NQA2B kho tài liệu bách khoa

NUMERICALS-ANSWERS QUESTIONS A 220 V, 100 W bulb is connected to a 110 V source Calculate the power consumed by the bulb and the resistance of the bulb A 12 V storage battery of negligible internal resistance is connected across a 40 Ω resistor Calculate the heat energy produced in the resistor in hour An electric motor operating on a 50 V d.c supply draws a current of 12 A If the efficiency of the motor is 30%, estimate the resistance of the winding of the motor 10 MW power is transmitted through a 10 km line at 200 kV If the resistance of the lines is 0.1 Ω per km, calculate the power loss in the line A current of 5.0 A flows through an electric press of resistance 44 Ω Calculate the energy consumed by the press in minutes (AISSCE 1994) The emf of a Cu-Fe thermocouple varies with temperature θ of the hot junction as E (µ V) = 14 (AISSCE 1990) θ – 0.02 θ2 Determine the neutral temperature An electric bulb is marked 100 W, 230 V What is the energy produced by the bulb in 20 minutes when operated at 230 V ? (AISSCE 1990 C) An electric power station transmits power to a distant load through long cables Which of the two modes of transmission would result in lesser power wastage—transmission at (i) 20,000 V, (ii) 200 V Calculate the time taken by a kW heater to raise the temperature of kg of water by 40 °C 10 Two electric lamps have the ratings (100 W, 220 V) and (200 W, 220 V) Find the heat generated in each per second when they are connected in series across a 220 V source 11 Compare the amounts of heat produced per second in three conductors of the same material and length, having radii in the ratio 1: 2: 3, when connected in parallel across a source S Chand & Company Limited 12 Two bulbs of 60 W and 100 W are designed to operate at 220 V Find the ratio of their resistances 13 A dry cell of emf 1.6 V and internal resistance 0.10 ohm is connected to a resistor of resistance R ohm The current drawn from the cell is 2A (i) What is the voltage drop across R ? (ii) What is the energy dissipation in the resistor? (AISSCE Delhi 1993) 14 A battery of internal resistance r is connected across a pure resistive device of resistance R Show that the power output of the device is maximum when R = r 15 Two bulbs rated (60 W, 110 V) and (100 W, 110 V) are connected in series across an a.c supply of 220 V Which of the bulbs, if any, will fuse? 16 For a thermocouple the temperature of the cold junction is 30°C and the neutral temperature is 240° C Calculate the temperature of inversion 17 A current of 1.6 ampere is passing through a voltameter Find the number of Cu atoms deposited on the cathode per minute 18 A steady current of 10.0 A is maintained in a copper voltameter Find the time required to deposit 2.5 g of copper The relative atomic mass of copper is 63.5 and Faraday constant (F) is 96500 C mol–1 (AISSCE 1991) 19 Calculate the time required to deposite 250 milligram of copper in a copper voltameter using a current of 0.2 A Chemical equivalent of copper is 32 Use known value of Faraday constant 20 Calculate the time required to deposite 10.79 g of silver on the cathode of a silver voltameter by a current of 20 A ECE of silver is 0.001118 gC–1 21 A steady potential difference of 1.62 V is applied across two electrodes dipped in a CuCl2 solution It is found that 11.84 g of copper is deposited on the cathode in 20 minutes If the S Chand & Company Limited 22 23 24 25 back emf of the voltameter is 1.34 V, estimate the resistance of the Voltameter The ECE of copper is 3.29 × 10–7 kg/C A copper voltameter is connected in series with a coil of resistance 50 ohm A steady current is passed through the circuit for 10 minutes, and it deposits g of copper How many joules of heat energy are generated in the coil per second ECE of copper is 3.3 × 10–7 kgC–1 A steady current of 10.0 A is passed through a water voltameter for minutes