Page 1 33.1 CHAPTER – 33 THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT 1. i = 2 A, r = 25 ?, t = 1 min = 60 sec Heat developed = i 2 RT = 2 × 2 × 25 × 60 = 6000 J 2. R = 100 ?, E = 6 v Heat capacity of the coil = 4 J/k ?T = 15°c Heat liberate ? Rt E 2 = 4 J/K × 15 ? t 100 6 6 ? ? = 60 ? t = 166.67 sec = 2.8 min ? 3. (a) The power consumed by a coil of resistance R when connected across a supply v is P = R v 2 The resistance of the heater coil is, therefore R = P v 2 = 500 ) 250 ( 2 = 125 ?? (b) If P = 1000 w then R = P v 2 = 1000 ) 250 ( 2 = 62.5 ? 4. ƒ = 1 × 10 –6 ?m P = 500 W E = 250 v (a) R = P V 2 = 500 250 250 ? = 125 ? (b) A = 0.5 mm 2 = 0.5 × 10 –6 m 2 = 5 × 10 –7 m 2 R = A l ƒ = l = ƒ RA = 6 7 10 1 10 5 125 ? ? ? ? ? = 625 × 10 –1 = 62.5 m (c) 62.5 = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10 –3 × n ? n = 3 10 4 14 . 3 2 5 . 62 ? ? ? ? n = 14 . 3 8 10 5 . 62 3 ? ? ? ˜ 2500 turns ? 5. V = 250 V P = 100 w R = P v 2 = 100 ) 250 ( 2 = 625 ?? Resistance of wire R = A l ƒ = 1.7 × 10 –8 × 6 10 5 10 ? ? = 0.034 ? ? The effect in resistance = 625.034 ? ? The current in the conductor = R V = ? ? ? ? ? ? 034 . 625 220 A ? The power supplied by one side of connecting wire = 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? The total power supplied = 2 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? = 0.0084 w = 8.4 mw 6. E = 220 v P = 60 w R = P V 2 = 60 220 220 ? = 3 11 220 ? ? (a) E = 180 v P = R V 2 = 11 220 3 180 180 ? ? ? = 40.16 ˜ 40 w 10 cm Page 2 33.1 CHAPTER – 33 THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT 1. i = 2 A, r = 25 ?, t = 1 min = 60 sec Heat developed = i 2 RT = 2 × 2 × 25 × 60 = 6000 J 2. R = 100 ?, E = 6 v Heat capacity of the coil = 4 J/k ?T = 15°c Heat liberate ? Rt E 2 = 4 J/K × 15 ? t 100 6 6 ? ? = 60 ? t = 166.67 sec = 2.8 min ? 3. (a) The power consumed by a coil of resistance R when connected across a supply v is P = R v 2 The resistance of the heater coil is, therefore R = P v 2 = 500 ) 250 ( 2 = 125 ?? (b) If P = 1000 w then R = P v 2 = 1000 ) 250 ( 2 = 62.5 ? 4. ƒ = 1 × 10 –6 ?m P = 500 W E = 250 v (a) R = P V 2 = 500 250 250 ? = 125 ? (b) A = 0.5 mm 2 = 0.5 × 10 –6 m 2 = 5 × 10 –7 m 2 R = A l ƒ = l = ƒ RA = 6 7 10 1 10 5 125 ? ? ? ? ? = 625 × 10 –1 = 62.5 m (c) 62.5 = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10 –3 × n ? n = 3 10 4 14 . 3 2 5 . 62 ? ? ? ? n = 14 . 3 8 10 5 . 62 3 ? ? ? ˜ 2500 turns ? 5. V = 250 V P = 100 w R = P v 2 = 100 ) 250 ( 2 = 625 ?? Resistance of wire R = A l ƒ = 1.7 × 10 –8 × 6 10 5 10 ? ? = 0.034 ? ? The effect in resistance = 625.034 ? ? The current in the conductor = R V = ? ? ? ? ? ? 034 . 