Super 20 Combined Class 12 for Tamil Nadu Board Science Stream

(turned ON). The increase in collector current (IC) increases the voltage drop across RC, 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 (Vin = 0V), decreases the base current (IB) and in turn decreases the collector current (IC). The transistor will move into the cut-off region (turned OFF). The decrease in collector current (IC) decreases the drop across RC, thereby increasing the output voltage, close 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. PART - IV Answer all the questions. [5 × 5 = 25] 34. (a) Obtain the expression for the induced emf by changing relative orientation of the coil with the magnetic field (Graph not necessary) Consider a rectangular coil of N turns kept in a uniform magnetic cos t field B figure (a). The coil rotates in anti-clockwise direction with an angular velocity ω about an axis, perpendicular to the field. At time t = 0, the plane of the coil is perpendicular to the B field and the flux linked with the coil has its maximum value Фm = NBA (where A is the area of the coil). sin t In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is The coil has rotated Фm cos ωt, a component of Фm normal to the plane of the coil. through an angle θ = ωt The component parallel to the plane (Фm sin ωt) has no role in electromagnetic induction. Therefore, the flux linkage at this deflected position is NFB = NFm cos wt. According to Faraday’s law, the emf induced at that instant is e =  d  N B    d  N m cos t  dt dt = –NFm(–sin wt)w = NFmw sin wt When the coil is rotated through 90° from initial position, sin ωt = 1. Then the maximum value of induced emf is em = NFmw = NBAw since Fm = BA Therefore, the value of induced emf at that instant is then given by e = em sin wt It is seen that the induced emf varies as sine function of the time angle ωt. The graph between induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf. C-10 Physics – XII

[OR] 34. (b) Derive the mirror equation and the equation for lateral magnification. The mirror equation: • The mirror equation establishes a relation among object distance u, image distance v and focal length f for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C. • Let us consider three paraxial rays from point B on the object. • The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PB´. The third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path. • The three reflected rays intersect at the point B′. A perpendicular drawn as A′ B′ to the principal axis is the real, inverted image of the object AB. As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠ B′PA′ . The triangles ΔBPA and ΔB′PA′ are similar. Thus, from the rule of similar triangles, A'B' = PA ' ...(1) AB PA The other set of similar triangles are, ΔDPF and Δ B′A′ F. (PD is almost a straight vertical line) D A'B' = A'F B A PD PF A' C FP As, the distances PD = AB the above equation becomes, E B' A'B' = A'F ...(2) AB PF Mirror equation From equations (1) and (2) we can write, PA ' A 'F = PA PF As, A′ F = PA′ – PF, the above equation becomes, PA ' PA '− PF = ...(3) PA PF We can apply the sign conventions for the various distances in the above equation. PA = −u, PA' = −v, PF = − f All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes, −v = −v − (− f ) −u − f On further simplification, v= v − f ; v= v −1 u f uf Sample Paper - 1 C-11

Dividing either side with v, 1= 1 − 1 u fv After rearranging, 1 + 1 =1 ...(4) vu f The above equation (4) is called mirror equation. Lateral magnification in spherical mirrors: The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis. magnification (m) = height of the image (h ') height of the object (h) m = h ' ...(5) h Applying proper sign conventions for equation (1), A 'B' PA ' = AB PA A'B' = −h', AB = h, PA' = −v, PA = −u −hh ' = −v −u On simplifying we get, m = h' = − v ...(6) h u Using mirror equation, we can further write the magnification as, =hh − f v =f u ...(7) f f 35. (a) Deduce the expression for the force between two long parallel, current-carrying conductors. Force between two long parallel current r zr B1 carrying conductors: Oy Let two long straight parallel current carrying x conductors separated by a distance r be kept F2 B1 in air medium as shown in Figure. Let I1 and I2 be the electric currents passing through I1 I2 the conductors A and B in same direction (i.e. along z - direction) respectively. The net I1 I2 magnetic field at a distance r due to current I1 Two long straight parallel wires in conductor A is B1 = µI (− i ) = − µI i πr πr C-12 Physics – XII

From thumb rule, the direction of magnetic field is perpendicular to the plane of the paper and inwards i.e. along negative direction. Let us consider a small elemental length dl in conductor B at which the magnetic field is present. From equation dF (I dl B), Lorentz force on the element dl of conductor B is       d F =I2d l × B1 µ0 I1 (k × i) µ0 I1I 2 dl j ( ) =−I2dl 2π r =− 2π r Therefore the force on dl of the wire B is directed towards the conductor A. So the element of length dl in B is attracted towards the conductor A. Hence the force per unit length of the conductor B due to current in the conductor A is F =− µ I1I2 j l 2πr Similarly, the net magnetic induction due to current I2 (in conductor B) at a distance r in the elemental length dl of conductor A is B2 µ I2 i 2πr From the thumb rule, direction of magnetic field is perpendicular to the plane of the paper and outwards i.e., along positive direction. Hence, the magnetic force acting on element dl of the conductor A is z ( )  I2  I1I2dl Oy dF = I1dl⋅ B2 I1dl 2r k×i = 2r j x Therefore the force on dl of conductor A is directed towards the I1 I2 conductor B. So the length dl is attracted towards the conductor B dl as shown in Figure. B2 • F F × B1 The force acting per unit length of the conductor A due to the current in conductor B is r F µ I1I2 j Current in both the conductors are in l 2πr the same direction - attracts each other Thus the force between two parallel current The force between two parallel current carrying conductors is attractive if they carry carrying conductors is repulsive if they carry current in the same direction. current in opposite directions. Current in the same direction Current in the opposite direction F Current F F F Current Current Two parallel conductors carrying current in same direction experience an attractive force Two parallel conductors carrying current in opposite direction experience a repulsive force [OR] Sample Paper - 1 C-13

35. (b) Write down Maxwell equation in integral form. Maxwell’s equations in integral form: Electrodynamics can be summarized into four basic equations, known as Maxwell’s equations. These equations are analogous to Newton’s equations in mechanics. Maxwell’s equations completely explain the behaviour of charges, currents and properties of electric and magnetic fields. So we focus here only in integral form of Maxwell’s equations: (i) First equation is nothing but the Gauss’s law. It relates the net electric flux to net electric enclosed in a surface. as ∫charge Mathematically, it is expressed E.dA = Qenclosed  s ε Where E is the electric field and Qenclosed is the charge enclosed. This equation is true for both discrete or continuous distribution of charges. It also indicates that the electric field lines start from positive charge and terminate at negative charge. This implies that the electric field lines do not form a continuous closed path. In other words, it means that isolated positive charge or negative charge can exist. (ii) Second equation has no name. But this law is similar to Gauss’s law in electrostatics. So this law can also be called as Gauss’s law in magnetism. The surface integral of magnetic field over a closed surface is z∫eBro..dMAa=th0ematically,  s where B is the magnetic field. This equation implies that the magnetic lines of force form a continuous closed path. In other words, it means that no isolated magnetic monopole exists. (iii) Third equation is Faraday’s law of electromagnetic induction. This law relates electric field with the changing magnetic flux∫l wEh.diclh=is−mddtaΦthBematically written as  where E is the electric field. This equation implies that the line integral of the electric field around any closed path is equal to the rate of change of magnetic flux through the closed path bounded by the surface. (iv) Fourth equation is modified Ampere’s circuital law. This is also known as ampere-Maxwell’s law. This law relates the magnetic field around any closed path to the conduction current and displac∫emB .ednlt=cuµr0riecn+t µth0rεo0udgdth∫stEha.dtAp ath. l  Where B is the magnetic field. This equation shows that both conduction and also displacement current produces magnetic field. These four equations are known as Maxwell’s equations in electrodynamics. C-14 Physics – XII

36. (a) Describe Davisson – Germer experiment which demonstrated the wave nature of electrons. L.T. Davisson - Germer experiment: • De Broglie hypothesis of matter waves F Electron gun was experimentally confirmed by Clinton H.T. in aluminium Davisson and Lester Germer in 1927. They demonstrated that electron beams diaphragms are diffracted when they fall on crystalline Aluminium cylinder solids. • Since crystal can act as a three-dimensional diffraction grating for matter waves, the Incident electron waves incident on crystals are beam diffracted off in certain specific directions. Electron • The filament F is heated by a low tension detector (L.T.) battery. Electrons are emitted from the hot filament by thermionic emission. Scattered They are then accelerated due to the θ beam potential difference between the filament and the anode aluminium cylinder by a high Ni crystal tension (H.T.) battery. Experimental set up of • Electron beam is collimated by using two Davisson – Germer experiment thin aluminium diaphragms and is allowed to strike a single crystal of Nickel. • The electrons scattered by Ni atoms in different directions are received by the electron detector which measures the intensity of scattered electron beam. • The detector is rotatable in the plane of the paper so that the angle φ between the incident beam and the scattered beam can be changed at our will. Inte ted V = 54V • The intensity of the scattered electron beam is electron beam measured as a function of the angle θ. • From the graph shows the variation of intensity of the scattered electrons with the angle θ for the accelerating voltage of 54V. For a given accelerating voltage V, the scattered wave 50° shows a peak or maximum at an angle of 0° 30° 60° 90° 50° to the incident electron beam. This peak θ in intensity is attributed to the constructive interference of electrons diffracted from Variation of intensity of di racted electron beam with the angle θ various atomic layers of the target material. Sample Paper - 1 C-15

• From the known value of interplanar spacing of Nickel, the wavelength of the electron wave has been experimentally calculated as 1.65Å. • The wavelength can also be calculated from de Broglie relation for V = 54 V from equation. =λ 12=.27 Å 12.27 V 54 λ =1.67 Å • This value agrees well with the experimentally observed wavelength of 1.65Å. Thus this experiment directly verifies de Broglie’s hypothesis of the wave nature of moving particles. [OR] 36. (b) (i) Derive an expression for the orbital energy of an electron in hydrogen atom using Bohr theory. (ii) An electron in Bohr's hydrogen atom has an energy of – 3.4 eV. What is the angular momentum of the electron? (i) The energy of an electron in the nth orbit Since the electrostatic force is a conservative force, the potential energy for the nth orbit is Un = 4π1ε0 (+Ze)(−e) = −1 Ze2 rn 4πε0 rn = – 1 Z 2me4  rn = ε0h2 n2  4ε02 h2n2 πme2 Z  The kinetic energy for the nth orbit is =KEn 12=mvn2 me4 Z2 8ε02h2 n2 This implies that Un = –2 KEn. Total energy in the nth orbit is En = KEn + Un = KEn – 2KEn = – KEn En = − me4 Z2 8ε02h2 n2 For hydrogen atom (Z = 1), En = − me4 1 joule ...(1) 8ε02h2 n2 where n stands for principal quantum number. The negative sign in equation (1) indicates that the electron is bound to the nucleus. Substituting the values of mass and charge of an electron (m and e), permittivity of free space ε0 and Planck’s constant h and expressing in terms of eV, we get 1 En = −13.6 n2 eV C-16 Physics – XII