Calculate the volume of hydrogen evolved at S.T.P Faraday constant F = 96500 C/mol Molecular weight of hydrogen is 2.016 A current of 10 A is passed for 77s through a voltameter so that 0.253 g of substance is deposited on the electrode Taking F = 96500 Cmol–1, calculate the atomic mass of the substance deposited What current will be required by a lamp marked 225 V, 40 W ? Find the resistance of such a lamp and the amount of heat produced in minutes 26 An electric heater consists of 20m length of manganin wire of 2.3 × 10 –7 m cross sectional area Calculate the wattage of the heater when the potential difference across the heater is 200 V Resistivity of manganin is 4.6 × 10–7 Ω m 27 A (100 W, 220 V) bulb is connected to a 110 V source Calculate the power consumed by the bulb 28 Two electric lamps have ratings (220 V, 40 W) and (220 V, 60W) Find the heat generated in each lamp per second when they are connected in series across a 220 V supply (AISSCE 1983) 29 An electric motor operating on a 100 V d.c supply draws a current of 15 A If the efficiency of the motor is 25%, calculate the resistance of the winding of the motor 30 In a given thermocouple the temperature of the cold junction is 25°C while the neutral temperature is 275°C Find the temperature of inversion S Chand & Company Limited 31 The thermo-emf of a copper constantan couple is 40 µ V/°C What is the smallest temperature difference that can be detected with this thermo couple and a galvanometer of 100 Ω capable of detecting a current as low as 10–6 A ? 32 A dry cell of potential difference 1.5 V and internal resistance 0.2 Ω is connected across a resistance in series with a very low resistance ammeter If the ammeter reading is 2.5 A, then calculate (a) the rate of chemical energy consumption of the cell, (b) the rate of energy dissipation inside the cell, and (c) the rate of energy dissipation in the resistance 33 In a voltameter containing CuSo4 as electrolyte, charge of 4.0 × 105 C flows Calculate the following : (a) Number of Cu++ ions liberated from the electrolyte (b) Number of electrons flowing in the external circuit 34 Atomic weight of silver is 108, its valency is and E.C.E is 1.18 × 10 –6 kg/C Calculate the following : (i) Kg-equivalent of silver (ii) Charge required to liberate one kg-equivalent of silver (iii) Electronic charge e, assuming that there are 6.0 × 1026 atoms in one kg atom of silver (iv) Number of ions of silver that would be displaced by coulomb of charge (v) Number of ions of copper that would be displaced by coulomb of charge 35 How much time will be required to electroplate 3.00 g of silver on a vessel of brass if 15.0 A current be used E.C.E of copper is 1.118 × 10–6 kg/coulomb? 36 E.C.E of silver is 1.118 × 10–6 kg/ coulomb Its atomic weight is 107.9 and valency Assuming the avogadro number to be 6.06 × 1026 per kg atom, calculate the charge on a Ag ion in AgNo3 solution S Chand & Company Limited 37 The atomic weights of silver and copper are 108 and 63 respectively If E.C.E of copper is 3.29 × 10–7 kg/C, calculate the E.C.E of silver 38 Three equal resistance, connected in series across a source of e.m.f., together dissipates 20 W of power What would be the power dissipated if the same resistances are connected in parallel across the same source ? 39 An electric motor operates on a 50 V supply and draws a current of 12 A If the motor yields a mechanical power of 150 W, what is the percentage efficiency of the motor ? (NCERT Book) 40 A heating element is marked 210 V, 630 W What is the current drawn by the element when connected to a 210 V d.c mains ? What is the resistance of the element ? (NCERT Book) 41 A battery of emf E and internal resistance r is connected across a pure resistive device of resistance R (a) Show that the power output of the device is maximum when there is a perfect ‘matching’ between the external resistance and the source