625 220 A ? The power supplied by one side of connecting wire = 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? The total power supplied = 2 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? = 0.0084 w = 8.4 mw 6. E = 220 v P = 60 w R = P V 2 = 60 220 220 ? = 3 11 220 ? ? (a) E = 180 v P = R V 2 = 11 220 3 180 180 ? ? ? = 40.16 ˜ 40 w 10 cm Thermal & Chemical Effects of Electric Current 33.2 (b) E = 240 v P = R V 2 = 11 220 3 240 240 ? ? ? = 71.4 ˜ 71 w 7. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v The resistance of bulb R = P V 2 = 100 ) 220 ( 2 = 484 ? (a) For minimum power consumed V 1 = 220 – 1% = 220 – 2.2 = 217.8 ? i = R V 1 = 484 8 . 217 = 0.45 A Power consumed = i × V 1 = 0.45 × 217.8 = 98.01 W (b) for maximum power consumed V 2 = 220 + 1% = 220 + 2.2 = 222.2 ? i = R V 2 = 484 2 . 222 = 0.459 Power consumed = i × V 2 = 0.459 × 222.2 = 102 W 8. V = 220 v P = 100 w R = P V 2 = 100 220 220 ? = 484 ? P = 150 w V = PR = 22 22 150 ? ? = 150 22 = 269.4 ˜ 270 v ? 9. P = 1000 V = 220 v R = P V 2 = 1000 48400 = 48.4 ? Mass of water = 1000 100 1 ? = 10 kg Heat required to raise the temp. of given amount of water = ms ?t = 10 × 4200 × 25 = 1050000 Now heat liberated is only 60%. So % 60 T R V 2 ? ? = 1050000 ? T 100 60 4 . 48 ) 220 ( 2 ? ? = 1050000 ? T = 60 1 6 10500 ? nub = 29.16 min. ? 10. Volume of water boiled = 4 × 200 cc = 800 cc T 1 = 25°C T 2 = 100°C ? T 2 – T 1 = 75°C Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg Q(heat req.) = MS ?? = 0.8 × 4200 × 75 = 252000 J. 1000 watt – hour = 1000 × 3600 wattsec = 1000× 3600 J No. of units = 3600 1000 252000 ? = 0.07 = 7 paise (b) Q = mS ?T = 0.8 × 4200 × 95 J No. of units = 3600 1000 95 4200 8 . 0 ? ? ? = 0.0886 ˜ 0.09 Money consumed = 0.09 Rs = 9 paise. 11. P = 100 w V = 220 v Case I : Excess power = 100 – 40 = 60 w Power converted to light = 100 60 60 ? = 36 w Case II : Power = 484 ) 220 ( 2 = 82.64 w Excess power = 82.64 – 40 = 42.64 w Power converted to light = 100 60 64 . 42 ? = 25.584 w Page 3 33.1 CHAPTER – 33 THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT 1. i = 2 A, r = 25 ?, t = 1 min = 60 sec Heat developed = i 2 RT = 2 × 2 × 25 × 60 = 6000 J 2. R = 100 ?, E = 6 v Heat capacity of the coil = 4 J/k ?T = 15°c Heat liberate ? Rt E 2 = 4 J/K × 15 ? t 100 6 6 ? ? = 60 ? t = 166.67 sec = 2.8 min ? 3. (a) The power consumed by a coil of resistance R when connected across a supply v is P = R v 2 The resistance of the heater coil is, therefore R = P v 2 = 500 ) 250 ( 2 = 125 ?? (b) If P = 1000 w then R = P v 2 = 1000 ) 250 ( 2 = 62.5 ? 