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COUNSELLING CODE P.B. College of Engineering 1222 Campus Address : Administration Office : Irungattukottai, Chennai – Bangalore National Highway Sriperumbudur Taluk, Kanchipuram Dist., Chennai – 602 117. Tel. : 044-27198033, 9940026861, 9600112733 E-mail : [email protected] Website : www.pbce.in 3/117, Brahmin Street, Karambakkam, Porur, Chennai – 600 116. Tel. : 044-24766386, Telefax : 044-24766941 Cell : 9600113031, 9962206098 MODEL QUESTION PAPER – II PHYSICS Note : i) Answer all the questions ii) Choose and Write the correct answer iii) Each question carries one mark. PART – I 1. The unit of electric field intensity is ____ (a) NC-1 (b) NC (c) Vm-1 (d) Vm Answer : (c)Vm-1 2. A capacitor of Capacitance 6F is connected to a 100 V battery. The energy Stored in the Capacitor is _____ (a) 30J (b) 3J (c) 0.03J (d) 0.06J Answer : (c) 0.03J 3. The ratio of electric potential at points 10 cm and 20 cm from the centre of an electric dipole along its axial line is (a) 1 :2 (b) 2 :1 (e) 1 :4 (d) 4 :1 Answer : (d) 4 :1 4. A dielectric will have (a) Free electrons (b) No electrons (c) Tightly bound electron (d) one electron in the outer orbit. Answer : (c) Tightly bound electron 5. A toaster operating 240V has a resistance 120. The power is (a) 400W (b) 2 W (c) 480 W (d) 240 W. Answer : (d) 240 W. 6. Phosphor – bronze wire is used for suspension in a moving coil galvanometer, because it has (a) high Conductivity (b) high resistivity (c) large couple per unit twist (d) Small couple per unit twist. Answer : (d) Small couple per unit twist. 7. The direction of force on current carrying conductor placed in a magnetic field is given by (a) Fleming’s left hand rule (c) End rule (b) Fleming’s Right hand rule . (d) Right hand Palm rule. Answer : (a) Fleming’s left hand rule 1 COUNSELLING CODE P.B. College of Engineering 8. 1222 If the flux associated with a coil varies at the rate of 1Wb/minute then the induced e.m.f. is ___ (a) IV (b) 1/60V (c) 60 V (d) 0.60 V. Answer : (b) 1/60V 9. A power of 11,000W is transmitted at 220V. The current through live wires is. (a) 50A (b) 5A (c) 500A (d) 0.5A Answer : (a) 50A 10. The Unit Henry can also be written as P1 (a) VsA-1 (b) WbA-1 (c) S (d) all the above Answer : (d) all the above 11. Q factor has values lying between ____for normal frequencies. (a) 0 to 10 (b) 10 to 50 (c) 50 to 100 (d) 10 to 100. Answer : (d) 10 to 100. 12. When a drop of water is introduced between the glass plate and plano convex lens in Newton rings system The rings. (a) Contracts (b) expands (c) remains same (d) first expands, then contracts Answer : (a) Contracts 13. Electric filament lamp gives rise to (a) line spectrum (b) Continuous Spectrum (c) Continuous absorption spectrum (d) line absorption spectrum. Answer : (b) Continuous Spectrum 14. In Newton’s ring experiment, the radii of mth and (m+4)th dark rings are respectively 5 mm and7 mm. What is the Value of m ? (a) 2 (b) 4 (c) 8 (d) 10. Answer : (d) 10. 15. Ratio of intensities of two waves 4 :1 then, the ratio of their amplitudes. (a) 2:1 (b) 1:2 (c) 4:1 (d) 1:4 Answer : (a) 2:1 16. The ratio of the radii of the first three Bohr orbits is ________ (a) 1 : ½ : 1/3 (b) 1 : 2 :3 (c) 1 : 4 : 9 (d) 1 : 8 : 27 Answer : (c) 1 : 4 : 9 17. Maser Materials are (a) diamagnetic ions (b) Paramagnetic ions. (c) Ferro magnetic ions (d) non-magnetic ions. Answer : (b) Paramagnetic ions. 2 COUNSELLING CODE P.B. College of Engineering 1222 18. In Somerfield atom model, for principal quantum number n=3, which of the following subshells represents circular orbit ? (a) 2S (b) 2P (c) 2d (d) None of these. Answer : (b) 2P 19. The Value of Rydberg’s Constant is __________ (a) 1.094 x 10-7m-1 (b) 1.094 x 107m+1 (c) 1.094 x 107 m-1 (d) 1.094 x 10-7 m1 Answer : (c) 1.094 x 107 m-1 20. If 1 kg of substance is fully converted to energy, the energy produced is. (a) 9x1016J (b) 9x1024J (c) 1J (d) 3x108J Answer : (a) 9x1016J 21. The number of debroglie waves of an electron in the nth orbit of an atom is (a) n (b) n-1 (c) n+1 (d) 2n Answer : (a) n 22. The radio isotope used in agriculture is (a) 15P31 Answer : (a) (b) 15P32 (c) 11Na23 (d) 11Na24 15P31 23. The mean life of radon is 5.5 days. It half life is (a) 8 days (b) 2.8 days (c)0.38 days (d) 3.8 days. Answer : (d) 3.8 days. P2 24. Which of the following belongs to Baryon group? (a) Photon (b) Electron (c) Pion (d) Proton. Answer : (d) Proton. 25. The half – life Period of N13 is 10.1 minute .Its life time is (a) 5.05 minute (b) 20.2 minute (c) Answer : (d) infinity 10.1 minute 0.6931 (d) infinity 26. The colour of light emitted by a LED depends on (a) its reverse bias (b) amount of forward current (c) Its forward bias (d) type of semiconductor material. Answer : (d) type of semiconductor material. 