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Sample Question Paper – III Time : Hours Maximum Marks :70 General Instructions (i) (ii) All questions are compulsory There is no overall choice However, an internal choice has been provided in one question of two marks, one question of three marks and one question of five marks You have to attempt only one of the choices in such questions (iii) Question number to are very short answer questions, carrying mark each (iv) Question number to 12 are short answer questions, carrying marks each (v) Question number 13 to 24 are also short answer questions, carrying marks each (vi) Question number 25 to 27 are long answer questions, carrying marks each (vii) Use of calculators is not permitted However, you may use Log Tables, if necessary (viii) You may use the following physical constants wherever necessary: c = × 108 ms–1 h = 6.6 × 10–34 Js e = 1.6 × 1019 C à0 = ì 107 T mA–1 State any two properties of electromagnetic waves Why are infrared waves preferred for optical fibre communication? Two electric bulbs are rated as 220V-100W & 220V-60W Which one of these has greater resistance and why? S Chand & Company Limited 10 11 12 In which direction will a freely suspended magnet allign itself when placed at the magnetic pole of the earth? A graph is plotted between the maximum K.E of emitted photoelectrons & the frequency of incident radiations Which physical constant can be determined from the slope of this graph? Define the term resistivity Write an expression for the resistivity of a conductor in terms of relaxation time An ac generator has a coil of N turns each of area A, rotating with angular speed ω in a uniform magnetic field B (i) What is the maximum emf between its slip rings? (ii) What is the flux associated with the coil when the emf across it is zero? Draw the energy level diagrams for (i) n-type & (ii) p-type semiconductors Mark the donor and acceptor levels in the diagrams The refractive index of water is 4/3 Obtain the value of the semi vertical angle of the cone within which the entire outside view would be confined for a fish under water Draw an appropriate ray diagram When an uncharged body B is charged by induction, using a charged body A, the charge on A does not change Explain how this fact is in accordance with the principle of conservation of charge A coil of n turns and radius R carries a current I It is unwound and rewound to make another coil of radius R/2, current remaining the same Calculate the ratio of the magnetic moments of the new coil and the original coil What will be the effect on the interference fringes in Young’s double slit experiment when, (i) the monochromatic source is replaced by another monochromatic source of shorter wavelength (ii) monochromatic source is replaced by a source of white light Justify your answer in each case S Chand & Company Limited 13 14 15 16 17 OR The polarisation of a beam of light by reflection is best achieved when the reflected and refracted rays are at right angles to each other Show that the polarising angle of incidence is then given by ip = tan–1 µ State any two properties of the electric lines of force Draw the lines of force in each of the following cases:(a) An isolated point charge (i) q > (ii) q < (b) Between the plates of a charged parallel plate capacitor A bar magnet is placed in a unifrom magnetic field with its magnetic moment making an angle θ witht he field (i) Write an expression for the torque acting on the magnet and hence define its magnetic moment (ii) Write an expression for the potential energy of the magnet in this orientation Hence obtain the orientation for which this energy becomes minimum Define the term binding energy of a nucleus Which of the two, 235 or 144 56 Ba, has a higher value 92 U of B.E./nucleon? Give the formula for B.E./nucleon for a nucleus of mass number A and atomic number Z Name the nucleus of low atomic number that is much more stable than its immediate neighbours Two charges of magnitude +25 nC and –9nC are located at points A(1,2) and B(5,2) respectively Find the magnitude of electric field due to these charges at the point C(5,5) All distances are measured in metres How does one remove the defect of polarisation in a (i) Daniel cell (ii) Leclanche cell S Chand & Company Limited 18 19 Write the chemical equation for charging in (i) Lead accumulator (ii) Alkali accumulator A small town with a demand of electric power 800 kW at 220 V is situated 15 km away from an electric generating station The resistance of the two wire line carrying power is 0.5 Ω/km The town gets power from the line through a 4000 V – 220 V step down transformer at a substation in the town (i) Estimate the line power