4. ƒ = 1 × 10 –6 ?m P = 500 W E = 250 v (a) R = P V 2 = 500 250 250 ? = 125 ? (b) A = 0.5 mm 2 = 0.5 × 10 –6 m 2 = 5 × 10 –7 m 2 R = A l ƒ = l = ƒ RA = 6 7 10 1 10 5 125 ? ? ? ? ? = 625 × 10 –1 = 62.5 m (c) 62.5 = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10 –3 × n ? n = 3 10 4 14 . 3 2 5 . 62 ? ? ? ? n = 14 . 3 8 10 5 . 62 3 ? ? ? ˜ 2500 turns ? 5. V = 250 V P = 100 w R = P v 2 = 100 ) 250 ( 2 = 625 ?? Resistance of wire R = A l ƒ = 1.7 × 10 –8 × 6 10 5 10 ? ? = 0.034 ? ? The effect in resistance = 625.034 ? ? The current in the conductor = R V = ? ? ? ? ? ? 034 . 625 220 A ? The power supplied by one side of connecting wire = 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? The total power supplied = 2 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? = 0.0084 w = 8.4 mw 6. E = 220 v P = 60 w R = P V 2 = 60 220 220 ? = 3 11 220 ? ? (a) E = 180 v P = R V 2 = 11 220 3 180 180 ? ? ? = 40.16 ˜ 40 w 10 cm Thermal & Chemical Effects of Electric Current 33.2 (b) E = 240 v P = R V 2 = 11 220 3 240 240 ? ? ? = 71.4 ˜ 71 w 7. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v The resistance of bulb R = P V 2 = 100 ) 220 ( 2 = 484 ? (a) For minimum power consumed V 1 = 220 – 1% = 220 – 2.2 = 217.8 ? i = R V 1 = 484 8 . 217 = 0.45 A Power consumed = i × V 1 = 0.45 × 217.8 = 98.01 W (b) for maximum power consumed V 2 = 220 + 1% = 220 + 2.2 = 222.2 ? i = R V 2 = 484 2 . 222 = 0.459 Power consumed = i × V 2 = 0.459 × 222.2 = 102 W 8. V = 220 v P = 100 w R = P V 2 = 100 220 220 ? = 484 ? P = 150 w V = PR = 22 22 150 ? ? = 150 22 = 269.4 ˜ 270 v ? 9. P = 1000 V = 220 v R = P V 2 = 1000 48400 = 48.4 ? Mass of water = 1000 100 1 ? = 10 kg Heat required to raise the temp. of given amount of water = ms ?t = 10 × 4200 × 25 = 1050000 Now heat liberated is only 60%. So % 60 T R V 2 ? ? = 1050000 ? T 100 60 4 . 48 ) 220 ( 2 ? ? = 1050000 ? T = 60 1 6 10500 ? nub = 29.16 min. ? 10. Volume of water boiled = 4 × 200 cc = 800 cc T 1 = 25°C T 2 = 100°C ? T 2 – T 1 = 75°C Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg Q(heat req.) = MS ?? = 0.8 × 4200 × 75 = 252000 J. 1000 watt – hour = 1000 × 3600 wattsec = 1000× 3600 J No. of units = 3600 1000 252000 ? = 0.07 = 7 paise (b) Q = mS ?T = 0.8 × 4200 × 95 J No. of units = 3600 1000 95 4200 8 . 0 ? ? ? = 0.0886 ˜ 0.09 Money consumed = 0.09 Rs = 9 paise. 11. P = 100 w V = 220 v Case I : Excess power = 100 – 40 = 60 w Power converted to light = 100 60 60 ? = 36 w Case II : Power = 484 ) 220 ( 2 = 82.64 w Excess power = 82.64 – 40 = 42.64 w Power converted to light = 100 60 64 . 42 ? = 25.584 w Thermal & Chemical Effects of Electric Current 33.3 ?P = 36 – 25.584 = 10.416 Required % = 100 36 416 . 10 ? = 28.93 ˜ 29% ? 12. R eff = 1 8 12 ? = 2 5 i = ? ? 