27. The following arrangement performs the logic function of A Y B (a) AND gate (b) OR gate . (c) NAND gate (d) Ex-OR gate. Answer : (a) AND gate 3 COUNSELLING CODE P.B. College of Engineering 1222 28. Bark hausen Condition for maintenance of oscillation (a) =1/A (b) AB = (c) A= (d) A = ½ Answer : (a) =1/A 29. The audio frequency range is (a) 20HZ to 200HZ (b) 20HZ to 2000HZ (c) 20Hz to 200000HZ (d) 20 HZ to 20000 HZ Answer : (d) 20 HZ to 20000 HZ 30. Videocon Camera tube works on the Principle of (a) Photo Conductivity (b) thermoelectric effects. (c) themionic emission (d) seeback effect Answer : (a) Photo Conductivity PART – II Note : Answer any Fifteen Questions. 15x3=45 31. Define electric flux. Give its unit : Solution : The electric flux is defined as the total number of electric lines of force, crossing through the given area. The electric flux d through the area is d = = Eds cos . Its unit is Nm2C-1 32. Calculate the potential at a point due to a charge 4 x 10-7C located at 0.09 m away from it. Solution : V = = 1 4 q r 9 x 109 x 4 x 10-7 9 x 10-2 = 4 x 104 V 33. Define temperature coefficient of resistance. Solution : The Temperature coefficient of resistance is defined as the ratio of increase in resistance per degree rise in temperature to its resistance at 0oC o Its Unit is per C 4 COUNSELLING CODE P.B. College of Engineering 1222 34. Two wires of same material and length have resistance 5 and 10 respectively. Find the ratio of radii of the two wires. Solution : R1 = 5 ,R2 = 10 Radius of 1st wire = r1, radius of 2nd wire = r2 specific Resistance = Length = l l A l = r12 (A = r 2) R = R1 R2 R1 = R2 = r2 r1 : r2 r22 r12 r22 r1 l ,R2 = = R1 10 5 = 2 2:1 = 35. From the following network, find the effective resistance between A and B. R1=15 A B R2=15 Solution : R1=15 A B R2=15 1 RP 1 = = = RP = R1 1 15 1 + + R2 1 15 2 15 15 2 = 7.5 ohm. 5 COUNSELLING CODE P.B. College of Engineering 1222 36. Find the magnetic induction at a point, 10 cm from a long straight wire carrying current of 10A. I = 10A, a = 10cm = 10 x 10-2m, B=? Solution : B = oI 2a xx = 2 x x B = 2 x 10-5 T 37. State Fleming’s left hand rule. Solution : The fore finger, the middle finger and the thumb of the left hand are stretched in mutually perpendicular directions. If the fore finger points in the direction of the magnetic field, the middle finger points in the direction of the current, then the thumb points in the direction of the force of the conductor. 38. What is Capacitive reactance ? Solution : (i) capacitive reactance is the resistance offered to the flow of current by a capacitor. Xc = c = vc its unit is ohm. (ii) In a d.c. circuit frequency = O, Xc = ∞It blocks D.C. (iii) For an a.c. the capacitive reactance varies inversely as the frequency of a.c and also inversely as the capacitance of the Capacitor. 39. Why does the sky appear blue in colour ? Solution : At Sunrise the blue appearance of the sky is due to scattering of sunlight by the atmosphere. According to Rayleigh’s Scattering law, blue light to a greater extent than red light. This Scattered radiation causes the sky to appear blue. 40. The refractive index of a medium is 3. Calculate the angle of refraction if the unpolarised light is incident on it at the polarizing angle of the medium. = 3. ; r = ? Solution : According to Brewster’s law = tan ip 6 COUNSELLING CODE P.B. College of Engineering = 3 1222 or Angle of refraction, ip = tan-1 3 = 600 r = 90-ip r = 90-60 r = 30o 41. Write the principle of Millikan’s oil drop experiment. Solution : (i) It is based on the study of the motion of uncharged oil drop under free fall due to gravity and charged oil drop in a uniform electric field. (ii) By adjusting uniform electric field suitably, a charged oil drop can be made to move up or down or even kept balanced in the field of view for sufficiently long time and a series of observations can be made. 42. Write any three medical applications of LASER. Solution : (i) Micro surgery has become possible due to narrow angular spread of the laser beam. (ii) The laser beams are used in endoscopy. (iii) It can also be used for the treatment of human and animal cancer. 43. State the postulates of special theory of relativity. Solution : (i) The laws of Physics are the same in all inertial frames of reference. (ii) The velocity of light in free space is a constant in all the frames of reference. 44. Define critical mass and critical size. Solution : (i) Critical size of a system containing a fissile material is defined as the minimum size in which atleast one neutron is available for further fission reaction. (ii) The mass of the fissile material at the critical size is called critical mass. (iii) The chain reaction is not possible, if the size is less than the critical size. 45. Write the proton – proton cycle that takes place in sun and stars. Solution : 1 H 1 + 1H 1 → 1 H 1 + 1H 2→ 2 1 H2 + 1e0 + (emission of positron and neutrino) He3 + + (emission of gamma rays) 22He3 → 2He4 + 21H1 The reaction cycle is written as 4 H 1→ 1 2 He4 + 21e0 + 2 + energy (26.7 Mev) 7 COUNSELLING CODE P.B. College of Engineering 1222 This four protons fuse together to form an alpha particle and two positrons with a release of large amount of energy. 