resistance, i.e., R = r (b) Determine this maximum output power 42 For a copper-iron and a chromel - alumel thermocouple plots between thermoelectric emf and the temperature θ of the hot junction are found to satisfy approximately the parabola equation : E = α θ + β θ2 with α = 14µ V/°C  2 β = – 0.04 µ V/°C  for copper - iron,  for chromel-alumel  β = + 0.002µ V/ °C  Which of the two thermocouples would you use to measure temperature in the range of about 500°C to 600°C ? (NCERT Book) and α = 41µ V/°C S Chand & Company Limited 43 A series battery of lead accumulators, each of emf 2.0 V and internal resistance 0.25 Ω, is charged by 230 V d.c mains To limit the charging current, a series resistance of 53 Ω is used in the charging circuit What is (a) the power supplied by the mains, (b) the power dissipated as heat ? (NCERT Book) 44 A steady current is passed for a certain time through three voltameters connected in series: a copper voltameter (Cu electrodes in CuSo4), a silver voltameter (Ag electrodes in AgNO3) and an iron voltameter (Fe electrodes in FeCl3) The mass of Cu dissolved off the anode of the first voltameter is found to be 437.1 g Predict the masses of silver and iron deposited on the respective cathodes of the other two voltameters Relative atomic masses of Cu, Ag and Fe are 63.54, 107.9, 55.85 respectively (NCERT Book) 45 The potential difference across the terminals of a battery of emf 12 V and internal resistance 2Ω drops to 10 V when it is connected to a silver voltameter Calculate the silver deposited at the cathode in half an hour Atomic weight of silver is 107.9 g/mol–1 (AISSCE 1998) 46 A silver and a copper voltameter are connected in series with a 12.0 V battery of negligible internal resistance 0.806 g of silver is deposited in half an hour in the silver voltameter Calculate (i) the magnitude of the current flowing in the circuit, (ii) the mass of copper deposited in the copper voltameter during same period (E.C.E of silver = 1.12 × 10–8 kg C–1 E.C.E of (AISSCE 1998) copper = 6.6 × 10–7 kg C–1) 47 A copper voltameter is connected in series with a coil of resistance 100 Ω A steady current flowing in the circuit for 10 minutes gives a deposit of 0.1g of copper in the voltameter Calculate the heat generated in the coil (E.C.E of copper = 3.3 × 10–7 kg C–1) 48 A copper voltameter is in series with a heater coil of resistance 0.1 ohm A steady current flows in the circuit for 20 minutes, and a mass of 0.99 gm of copper is deposited at the cathode If E.C.E for copper is 0.00033 gC–1 Calculate the heat generated in the coil (AISSCE Delhi 2000) S Chand & Company Limited ANSWERS R e s is ta n c e R = Here ∴ V P V = 220 V P = 100 W R= 220 × 220 100 P = V2/R Power consumed = 110 × = 484Ω 110 484 = 25W V = 12 V R = 40 Ω t = hr = 3600 s Heat produced H = V2 R t= 12 ×12 40 × 3600 = 13kJ V = 50 V I = 12 A The power drawn from the source P = VI = 50 ×12 = 600 W Since efficiency of the motor is 30% the power consumed by the winding of the motor, P′ = 70% of P 600 × 70 100 = 420 W S Chand & Company Limited R= resistance of motor winding P′ I = P = 10 MW = 10 ×106 W V = 200 kV = 200 × 103 = × 105 V R = 10km × 0.1 Power loss Ω km 420 12×12 = 2.9Ω = 1Ω = I2 R P =  R V   10×106  =  ×1  2×10  = 2500W = 2.5 kW i = A, R = 44 Ω, t = × 60 s Energy consumed H= i2 Rt = × × 44 × × 60 = 3.3 × 105 J Given E = 14 θ – 0.02 θ2 At neutral temperature, E is maximum Therefore, S Chand & Company Limited dE dθ or =0 14 – 2(0.02)θ n = 14 = 350 °C 0.04 P = 100 W, t = 1200 s Energy produced by the bulb or θn = U = Pt = 100 × 1200 = 1.2 × 105 J Let I1 and I2 be the currents when power transmission takes place at 20,000 V and at 200 V, respectively If P is the power of the generator then P I1 = 20, 000 and I2 = P/200 Thus I1 < I2 Now Power loss = I2R Since I1 < I2, power loss is less in the first case, i.e., at 20,000 V Let the required time be t Heat produced