2 / 5 6 = 5 12 Amp. i ? 6 = (i – i ?)2 ? i ? 6 = i 2 2 5 12 ? ? 8i ? = 5 24 ? i ? = 8 5 24 ? = 5 3 Amp i – i ? = 5 3 5 12 ? = 5 9 Amp (a) Heat = i 2 RT = 60 15 2 5 9 5 9 ? ? ? ? = 5832 2000 J of heat raises the temp. by 1K 5832 J of heat raises the temp. by 2.916K. (b) When 6 ? resistor get burnt R eff = 1 + 2 = 3 ?? i = 3 6 = 2 Amp. Heat = 2 × 2 × 2 ×15 × 60 = 7200 J 2000 J raises the temp. by 1K 7200 J raises the temp by 3.6k 13. ? = 0.001°C a = – 46 × 10 –6 v/deg, b = –0. 48 × 10 –6 v/deg 2 Emf = a BlAg ? +(1/2) b BlAg ? 2 = – 46 × 10 –6 × 0.001 – (1/2) × 0.48 × 10 –6 (0.001) 2 = – 46 × 10 –9 – 0.24 × 10 –12 = – 46.00024 × 10 –9 = – 4.6 × 10 –8 V ? 14. E = a AB ? + b AB ? 2 a CuAg = a CuPb – b AgPb = 2.76 – 2.5 = 0.26 ?v/°C b CuAg = b CuPb – b AgPb = 0.012 – 0.012 ?vc = 0 E = a AB ? = (0.26 × 40) ?V = 1.04 × 10 –5 V 15. ? = 0°C a Cu,Fe = a Cu,Pb – a Fe,Pb = 2.76 – 16.6 = – 13.8 ?v/°C B Cu,Fe = b Cu,Pb – b Fe,Pb = 0.012 + 0.030 = 0.042 ?v/°C 2 Neutral temp. on – b a = 042 . 0 8 . 13 °C = 328.57°C 16. (a) 1eq. mass of the substance requires 96500 coulombs Since the element is monoatomic, thus eq. mass = mol. Mass 6.023 × 10 23 atoms require 96500 C 1 atoms require 23 10 023 . 6 96500 ? C = 1.6 × 10 –19 C (b) Since the element is diatomic eq.mass = (1/2) mol.mass ? (1/2) × 6.023 × 10 23 atoms 2eq. 96500 C ? 1 atom require = 23 10 023 . 6 2 96500 ? ? = 3.2 × 10 –19 C 17. At Wt. At = 107.9 g/mole I = 0.500 A E Ag = 107.9 g [As Ag is monoatomic] Z Ag = f E = 96500 9 . 107 = 0.001118 M = Zit = 0.001118 × 0.5 × 3600 = 2.01 1 ? ? i ii ? 6 6 ? ? 2 ? ? Page 4 33.1 CHAPTER – 33 THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT 1. i = 2 A, r = 25 ?, t = 1 min = 60 sec Heat developed = i 2 RT = 2 × 2 × 25 × 60 = 6000 J 2. R = 100 ?, E = 6 v Heat capacity of the coil = 4 J/k ?T = 15°c Heat liberate ? Rt E 2 = 4 J/K × 15 ? t 100 6 6 ? ? = 60 ? t = 166.67 sec = 2.8 min ? 3. (a) The power consumed by a coil of resistance R when connected across a supply v is P = R v 2 The resistance of the heater coil is, therefore R = P v 2 = 500 ) 250 ( 2 = 125 ?? (b) If P = 1000 w then R = P v 2 = 1000 ) 250 ( 2 = 62.5 ? 4. ƒ = 1 × 10 –6 ?m P = 500 W E = 250 v (a) R = P V 2 = 500 250 250 ? = 125 ? (b) A = 0.5 mm 2 = 0.5 × 10 –6 m 2 = 5 × 10 –7 m 2 R = A l ƒ = l = ƒ RA = 6 7 10 1 10 5 125 ? ? ? ? ? = 625 × 10 –1 = 62.5 m (c) 62.5 = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10 –3 × n ? n = 3 10 4 14 . 3 2 5 . 62 ? ? ? ? n = 14 . 3 8 10 5 . 62 3 ? ? ? ˜ 2500 turns ? 5. V = 250 V P = 100 w R = P v 2 = 100 ) 250 ( 2 = 625 ?? Resistance of wire R = A l ƒ = 1.7 × 10 –8 × 6 10 5 10 ? ? = 0.034 ? ? The effect in resistance = 625.034 ? ? The current in the conductor = R V = ? ? ? ? ? ? 