46. Define input impedance of a transistor is connected in common emitter mode. Solution : The Input impedance of the transistor is defined as the ratio of small change in base emitter voltage to the corresponding change in base current at a given VCE Input impedance ri = V I ohm VCE 47. What is an extrinsic semi conductor ? Solution : An extrinsic semiconductor is one in which an impurity with a valency higher or lower than the valency of the pure semiconductor is added, so as to increase the electrical conductivity of the semiconductor. Types ; P-type, N-type. 48. What are the advantages of negative feedback ? Solution : (i) Highly established gain (ii) Reduction in the noise level (iii) Increased band width (iv) Increased input impedance and decreased output impedance (v) Less distortion 49. Draw the circuit configuration of NPN transistor in common collector (c.c.) mode. Solution : VBB = input voltage source VCC = output voltage source E = Emitter B = Base C = Collector 50. What is Directivity of an Antenna ? Solution : Directivity is the ability of the antenna to concentrate the electromagnetic waves in the most desired directions during transmission or to have maximum reception from most preferred directions during reception. 8 COUNSELLING CODE P.B. College of Engineering 1222 PART – III Note : (i) Answer Q. No 56 Compulsorily (ii) Answer any six of the remaining questions (iii) Draw Diagrams wherever necessary. 51. Derive an expression for the capacitance of a parallel plate capacitor with a dielectric medium. Solution : Consider a parallel plate capacitor having two conducting plates X and Y each of area A, separated by a distance d apart. X is given a positive charge, so that the surface charge density on it is and Y is earthed. Let a dielectric slab of thickness t and relative permittivity r be introduced +σ between the plates. AIR X d Thickness of dielectric Slab = t Thickness of air gap = (d-t) DIELECTRIC t AIR Y t<d Electric field at any point in the air between the plates, E = Electric field at any point, in the dielectric slab. E΄ = r The total potential difference between the plates is the work done in crossing unit positive charge from one plate to another in the field E over a distance (d-t) and in the field E’ over a distance t, then. V = E(d-t) + E΄t = = (d-t) + [(d-t) + t r t r ] The charge on the plate X, q = A Hence the capacitance of the capacitor is C = q v A = A = (d-t) + t r (d-t) + t r 9 COUNSELLING CODE P.B. College of Engineering 1222 Effect of dielectric: In capacitors, the region between the two plates is filled with dielectric like mica or oil. The Capacitance of the air filled capacitor, C = A d The Capacitance of the dielectric filled capacitor , C’ = C’ C = r r A d C’ = r C or 52. Explain the determination of the internal resistance of a cell using voltmeter. Solution : The Internal resistance of a cell : The electric current in an external circuit flows from the positive terminal to the negative terminal of the cell, through different circuit elements. In order to maintain continuity, the current has to flow through the electrolyte of the cell, from its negative terminal to the positive terminal. During this process of flow of current inside a cell, a resistance is offered to current flow by the electrolyte of the cell. A freshly prepared cell has low internal resistance and this increases with aging. Determination of internal Resistance : The circuit connections are made as shown, With key k open, the emf of cell E of is found by connecting a high resistance voltmeter across it. Since the high resistance voltmeter draws only a very feeble current for defection, the circuit may be considered as an open circuit. Hence the voltmeter reading gives the emf of the cell. A small value of resistance R is included in the external circuit and key k is closed. The potential difference R is equal to the potential difference across cell (v) The potential drop across R V = IR (or) IR = V …………….. 1 Due to Internal resistance r of the cell Then V = E – Ir (or) Ir = E-V …………….. 