For the first orbit (ground state), the total energy of electron is E1= – 13.6 eV. For the second orbit (first excited state), the total energy of electron is E2 = –3.4 eV. For the third orbit (second excited state), the total energy of electron is E3 = –1.51 eV and so on. Notice that the energy of the first excited state is greater than the ground state, second excited state is greater than the first excited state and so on. Thus, the orbit which is closest to the nucleus (r1) has lowest energy (minimum energy compared with other orbits). So, it is often called ground state energy (lowest energy state). The ground state energy of hydrogen (–13.6 eV ) is used as a unit of energy called Rydberg (1 Rydberg = –13.6 eV ). The negative value of this energy is because of the way the zero of the potential energy is defined. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes. (ii) The angular momentum of an electron when energy is – 3.4 eV. n = 2 Ln = nh ⇒ L=2 2=h 6.6 ×10−34 2π 2π 3.14 L2 = 2.1 Kg m2s-1 37. (a) State and prove De Morgan’s First and Second theorems. De Morgan’s First Theorem: The first theorem states that the complement of the sum of two logical inputs is equal to the product of its complements. A B A+B A + B A B A. B Proof The Boolean equation for NOR gate is Y = A + B . 0 0 0 1 1 1 1 The Boolean equation for a bubbled AND gate is 01 1 0 10 0 Y = A. B . 10 1 0 01 0 11 1 0 00 0 Both cases generate same outputs for same inputs. It can be verified using the following truth table. From the above truth table, we can conclude A + B = A. B . Thus De Morgan’s First Theorem is proved. It also says that a NOR gate is equal to a bubbled AND gate. The corresponding logic circuit diagram is shown in figure. AA Y Y BB NOR gate equals bubbled AND gate De Morgan’s Second Theorem: The second theorem states that the complement of the product of two inputs is equal to the sum of its complements. Proof The Boolean equation for NAND gate is Y = AB Sample Paper - 1 C-17

The Boolean equation for bubbled OR gate is Y = A + B . A and B are the inputs and Y is the output. The above two equations produces the same output for the same inputs. It can be verified by using the truth table. A B A.B A.B A B A + B 00 0 1 1 1 1 01 0 1 1 0 1 10 0 1 0 1 1 11 1 0 0 0 0 From the above truth table we can conclude A.B = A + B . Thus De Morgan’s Second Theorem is proved. It also says, a NAND gate is equal to a bubbled OR gate. The corresponding logic circuit diagram is shown in figure. A A Y B YB NAND gate equals bubbled OR gate [OR] 37. (b) Find out the phase relationship between voltage and current in a pure inductive circuit. AC circuit containing only an inductor: Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source. The alternating voltage is given by the equation. v = Vm sin wt ...(1) The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by Back emf, ε = −L di dt By applying Kirchoff’s loop rule to the purely inductive circuit, we get v + e = 0 L Vm sin wt = L di dt ~ di Vm = L sin wt dt v = Vm sin t sin t dt Integrating both sides, we get AC circuit with inductor constant  i = Vm = Vm (–cos wt) + L Lω The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero. C-18 Physics – XII

  cost  sin     2  t       sin t   2 i = Vm sin  t     (or) i = Im sin  t    ...(2) Lω  2   2  where Vm = Im, the peak value of the alternating current in the circuit. From equation (1) and Lω current lags behind the applied voltage by π in an inductive circuit. This (2), it is evident that 2 fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90°. Vm ωL Inductive reactance XL: The peak value of current Im is given by Im = . Let us compare this equation with Im quantity same role as the = Vm from resistive circuit. The ωL plays the R resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance (XL). It is measured in ohm. XL = wL The inductive reactance (XL) varies directly as the frequency. XL = 2pf L...(3) where f is the frequency of the alternating current. For a steady current, f = 0. Therefore, XL = 0. Thus an ideal inductor offers no resistance to steady DC current. vi v Im sin ( t– ) Vm sin t Ai v 0 t t 0i B OA = Vm OB = Im Phasor diagram and wave diagram for AC circuit with L 38. (a) State Gauss Law in electrostatics. Obtain the expression for electric field due to an infinitely long charged wire. Definition: Gauss law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux ΦE through the closed surface is  ΦE = E  d A  Qencl 0 Sample Paper - 1 C-19

Electric field due to an infinitely long charged wire: Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A1 and A2 on the wire which are at equal distances from the point P. The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry. A2 r EQ R P r PE S A1 Electric field due to infinite long charged wire Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this c=lo∫sEed. surface is ΦE dA      ...(1)   =ΦE ∫ E.d A + ∫ E.dA+ ∫ E.d A ...(2) Curved Top Bottom surface surface surface    It is seen that forthe curved surface, E is parallelto A and E.d A = E dA. For the top and bottom surface, E is perpendicular to A and E.d A = 0 Substituting these values in the equation (2) and applying Gauss law to the cylindrical surfaces, we have ∫ =ΦE =EdA Qencl ...(3) ε0 Curved surface Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration and Qencl is given by Qencl = λL where λ is the linear charge density. ∫ E dA = λL ...(4) ε0 Curved surface C-20 Physics – XII

Here ∫ dA = total area of the curved surface = 2πrL. Substituting this in equation (4), Curved surface we get E.2πrL =λL (or) E = 1 λ ...(5) ε0 2πε0 r  1 λ r  ...(6) In vector form, E = 2πε0 r 1 n The electric field due to the infinite charged wire depends on rather 1 r than r2 for a point charge. Equation (6) indicates that the electric field is always along the r perpendicular direction ( 0 then E points perpendicular outward ( r ) to wire. In fact, if λ > λ < 0, then E points n r ) from the wire and if perpendicular inward (- r ). n Cylindrical Gaussian surface [OR] 38. (b) Explain the determination of the internal resistance of a cell using voltmeter. Determination of internal resistance: The emf of cell ε is ++ ε –– r measured by connecting a high resistance voltmeter across ε r it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance R is (a) ++ VV –– included in the circuit and current I is established in the (a) circuit. The potential difference across R is equal to the Voltmeter potential difference across the cell (V). Voltmeter The potential drop across the resistor R is ++ ε –– r ε r V = IR ...(1) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ε. It is because, I ++ VV –– I R certain amount of voltage (Ir) has dropped across the R internal resistance r. Then V = ε – Ir Ir = ε – V ...(2) (b) V (b) resistanVce Dividing equation (2) by equation (1), we get of the cell Internal Ir =   V Internal resistance of the cell IR V r =   V R ...(3) V Since ε, V and R are known, internal resistance r can be determined. Sample Paper - 1 C-21

4SAMPLE PAPER – Time: 3 Hours (UNSOLVED) Maximum Marks: 70 PART - I [15 × 1 = 15] Answer all the questions. Choose the correct answer. 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 2. In Joule’s heating law, when I and t are constant, if the H is taken along the y axis and I2 along the x axis, the graph is .......................... . (a) straight line (b) parabola (c) circle (d) ellipse 3. Electromotive force is most closely related to ................ . (a) electric field (b) magnetic field (c) potential difference (d) mechanical force 4. A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an axis perpendicular to its plane passing through the center with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the axis of rotation ................ . (a) 1 σωπ BR (b) 1 σωπ BR2 (c) 1 σωπ BR3 (d) 1 σωπ B R4 4 44 4 5. 2π02 H inductor is connected to a capacitor of capacitance C. The value of C in order to impart maximum power at 50 Hz is ................ . (a) 50 μF (b) 0.5 μF (c) 500 μF (d) 5 μF 6. In a LCR AC circuit off resonance, the current ................ . (a) is always in phase with the voltage (b) always lags behind the voltage (c) always leads the voltage (d) may lead or lag behind the voltage 7. If the amplitude of the magnetic field is 3 × 10−6 T, then amplitude of the electric field for a electromagnetic waves is ................ . (a) 100 V m−1 (b) 300 V m−1 (c) 600 V m−1 (d) 900 V m−1 8. An object is placed in front of a convex mirror of focal length of f and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. (a) 2f and c (b) c and ∞ (c) f and O (d) None of these 9. Which of the following is used in optical fibres? (a) Total internal reflection (b) Diffraction (c) Refraction (d) Scattering C-56

10. Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of emitted electrons will be ..................... . (a) 1:4 (b) 1:3 (c) 1:1 (d) 1:9 11. The mass of a 7 Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The 3 binding energy per nucleon of 7 Li nucleus is nearly ................ . 3 (a) 46 MeV (b) 5.6 MeV (c) 3.9 MeV (d) 23 MeV 12. The allowed energy for the particle for a particular value of n is proportional to ................ . (a) a–2 (b) a–3/2 (c) a–1 (d) a2 13. When a transistor is fully switched on, it is said to be ............. . (a) Shorted (b) Saturated (c) Cut-off (d) Open 14. The output transducer of the communication system converts the radio signal into ............. . (a) Sound (b) Mechanical energy (c) Kinetic energy (d) None of the above 15. The materials used in Robotics are ................ . (a) Aluminium and silver (b) Silver and gold (c) Copper and gold (d) Steel and aluminium PART - II Answer any six questions in which Q. No 18 is compulsory. [6 × 2 = 12] 16. What is meant by electrostatic energy density? 17. What is superconductivity? 18. The repulsive force between two magnetic poles in air is 9 × 10–3 N. If the two poles are equal in strength and are separated by a distance of 10 cm, calculate the pole strength of each pole. 19. What is meant by mutual induction? 20. What is the reason for reddish appearance of sky during sunset and sunrise? 21. Give some important uses of photo-cells. 22. What is half-life of nucleus? Give the expression. 23. Draw the output waveform of a full wave rectifier. 24. Define bandwidth? PART - III Answer any six questions in which Q.No. 25 is compulsory. [6 × 3 = 18] 25. A parallel plate capacitor has square plates of side 5 cm and separated by a distance of 1 mm. (a) Calculate the capacitance of this capacitor. (b) If a 10 V battery is connected to the capacitor, what is the charge stored in any one of the plates? (The value of ε0 = 8.85 × 10–12 Nm2 C–2). Sample Paper - 4 C-57