034 . 625 220 A ? The power supplied by one side of connecting wire = 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? The total power supplied = 2 034 . 0 034 . 625 220 2 ? ? ? ? ? ? ? ? = 0.0084 w = 8.4 mw 6. E = 220 v P = 60 w R = P V 2 = 60 220 220 ? = 3 11 220 ? ? (a) E = 180 v P = R V 2 = 11 220 3 180 180 ? ? ? = 40.16 ˜ 40 w 10 cm Thermal & Chemical Effects of Electric Current 33.2 (b) E = 240 v P = R V 2 = 11 220 3 240 240 ? ? ? = 71.4 ˜ 71 w 7. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v The resistance of bulb R = P V 2 = 100 ) 220 ( 2 = 484 ? (a) For minimum power consumed V 1 = 220 – 1% = 220 – 2.2 = 217.8 ? i = R V 1 = 484 8 . 217 = 0.45 A Power consumed = i × V 1 = 0.45 × 217.8 = 98.01 W (b) for maximum power consumed V 2 = 220 + 1% = 220 + 2.2 = 222.2 ? i = R V 2 = 484 2 . 222 = 0.459 Power consumed = i × V 2 = 0.459 × 222.2 = 102 W 8. V = 220 v P = 100 w R = P V 2 = 100 220 220 ? = 484 ? P = 150 w V = PR = 22 22 150 ? ? = 150 22 = 269.4 ˜ 270 v ? 9. P = 1000 V = 220 v R = P V 2 = 1000 48400 = 48.4 ? Mass of water = 1000 100 1 ? = 10 kg Heat required to raise the temp. of given amount of water = ms ?t = 10 × 4200 × 25 = 1050000 Now heat liberated is only 60%. So % 60 T R V 2 ? ? = 1050000 ? T 100 60 4 . 48 ) 220 ( 2 ? ? = 1050000 ? T = 60 1 6 10500 ? nub = 29.16 min. ? 10. Volume of water boiled = 4 × 200 cc = 800 cc T 1 = 25°C T 2 = 100°C ? T 2 – T 1 = 75°C Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg Q(heat req.) = MS ?? = 0.8 × 4200 × 75 = 252000 J. 1000 watt – hour = 1000 × 3600 wattsec = 1000× 3600 J No. of units = 3600 1000 252000 ? = 0.07 = 7 paise (b) Q = mS ?T = 0.8 × 4200 × 95 J No. of units = 3600 1000 95 4200 8 . 0 ? ? ? = 0.0886 ˜ 0.09 Money consumed = 0.09 Rs = 9 paise. 11. P = 100 w V = 220 v Case I : Excess power = 100 – 40 = 60 w Power converted to light = 100 60 60 ? = 36 w Case II : Power = 484 ) 220 ( 2 = 82.64 w Excess power = 82.64 – 40 = 42.64 w Power converted to light = 100 60 64 . 42 ? = 25.584 w Thermal & Chemical Effects of Electric Current 33.3 ?P = 36 – 25.584 = 10.416 Required % = 100 36 416 . 10 ? = 28.93 ˜ 29% ? 12. R eff = 1 8 12 ? = 2 5 i = ? ? 2 / 5 6 = 5 12 Amp. i ? 6 = (i – i ?)2 ? i ? 6 = i 2 2 5 12 ? ? 8i ? = 5 24 ? i ? = 8 5 24 ? = 5 3 Amp i – i ? = 5 3 5 12 ? = 5 9 Amp (a) Heat = i 2 RT = 60 15 2 5 9 5 9 ? ? ? ? = 5832 2000 J of heat raises the temp. by 1K 5832 J of heat raises the temp. by 2.916K. (b) When 6 ? resistor get burnt R eff = 1 + 2 = 3 ?? i = 3 6 = 2 Amp. Heat = 2 × 2 × 2 ×15 × 60 = 7200 J 2000 J raises the temp. by 1K 7200 J raises the temp by 3.6k 13. ? = 0.001°C a = – 46 × 10 –6 v/deg, b = –0. 