2 10 COUNSELLING CODE P.B. College of Engineering dividing 2 by 1 Ir IR = r 1222 E-V V E-V V = R. Thus r can be calculated by knowing E, V and R. 53. In the given network, calculate the effective resistance between points A and B. 10 5 5 10 5 10 B A 5 5 10 10 5 10 Solution : The network has three individual units The simplified form of one unit is The equivalent resistance of one unit is 1 Rp = 1 R1 + 1 R2 = 1 15 + 1 15 Rp = 7.5 Each Unit has a resistance of 7.5 . The total network reduces to A The combined resistance between A and B R = R’ + R’ = R’ = 7.5 + 7.5 + 7.5 R = 22.5 11 COUNSELLING CODE P.B. College of Engineering 1222 54. A moving Coil galvanometer of resistance 20 produces fullscale deflection for a current of 50mA. How will you convert the galvanometer into (i) an ammeter of range 20A and (ii) a voltmeter of range 120 volt ? Data G = 20 Ig = 50 x 10-3 A I = 20A S = ? ,V = 120V, R = ? Solution : (i) S = G S = S = Ig I-Ig 20 x 50 x10-3 20 – 50 x10-3 20 x 0.05 = 20 – 0.05 20 x 0.05 19.95 = 0.05 A Shunt of 0.05 Should be connected in parallel with galvanometer. (ii) R = V Ig R = 120 – 20 50 x 10-3 -G = 2400 – 20 = 2380 A resistance of 2380 should be connected in series with galvanometer. 55. Obtain the phase relation between current and voltage in an a.c. circuit with inductor only. Solution : Let an alternating source of emf be applied to a pure inductor of inductance L. The inductor has a negligible resistance and is wound on a laminated iron core. Due to an alternating emf that is applied to the inductive coil. a self induced emf is generated which opposes the applied voltage. (eg) Choke coil. The instantaneous value of applied emf is given by, e = Eo sin t ……….. (1) di Induced emf e’ = - L / dt where L is the self inductance of the coil. In an ideal inductor circuit induced emf is equal and opposite to the applied voltage. 12 COUNSELLING CODE P.B. College of Engineering 1222 Therefore e = -e΄ E e΄ = – di = –L Eo L di dt sin t dt Integrating both the sides i = = i = Eo L Eo L Eo L sin t dt – cos t =– sin ( t – ) 2 i = Io. sin(t – where Io = Eo cos t L Eo L 2 ) ……… (2) . Here, L is the resistance offered by the coil. It is called inductive reactance. Its unit is ohm. From equations (1) and (2) it is clear that in an a.c. circuit containing a pure inductor the current i lags behind the voltage e by the phase angle of /2. Conversely the voltage across L leads the current by the phase angle of /2. This fact is presented graphically in figure (b) Fig. (c) represents the phasor diagram of a.c. circuit containing only L. Inductive reactance XL = L = 2vL, where v is the frequency of the a.c. supply. For d.c. = 0 ; XL = 0 Thus a pure inductor offers zero resistance to d.c. But in an a.c. circuit the reactance of the coil increases with increase in frequency. 13 COUNSELLING CODE P.B. College of Engineering 1222 56. A plane transmission grating has 5000 lines / cm. Calculate the angular separation in second order spectrum of redline 7070Å and Blue line 5000 Å N = 5000 lines / cm = 5000 x 102 lines /m = 5 x 105 lines/m m = 2 ; R = 7070Å = 7.07 x10-7m B = 5000 Å = 5 x 10-7 m sin R = NmR = 5 x105 x 2 x 7.07 x 10-7 = 0.707 R = 450 sinB = NmB = 5 x105 x 2 x 5 x 10-7 = 0.5 B = 300 Angular Separation = R - B = 450 - 300 = 150 (or) A 500mm long tube containing 60 C.C. of sugar solution produces a rotation of 9° when paced in a polarimeter. If the specific rotation is 60°, Calculate the quantity of sugar contained in the solution. l = 500mm = 5 dm = 90 ;S = 600;V = 60cc; m= ? S= lxC = ; m= l x (m/v) v I x S= 9 x 60 5 x 60 m = 1.8 g 57. Describe laue experiment. What are the facts established by it. Von laue suggested that a crystal can act as a three dimensional grating for an X ray beam. X rays from the X – ray tube is collimated into a fine beam by two slits S1, and S2. The beam is now allowed to pass through a zinc sulphide (ZnS) crystal. The emergent rays are made to fall on a photographic plate P. 14 COUNSELLING CODE P.B. College of Engineering 1222 The diffraction patterns so obtained consist of a central spot 0 and a series of spots arranged in a definite pattern about O. The central Spot is due to the direct beam, whereas the regularly arranged spots are due to the diffraction pattern from the atoms of the various crystal planes. These spots are known as laue spots. The Laue experiment has established following two important facts. (i) x rays are electromagnetic waves of extremely short wave length. (ii) The atoms in a crystal are arranged in a regular three dimensional lattice. 58. Obtain Einstein’s photoelectric equation. In 1905, Albert Einstein, successfully applied quantum theory of radiation to photo electric effect. According to Einstein, the emission of photo electron is the result of the infraction between a single photon of the incident radiation and an electron in the metal. When a photon of energy h is incident on a metal surface, it’s energy is used up in two ways. (i) A part of the energy of the photon is used in extracting the electron from the surface of metal, Since the electrons in the metal are bound to the nucleus. This energy W spent in releasing the photo electron is known as photo electric work function of the photo metal. The work function of a photo metal is defined as the minimum amount of energy required to liberate an electron from the metal surface. (ii) The remaining energy of the photon is used to import kinetic energy to the liberated electron. If m is the mass of an electron and v, its velocity then. Energy of the Incident photon = Work function + kinetic energy of the electron h = W + 1 mv2 2 If the electron does not lose energy by internal collisions, as it escapes from the metal, the entire energy (h –w) will be exhibited as the kinetic energy of the electron. Thus, (h-w) represents the maximum kinetic energy of the ejected photo electron. If Vmax is the maximum velocity with which the photo electron can be ejected, then h = W+ 1 2 m 2 max …………..