26. What is Thomson effect? 27. Is an ammeter connected in series or parallel in a circuit? Why? 28. What are LC oscillations? 29. A pulse of light of duration 10−6 s is absorbed completely by a small object initially at rest. If the power of the pulse is 60 × 10−3 W, calculate the final momentum of the object. 30. What is critical angle and total internal reflection? 31. In alpha decay, why the unstable nucleus emits 4 He nucleus? Why it does not emit four 2 separate nucleons? 32. Define skip distance. 33. What are sub atomic particles? PART - IV Answer all the questions. [5 × 5 = 25] 34. (a) Obtain the expression for capacitance for a parallel plate capacitor. [OR] (b) Find the magnetic field due to a long straight conductor using Ampere’s circuital law. 35. (a) How will you induce an emf by changing the area enclosed by the coil? [OR] (b) Write short notes on (i) Infrared radiation (ii) Ultraviolet radiation (iii) Gamma radiation. 36. (a) Derive the equation for thin lens and obtain its magnification. [OR] (b) Describe briefly Davisson – Germer experiment which demonstrated the wave nature of electrons. 37. (a) Discuss the alpha decay process with example. [OR] (b) Explain the formation of PN junction diode. Discuss its V–I characteristics. 38. (a) Write a short notes on the prisms making use of total internal reflection. [OR] (b) Elaborate any two types of Robots with relevant examples C-58 Physics – XII

One-mark & Two-mark Solutions for Unsolved Sample Papers Sample Paper-4 PART - I 1. (c) uniformly charged infinite plane 2. (a) straight line 3. (c) potential difference 4. (d) 1 σwπ BR4 4 5. (d) 5 μF 6. (d) may lead or lag behind the voltage 7. (d) 900 V m−1 8. (d) None of these Hint: There is no maximum minimum object distance for convex mirror to form real and inverted image. 9. (a) Total internal reflection Hint : The working of optical fibres is based on total internal reflection. 10. (b) 1:3 =Hint: v1max =2(hυ - φ) 2=(0.9 - 0.6) 2(0.3) m m m = v2max =2(hυ - φ) 2=(3.3 - 0.6) 2(2.7) m m m v1max =2(0.3) × m =0.3 =1 v2max m 2(2.7) 2.7 3 (v1 : v2)max = 1 : 3 11. (b) 5.6 MeV Hint : If m = 1 u, C = 3 × 108 ms–1 then, E = 931 MeV 1 u = 931 Mev Binding energy = 0. 042 × 931 = 39. 10 MeV B.E 39.10 Binding energy per nucleon = A = 7 = 5.58 = 5.6 MeV 12. (a) a–2 λ or λ =2a Hint: For the standing wave, a = n 2n P=2 n2h2 P= h= nh ;=E 2m 2a2m ; E ∝ a–2 λ 2a C-77

13. (b) Saturated 14. (a) Sound 15. (d) Steel and aluminum PART - II U Volume 16. The energy stored per unit volume of space is defined as energy density uE = From equation, uE =1 ( ε0A ) ( Ed )2= 1 ε0 ( Ad ) E2  or  uE= 1 ε0E2 2 d 2 2 17. The ability of certain metals, their compounds and alloys to conduct electricity with zero 18. TreThsheiseftaomnrcacegenbaiettutvwdeeeryeonflottwwheotefpomorclpeeesriaastrueFrge=sivkiesqncmabArlyq2lemdBFs=upkeqrcmoArnq2dmuBctivity. Given : F = 9 × 10–3N, r = 10 cm = 10 × 10–2 m Therefore, 9 ×10-3 = 10-7 × qm2 ⇒ qm = 30NT-1 (10 ×10-2 )2 19. When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction. 20. During sunrise or sunset, the sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the sun appears red. 21. Applications of photo cells: • Photo cells have many applications, especially as switches and sensors. • Automatic lights that turn on when it gets dark use photocells, as well as street lights that switch on and off according to whether it is night or day. • Photo cells are used for reproduction of sound in motion pictures and are used as timers to measure the speeds of athletes during a race. 22. the half-life T1/2 is defined as the time required for the number of atoms initially present to reduce to one half of the initial amount. T=1/ 2 ln 2 0.6931 = λ λ 23. Output waveform of a full wave rectifier: Vpeak Input voltage 0 Time −Vpeak Output voltage 0 (c) Time C -7 8 Input and output waveforms Physics – XII

24. The frequency range over which the baseband signals or the information signals such as voice, music, picture, etc. is transmitted is known as bandwidth. One-mark & Two-mark Solutions C-79

CHEMISTRY QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to be Total Marks answered Multiple Choice Questions 1 15 15 Very short answers: 2 6 12 (Totally 9 questions will be given. Any one question should be answered compulsorily.) Short answers: 3 6 18 (Totally 9 questions will be given. Any one question should be answered compulsorily.) Essay type 55 25 Total 70 Practical Marks 20 Internal Assessment 10 100 Total Marks S.No. Weightage of Marks Weightage 1. 30% 2. Purpose 40% 3. Knowledge 20% 4. Understanding 10% Application Skill/Creativity D-1

Time: 3 Hours 1SAMPLE PAPER – Maximum Marks: 70 (SOLVED) PART - I Answer all the questions. Choose the correct answer. [15 × 1 = 15]  [Answers are in Bold] 1. The following set of reactions are used in refining Zirconium Zr (impure) + 2I2 523 K ZrI4 ZrI4 1800K Zr (pure) + 2I2 This method is known as .................... . (a) Liquation (b) Van Arkel process (c) Zone refining (d) Mond’s process 2. The stability of +1 oxidation state increases in the sequence ............................ . (a) Al < Ga < In < Tl (b) Tl < In < Ga < Al (c) In < Tl < Ga < Al (d) Ga < In < Al < Tl 3. Assertion : Bond dissociation energy of fluorine is greater than chlorine gas. Reason: Chlorine has more electronic repulsion than fluorine. (a) Both assertion and reason are true and reason is the correct explanation of assertion (b) Both assertion and reason are true but reason is not the correct explanation of assertion (c) Assertion is true but reason is false (d) Both assertion and reason are false 4. Which of the following pair has d10 electrons? (d) Mn2+ , Fe3+ (a) Ti3+ , V4+ (b) Co3+ , Fe2+ (c) Cu+ , Zn2+ 5. Which is used for the separation of lanthanides, in softening of hard water and also in removing lead poisoning? (a) [Ni (CO)4] (b) EDTA (c) [Ni(DMG)2] (d) Ti Cl4 + AI (C2H5)3 6. The yellow colour in NaCl crystal is due to .................... . (a) excitation of electrons in F centers (b) reflection of light from Cl– ion on the surface (c) refraction of light from Na+ ion (d) all of the above 7. For a reaction Rate = k[acetone]3/2 then unit of rate constant and rate of reaction respectively is .................... . (b) (mol–1/2 L1/2 s–1) , (mol L–1 s–1) (a) (mol L–1 s–1),(mol–1/2 L1/2 s–1) (c) (mol1/2 L1/2 s–1) , (mol L–1 s–1) (d) (mol L s–1) , (mol1/2 L1/2 s) Sol. Rate = k [A]n Rate = −d[A] ; unit of rate = mol L−1 = mol L–1s–1 ( )dt s mol L−1s−1 ( ) unit of rate constant = mol L−1 n = mol1– n Ln – 1 s– 1 In this case, rate = k [Acetone]3 2 n = 3/2 mol1−(3 2)L(3 2)−1s−1 ⇒ mol−(1 2)L(1 2)s−1 D-3

8. Arrange the acids (i) H2SO3 (ii) H3PO3 and (iii) HClO3in the decreasing order of acidity. (a) (i) > (iii) > (ii) (b) (i) > (ii) > (iii) (c) (ii) > (iii) > (i) (d) (iii) > (i) > (ii) Sol. Acidity is directly proportional to oxidation number. As the oxidation number of S, P and Cl in H2SO3 , H3PO3 and HClO3 is +4, +3, +5 respectively. So decreasing order of acidity will be (iii) > (i) > (ii) 9. The conductivity of strong electrolyte is ............ . (a) increase on dilution slightly (b) decrease on dilution (c) does not change with dilution (d) depend upon density of electrolyteitself 10. Which one of the is not a surfactant? (a) CH3–(CH2)15 – –(CH3)2CH2Br (b) CH3–(CH2)15 – NH2 (c) CH3–(CH2)16 – CH2OSO2– Na+ (d) OHC–(CH2)14 – CH2–COO– Na+ 11. Which compound has the highest boiling point? (a) Acetone (b) Diethyl ether (c) Methanol (d) Ethanol 12. The reagent used to distinguish between acetaldehyde and benzaldehyde is ..................... . (a) Tollens reagent (b) Fehling’s solution (c) 2,4 – dinitrophenyl hydrazine (d) semicarbazide 13. CH3CH3Br aq NaOH A KMnO4 /H+ B NH3 C Br2/NaOH D. ‘D’ is ................ . Δ Δ Δ (a) bromomethane (b) α - bromo sodium acetate (c) methanamine (d) acetamide Hint: CH3CH3Br aq NaOH CH3CH2– OH KMnO4 CH3– COOH NH3 CH3CONH2 Δ Br2/NaOH CH3–NH2 14. Which one of the reagent does not react with glucose? (a) Acetic anhydride (b) Tollen’s reagent (c) Sodium bi sulphite (d) Bromine water 15. The rate of decomposition of ozone drops sharply in ................. . (a) acidic medium (b) alkaline medium (c) neutral medium (d) Ether medium PART - II Answer any six questions. Question No. 23 is compulsory. [6 × 2 = 12] 16. What is the role of Limestone in the extraction of Iron from its oxide Fe2O3? In the extraction of iron, a basic flux limestone is used. Limestone decomposes to form CaO which reacts with silica gangue present in the iron ore is acidic in nature to form calcium silicate (slag). CaCO3 CaO + CO2 CaO + SiO2 CaSiO3 Flux Gangue Slag 17. Mention the uses of the potash alum. (i) It is used for purification of water (ii) It is also used for water proofing and textiles D-4    Chemistry – XII