48 × 10 –6 v/deg 2 Emf = a BlAg ? +(1/2) b BlAg ? 2 = – 46 × 10 –6 × 0.001 – (1/2) × 0.48 × 10 –6 (0.001) 2 = – 46 × 10 –9 – 0.24 × 10 –12 = – 46.00024 × 10 –9 = – 4.6 × 10 –8 V ? 14. E = a AB ? + b AB ? 2 a CuAg = a CuPb – b AgPb = 2.76 – 2.5 = 0.26 ?v/°C b CuAg = b CuPb – b AgPb = 0.012 – 0.012 ?vc = 0 E = a AB ? = (0.26 × 40) ?V = 1.04 × 10 –5 V 15. ? = 0°C a Cu,Fe = a Cu,Pb – a Fe,Pb = 2.76 – 16.6 = – 13.8 ?v/°C B Cu,Fe = b Cu,Pb – b Fe,Pb = 0.012 + 0.030 = 0.042 ?v/°C 2 Neutral temp. on – b a = 042 . 0 8 . 13 °C = 328.57°C 16. (a) 1eq. mass of the substance requires 96500 coulombs Since the element is monoatomic, thus eq. mass = mol. Mass 6.023 × 10 23 atoms require 96500 C 1 atoms require 23 10 023 . 6 96500 ? C = 1.6 × 10 –19 C (b) Since the element is diatomic eq.mass = (1/2) mol.mass ? (1/2) × 6.023 × 10 23 atoms 2eq. 96500 C ? 1 atom require = 23 10 023 . 6 2 96500 ? ? = 3.2 × 10 –19 C 17. At Wt. At = 107.9 g/mole I = 0.500 A E Ag = 107.9 g [As Ag is monoatomic] Z Ag = f E = 96500 9 . 107 = 0.001118 M = Zit = 0.001118 × 0.5 × 3600 = 2.01 1 ? ? i ii ? 6 6 ? ? 2 ? ? Thermal & Chemical Effects of Electric Current 33.4 18. t = 3 min = 180 sec w = 2 g E.C.E = 1.12 × 10 –6 kg/c ? 3 × 10 –3 = 1.12 × 10 –6 × i × 180 ? i = 180 10 12 . 1 10 3 6 3 ? ? ? ? ? = 2 10 72 . 6 1 ? ˜ 15 Amp. 19. L 4 . 22 H 2 ? 2g 1L ? 4 . 22 2 m = Zit 4 . 22 2 = T 5 96500 1 ? ? ? T = 5 96500 4 . 22 2 ? = 1732.21 sec ˜ 28.7 min ˜ 29 min. 20. w 1 = Zit ? 1 = 3600 5 . 1 2 96500 3 mm ? ? ? ? ? mm = 3600 5 . 1 2 96500 3 ? ? ? = 26.8 g/mole 2 1 E E = 2 1 w w ? ? ? ? ? ? ? 3 mm 9 . 107 = 1 w 1 ? w 1 = 8 . 26 3 9 . 107 ? = 12.1 gm 21. I = 15 A Surface area = 200 cm 2 , Thickness = 0.1 mm Volume of Ag deposited = 200 × 0.01 = 2 cm 3 for one side For both sides, Mass of Ag = 4 × 10.5 = 42 g Z Ag = F E = 96500 9 . 107 m = ZIT ? 42 = T 15 96500 9 . 107 ? ? ? T = 15 9 . 107 96500 42 ? ? = 2504.17 sec = 41.73 min ˜ 42 min 22. w = Zit 2.68 = 60 10 i 96500 9 . 107 ? ? ? ? I = 6 9 . 107 965 68 . 2 ? ? = 3.99 ˜ 4 Amp Heat developed in the 20 ? resister = (4) 2 × 20 × 10 × 60 = 192000 J = 192 KJ ? 23. For potential drop, t = 30 min = 180 sec V i = V f + iR ? 12 = 10 + 2i ? i = 1 Amp m = Zit = 60 30 1 96500 9 . 107 ? ? ? = 2.01 g ˜ 2 g 24. A= 10 cm 2 × 10 –4 cm 2 t = 10m = 10 × 10 –6 Volume = A(2t) = 10 × 10 –4 × 2 × 10 × 10 –6 = 2 × 10 2 × 10 –10 = 2 × 10 –8 m 3 Mass = 2 × 10 –8 × 9000 = 18 × 10 –5 kg W = Z × C ? 18 × 10 –5 = 3 × 10 –7 × C ? q = 7 5 10 3 10 18 ? ? ? ? = 6 × 10 2 V = q W = ? W = Vq = 12 × 6 × 10 2 = 76 × 10 2 = 7.6 KJ ? ? ? ? ? 20 < >Read More
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