(2) This equation is known as Einstein’s photo electric equation. 15 COUNSELLING CODE P.B. College of Engineering 1222 When the frequency () of the incident radiation is equal to the threshold frequency (o) of the metal surface, kinetic energy of the electron is zero. Then equation (2) becomes, ho = W Substituting the value of W in equation (2) we get, h – ho = 1 2 mmax 2 (or) h(-o) = 1 m2max 2 This is another form of Einstein’s photo electric equation. 59. How fast would a rocket have to go relative to an observer for its length to be corrected to 99% of its length at rest ? Data : = 99% = 99/100 ; v = ? Solution : l = lo 1 – l = 99 lo 100 l lo = 99 100 99 = 100 1 – v2 c2 v2 c2 v = 0.141c v = 0.141 x 3 x 108 v = 0.423 x 108 ms-1 60. The binding energy per nucleon for 6C12 Nucleus is 7.68 Mev and that for 6C13 is 7.47 Mev, Calculate the energy required to remove a neutron from 6C13 nucleus. Data : Binding energy per nucleon of 6C12 = 7.68 MeV Binding energy of neutron = ? Working : We can write the reaction as 6C13 → 6C12 + 0n1 Total binding energy of 6C13 = 7.47 x 13 = 97.11MeV Total binding energy of 6C12 = 7.68 x12 = 92.16 MeV Total binding energy of the reactant = Total binding energy of the product 97.11 MeV = 92.16 MeV + Binding energy of a neutron 16 COUNSELLING CODE P.B. College of Engineering 1222 Binding energy of a neutron = 97.11-92.16 = 4.95 MeV 61. With the circuit diagram, explain voltage divider biasing of a transistor. Voltage divider bias : In this method, two resistances R1, and R2 are connected across the supply voltage Vcc and provide biasing. The emitter resistance RE provides stabilization. The voltage drop across R2 forward biases the base emitter junction. This causes the base current and hence collector to flow in zero signal conditions. The stabilization provided by RE : Since b is very Sensitive to temperature changes, the collector current IC increases with rise in temperature, Consequently, it can be seen that IE increases. This will cause the voltage drop across emitter resistance RE to increase. The voltage drop across R2 = VBE + VRE. As voltage drop across R2 is independent of IC, VBE decreases. This decreases IB and the reduced value of IB tends to bring back IC to the original value. Hence any variation of b will have no effect on the operating point. 62. With the help of block diagram, explain the operation of an FM super heterodyne receiver. An FM receiver is a super heterodyne type like a typical AM receiver. The functional block diagram of any FM receiver is shown in the figure. 17 COUNSELLING CODE P.B. College of Engineering 1222 The RF section, selects the modulated signals and is amplified. It is then fed into the mixer and local oscillator. Here the frequency of the modulated signal is changed to intermediate frequency. For FM receivers, this IF is 10.7 MHz. The intermediate frequency wave is amplified using IF amplifier and then its amplitude is maintained constant using a limiter. The output of this section is applied to the FM detector which demodulates the modulated wave. The AF signal from the FM detector is then passed on through a de – emphasis network, where the various frequencies attain their original power distribution. Finally it is fed into the loud speaker after performing AF amplification. PART – IV Note : (i) Answer any four question in detail (ii) Draw diagrams wherever necessary. 63. What is an electric dipole ? Derive an expression for the electric field due to an electric dipole at a point on its axial line. Two Equal and opposite charges separated by a very small distance constitute an electric dipole. Some examples are water, ammonia, and chloroform molecules. Derivation : AB is an electric dipole of two point charges – q and + q separated by a small distance 2d. P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole. The electric field at the point P due to + q placed at B is , E1 = 1 q 4 (r-d)2 (along BP) The electric field at the point P due to – q placed at A is E2 = 1 4 . q (r+d)2 (along PA) E1 and E2 act in opposite direction. Therefore the magnitude of resultant electric field (E) acts in the direction of the vector with a greater magnitude. The resultant electric field at P is, E = E1 + (–E2) 18 COUNSELLING CODE P.B. College of Engineering E 1 = = = 4 . q – (r+d)2 q 1q 4 (r-d)2 1 4rd 4 (r2-d2)2 . 