(iii) It is used in dyeing, paper and leather tanning industries (iv) It is employed as a styptic agent to arrest bleeding. 18. Why Gd3+ is colourless? Gd – Electronic Configuration : [Xe] 4f 7 5d1 6s2 Gd3+ – Electronic Configuration : [Xe] 4f 7 In Gd3+, no electrons are there in outer d-orbitals. d – d transition is not possible. So it is colourless. 19. Ionic solids conduct electricity in molten state but not in solid state. Explain. In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free to move but remain held together by strong electrostatic forces of attraction, so they cannot conduct electricity in the solid state. 20. What is meant by conjugate acid-base pair? Find the conjugate acid / base for the Afonllaocwidin-bgasspeepcaieirs:whHicNhOd2if,fCerHs –b,yHaCplOro4to, nOHon–l,yC(OH3A2–,S2–A– + H+) is known as conjugate acid-base pair. Conjugate acid: HCN, H2O, HCO3–, HS – . Conjugate base: NO2–, ClO4–, O2– . 21. NH3, CO2 are readily adsorbed where as H2, N2 are slowly adsorbed. Give reason. • The nature of adsorbate can influence the adsorption. Gases like NH3, CO2 are easily liquefiable as have greater Van der Waals forces of attraction and hence readily adsorbed due to high critical temperature. • But permanent gases like H2, N2 can not be easily liquefied and having low critical temperature and adsorbed slowly. 22. Alcohol can act as Bronsted base. Prove this statement. Alcohols can also act as a Bronsted bases. It is due to the presence of unshared electron pairs on oxygen which make them to accept proton. So proton acceptor are Bronsted bases. i.e., alcohols are Bronsted bases. CH3 CH2 O•• H + H O H CH3 CH2 O H + OH H 23. Arrange the following in decreasing order of basic strength CH3CH2NH2 , O2N NH2 , NH2 , CH3– NH2 (i) Aliphatic amines are more basic than aromatic amines. Therefore CH3CH2NH2 and CH3NH2 are more basic. Among the ethylamine and methylamine, ethylamine was experienced more +I effect than methylamine and hence ethylamine is more basic than methylamine. (ii) Nitrogroup has a powerful electron withdrawing group and they have both –R effect as well as –I effect. As a result, all the nitro anilines are weaker bases than aniline. In P-nitroaniline ( O2N NH2 ) both –R effect and –I effect of the NO2 group decrease the basicity. Sample Paper - 1    D-5

(iii) Therefore decreasing order of basic strength is, CH3CH2NH2 > CH3NH2 > NH2 > O2N NH2 Ethylamine > Methylamine > Aniline > p - nitro aniline 24. Give one example for each of the following: (i) icosogens (ii) tetragen (iii) pnictogen (iv) chalcogen (i) Icosogens: (a) Boron, (b) Aluminium, (c) Gallium (ii) Tetragen: (a) Carbon, (b) Silicon, (c) Germanium (iii) Pnictogen: (a) Nitrogen, (b) Phosphorus, (c) Bismuth (iv) Chalcogen: (a) Oxygen, (b) Sulphur, (c) Selenium PART-III [6 × 3 = 18] Answer any six questions. Question No. 30 is compulsory. 25. Give the uses of silicones. Uses of silicones: (i) Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc. (ii) They are used for making water proofing clothes (iii) They are used as insulting material in electrical motor and other appliances (iv) They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals. 26. Complete the following reactions. 1. NaCl + MnO2 + H2SO4 2. I2 + S2O32− 3. P4 + NaOH + H2O 1. 4NaCl + MnO2 + 4H2SO4 Cl2 + MnCl2 + 4 NaHSO4 + 2H2O 2. I2 + 2S2O32− S4O62– + 2I– 3. P4 + 3NaOH + 3H2O 3NaH2PO2 + PH3 ↑ 27. Zn, Cd, Hg belong to d-block elements even though they do not have partially filled d-orbitals. Give reason. (i) Zn, Cd, Hg belong to d-block elements even though they do not have partially filled d-orbitals either in their elemental state or in their normal oxidation states. (ii) However they are treated as transition elements, because their properties are an extension of the properties of the respective transition elements. 28. Why ionic crystals are hard and brittle? The ionic compounds are very hard and brittle. In ionic compounds the ions are rigidly held in a lattice because the positive and negative ions are strongly attracted to each other and difficult to separate. But the brittleness of a compound is now easy to shift the position of atoms or ions in a lattice. If we apply a pressure on the ionic compounds the layers shifts slightly. The same charged ions in the lattice comes closer. A repulsive forces arises between same charged ions, due to this repulsions the lattice structure breaks down chemical bonding. D-6    Chemistry – XII

29. Paracetamol is prescribed to take once in 6 hours. Justify this statement. (i) Paracetamol is a well known antipyretic and analgesic that is prescribed in cases of fever and body pain. (ii) Paracetamol has a half life of 2.5 hours within the body. (i.e) the plasma concentration of the drug is halved after 2.5 hours. So after 10 hours (4 half lives), only 6.25% of drug remains. Based on this, the dosage and frequency will be decided. (iii) In the case of paracetamol, it is usually prescribed to take once in 6 hours. 30. A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g. Calculate the equivalent mass of the metal. Given, I = 0.15 amperes t = 150 mins ⇒ t = 150 × 60 sec ⇒ t = 9000 sec Q = It ⇒ Q = 0.15 × 9000 coulombs ⇒ Q = 1350 coulombs Hence, 135 coulombs of electricity deposit is equal to 0.783g of metal. ∴ 96500 coulombs of electricity, 0.783 × 96500 = 55.97 g of metal 1350 Hence equivalent mass of the metal is 55.97 31. What happens when (i) 2 – Nitropropane boiled with HCl (ii) Nitrobenzen electrolytic reduction in strongly acidic medium. (i) 2 – Nitropropane boiled with HCl: 2–nitropropane upon hydrolysis with boiling HCl give a ketone (2–propanone) and nitrous oxide. CH3 CH − NO2 Boiling HCl CH3 C = O + N2O + H2O CH3 CH3 (2-propanone) (Acetone) (nitrous oxide) (ii) Nitrobenezen electrolytic reduction in strongly acidic medium: Electrolytic reduction of nitrobenzene in weakly acidic medium gives aniline but in strongly acidic medium, it gives p-aminophenol obviously through the acid - catalysed rearrangement of the initially formed phenylhydroxylamine. NO2 NHOH NH2 Electrolytic reduction rearrangement Strongly acidic medium (Nitro benzene) (Phenyl hydroxylamine) OH 32. Write a short note on peptide bond. (P - amino phenol) (i) The amino acids are linked covalently by peptide bonds. (ii) The carbonyl group of the first amino acid react with the amino group of the second amino acid to give an amide linkage (–CONH) between these aminoacids. This amide linkage is called peptide bond. Sample Paper - 1    D-7

H2N − CH2 − COOH + H2N − CH − COOH −H2O H2N − CH2 − C − N − CH − COOH || | | (Glycine) CH3 O H CH3 (Alanine) (Peptide Bond) (Compound is Glycylalanine) (iii) The resulting compound is called a dipeptide. Because, two amino acids are inovlved for getting one peptide bond. (iv) If large number of amino acids combined through peptide bond, the resulting giant molecule is called a protein. (v) The amino end of the peptide is known as N-terminal, while the carboxy end is called C - terminal. 33. Cr and Cu are more stable. Give reason. (i) The electronic configuration of Cr and Cu are [Ar] 3d5 4s1 and [Ar] 3d10 4s1 respectively. The extra stability of half filled and fully filled d orbitals, as already explained in XI STD, is due to symmetrical distribution of electrons and exchange energy. (ii) The extra stability of half filled and fully filled d orbitals is due to symmetrical distribution of electrons and exchange energy. (iii) When the d-orbitals are considered together, they will constitute a sphere. So the half filled and fully filled configuration leads to complete symmetrical distribution of electron density. (iv) On the other hand, an unsymmetrical distribution of electron density will result in building up of a potential difference. To decrease this and to achieve a tension free state with lower energy, a symmetrical distribution is preferred. Answer all the questions. PART - IV [5 × 5 = 25] 3 4.(a) (i) What is smelting?(2) (ii) How will you prepare potash alum? (3) [OR] (b) (i) Give the balanced equation for the reaction between chlorine with cold NaOH.(3) (ii) Nitrogen exists as diatomic molecule and Phosphrus as P4 . Why?(2) (a) (i) Smelting is a process of reducing the roasted metallic oxide to metal in molten condition. In this process, impurities are removed by the addition of flux as slag. (ii) The alunite the alum stone is the naturally occurring form and it is K2SO4. Al2(SO4)3.4Al(OH)3. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation. K2SO4.Al2(SO4)3.4Al(OH)3 + 6H2SO4 → K2SO4 + Al2(SO4)3 + 12H2O K2SO4 + Al2(SO4)3 + 24H2O → K2SO4.Al2(SO4)3.24H2O [OR] (b) (i) Reaction between chlorine with cold NaOH: Cl2 + H2O HCl + HOCl D-8    Chemistry – XII

HCl + NaOH NaCl + H2O HOCl + NaOH NaOCl + H2O Overall reaction, Cl2 + 2 NaOH NaOCl + NaCl + H2O (cold) (Sodium hypochlorite) (Sodium chloride) Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite. (ii) Nitrogen has a triple bond between its two atoms because of its small size and high electronegativity. Phosphorus P4 has single bond, that is why it is tetra–atomic. 3 5.(a) (i) What are Wade’s Rule?(2) (ii) Compare Lanthanoids and Actinoids. (3) [OR] (b) (i) In the complex, [Pt(NO2)(H2O)(NH3)2]Br , identify the following(3) 1. Central metal atom/ion 2. Ligand(s) and their types 3. Coordination entity 4. Oxidation number of the central metal ion 5. Coordination number (2) (ii) Cu+, Zn2+, Sc3+, Ti4+ are colourless. Prove this statement. (a) (i) Wade’s rules are used to rationalize the shape of borane clusters by calculating the total number of skeletal electron pairs (SEP) available for cluster bonding. (ii) Lanthanoids Actinoids 1. Differentiating electron enters in 4f orbital Differentiating electron enters in 5f orbital 2. Binding energy of 4f orbitals are higher Binding energy of 5f orbitals are lower 3. They show less tendency to form complexes They show greater tendency to form complexes 4. Most of the lanthanoids are colourless Most of the actinoids are coloured. E.g : U3+ (red), U4+ (green), UO22+ (yellow) 5. They do not form oxo cations They do form oxo cations such as UO22+ NpO22+ etc. 6. Besides +3 oxidation states lanthanoids show Besides +3 oxidation states actinoids show higher +2 and +4 oxidation states in few cases. oxidation states such as +4, +5, +6 and +7. [OR] (b) (i) [Pt(NO2)(H2O)(NH3)2]Br – Pt2+ 1. Central metal ion 2. Ligands and their types – NO2 - Mono dendate ligand, H2O and NH3 - neutral monodendate ligand – [Pt(NO2)(H2O)(NH3)2]+ 3. Coordination entity 4. Oxidation number of the – x +1(–1)+1(0)+1(0) = +1 central metal ion x–1=+1 x = +2 5. Coordination number – 4 (ii) 1. Cu+, Zn2+ have d10 configuration and Sc3+, Ti4+ have d1 configuration. 2. d - d transition is not possible in the above complexes. So they are colourless. Sample Paper - 1    D-9

