii) f(2) = 2² – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x – 1) = (x – 1)² – 5(x – 1) + 6
= x² – 2x + 1 – 5x + 5 + 6
= x² – 7x + 12

Question 4.
A graph representing the function f(x) is given in it is clear that f(9) = 2.

(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
Answer:
(a) f (0) = 9
(b) f (7) = 6
(c) f (2) = 6
(d) f(10) = 0

(ii) For what value of x is f(x) = 1 ?
Answer:
When f(x) = 1 the value of x is 9.5

(iii) Describe the following
(i) Domain
(ii) Range.
Answer:
Domain = {0, 1, 2, 3,… .10}
= {x / 0 < x < 10, x ∈ R}
Range = {0,1,2,3,4,5,6,7,8,9}
= {x / 0 < x < 9, x ∈ R}

(iv) What is the image of 6 under f?
Answer:
The image of 6 under f is 5.

Question 5.
Let f (x) = 2x + 5. If x ≠ 0 then find
\(\frac{f(x+2)-f(2)}{x}\)
Answer:
f(x) = 2x + 5
f(x + 2) = 2(x + 2) + 5
= 2x + 4 + 5
= 2x + 9

Question 6.
A function/is defined by f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
(ii) find x such that f(x) = 0.
(iii) find x such that/ (A:) = x.
(iv) find x such that fix) =/(l – x).
Answer:
(i) f(x) = 2x – 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = 2 – 3 = -1

(ii) f(x) = 0
2x – 3 = 0
2x = 3
x = \(\frac { 3 }{ 2 } \)

(iii) f(x) = x
2x – 3 = x
2x – x = 3
x = 3

(iv) f(1 – x) = 2(1 – x) – 3
= 2 – 2x – 3
= – 2x – 1
f(x) = f(1 – x)
2x – 3 = – 2x – 1
2x + 2x = 3 – 1
4x = 2
x = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

Question 7.
square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.

Solution:

After cutting squares we will get a cuboid,
length of the cuboid (l) = 24 – 2x
breadth of the cuboid (b) = 24 – 2x
height of the cuboid (h) = 2x
Volume of the box = Volume of the cuboid
V = (24 – 2x)(24 – 2x) (x)
= (24 – 2x)² (x)
= (576 + 4x² – 96x) x
= 576x + 4x³ – 96x²
V = 4x³ – 96x² + 576x
V(x) = 4x³ – 96x² + 576x

Question 8.
A function f is defined by f(x) = 3 – 2x. Find x such that f(x²) = (f (x))².
Answer:
f(x) = 3 – 2x
f(x²) = 3 – 2 (x²)
= 3 – 2x²
(f (x))² = (3 – 2x)²
= 9 + 4x² – 12x
But f(x²) = (f(x))²
3 – 2 x² = 9 + 4x² – 12x
-2x² – 4x² + 12x + 3 – 9 = 0
-6x² + 12x – 6 = 0
(÷ by – 6) ⇒ x² – 2x + 1 = 0
(x – 1) (x – 1) = 0
x – 1 = 0 or x – 1 = 0
x = 1
The value of x = 1

Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.
Solution:
Speed = \(\frac{\text { distance covered }}{\text { time taken }}\)
⇒ distance = Speed × time
⇒ d = 500 × t [ ∵ time = t hrs]
⇒ d = 500 t

Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as y = ax + b , where a, b are constants.
(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Length ‘x’ of forehand (in cm)	Height y (in inches)
35	56
45	65
50	69.5
55	74

Answer:
The relation is y = 0.9x + 24.5
(i) Yes the relation is a function.
(ii) When compare with y = ax + b
a = 0.9, b = 24.5
(iii) When the forehand length is 40 cm, then height is 60.5 inches.
Hint: y = 0.9x + 24.5
= 0.9 × 40 + 24.5
= 36 + 24.5
= 60.5 feet

(iv) When the height is 53.3 inches, her forehand length is 32 cm
Hint: y = 0.9x + 24.5
53.3 = 0.9x + 24.5
53.3 – 24.5 = 0.9 x
28.8 = 0.9 x
x = \(\frac { 28.8 }{ 0.9 } \)
x = 32 cm

Representation of functions

A function may be represented by

(a) Set of ordered pairs
(b) Table form
(c) Arrow diagram
(d) Graphical form

Vertical line test
A curve drawn in a graph represents a function, if every vertical line intersects the curve in at most one point.