4 q along BP (r+d)2 1q – q 1222 along BP (r+d)2 along BP If the point P is for away from the dipole, then d<<r E = E = q 4 1 4 . 4rd = r4 2P . q . 4 4d r3 along BP r3 [ Electric dipole moment p = q x 2d] E acts in the direction of dipole moment. 64. Apply Biot – Savart law, obtain an expression for the magnetic induction at a point due to infinitely long straight conductor carrying current. XY is an infinitely long straight conductor carrying current I . P is a point at a distance a from the conductor. AB is a small element of length dl. is the angle between the current element Idl and the line joining the element dl and the point P. According to Biot – Savart law, the magnetic induction at the point P due to the current element Idl is dB = Idl.sin 4 r2 ……………….(1) AC is drawn perpendicular to BP from A OPA = APB =d AC AC In ABC, sin AB dl AC = dl sin ……………. (2) From APC, AC = rd ……………. (3) from equation (2) and (3) rddl.sin ................(4) 19 COUNSELLING CODE P.B. College of Engineering 1222 substituting equation (4) in equation (1) dB = Ird 4 r2 = Id 4 r ………………….. (5) a In OPA, cos = r= r a cos substituting eqn. (6) in eqn. (5) dB = 1 . 4 cos d a The total magnetic induction at P due to the conductor XY is B = dB = = 4a cos d [ sin 1 + sin 2 ] 4a for infinitely long conductor, 1290° B= 2a If the conductor is placed in a medium of permeability, B= l 2a . 65. Discuss with theory the method of inducing e.m.f. in a coil by changing its orientation with respect to the direction of the magnetic field. PQRS is a rectangular coil of N turns and area A placed in a uniform magnetic field B. The coil is rotated with an angular velocity in the clockwise direction about an axis perpendicular to the direction of the magnetic field. Suppose, initially the coil is in vertical position, so that the angle between normal to the plane of the coil and magnetic field is zero. After a time t, let be the angle through which the coil is rotated. If is the flux linked with the coil at this instant , then cos(since = t) The induced emf is e = – d dt = – NBA d dt cos (t) 20 COUNSELLING CODE P.B. College of Engineering e 1222 = NBA sint ……………. (1) The maximum value of the induced emf is Eo = NAB INDUCED EMF BY CHANGING THE ORIENTATION OF THE COIL Hence, the induced emf can be represented as e – Eo sint. The induced emf e varies sinusoidally with time t and the frequency being cycles per second = (i) 2 When t = 0, the plane of the coil is perpendicular to the field B and hence e = 0. (ii) When t = /2, the plane of the coil is parallel to B and hence e = E0. (iii) When t = , the plane of the coil is at right angle to B and hence e = 0. (iv) When t = 3, the plane of the coil is again parallel to B and the induced emf is e = 0. (v) When t = 2, the plane of the coil is again perpendicular to B and hence e = 0. If the ends of the coil are connected to an external circuit through a resistance R, current flows through the circuit, which is also sinusoidal in nature. 21 COUNSELLING CODE P.B. College of Engineering 1222 66. Derive an expression for band width of interference fringes in young double slit experiment. Expression for bandwidth : Let d be the distance between two coherent sources A and B of wavelength . A screen XY is placed parallel to AB at a distance D from the coherent sources. C is the midpoint of AB. O is a point on the screen equidistant from A and B. P is a point at a distance x from O. Waves from A and B meet at P in phase or out of phase depending upon the path difference between two waves. Draw AM perpendicular to BP The path difference = BP – AP AP = MP = BP – AP = BP – MP = BM In right angled ABM, BM = d sin If is small, sin = The path difference = .d In right angled triangle COP, tan = OP CO = x D For small values of , tan = xd The path difference = D Bright fringes : By the principle of interference, condition for constructive interference is the path difference = n xd = n D Where n = 0,1,2…indicate the order of bright fringes. 22 COUNSELLING CODE P.B. College of Engineering x = 1222 D d n This equation gives the distance of the nth bright fringe from the point O. Dark fringes : By the principle of interference, condition for destructive interference is the path difference = (2n – 1) d Where n = 1,2,3…. indicate the order of the dark fringes. x= D d (2n – 1) d This equation gives the distance of the nth dark fringe from the point O. Thus, on the screen alternate dark and bright bands are seen on either side of the central bright band . Band width () : The distance between any two consecutive bright or dark bands is called bandwidth. The distance between (n+1)th and nth order consecutive bright friges from O is given by X(n + 1) –xn = D d (n+1) - D n = D d d Bandwidth, = D d Similarly, it can be proved that the distance between two consecutive dark bands is also equal to D . Since bright and dark fringes are of same width, they d are equi-spaced on either side of central maximum. 