36.(a) (i) What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure? (2) (ii) Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation log k = log A Ea R  1  2.303  T  1 a straight Where Ea is the activation energy. When a graph is plotted for log k Vs T line with a slope of – 4000K is obtained. Calculate the activation energy. (3) [OR] (b) (i) Ksp of Ag2CrO4 is 1.1 × 10–12. What is solubility of Ag2CrO4 in 0.1M K2CrO4 .(3) (ii) Identify the Lewis acid and the Lewis base in the following reactions. (2) (1) CaO + CO2 → CaCO3 CH3 Cl (2) CH3 – O – CH3 + AlCl3 O Al Cl CH3 Cl (a) (i) 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water. NaOH(aq) + CH3COOH(aq) CH3COONa(aq) + H2O(l) 2. Coordination number of atoms in a bcc structure is 8 (ii) log k = log A – Ea R  1  2.303  T  y = c + mx m = – Ea 2.303 R Ea = – 2.303 R m Ea = – 2.303 × 8.314 J K–1 mol–1 × (− 4000 K) Ea = 76,589 J mol–1 Ea = 76.589 kJ mol−1 [OR] (b) (i) A g2CrO4 2Ag+ + CrO42– x 2x x x is the solubility of Ag CrO in 0.l M K CrO 24 24 K2CrO4 2K+ + CrO42– 0.l M 0.2 M 0.1 M [Ag+] = 2x [CrO42– ] = ( x + 0.l) ≈ 0.1 ∴ x << 0.1 Ksp = [Ag+]2 [CrO42–] D-10    Chemistry – XII

1.1 × 10–12 = (2x)2(0.l) 1.1 × 10–12 = 0.4x2 x2 = 1.1 × 10−12 ⇒ x = 1.1 × 10−12 0.4 0.4 x = 2.75 ×10−12 x = 1.65 × 10–6 M (ii) 1. CaO + CO2 → CaCO3 (a) CaO - Lewis base ; All metals oxides are Lewis bases (b) CO2 - Lewis acid ; CO2 contains a polar double bond. H3C 2. CH3 – O – CH3 + AlCl3 O AlCl3 H3C (a) CH3 – O – CH3 - Lewis base ; Electron rich species (b) AlCl3 - Lewis acid ; AlCl3 is electron deficient molecule. 37.(a) (i) What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?  (3) (ii) What is meant by catalyst poison? (2) [OR] (b) (i) How would you calculate the solubility of sparingly soluble salt using Kohlrausch’s law?  (3) (ii) Formic acid act as reducing agent. Prove this statement. (2) (a) (i) Lyophilic Sols: Colloidal sols directly formed by mixing substances like gums, gelatin, starch, rubber, etc. with a suitable liquid (The dispersion medium) are lyophilic sols. An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation) the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. These sols are quite stable and cannot be easily coagulated. Lyophobic sols: These colloidal sols can only be prepared by some special methods. These sols are readily precipitated on the addition of small amount of electrolytes, by heating or by shaking and hence are not stable. Hydrophobic sols are water hating. They are formed by indirect method. These sols are irreversible sols. These sols are readily precipitated by the addition of small amount of electrolytes, by heating or by shaking and hence are not stable. (ii) 1. The substances when added to a catalysed reaction decreases or completely destroys the activity of a catalyst are often known as catalytic poisons. 2. In the reaction 2SO2 + O2 → 2SO3 with Pt catalyst, the catalyst poison is AS2O3. [OR] Sample Paper - 1    D-11

(b) (i) 1. Substances like AgCl, PbSO4 are sparingly soluble in water. The solubility product can be determined using conductivity experiments. 2. Let us consider AgCl as an example AgCl(s)  Ag+ + Cl– Ksp = [Ag+] [Cl–] 3. Let the concentration of [Ag+] be ‘C’ mol L–1. If [Ag+] = C, then [Cl– ] is also equal to C mol L–1. ∴ Ksp = C.C ⇒ Ksp = C2. 4. The relationship between molar conductance and equivalent conductance is κ × 10−3 (or) C = κ × 10−3 C mol L−1 Λ° =( )Λ° Substitute the concentration value in the relation Ksp = C2 2 Ksp = κ × 10−3   Λ°   (ii) 1. Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent. O O H C OH H C OH Aldehyde group Carboxylic acid group 2. Formic acid reduces Tollen’s reagent (ammonical silver nitrate solution) to metallic silver. HCOO- + 2Ag+ + 3OH - → S2iAlvge r m+ i r r CorO32 - (Tollen’s reagent) + 2H2O 3. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous ions. H COO - + 2Cu2+ + 5OH - → RCedu2pOre↓c i p i t a +te CO3 2 - + 3H2O (Fehlings solution) (Blue) 38.(a) (i) What are the uses of nitrobenzene? (2) (ii) Define (i) Epimers (ii) Epimerisation. (3) [OR] (b) (i) CH3CONH2 is a weaker base than CH3CH2NH2. Why?(2) (ii) Write about the classification of organic nitro compounds.(3) (a) (i) 1. Nitro benzene is used to produce lubricating oils in motors and machinery. 2. It is used in the manufacture of dyes, drugs, pesticides, synthetic rubber, aniline and explosives like TNT, TNB. D-12    Chemistry – XII

(ii) 1. Sugar differing at an asymmetric centre is known as epimers. 2. The process by which one epimer is converted into other is called epimerisation and it requires the enzyme epimersase. 3. Galactose is converted to glucose by this manner in our body. C2 - epimers C4 - epimers CHO 1CHO CHO HO − C − H H − C − ΟH 2 H − C − ΟH HO − C − H HO −3C − H HO − C − H H − C − OH HO − C − H 4 H − C − OH H − C − OH H −5C − OH H − C − OH CH2OH 6 CH2OH CH2OH D - mannose D - glucose D - galactose Glucose and mannose are epimers at C2 carbon and glucose and galactose are epimers at C4 carbon [OR] (b) (i) Due to resonance, the lone pair of electrons on the nitrogen atom in CH3CONH2 is delocalised over the keto group. There is no such effect in CH3CH2NH2. Due to reduction in electron density on N-atom of CH3CONH2, it is a weaker base than CH3CH2NH2. (ii) Nitro compounds (Organic compounds containing -NO2 group) Aliphatic nitro compounds Aromatic nitro compounds Nitroalkanes Alkylnitrites N=O Nitro arenes Aryl nitroalkanes CH3CH2 NO2 CH3CH2 O C6H5 CH2 NO2 Nitroethane ethylnitrite -O N+ Onitrobenzene Primary (1 ) nitroalkane Secondary (2 ) nitroalknane Tertiary (3 ) nitroalknane Example: Example: Example: H CH3 CH3 CH3 C NO2 CH3 C NO2 CH3 C NO2 H H CH3 Nitroethane 2-nitropropane 2-methyl-2-nitropropane Sample Paper - 1    D-13

4SAMPLE PAPER – Maximum Marks: 70 Time: 3 Hours (UNSOLVED) PART - I Answer all the questions. [15 × 1 = 15] Choose the most suitable answer from the given four alternatives and write the option code and the corresponding answer. 1. Consider the following statements. (i) All ores are minerals but all minerals are not ores. (ii) Bauxite is an ore of aluminium while clay is not. (iii) Extraction of aluminium form clay is profitable one. Which of the above statement(s) is / are not correct? (a) (i) only (b) (ii) only (c) (iii) only (d) (i), (ii) and (iii) 2. The compound that is used in nuclear reactors as protective shields and control rods is ....... . (a) Metal borides (b) metal oxides (c) Metal carbonates (d) metal carbide 3. An element belongs to group 15 and 3 rd period of the periodic table, its electronic configuration would be .................... . (a) 1s2 2s2 2p4 (b) 1s2 2s2 2p3 (c) 1s2 2s2 2p6 3s2 3p2 (d) 1s2 2s2 2p6 3s2 3p3 4. Which one of the following is more basic in nature? (a) La(OH)3 (b) Ce(OH)3 (c) Gd(OH)3 (d) Lu(OH)3 5. Formula of tris(ethane-1,2-diamine)iron(II)phosphate (a) [Fe(CH3– CH(NH2)2)3] (PO4)3 (b) [Fe(H2N – CH2– CH2 –NH2)3] (PO4) (c) [Fe(H2N – CH2 – CH2 – NH2)3](PO4)2 (d) [Fe(H2N – CH2 – CH2 – NH2)3](PO4)2 6. Which one of the following is an example for molecular crystals? (a) Diamond (b) Silica (c) Glass (d) Naphthalene 7. ; This reaction follows first order kinetics. The rate constant at particular temperature is 2.303 × 10−2 hour−1. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (log 2 = 0.3010) (a) 0.125 M (b) 0.215 M (c) 0.25 × 2.303 M (d) 0.05 M 8. Conjugate base for bronsted acids H2O and HF are .................. . (a) OH – and H2FH+ , respectively (b) H3O+ and F – , respectively (c) OH – and F – , respectively (d) H3O+ and H2F + , respectively 9. The salt bridge used in Daniel cell contains .......... . (a) Na2SO4 + NaCl (b) Agar-Agar gel + Na2SO4 (c) Silica gel + CuSO4 (d) ZnSO4 + CuSO4 10. Consider the following statements: (i) A catalyst needed in very small quantity (ii) A catalyst can initiate a reaction D-38