by heater = Pt Heat required by water = mc∆θ So, Pt = mc∆θ mc∆θ t= or P S Chand & Company Limited here m = kg, c = 4.2 × 103 J kg– °C–1, ∆θ = 40 °C, P = kW = 2000 W So, t= 1× 4.2 ×10 × 40 2000 = 84s 10 Let R1 and R2 be the resistances of the two lamps Then R1 = R2 = V2 P1 V = P2 = 220 × 220 100 220 × 220 200 = 484 Ω = 242 Ω Total resistance of the two lamps in series = R1 + R2 = 726 Ω V 200 current I = = A R 726 Heat produced per second in the first bulb  220  = I R1 =   × 484 = 44.4 J  726  2  220   × 242  726  Heat produced per secondin thesecond bulb =  = 22.2 J S Chand & Company Limited Here m = 2.5 g, p = 2, F = 96500 C mol–1 M = 63.5, i = 10A ∴ t= 2.5 × × 96500 63.5 × 10 E m = zit = 19 F it t= or = 760 s mF Ei Here m = 250 mg = 250 × 10 – g, i = 0.2 A F = 96500 C mol–1, E = 32 –3 ∴ t= 250 ×10 × 96500 32 × 0.2 = 3770 = 63minutes 20 According to Faraday’s first law, m = zit or, t= m zi Here, z = 0.001118 gC–1, i = 20 A, m = 10.79 g ∴ t= 10.79 0.001118 × 20 = 457.2s S Chand & Company Limited 21 t = 20 = 20 × 60 = 1200 s, m = 11.84 g z = 3.29 × 10–7 kg C–1, F = 96500 Cmol–1 Applied emf = 1.62 V Back emf = 1.43 V So, net emf in the circuit = 1.62 – 1.34 = 0.28 V Now m = zit or i= m = 11.84 = 29.99A zt 3.29 ×10 –4 ×1200 [z = 3.29 × 10–7 Kg C–1 = 3.29 × 10–4 gC–1] Resistance of voltameter = Net emf in the circuit I 0.28 = = 0.009336 29.99 = 9.34 × 10 –3 Ω 22 t = 10 = 10 × 60 s m=1g z = 3.3 × 10–7 kgC–1 = 3.3 × 10–4 gC–1 From Faraday’s first law, m = zit i= m zt = = 5.05 A 3.3 × 10 × 10 × 60 Now heat produced per second is H = i2R or –4 S Chand & Company Limited = (5.05) × 50 = 1.28 × 10 J –1 23 i = 10A, F = 96500 Cmol , M = 2.016 t = 5min = × 60 = 300 s Massof hydrogen evolved m= Mit pF = 2.016 ×10 × 300 × 96500 = 0.0313g = Volumeof hydrogen 22.4 × 0.0313 2.018 =0.347litres 24 i = 10 A; t = 77 s ; m = 0.253 g F = 96500 Cmol–1, p = From Faraday's first law, m = zit = or M = mit pF mpF 0.253 ×2 × 96500 = 10 × 77 it = 63.4g 25 We have P= V R S Chand & Company Limited or R= V = P 225× 225 40 = 1265.6 Ω = 1.27 kΩ Also P = VI or I= P  40  = A V  225  Heat produced = I Rt 225× 225  40  × 5× 60  × 40  225  = =12000Joule= 12 kJ 26 We know that R = ρl/A Here l = 20 m, A = 2.3 × 10–7 m2, V = 200 V and ρ = 4.6 × 10–7 Ω m ∴ R= Power P= = 4.6 × 10 –7 × 20 2.3 × 10 V –7 = 40 Ω R 200 × 200 40 = 1000 W S Chand & Company Limited 27 R= V P = 220 × 220 100 Power consumed = = 484 Ω 110 × 110 = 25 W 484 28 Let R1 and R2 be the resistances of the two lamps Then R1 = R2 = V P1 V = = 220 × 220 = 1210 Ω 40 220 × 220 = 2420 Ω 60 P2 Total resistance inseries R = R1 + R2 = 1210 + 2420 = 6050 Ω Current through the lamps V 220 × I= = = A R 6050 55 Heat produced per second in the first lamp = I R1 = × ×1210 = 14.4 J 55 55 Heat produced per second in the second lamp S Chand & Company Limited = I R2 = × × 2420 = 9.6 J 55 55 29 We have P = VI Here V = 100 V and I = 15A Power drawn from source, P = 100 × 15 = 1500 W Since efficiency is 25%, the power dissipated as heat is P ′ = 1500 × 75 100 = 1125W Now, P′ = I R or R= P′ = 1125 15 ×15 I ° ° 30 θi = 25 C, θn = 275 C =5Ω θ c + θi We know that θn = or 275 = (25 + θ i ) / or 550 = 25+θ i or θi = 525°C 31 If t is the smallest temperature difference that can be detected, then 40 × 10–6 t = 100 × 10–6 ⇒ t = 2.5 °C S Chand & Company Limited 32 V = 1.5 V, r = 0.2 Ω, I = 2.5 A (a) Rate of chemical energy consumption of the cell = V × I = 1.5 × 2.5 = 3.75 W (b) Rate of energy dissipation inside the cell = I2 r = (2.5)2 × 0.2 = 1.25 W (c) Rate of energy dissipation inside the resistance = VI – I2r = 3.75 – 1.25 = 2.50 W 33 q = × 105 C, e = 1.6 × 10–19 C (a) If N1 is the number of copper ions liberated then N1 × 2e = q or N1 = q = × 10 –19 = 1.25 × 10 24 2e ×1.6 × 10 (b) The number of electrons flowing in the external circuit = N1 = × 1.25 × 1024 = 2.5 × 1024 34 (i) Chemical equivalent of silver = ∴ atomic weight valency Kg-equivalent