67. State Bohr’s Postulates. Obtain an expression for the radius of the n th orbit of hydrogen atom. (i) An electron cannot revolve round the nucleus in all possible orbits. The electrons can revolve round the nucleus only in those allowed or permissible orbits for which h the angular momentum of the electron is an integral multiple of (where h is 2 Planck’s Constant = 6.626 x 10-34 Js). These orbits are called stationary orbits or non-radiating orbits and an electron revolving in these orbits does not radiate any energy. If m and are the mass and velocity of the electron in a permitted orbit of radius r nh then angular momentum of electron =mr = , Where n is called principal 2 23 COUNSELLING CODE P.B. College of Engineering 1222 quantum number and has the integral values 1,2,3… This is called Bohr’s quantization condition. (iii) An atom radiates energy, only when an electron jumps from a stationary orbit of higher energy to an orbit of lower energy. If the electron jumps from an orbit of energy E2 to an orbit of energy E1, a photon of energy h = E2- E1 is emitted. This condition is called Bohr’s frequency condition. Radius of the nth Orbit (rn) : Consider an atom whose nucleus has a positive charge Ze, where Z is the atomic number that gives the number of protons in the nucleus and e, the charge of the electron which is numerically equal to that of proton. Let an electron revolve around the nucleus in the nth orbit of radius rn By Coulomb’s law, the electrostatic force of attraction between the nucleus 1 (Ze) (e) and the electron = . ………………(1) rn2 4 Where 0 is the permittivity of the free space. Since, the electron revolves in a circular orbit, it experiences a centripetal force, 2 mn = rn mrnn2 ………………(2) where m is the mass of the electron, n and n are the linear velocity and angular velocity of the electron in the nth orbit respectively. The necessary centripetal force is provided by the electrostatic force of attraction. For equilibrium, from equations (1) and (2), 1 4 1 4 From equation (4), Ze2 . rn2 . Ze2 rn2 = mn2 rn = mrnn2 ………………(3) ………………(4) Ze2 n2 = 4mrn3 ………………(5) The angular momentum of an electron in nth orbit is, L = mnrn = mrn2n ………………(6) By Bohr’s first postulate, the angular momentum of the electron 24 COUNSELLING CODE P.B. College of Engineering L= 1222 nh ………………(7) 2 where n is an integer and is called as the principal quantum number. From equations, (6) and (7), nh mrn2n = 2 nh (or) n = 2 mrn2 squaring both sides, n2 = n2h2 42 m2rn4 ……………………..(8) From equations (5) and (8), n2h2 Ze2 4mrn3 (or) = 42 m2r 4 n rn = n2h2 mZe2 …………………….(9) From equation (9), it is seen that the radius of the nth orbit is proportional to the square of the principal quantum number. Therefore, the radii of the orbits are in the ratio 1 : 4 : 9…………. For hydrogen atom, Z = 1 From equation (9) rn = n2h2 me2 Substituting the known values in the above equation we get, rn = n2 x 0.53 Å If n = 1, r1 = 0.53 Å This is called Bohr radius. 68. Obtain an expression for the amount of the radioactive substance present at any moment. Obtain the relation between half –life period and decay constant. Radioactive law of disintegration : Rutherford and Soddy found that the rate of disintegration is independent of physical and chemical conditions. The rate of disintegration at any instant is directly proportional to the number of atoms of the element present at that instant. This is known as Radioactive law of disintegration. 25 COUNSELLING CODE P.B. College of Engineering 1222 Let N0 be the number of radioactive atoms present initially and N, the number of atoms at a given instant t. Let dN be the number of atoms undergoing disintegration in a small interval of time dt. Then the rate of disintegration is _ dN N dt dN =–N ………………… (1) dt where is a constant known as decay constant or disintegration constant. The negative sign indicate that N decreases with increase in time. Equation (1) can be written as dN = - dt dt Integrating, logeN = - t + C ………………… (2) where C is a constant of integration, At t = 0, N = N0 logeN0 = C Substituting for C, equation (2) becomes, logeN = - t + logeN0 loge N N0 N N0 =- t = e- t N = N0e- t ………………… (3) Equation (3) shows that the number of atoms of a radioactive substance decreases exponentially with increase in time. 26 COUNSELLING CODE P.B. College of Engineering 1222 Initially the disintegration takes place at a faster rate. As time increases, N gradually decreases exponentially. Theoretically, an infinite time is required for the complete disintegration of all the atoms. Half life period : The half period of a radioactive element is