(iii) Catalyst are highly specific in nature Which of the above statement is / are not correct? (a) i & iii (b) ii & iii (c) iii only (d) ii only 11. An alcohol (x) gives blue colour in victormayer’s test and 3.7g of X when treated with metallic sodium liberates 560 mL of hydrogen at 273 K and 1 atm pressure what will be the possible structure of X? (a) CH3 CH (OH) CH2CH3 (b) CH3 – CH (OH) – CH3 (c) CH3 – C (OH) – (CH3)2 (d) CH3– CH2– CH (OH) – CH2 – CH3 12. In which case chiral carbon is not generated by reaction with HCN O OH (a) (b) O OO (c) (d) Ph OH Ph 13. When aniline reacts with acetic anhydride the product formed is ..................... . (a) o – aminoacetophenone (b) m-aminoacetophenone (c) p – aminoacetophenone (d) acetanilide 14. Among the following the achiral amino acid is .................. . (a) 2-ethylalanine (b) 2-methylglycine (c) 2-hydroxymethylserine (d) Tryptophan 15. Which one of the complex salt is acting as a photo sensitiser in photosynthesis process? (a) Wilkinson’s compound (b) Cobalamine (c) Chlorophyll (d) Haemoglobin PART - II Answer any six questions. Question No. 22 is compulsory. [6 × 2 = 12] 16. List out the common refining methods. 17. Mention the use of boric acid. 18. Which metal in the 3d series exhibits +1 oxidation state most frequently and why? 19. How many lattice points are there in one unit cell of each of the following lattice? (i) Face-centred cubic (ii) Face-centred tetragonal (iii) Body-centered 20. The aqueous solution of sugar does not conduct electricity whereas when sodium chloride is added to water, it conducts electricity. Justify this statement. 21. What are the limitations of Freundlich adsorption isotherm? 22. Writes the chemical equation for Williamson synthesis of 2- ethoxy - 2- methyl pentane starting from ethanol and 2 - methyl pentan -2-ol. 23. Aniline is basic in nature. Justify this statement. 24. What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed? Sample Paper - 4    D-39

PART-III Answer any six questions. Question No. 31 is compulsory. [6 × 3 = 18] 25. Mention the properties of silicones. 26. Explain about the variation of melting point among the transition metal series. 27. The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH. 28. Describe the action of active centres present in the catalyst. 29. The electrochemical cell reaction of the Daniel cell is Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu (s) What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10? 30. What are the factors which influence the adsorption of a gas on a solid? (A) H3O+ 31. Identify A, B and C. Benzyl bromide NaCN (C) THF Mg ether (B) (i) CO2 (ii) H3O+ 32. Explain about the cyclic structure of Glucose. 33. How will you distinguish between primary secondary and tertiary alphatic amines. Answer all the questions. PART - IV [5 × 5 = 25] 34. (a) (i) Bleaching action of chlorine is permanent. Justify this statement. (3) (ii) What is Royal water? (2) [OR] (b) (i) What is lanthanide contraction and what are the effects of lanthanide contraction? (3) (ii) How many series are in d-block elements? What are they? (2) 3 5.(a) (i) Write the postulates of Werner’s theory. (3) (ii) What is coordination entity? Give example. (2) [OR] (b) (i) Distinguish between hexagonal close packing and cubic close packing.(2) (ii) What are molecular solids? Explain their classification with suitable examples. (3) 36.(a) Derive an expression for Ostwald’s dilution law.(5) [OR] (b) (i) For strong electrolytes the molar conductivity increases on dilution and reaches a maximum value at infinite dilution. Justify this statement. (2) (ii) A solution of silver nitrate is electrolysed for 20 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode. (3) 3 7.(a) (i) Write a note about nano catalyst. (3) (ii) How would you distinguish natural honey from artificial honey?  (2) [OR] (b) An organic compound (A) of molecular formula C2H6O on reaction with conc.H2SO4 at D-40    Chemistry – XII

443 K gives an unsaturated hydrocarbon (B). (B) on reaction with Baeyer’s reagent produces (C) of molecular formula C2H6O2. (C) on reaction with anhydrous ZnCl2 produces (D) of molecular formula C2H4O. (D) reduces Tollen’s reagent. Identify A, B, C and D and explain the reactions involved.(5) 38.(a) (i) Arrange the following in increasing order of basic strength(3) (a) aniline, p- toludine and p - nitroaniline (b) C6H5NH2 , C6H5NHCH3 , C6H5NH2 , p-Cl-C6H4– NH2 (3) (ii) Convert Ethanamine into Methanamine. (2) [OR] (b) (i) Identify X and Y. CH3MgBr X H3O+ Y  (2) CH3COCH2CH2COOC2H5 (ii) Explain the action of the following reagents with acetyl chloride? 1. Pd/BaSO4 2. LiAlH4(3) Sample Paper - 4    D-41

One-mark & Two-mark Solutions for Unsolved Sample Papers Sample Paper- 4 PART - I 1. (c) (iii) only 2. (a) Metal borides 3. (d) 1s2 2s2 2p6 3s2 3p3 4. (a) La(OH)3 5. (d) [Fe(H2N – CH2 – CH2 – NH2)3](PO4)2 6. (d) Naphthalene 7. (b) 0.215 M k =  2.303 log  [[AA0]]  t   ⇒ 2.303 × 10–2 hour–1 =  2.303  log  0.25  1806 min   [A]  ⇒  2.303 × 10−2 hour −1 ×1806 min  = log  0.25 ⇒  1806 × 10−2  = log  0.25  2.303   [A]   60   [A]  ⇒ 0.301 = log  0.25 ⇒ log 2 = log  0.25 ⇒ 2 =  0.25  [A]   [A]   [A]  [A] =  0.25 = 0.125 M  2  8. (c) OH – and F – , respectively H2O + H2O H3O+ + OH– acid 1 base 1 acid 2 base 2 HF + H2O H3O+ + F – acid 1 base 1 acid 2 base 2 ∴ Conjugate bases are OH– and F– respectively 9. (b) Agar-Agar gel + Na2SO4 10. (d) ii only 11. (a) CH3 CH (OH) CH2CH3 Hint : 2 R – OH + Na → 2 RONa + 2H2 ↑ 2 moles of alcohol gives 1 mole of H2 which occupies 22.4L at 273K and 1 atm ∴ number of moles of alcohol = 2 moles of R − OH × 560 mL = 0.05 moles 22.4L of H2 ∴ number of moles = mass ⇒ molar mass = 3.7 = 74 g mol–1 molar mass 0.05 General formula for R – OH Cn H2n + 1 – OH D-66

∴ n(12) + (2n + 1) (1) + 16 + 1 = 74 14n = 74 – 18 = 56 ∴ n = 56 = 4 4 The 2° alcohol which contains 4 carbon is CH3 CH(OH)CH2 CH3 O 12. (a) Hint: 13. (d) acetanilide 14. (c) 2-hydroxymethylserine 15. (c) Chlorophyll PART - II 16. Common refining methods : (i) Distillation (ii) Liquation (iii) Electrolytic refining (iv) Zone refining (v) Vapour phase method 17. i. Boric acid is used in the manufacture of pottery glazes, glass, enamels and pigments. ii. It is used as an antiseptic and as an eye lotion. iii. It is also used as a food preservative. 18. i. The first transition metal copper exhibits only +1 oxidation state. ii. It is unique in 3d series having a stable +1 oxidation state. Cu (Z = 29) Electronic configuration is [Ar] 3d10 4s1 iii. So copper element only can have +1 oxidation state. 19. i. Lattice points in face-centred cubic lattice = 8 (at corners) + 6 (at the face centre) = 14 ii. Face centred tetragonal = 8 (at corners) + 6 (at the face centre) = 14 iii. Lattice points in body-centred cube = 8 (at corners) +1 (at the body centre) = 9 20. • Sugar is a non electrolyte and when it dissolves in water, there will be no ionisation takes place. If there is no free ions, it does not conduct electricity. • When sodium chloride is added to water, it is completely ionised to give Na+ ions and Cl– ions. Due to the presence of ions, they will be possibility of electrical conductance. Because ions are carriers of electric current. 21. i. The Freundlich adsorption isotherm is purely empirical and valid over a limited pressure range. ii. The values of constants ‘k’ and ‘n’ also found vary with temperature. No theoritical explanations were given. 22. Williamson synthesis: HBr / Δ (or) PBr3 Step 1: CH3CH2OH CH3 – CH2 –Br + H2O (Ethanol) (Ethyl bromide)  Answers    D-67

Step 2: OH Na - metal ONa 2 CH3 – C – CH2 – CH2 – CH3 2 CH3 – C – CH2 – CH2 – CH3 + H2 ↑ CH3 CH3 (2-methyl-pentan-2-ol) (Sodium alkoxide) Step 3: O – Na O – CH2 – CH3 CH3 – C – CH2 – CH2 – CH3 + CH3 – CH2 – Br Δ CH3 – C – CH2 – CH2 – CH3 + NaBr CH3 CH3 (2-methyl-pentan-2-ol) (2-ethoxy-2methyl pentane) Williamsons synthesis occurs by SN2-mechanism and primary alkyl halides are more reactive in SN2 reactions. Therefore ethanol is converted into ethyl bromide. 23. The lone pair of electrons on nitrogen atom in aniline makes it base. Aniline reacts with mineral acids to form salt. C6H5 NH3Cl – C6H5 NH2 + HCl Aniline Anilinium chloride 24. On mixing Fe(OH)3 positive sol and As2S3 negative sol, mutual coagulation occurs which causes precipitation. When these sol got mixed with each other, due to Fe3+ and S2– ions neutralisation of charges will happen and precipitate will be formed. Fe(OH)3 + As2S3 → Fe2S3 + As(OH)3 D-68 Chemistry – XII

MATHEMATICS QUESTION PAPER DESIGN (Strictly based on Reduced Syllabus for 2022 Board Exams) Types of Questions Marks No. of Questions to Total be Answered Marks Part-I 1 20 20 Multiple Choice Questions Part-II 2 7 14 (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily) Part-III 3 7 21 (Totally 10 questions will be given. Answer any Seven. Any one question should be answered compulsorily) Part-IV 57 35 Total Marks 90 S.No. Weightage of Marks Weightage 1. 30% 2. Purpose 40% 3. Knowledge 20% 4. Understanding 10% Application Skill/Creativity E-1