losses in the form of heat (ii) Assuming negligible power losses due to leakage, find the power that must be supplied by the plant State Huygens’ principle and use it to construct refracted wave-front for refraction of a plane wavefront at a plane refracting surface Hence derive Snell’s Law 20 Compare the energy of an electron of de Broglie wavelength 1Å with that of an X-ray photon of the same wavelength 21 Differentiate between absolute magnitude and apparent magnitude of brightness of a star The apparent and absolute magnitudes of stars A and B are as under: star Apparent magnitude Absolute magnitude A -2 B -1 Which of the stars is nearer to earth? Give reason for your answer 22 Explain the working of a typical fibre optic communication link using a block diagram 23 A 50W - 100 V electric bulb is to be used on a 200V-50Hz ac supply Calculate the inductance of the lamp so that it may glow with its normal brightness (Take π ≈ 3) S Chand & Company Limited 24 What will be the required height of a TV tower which can cover the population of 60.3 lakhs if average population density around the tower is 1000 km–2 Radius of earth = 6.4 × 106m OR 24 How we make the choice of a communication channel? A message signal has a bandwidth of MHz Suggest a possible communication channel for its transmission 25 Explain the function of the (i) collimator (ii) telescope in a spectrometer With the help of an appropriate ray diagram discuss how we use a spectrometer for finding the refracting angle of a prism Write the formula used for determination of refractive index of the material of the prism using the spectrometer 26 Define the term potential gradiant Using this concept, explain the method for comparison of emfs of two primary cells using a potentiometer Establish the relation used Write two possible causes of potentiometer giving only one sided deflection in this method 27 With the help of a block diagram, explain the concept of feedback used in an oscillator Explain by drawing a circuit diagram, the working of a transistor as an oscillator OR Draw a circuit diagram of a common emitter amplifier using npn transistor Write an expression for the current gain Show input and output voltages graphically and explain the phase relation between them ANSWERS (i) Transverse nature ½ (ii) Can propagate through vacuum or Any two other properties S Chand & Company Limited ½ For infrared wave lengths, fiber attenuation is minimum 60 W bulb has greater resistance, since, V2 1 R= , Hence R ∝ for a given voltage P P It will align itself normal (vertical) to the surface of the earth Planck’s constant (h) can be determined from the slope of this graph Resistance offered by unit length and unit area of cross section is resistivity m ρ= ne τ (i) Emax = NBAω (ii) flux = NBA n type = Conduction band Donor level Valence band p type = Conduction band Acceptor level Valence band S Chand & Company Limited ½ ½ 1 Diagram ic q = 2ic à= ẵ sin ic ½ = sin ic sin ic = ic = 49° = semi-vertical angle 10 ½+½ During charging of B by induction, the charged object A simply helps B to share charges with the earth It is, therefore, the earth that provides/takes away charge from B and there is no violation of the principle of charge conservation S Chand & Company Limited 11 L = n.2 πR = n ′.2 π Hence n1 = 2n ½ M = n (π R2) I ½ πR2 πR2 M′ = n′ I I = n ½ M′ = M ∴ 12 R (i) ½ The fringe width decreases as β = λD d (ii) Central fringe will be white surrounded by coloured fringes on both side of the central fringe This is because the fringe width depends on the wavelength of light ½+½ OR 12 Ray diagram angle of reflection + angle of refraction = 90° ∴ angle of refraction = 90º – angle of reflection = 90º – angle of incidence (∠ i = ∠ r) = 90º – ip By Snell’s Law µ= ½ sin ip sin i = sin r sin(90 − ip ) S Chand & Company Limited ẵ = tan ip or 13 ip = tan1à ẵ Any two properties q>0 14 (i) → → q< → τ =M ×B τ = MB sin θ M= τ B sin θ if B = tesla and sin θ = 1, M = τ ½ Hence the magnetic moment is the torque experienced by the magnet when it is placed normally in a magnetic field of uniform unit strength, i.e, tesla S Chand & Company Limited → → (ii) U = − M B or U = − MB cos θ P.E will be minimum when 15 θ = 0, (Umax = – M B) Hence the P.E is minimum when the magnetic moment of the magnet is aligned parallel to the external uniform magnetic field Definition of B.E 144 56 Ba has higher B.E./nucleon ½ Formula Zmp + ( A − Z )mn − M − c B.E./nucleon = A Name – He (the alpha particle) ½ 16 E1 C (5,5) 3m Y E2 4m A (1,2) B (5,2) X S Chand & Company Limited → E1 = q 12 π ε (5) × 109 × = → 25 × 10 −9 25 N C ẵ E2 = q ì 109 (9) × 10 −9 22 = π ε (3) = −9 N C → E ½ = → → → → 2 E1 + E2 + E1 E2 cos θ = 92 + 92 + 2(9) (9) cos (π/2 + α) = 81 + 81 + (+81) (–sinα) = 162 – 162 × 3/5 = 162 – 97.2 = 64.8 → E = 64.8 = 8.05 NC −1 17 Daniel cell — ½ copper sulphate acts as the depolariser/H+ ions interact with CuSo4 and form sulphuric acid and release Cu++ ions ½ S Chand & Company Limited Leclanche cell —Manganese dioxide acts as the depolariser / 2H++ MnO2→ Mn2O3 + H2O + units of positive charge ½ Lead accumulator : At the cathode Pb So4 + 2H+ + 2e– → Pb + H2So4 At the anode Pb So4 + 2H2O → PbO2 + 2H2SO4 + 2e– ½+½ Alkali accumulator: 18 At the cathode Fe (OH)2 + 2K+ + 2e– → Fe + 2KOH At the anode Ni (OH)2 + 2OH– → Ni(OH)4 + 2e– ½+½ Length of wire = 15 × = 30 km Resistance of wire = ì 30 = 15 ẵ P = Erms × Irms I rms = P × 105 = = 200A Erms 4000 ½ (a) Line power loss = I2rms R = 600 kW (b) Plant must supply 600 kW + 800 kW = 1400 kW 19 Statement of Huygens’ principle Diagram with complete idea of formation of secondary/refracted wavefront S Chand & Company Limited sin i = constant sin r Proof of 20 for the electron, λ = Also K.E = h h or p = λ p ½ p2 2m ∴ Electron energy = h2 2mλ For the X-Ray photon, energy = hν = hc λ = h 2mλc 6.6 × 10 −34 × 10 −10 × × 9.1 × 10 −31 × × 108 = / 82.4 21 ẵ Electron energy h2 = ì X-ray Photon energy 2mλ hc ∴ = 1½ Absolute Magnitude: It is the magnitude of brightness of a star observed from a distance of 10 parsec ½ Apparent magnitude: It is magnitude of brigthness of a star as observed from the earth Star B is nearer because it is less bright but appears brighter than A S Chand & Company Limited 22 Working: The input analog signal is first sampled and modulated using PCM which produces discrete pulses & gives coded stream of bit (0´S & 1´S) representing the signal Corresponding to this stream of bits, the optical beam from a source is modulated This modulated signal is propagated in medium called optical fibre At the receiver, an optical detector detects the signal which in turn, is converted back to electrical signal This is decoded and original analog signal is reconstructed Transmitter Input Signal Optical source Modulation Receiver Demodulation Optical Detector 23 I rms = Output Signal 50 = A 100 ½ S Chand & Company Limited voltage across L (Veff ) − (V R ) VL = = 100 V VL 100 = = 200 I rms X L = 2πfL XL = XL = 2πf = 1.154 H L= 24 πr2 × density = Population covered r π ×1000 = 60.3×105 10 r2 = 60.3×105 ×103 π = 1.92 ×109 m2 r = 2Rh 1.92 ×109 r2 = 2R × 6.4 ×106 = 150 m h= S Chand & Company Limited OR 24 The communication channel is determined both by the nature of the signal and the nature of medium (i) For guided media, the medium itself is important ½ (ii) For unguided media, frequency band of signal is important Possible communication channel (any one) ½ Co-axial cable, AM radio 25 Function of collimator: To obtain a parallel beam of light coming from a source ½ Function of telescope: To obtain image from rays refracted from the prism ½ Ray diagram FROM COLLIMATOR A PE CO S LE TE 2A PRISM TE LE SC OP E Explanation with relevant formula for finding the angle of prism Relevent formula, µ = sin (A + δm) / 2) / sin (A/2) ½ S Chand & Company Limited 26 Definition of potential gradient Circuit diagram 1½ Explanation of comparison of emf’s using potentiometer with formula Causes (i) All the positive terminals may not be connected to the same terminal of the potentiometer (ii) Emf of main supply producing the potential gradient may be less than the emf of cells ½+½ 27 Block diagram Concept of feedback Circuit diagram of oscillator 1½ Working of an oscillator 1½ Input Output Amp Feedback Network OR S Chand & Company Limited Ic C Ib B RL E Vo Vi Input signal Ie Output signal Ic Vcc Vbe V1 t V0 t Explanation: Output voltage differs by a phase angle of π with respect to the input voltage Vo = Vcc – IC RL During the half of cycle of input voltage emitter current Ie increases and thus Ic increases Vo decreases (according to the above equation) Increase in input signal voltage decreases output voltage Vo Output collector voltage and input signal voltage are out of phase in common emitter amplifier S Chand & Company Limited ... = 2πf = 1.154 H L= 24 πr2 × density = Population covered r π ×1000 = 60 .3? ?105 10 r2 = 60 .3? ?105 ×1 03 π = 1.92 ×109 m2 r = 2Rh 1.92 ×109 r2 = 2R × 6.4 ×106 = 150 m h= S Chand & Company... particle) ½ 16 E1 C (5,5) 3m Y E2 4m A (1,2) B (5,2) X S Chand & Company Limited → E1 = q 12 π ε (5) × 109 × = 25 ì 10 25 N C ẵ E2 = q × 109 (9) × 10 −9 22 = π ε (3) = −9 N C → E ½ = → → →... Fe + 2KOH At the anode Ni (OH)2 + 2OH– → Ni(OH)4 + 2e ẵ+ẵ Length of wire = 15 ì = 30 km Resistance of wire = × 30 = 15 ẵ P = Erms ì Irms I rms = P × 105 = = 200A Erms 4000 ½ (a) Line power loss
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