Types of function

1. One – One function (injection)

A function f: A → B is called one-one function if distinct elements of A have a distinct image in B.

2. Many – One function

A function f: A → B is called many-on function if two or more elements of A have same image in B.

3. Onto function (surjection)

A function f: A → B is said to be on to function if the range of f is equal to the co-domain of f.

4. Into function

A function f: A → B is called an into function if there exists at least one element in B which is not the image of any element of A.

5. Bijection

A function f: A → B is both one – one and onto, then f is called a bijection from A to B.

Horizontal line test
A function represented in a graph is one – one, if every horizontal line intersects the curve in at most one point.

Special cases of function

1. Constant function

A function f: A → B is called a constant function if the range of f contains only one element.

2. Identity function

A function f : A → A defined by f(x) = x for all x ∈ A is called an identity function on A and is denoted by I_A.

3. Real valued function
A function f: A → B is called a real valued function if the range of f is a subset of the set of all real numbers R.

Students can download Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 1.
Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B?
(i) R₁ = {(2,1), (7,1)}
(ii) R₂ = {(-1,1)}
(iii) R₃ = {(2,-1), (7, 7), (1,3)}
(iv) R₄ = {(7, -1), (0, 3), (3, 3), (0, 7)}
Answer:
A = {1,2,3,7} B = {3,0,-1, 7}
A × B = {1,2,3} × {3, 0,-1, 7}
A × B = {(1,3) (1,0) (1,-1) (1,7) (2,3) (2, 0)
(2, -1) (2, 7) (3, 3) (3,0) (3,-1)
(3, 7) (7, 3) (7, 0) (7,-1) (7, 7)}

(i) R₁ = {(2, 1)} (7, 1)
It is not a relation, there is no element of (2, 1) and (7, 1) in A × B

(ii) R₂ = {(-1),1)}
It is not a relation, there is no element of
(-1, 1) in A × B

(iii) R₃ = {(2,-1) (7, 7) (1,3)}
Yes, It is a relation

(iv) R₄ = {(7,-1) (0,3) (3, 3) (0,7)}
It is not a relation, there is no element of (0, 3) and (0, 7) in A × B

Question 2.
Let A = {1, 2, 3, 4,…,45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.
Solution:
A = {1, 2, 3, 4, . . . 45}, A × A = {(1, 1), (2, 2) ….. (45, 45)}
R – is square of’
R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}
R ⊂ (A × A)
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1, 4, 9, 16, 25, 36}

Question 3.
A Relation R is given by the set {(x, y)/y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.
Answer:
x = {0, 1, 2, 3, 4, 5}
y = x + 3
when x = 0 ⇒ y = 0 + 3 = 3
when x = 1 ⇒ y = 1 + 3 = 4
when x = 2 ⇒ y = 2 + 3 = 5
when x = 3 ⇒ y = 3 + 3 = 6
when x = 4 ⇒ y = 4 + 3 = 7
when x = 5 y = 5 + 3 = 8
R = {(0, 3) (1,4) (2, 5) (3, 6) (4, 7) (5, 8)}
Domain = {0, 1, 2, 3, 4, 5}
Range = {3, 4, 5, 6, 7, 8}

Question 4.
Represent each of the given relations by
(a) an arrow diagram
(b) a graph and
(c) a set in roster form, wherever possible.
(i) {(x,y) | x = 2y,x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}
(ii) {(x, y) | y = x + 3, x, y are natural numbers < 10}
Answer:
(i) x = {2, 3, 4, 5} y = {1, 2, 3, 4}
x = 2y
wheny y = 1 ⇒ x = 2 × 1 = 2
when y = 2 ⇒ x = 2 × 2 = 4
when y = 3 ⇒ r = 2 × 3 = 6
when y = 4 ⇒ x = 2 × 4 = 8

(a) Arrow diagram

(b) Graph

(c) Roster form R = {(2, 1) (4, 3)}

(ii) x = {1, 2, 3, 4, 5, 6, 7, 8, 9}
y = {1,2, 3, 4, 5, 6, 7, 8,9}
y = x + 3
when x = 1 ⇒ y = 1 + 3 = 4
when x = 2 ⇒ y = 2 + 3 = 5
when x = 3 ⇒ y = 3 + 3 = 6
when x = 4 ⇒ y = 4 + 3 = 7
when x = 5 ⇒ y = 5 + 3 = 8
when x = 6 ⇒ y = 6 + 3 = 9
when x = 7 ⇒ y = 7 + 3 = 10
when x = 8 ⇒ y = 8 + 3 = 11
when x = 9 ⇒ y = 9 + 3 = 12