defined as the time taken for one half the radioactive element to undergo disintegration. From the law of disintegration N = N0e -t Let T½ be the half life period. Then, at t = T½, N = N0 2 = N0e N0 2 -T½ loge2 = T½ T½ = loge2 = log102 x 2.3026 = 0.6931 The half life period is inversely proportional to its decay constant. For a radioactive substance, at the end of T½, 50% of the material remain unchanged. After another T½ i.e., at the end of 2T½, 25% remain unchanged. At the end of 3T½, 12.5% remain unchanged and so on. 69. Sketch the circuit of a colpit oscillator and explain its working. Colpitt’s oscillater : The circuit diagram of Colpitt’s oscillator is shown in figure. The resistance R1, R2 and RE provide the sufficient bias for the circuit. The frequency determining network is the parallel resonant circuits consisting of capacitors C1, C2 and the inductor L. The junction of C1 and C2 is earthed. The function of the capacitor C4 is to block d.c. and provide an a.c path from the collector to the tank circuit. The voltage developed across C1 provides the positive feedback for sustained oscillations. Working : When the collector supply voltage is switched on, a transient current is produced in the tank circuit and damped harmonic oscillation are produced. The oscillations across C1 are applied to the base emitter junction and appear in the amplified form in the collector circuit. If terminal 1 is at positive potential with respect to terminal 3 at any instant, then terminal 2 will be at negative potential with respect to 3, since 3 is grounded. Hence points 1 and 2 are 180 0 out of phase. The amplifier produce further phase shift 1800. Thus the total phase shift is 3600. In other words, energy supplied to the tank circuit is in phase with the oscillations and if A = 1, oscillations are sustained in the circuit. 27 COUNSELLING CODE P.B. College of Engineering 1222 The frequency of oscillations is given by f = 1 where C = 2 LC f= 1 (C1+C2) 2 LC1C2 C1C2 C1+C2 70. With the help of a block diagram, explain the function of a RADAR system. The Block diagram of a simple radar system is shown in the figure. This block diagram indicates that the radar system consists of both the transmitting and the receiving system. The transmitting system consists of a transmitter and a pulser. The receiving system consists of a receiver and an indicator. In most of the cases, a single antenna is used for both transmission and reception and this is achieved with the use of TR switch (Transmitter Receiver Switch). This switching arrangement is called as ‘duplexer’. This connects the antenna to the transmitter during transmission and to the receiver during reception. Moreover, this switch isolates the sensitive receiver from the damaging effects of the high power transmitter. The transmitter is essentially a high power magnetron oscillator which generates high power pulses. This transmitter is turned on and off with a periodic pulse from the pulser. Thus the transmitter generates periodic pulses of very short duration. These short pulses are fed to the antenna which radiates them into the space. The antenna is highly directional. 28 P.B. College of Engineering COUNSELLING CODE 1222 If the transmitted pulse hits any target, a weak echo signal returns to the same antenna. But, now the TR switch puts the antenna in contact with the receiver. This echo signal is amplified and demodulated by the superhet receiver. The sensitivity of the receiver is very high. The detected output is sent to the indicator. The indicator is a cathode ray tube. The CRT displays the original transmitted pulse as well as the detected echo pulse along a horizontal base line. The synchronizing pulse generated by the ‘timer’ is supplied to both transmitting and receiving systems. So, the indicator records the transmitted pulse as well as the returning pulse simultaneously. The returning echo pulse appears slightly displaced from the transmitted pulse and this displacement is a measure of the range of the target. 29 COUNSELLING CODE P.B. College of Engineering 1222 COURSE OFFERED : UG Courses : PG Courses : B.E. Aeronautical Engg. B.E. Automobile Engg. B.E. Computer Science & Engg. B.E. Civil Engg. B.E. Electronics & Communication Engg. B.E. Electrical & Electronics Engg. B.E. Mechanical Engg. B.Tech. Information Technology M.E. M.E. M.E. M.E. Aeronautical Engg. Computer Science & Engg. Communication & Networking Power Electronics & Drives GOLD MEDALIST - J. SUJATHA, M.E. (Aero) – UNIVERSITY RANK I ANNA UNIVERSITY RANK HOLDERS B. SUBASHREE YUVARAJ P.S. PRABHAJA V. THULASI STEPHEN B S. ANITHA D. PRASHANTH R. VISHNU PRIYA M. DEVAMBIKA B. NANCY PUSHPARANI U. SARASWATHI G. SARGUNARUBI G. SAI DIVYA G. GAJALAKSHMI OLIVER PRASANNA S. SHATHEESH KUMAR J. REVATHI JYOTHI LAKSHMI N. SIVARAM A. 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