1SAMPLE PAPER – Time: 3 Hours (SOLVED) Maximum Marks: 90 PART-I I. Choose the correct answer. Answer all the questions. [Answers are in bold] [20 × 1 = 20] 2 0 1 4 (d) –20 1. If A = 1 5 and B = 2 0 then | adj (AB) | = ................ . (a) – 40 (b) –80 (c) –60 2. in + in + 1 + in + 2 + in + 3 is .................. . (c) –1 (d) i (a) 0 (b) 1 z +1 ω ω2 1 = 0. 3. If ω = cis 2π , then the number of distinct roots of ω z + ω2 z+ω 3 ω2 1 (a) 1 (b) 2 (c) 3 (d) 4 4. sin−1 (cos x) = π − x is valid for .................... 2 (b) 0 ≤ x ≤ p (c) (a) –p ≤ x ≤ 0 π π (d) π 3π − 2 ≤ x ≤ 2 − 4 ≤ x ≤ 4 5. tan–1 x + cot–1 x = .................... (a) 1 (b) –p (c) π (d) p 2 6. The equation of the normal to the circle x2 + y2 − 2x − 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is .............. . (a) x + 2y = 3 (b) x + 2y + 3 = 3 (c) 2x + 4y + 3 = 0 (d) x – 2y + 3 = 0 7. The axis of the parabola x2 = 20y is ................. . (a) y = 5 (b) x = 5 (c) x = 0 (d) y = 0 ( ) ( ) 8. If a ××c = a ×  × c  for non-coplanar vectors a,  c then ..........    b  b b b, b 0 b a (a) parallel to (b) parallel to c (c) c parallel to a (d) a + + c = 9. The vector equation of a plane whose distance from the origin is p and perpendicular to a unit vectorn is ................... . (a) r ⋅ n = p (b)  ⋅ n = q (c)   p (d)  ⋅ n = p r r ×n= r 10. The point of inflection of the curve y = (x – 1)3 is ............. . (a) (0, 0) (b) (0, 1) (c) (1, 0) (d) (1, 1) E-3

11. The asymptote to the curve y2(1 + x) = x2 (1 – x) is ................ . (a) x = 1 (b) y = 1 (c) y = –1 (d) x = –1 12. The solution of a linear differential equation dx + Px = Q where P and Q are function of y, dy is ............... . ∫ (a) y (I.F) = (I.F) Q dx + c ∫(b) x (I.F) = (I.F) Q dy + c ∫ ∫ (c) y (I.F) = (I.F) Q dy + c (d) x (I.F) = (I.F) Q dx + c 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% ∫14. For any value of n∈, π ecos2 x cos3 [(2n + 1) x]dx is ............. . 0 (a) π 2 (b) π (c) 0 (d) 2 ∫ 15. If n is odd then π 2 sin n x dx is ............... . 0 (a) n ⋅ n − 2 ⋅ n − 4 ... π (b) n −1 ⋅ n − 3 ⋅ n − 5 ... 1 π n −1 n −3 n −5 2 n n−2 n−4 2 2 (c) n ⋅n−2⋅ n − 5 ... 3 ⋅1 (d) n − 1 ⋅ n − 3 ⋅ n − 5 ... 2 ⋅1 n −1 n −3 n − 4 2 n n − 2 n − 4 3 16. The solution of dy = 2y – x is ............. . dx (a) 2­x + 2y = c (b) 2­x – 2y = c (c) 1 − 1 =c (d) x+y=c 2x 2y equation y dy  d2y  17. If p and q are the order and degree of the differential dx + x3  dx2  + xy = cos x, when............. . (a) p < q (b) p = q (c) p > q (d) p exists and q does not exist 18. A pair of dice numbered 1, 2, 3, 4, 5, 6 of a six-sided die and 1, 2, 3, 4 of a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Then the number of elements in the inverse image of 7 is ........... . (a) 1 (b) 2 (c) 3 (d) 4 19. If in 6 trials, X is a binomial variate which follows the relation 9P(X = 4) = P(X = 2), then the probability of success is ..................... . (a) 0.125 (b) 0.25 (c) 0.375 (d) 0.75 20. Which one of the following statements has truth value F? (a) Chennai is in India or 2 is an integer (b) Chennai is in India or 2 is an irrational number (c) Chennai is in China or 2 is an integer (d) Chennai is in China or 2 is an irrational number E-4 Mathematics–XII

PART-II [7 × 2 = 14] II. Answer any seven questions. Question No. 30 is compulsory. 21. State the reason for cos−1 cos  − π   ≠ − π  6   6 Ans. We know cos(–θ) = cos θ so cos−1 cos  − π   = cos−1 cos π = π ≠ − π  6   6  6 6 22. Find the equations of the tangent and normal to the circle x2 + y2 = 25 at P(–3, 4) . Ans. Equation of tangent to the circle at P (x1 , y1) is xx1 + yy1 = a2 . That is, x(−3) + y(4) = 25 −3x + 4y = 25 Equation of normal is xy1 – yx1 = 0 That is, 4x + 3y = 0 23. Determine whether the three vectors 2i + 3j + k , i − 2j + 2k and 3i + j + 3k are coplanar.    Ans. Let a = 2i + 3 j + k  b = i− 2j + 2k  a, b, c c = 3i + j + 3k a b c we know that are coplanar if and only if =0  2 3 1 a b c = 1 −2 2 = 2 (–6 –2) –3 (3 – 6) + (1 + 6) = –16 + 9 + 7 = 0 313 ∴ Given vectors are coplanar 24. Find the intervals of monotonicities and hence find the local extremum for the following functions f (x) = x Ans. x − 5 x f − (x) = x 5 f ' (x) = ( x − 5)(1) − x (1) = x −5− x (x − 5)2 (x − 5)2 = ( x −5 <0 5) − 5)2 (when x ≠ f (x) is strictly decreasing on (–∞, 5) and (5, ∞) And there is no local extremum Sample Paper-1 E-5

 8 −4 25. If A = −5 3 , verify that A(adj A) = (adj A)A = |A|I2. Ans. A = 8 −4   3  −5 8 −4 |A| = −5 3 = 24 – 20 = 4 adj A = 3 4  5 8  8 −4  3 4  24 − 20 32 − 32  Now A(adj A) = 3  5 8  =    −5  −15 + 15 −20 + 24   = 4 0 = 4  1 0 = |A| I2...(1)    0   0 4   1  (adj A) A = 3 4 8 −4 =  24 − 20 −12 +12   −5 3    5 8   40 − 40 −20 + 24    = 4 0 = 4  1 0 = |A| I2...(2)  0 4   0   1  (1) = (2) ⇒ A(adj A) = (adj A)A = |A| I2 2π ∫ 26. Evaluate sin4 x cos3 x dx 0 Ans. f (x) = sin4x cos3x ∫ 2a f (x) dx = 0 if f (2a – x) = – f (x). 0 f (2π – x) = sin4 (2π – x) cos3 (2π – x) ∫ 2a ∫2 a = sin4x cos3x = f (x) 0 f (x) dx = 0 f (x) dx if f (2a – x) = f (x). π = ∫2 sin4 x cos3 xdx 0 Again f (x) = sin4x cos3x f (π – x) = sin4 (π – x) cos3 (π – x) f (π – x) = – sin4 (π – x) cos3 (π – x) = – f (x) = 0. 27. Solve the differential equation dy = 1 − y2 dx 1− x2 dy 1− y2 Ans. dx = 1− x2 dy = 1− y2 dx 1− x2 Separating the variables E-6 Mathematics–XII

∫ dy =∫ dx 1− y2 1− x2 sin–1 y = sin–1 x + c 28. Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function f (x) = k 200 ≤ x ≤ 600 . Find the value of k  otherwise  0 Ans. Given p.d.f. is f ( x) = k, 200 ≤ x ≤ 600 0, otherwise ∞ Since f (x) is a p.d.f, ∫ f (x) dx = 1 −∞ 600 i.e., k ∫ dx = 1 200 [ ]kx 600 = 1 200 k (600 – 200) = 1 400 k = 1 ⇒k= 1 400 29. Let * be defined on  by (a*b) = a + b + ab − 7 . Is * binary on  ? If so, find 3*  −7   15  Ans. a * b = a + b + ab − 7 . Now when a, b ∈ , then ab ∈  also a + b ∈ . So, a + b + ab ∈ . We know – 7 ∈ . So, a + b + ab – 7 ∈ . (i.e.) a * b ∈ . So, * is a binary operation on . Now, to find 3*  − 7  15 3*  − 7 = 3+  − 7 + (3)  − 7 − 7  15  15 15 = 3 − 7 − 21 − 7 = 45 − 7 − 21 −105 = − 88 15 15 15 15 30. Solve the equation : x4 − 14x2 + 45 = 0. Ans. The given equation is x4 − 14x2 + 45 = 0 Let x2 = y Sample Paper-1 E-7

y2 – 14y + 45 = 0 ⇒ (y – 9)(y – 5) = 0 y = 9, 5 ∴ x2 = 9, x2 = 5 x = ± 3, x = ± 5 ∴ The roots are ± 3, ± 5 PART-III III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21] 31. Solve the following system of linear equations, using matrix inversion method: 5x + 2y = 3, 3x + 2y = 5. 5 2 , X =  x  , B= 3 Ans. The matrix form of the system is AX = B, where A = 3 2  y  5 52 We find |A| = = 10 − 6 = 4 ≠ 0 3 2 So, A–1 exists and A−1 = 1 2 −2 . 4 −3 5 Then, applying the formula X = A−1B, we get  −4    x = 1 2 −2 3 = 1  6 −10  = 1 −4 =  4  = −1  y 4 −3 5 5 4 −9 + 25 4  16  16   4 So the solution is (x = −1, y = 4).  4  32. If z1 = 2 − i and z2 = −4 + 3i, find the inverse of z1 z2 and z1 z2 Ans. z1 = 2 – i, z2 = –4 + 3i (i) z1 z2 = (2 – i) (–4 + 3i) = (–8 + 6i + 4i – 3i2) = (–8 + 10i + 3) = (–5 + 10i) (z1z2)–1 = 1 ) = 1 × −5 −10i = −5 −10i = 5(−1− 2i) = −1 − 2i ( z1 z2 −5 +10i −5 −10i 25 +100 125 25 (ii) z1 = 2 − i × −4 − 3i = −8 − 6i + 4i + 3i2 = −8 − 2i − 3 = −11 − 2i z2 −4 + 3i −4 − 3i 16 + 9 25 25  z1 −1 = 25 2i = 25 × −11 + 2i = 25(−11+ 2i) = 25(−11+ 2i)  z2  −11 − −11− 2i −11+ 2i 121 + 4 125   = 15 (−11+ 2i) 33. If k is real, discuss the nature of the roots of the polynomial equation 2x2 + kx + k = 0, in terms of k. Ans. The given quadratic equation is 2x2 + kx + k = 0 a = 2, b = k, c = k E-8 Mathematics–XII