R = {(1,4) (2, 5) (3,6) (4, 7) (5, 8) (6, 9)}

(a) Arrow diagram

(b) Graph

(c) Roster form: R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

Question 5.
A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A₁, A₂, A₃, A₄ and A₅ were Assistants; C₁, C₂, C₃, C₄ were Clerks; M₁, M₂, M₃ were managers and E₁, E₂ were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.
Answer:
Assistants → A₁, A₂, A₃, A₄, A₅
Clerks → C₁, C₂, C₃, C₄
Managers → M₁, M₂, M₃
Executive officers → E₁, E₂

R = {00000, A₁) (10000, A₂) (10000, A₃) (10000, A₄) (10000, A₅)
(25000, C₁) (25000, C₂) (25000, C₃) (25000, C₄)
(50000, M₁) (50000, M₂) (50000, M₃) (100000, E₁) (100000, E₂)}

(a) Arrow diagram

Functions Definition
A relation f between two non – empty sets X and Y is called a function from X to Y if for each x ∈ X there exists only one Y ∈ Y such that (x, y) ∈ f
f = {(x, y) / for all x ∈ X, y ∈ f}
Note: The range of a function is a subset of its co-domain

Students can download Maths Chapter 1 Set Language Ex 1.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.4

Question 1.
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find
(i) (P∪Q)∪R
(ii) (P∩Q)∩S
(m) (Q∩S)∩R
Solution:
P = {1, 2, 5, 7, 9}; Q = {2, 3, 5, 9, 11}; R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
(i) P∪Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}
= {1, 2, 3, 5, 7, 9, 11}
(P∪Q)∪R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}
= {1, 2, 3, 4, 5, 7, 9, 11}

(ii) P∩Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}
= {2, 5, 9}
(P∩Q)∩S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}
= (2, 5}

(iii) Q∩S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}
= {2, 3, 5}
(Q∩S)∩R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}
= {3, 5}

Question 2.
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution:
P is a real number set
Q is a set of irrational number
∴ Q⊂P
P∪Q= Q∪P = P
∴ Union of sets is commutative.
P∩Q = Q∩P = Q
∴ Intersection of sets is commutative.

Question 3.
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.
Solution:
When union of sets is associative
A∪(B∪C) = (A∪B)∪C
(B∪C) = {m, n, q, s, t) ∪ {m, n, p, q, s}
= {m, n, p, q, s, t}
A∪(B∪C) = {p, q, r, s} ∪ {m, n, p, q, s, t}
= {m, n, p, q, r, s, t} ……..(1)
(A∪B) = {p, q, r, s} ∪ {m, n, q, s, t}
= {m, n, p, q, r, s, t}
(A∪B)∪C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}
= {m, n, p, q, r, s, t} ……….(2)
From (1) and (2) it is verified that A∪(B∪C) = (A∪B)∪C

Question 4.
Verify the associative property of intersection of sets for A = {-11, √2, √5, 7},
B = {√3, √5, 6, 13} and C = {√2, √3, √5, 9}.
Solution:
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9}
= {√3, √5}
A∩(B∩C) = {-11, √2, √5, 7} ∩ {√3, √5}
{√5} ………(i)
(A∩B) = {-11, √2, √5, 7} ∩ {√3, √5 ,6, 13}
= {√5}
(A∩B)∩C = {√5} n {√2, √3, √5, 9}
= {√5}……..(2)
From (1) and (2) it is verified that A∩(B∩C) = (A∩B)∩C

Question 5.
If A={ x : x = 2ⁿ, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Solution:
A = {x : x = 2ⁿ, n ∈ W and n < 4}
A = {1, 2, 4, 8}
B = {x : x = 2n, n ∈ N and n ≤ 4}
B = {2, 4, 6, 8}
C ={0, 1, 2, 5, 6}
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {2, 4, 6, 8} ∩ {0, 1, 2, 5, 6}
= {2, 6}
A∩(B∩C)= {1 ,2, 4, 8} ∩ {2, 6}
= {2}……….(1)
(A∩B) = {1, 2, 4, 8} ∩ {2, 4, 6, 8}
= {2, 4, 8}
(A∩B)∩c= {2, 4, 8} n {0, 1, 2, 5, 6}
= {2}………..(2)
From (1) and (2) we get A∩(B∩C) = (A∩B)∩C