of silver = 108 kg = 108 = 108 (ii) Since 1.118 × 10–6 kg of silver is liberated by coulomb of charge, the charge required to liberate 108 kg of silver S Chand & Company Limited = 108 1.118 ×10 (iii) We have e= or = 9.65 ×10 –7 coulomb –6 F = Ne F N = 9.65 ×10 ×10 = 1.6 × 10 26 –19 coulomb (iv) Number of silver ions displaced by coulomb of charge ×10 26 18 = 6.2 × 10 9.65 ×10 (v) Valency of copper is two So the number of Cu ions displaced = = 18 × 6.2 ×10 18 = 3.1 × 10 35 According to Faraday’s I law, m m = zIt or t = zI Here m = × 10–3 kg, z = 1.118 × 10–6 kg/coulombs I = 15.0 A ∴ t= 3×10 –3 1.118 ×10 –6 ×15 = 179 s S Chand & Company Limited 36 Chemical equivalent of silver = 107.9 =107.9 Charge required to liberate kg equivalent of silver = 107.9 1.118 × 10 –6 = 9.65 × 10 coulomb ∴ Chargeon a silver ion = 9.65 ×10 6.06 × 10 = 1.6 × 10 26 –19 coulomb 37 Valency of silver = Valency of copper = E.C.E of copper = 3.29 × 10– kg/coulomb 108 Equivalent weight of silver, E1 = Equivalent weight of copper, E = z1 We have or or z2 z1 –7 3.29 ×10 63 = = = 108 = 31.5 E1 E2 108 31.5 –6 z1 = 1.2×10 kg/coulomb S Chand & Company Limited 38 Let each resistance be of magnitude r Then total resistance in series is R = r + r + r = 3r V Power P= But P = 20 W V ∴ R = V 3r = 20 3r When the resistance are connected in parallel, the effective resistance is r R′ = Power P′ = V = 3V V    3r  = 9 R′ r = 20 × = 180 W 39 V = 50 V, I = 12 A Power drawn from source, P = VI = 50 × 12 = 600 W Output power = 150 W Output Power Efficiency of motor = × 100 Input Power 150 = ×100= 25% 600 S Chand & Company Limited 40 V = 210 V, P = 630 W We have P = V I P 630 or I = = = 3A V 210 P = Further, = V2 V2 or R = R P ( 210 )2 630 See Q 14 41 (a) = 70 Ω  E   R R+r P= (b) 2 E E  E  = = r r   r +r  4r 4r Pmax =  42 We have E = αθ + dE dθ βθ = α + βθ At neutral temperature dE dθ =0 S Chand & Company Limited or α + β θn = or θn = α/β 14 = 350 °C 0.04 A thermocouple cannot be used beyond the neutral temperature because a given emf corresponds to two different values of θ So this thermocouple can not be used to read temperature above 350°C (Actual neutral temperature is 250°) For chromel - alumel thermocouple, In the case of copper-iron thermocouple θ n = 41 = – 20500 °C 0.002 So this thermocouple can be used to measure any high temperature because it shows no maximum for θ > 43 EMF of the battery = × 2.0 = 12.0 V Internal resistance of the battery = × 0.25 = 1.5 Ω θn = – Charging Current = 230 – 12 = 218 53 + 1.5 54.5 (a) Power supplied by mains = 230 × 4.0 = 920 W (b) Power dissipated as heat I2 R = (4)2 × (54.5) = 872 W 44 Chemical equivalent of copper, E1 = 63.54 = 31.77 S Chand & Company Limited E2 = Chemical equivalent of silver, E3 = 107.9 = 107.9 55.85 = 18.617 Mass of copper dissolved off the anode, m1 = 473.1 g Chemical equivalent of iron, Let the masses of silver and copper deposited on the respective cathodes be m2 and m3 respectively We have m1 m2 E1 = E2 or 473.1 31.77 = m2 107.9 or m2 = 473.1×107.9 = 1606.8 g 31.77 m1 E1 Further, or m3 = m3 = E3 or 473.1 m3 = 31.77 18.62 473.1×18.617 = 277.27 g 31.77 45 The current flowing through the voltameter is i= E –V r = 12 – 10 = 1A S Chand & Company Limited Let the mass of silver deposited on the cathode be m Then m = zit = Mit pF , where M is the molecular weight, p is the valency and F is the Faraday constant Substituting the values m= 107.9 ×1×30 × 60 1× 96500 = 2.01 g 46 (i) m = zit −3 ⇒ i= 0.806 × 10 m = = 39.98 A zt 1.12 × 10 −8 × 30 × 60 (ii) Mass of copper deposited = zit = 6.6 × 10–7 × 39.98 × 30 × 60 = 47.5 g 47 From Faraday’s first law m = zit or i= m zt = 0.1×10 –3 –8 3.3 × 10 × 10 × 60 = 0.505A S Chand & Company Limited Heat produced in the coil = i2 Rt = (0.505)2 × 100 × 10 × 60 = 1.53 × 104 J 48 i= m zt = 0.99 0.00033 × 20 × 60 = 2.5 A Heat produced = I2 Rt = (2.5)2 × 0.1 × 20 × 60 = 750 J S Chand & Company Limited