∆ = b2 – 4ac = k2 – 4(2) k = k2 – 8k (i) If the roots are equal k2 – 8k = 0 ⇒ k(k –8) = 0 k = 0, k = 8 (ii) If the roots are real k2 – 8k > 0 k (k – 8) > 0 k ∈ (–∞, 0) ∪ (8, ∞) 08 (iii) If this roots are imaginary k2 – 8k < 0 ⇒ k ∈ (0, 8) 34. Solve the following system of linear equations by matrix inversion method: 2x + 5y = –2, x + 2y = −3  2 5  x  −2 Ans. The matrix form of the above equations is  1 2  y =  −3   (i.e) AX = B Where A = 2 5  ; X = x and B =  −2  1 2   y   −3 Now AX = B ⇒ X = A–1B To find A–1: A = 2 5  1 2  25 |A| = 1 2 = 4 – 5 = –1 ≠ 0 adj A = 2 −5     −1 2  A–1 = 1 (adj A) = 1 2 −5  =  −2 5 |A| −1  −1 2   1 −2  ∴ x = A–1B =  −2 5  −2 =  4 −15  =  −11  y   1   −3  −2 + 6  4   −2  ⇒ x = –11; y = 4 non-parametric and Cartesian equations of the ˆj + kˆ + s −iˆ + 2 Given a = ( ) ( ) ( ) 35. rF i=nd6tiˆh−e plane Ans. form of vector equation, ˆj + kˆ + t −5iˆ − 4 ˆj − 5kˆ .   6iˆ−  =  +  + k  5k ˆj + kˆ , b and c = −5i − 4 j − −i 2j  × c = i j k =  (−10 + 4) −  (5 + 5) +  (4 + 10) b −1 2 1 i j k  c = −5 −4 −5  b   × −6i −10 j + 14k Sample Paper-1 E-9

( ( )) (Nro−n-ap)a⋅ r−a6m(ire−t−r1ai0c)jf⋅o+br1m×4kcof vector equation = 0 = 0 [÷ by –2]   3i + 5 j − 7k ( )(r− a) ⋅ = 0  = a ⋅  3i + 5 j − 7k 3i + 5 j − 7k ( ) ( )r ⋅     3i + 5 j − 7k 6i − j + k 3i + 5 j − 7k ( ) ( )( )r ⋅ =    3i + 5 j − 7k 3i + 5 j − 7k ( ) ( )r ⋅ ⇒ r ⋅ =6 = 18 – 5 – 7 ( ) ( ) Cartesianxieq+uyajti+onzk ⋅  = 6 3i + 5 j − 7k 3x + 5y – 7z – 6 = 0 36. Find two positive numbers whose sum is 12 and their product is maximum. Ans. Let the two numbers be x, 12 – x. Their product p = x (12 – x) = 12x – x2 To find the maximum product. p' (x) = 12 – 2x p'' (x) = – 2 p' (x) = 0 ⇒ 12 – 2x = 0 ⇒ 2 = 12 ⇒ x = 6 at x = 6, p ''(x) = –2 = – ve ⇒ p is maximum at x = 6 when x = 6, 12 – x = 12 – 6 = 6 So the two numbers are 6, 6 37. Find the differential equation that will represent family of all circles having centres on the x-axis and the radius is unity. Ans. Equation of a circle with centre on x-axis and radius 1 unit is (x – a)2 + y2 = 1 ...(1) Differentiating with respect to x, 2 (x – a) + 2yy ' = 0 ⇒ 2 (x – a) = – 2yy ' (or) x – a = – yy ' ...(2) Substituting (2) in (1), we get, (– yy ')2 + y2 = 1 (i.e.,) y2 (y ')2 + y2 = 1 ⇒ y2 [1 + (y')2] = 1 E-10 Mathematics–XII

38. A six sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images. Ans. Six sided die marked ‘2’ on one face, ‘3’ on two faces and ‘4’ on three faces. When it is thrown twice, we get 36 sample points. ‘X’ denotes sum of the face numbers and the possible values of ‘X’ are 4, 5, 6, 7 and 8 For X = 4, the sample point is (2, 2) For X = 5, the sample points are (2, 3), (3, 2) For X = 6, the sample points are (3, 3), (2, 4), (4, 2) For X = 7, the sample points are (3, 4), (4, 3) For X = 8, the sample point is (4, 4) Values of X 4 5 6 7 8 Total Number of points in inverse images 1 2 3 2 1 9 39. Fill in the following table so that the binary operation * on A = {a,b,c} is commutative. * a bc ab b c ba ca c * a bc a b ca Ans. Given that the binary operation * is Commutative. b c ba c a ac To find a * b : a * b = b * a ( * is a Commutative) Here b * a = c. So a * b = c To find a * c : a * c = c * a ( * is a Commutative) c * a = a. (Given) So a * c = a To find c * b : c * b = b * c Here b * c = a. So c * b = a 40. Find the area of the region enclosed by y2 = x and y = x – 2. y Ans. The points of intersection of the parabola y2 = x and the line y = x – 2 x = y2 are (1, – 1) and (4, 2) (4, 2) y=x–2 x To compute the region [shown in figure] by integrating with respect to x, we would have to split the region into two parts, because the (1,–1) equation of the lower boundary changes at x =1. However if we integrate with respect to y no splitting is necessary. 2 Required area = ∫ f ( y) − g( y) dy −1 Sample Paper-1 E-11

∫2 + 2) − y2  dy =  y2 + 2y − y3  2  2 3  −1 = ( y −1 =  42 − 12 + (4 + 2) −  8 + 1  3 3 = 23 + 6 − 9 = 9 sq.units. 3 2 PART-IV IV. Answer all the questions. [7 × 5 = 35]  cosα 0 sin α    41. (a) If F(α) =  0 1 0  show that [F(α)]–1 = F(–α) − sin α 0 cosα Ans. Let A = F (α) So [F(α)]–1 = A–1  cosα 0 sin α    Now A =  0 1 0  − sin α 0 cosα cosα 0 sin α |A| = 0 1 0 − sin α 0 cosα Expanding the determinant - along R2 We get – 0( ) + 1[cos2 a + sin2 α] – 0( ) = 1 ≠ 0 So A–1 exists Now A–1 = 1 (adj A) = 11(adj A) = adj A |A| To Find adj A: adj A = (Aij)T  1 0 −0 0 +0 1 + 0 cos α − sin α   − sin α cosα 0  (Aij) =  0 sin α + cosα sin α − cosα 0  − 0 cos α − sin α cosα   − sin α 0     + 0 sin α − cosα sin α + cosα 0   1 0 00 0 1   +(cosα) −(0) +(sin α)   cosα 0 sin α    =  −(0) +(1) −(0)  =  0 1 0    +(− sin α) −(0) +(cosα)  − sin α 0 cosα E-12 Mathematics–XII

 cosα 0 − sin α ∴ adj A = (A–1) = (Aij)T =  0 1 0     sin α 0 cosα   cosα 0 − sin α (i.e) A–1 = [F(α)]–1 =  0 1 0  ...(1)    sin α 0 cosα   cosα 0 sin α    Given F(α) =  0 1 0  − sin α 0 cosα  cos(−α) 0 sin(−α)    So F(–α) =  0 1 0  − sin(−α) 0 cos(−α) cosα 0 − sin α   =  0 1 0   ...(2) sin α 0 cosα  (∴ cos (–θ) = cos θ and sin (–θ) = – sin θ) Here (1) = (2) ⇒ [ F(α)]–1 = F(–α) [OR] (b) If z = x + iy and arg  z−i  = π , then show that x2 + y2 + 3x − 3y + 2 = 0.  z + 2  4 Ans. arg  z−i  = π  z+2  4 We have arg  z1  = arg (z1) – arg (z2)  z2    arg (z – i) – arg (z + 2) = π ⇒ Let z = x + iy 4 arg (x + iy – i) – arg (x + iy + 2) = π ⇒ arg (x + i (y – 1)) – arg (x + 2 + iy) = π 4 4  y − 1 y   + x − − +  tan −1  y − 1 − tan −1  y  = π ⇒ tan–1  y 1 x 2  = π  x   + 2 4  × y x 1 4  ( y −1)( x + 2) − xy   x x+2  x(x + 2) − 1)  = π ⇒ tan–1  xy + 2 y − x − 2 − xy  = π tan–1  x(x + 2) + y( y  4  x2 + 2 x + y2 − y  4      x(x + 2)  Sample Paper-1 E-13

 2y − x−2  = tan π =1   4  x2 + y2 + 2x − y  2y – x – 2 = x2 + y2 + 2x – y x2 + y2 + 3x – 3y + 2 = 0 Hence proved. 42. (a) Solve the equation: 3x4 – 16x3 + 26x2 – 16x + 3 = 0. Ans. It is an even degree reciprocal equations Dividing (1) by x2 3 x2 3x2 – 16x + 26 – 16 + =0 x 3  x2 + 1 −16 x + 1  + 26 =0  x2  x Let y = x + 1 x 1 y2 – 2 = x2 + x2 ∴ 3(y2 – 2) – 16y + 26 = 0 3y2 – 6 – 16y + 26 = 0 3y2 – 16y + 20 = 0 (3y – 10) (y – 2) = 0 3y – 10 = 0 y – 2 = 0 3y = 10 y = 2 3(x + 1 ) = 10 x + 1 = 2 x x 3(x2 + 1) = 10x x2 + 1 = 2x 3x2 – 10x + 3 = 0 x2 – 2x + 1 = 0 (3x – 1) (x – 3) = 0 (x – 1)2 = 0 x = 1, 1 x = 1 and x = 3 3 1 The roots are 1, 1, 3 , 3 [OR] (b) A rod of length 1.2m moves with its ends always touching the coordinates axes. The locus of a point P on the rod, which is 0.3 m from the end in contact with x-axis is an ellipse. Find the eccentricity. Ans. From the diagram, (i) ∆le OAB be a right angle triangle. (ii) APDAaPnPDdBCPBC are corresponding angles. so corresponding angles are equal. E-14 Mathematics–XII