### 15-THERMAL _ CHEMICAL EFFECTS OF CURRENT _THERMOELECTRICITY.doc

• 1. THERMAL & CHEMICAL EFFECTS OF CURRENT &THERMOELECTRICITY HEATING EFFECT OF CURRENT, JOULE'S LAW (i) Whenever work is transformed into heat or heat into work, the quantity of work done is mechanically equivalent to the quantity of heat. (ii) Thus Work  Heat. or Work = a constant × heat or W = JH Where J is known as Joule's constant or mechanical equivalent of heat. It is defined as that mechanical work which produces unit calorie of heat. (iii) Expression for heat produced in a conductor due to current flow through it- A i R B (a) Let the potential difference between the points A and B of a conductor is V, on account of which a current i flows through it. (b) As potential difference is the work done per unit test charge  W = QV But Q = it  W = Vit But heat developed H = 2 W Vit i RT J J J   Where R is the resistance of wire. (c) If i i is in ampere, V is in volt and R is in ohm, then W = Vit Joule = Vit × 107 ergs.  Heat developed H = Vit J calories Vit 0.24 Vit cals. 4.2   = 0.24 i2Rt cals. JOULE'S LAWS OF HEATING EFFECTS OF CURRENT (i) The heat developed in a conductor is given by H = i2Rt Joule = 0.24 i2Rt cals (ii) The amount of heat developed in a conductor, in a given time, is directly proportional to the square of the current. i.e. 2 H i  , when R and t are constants. (iii) The amount of heat developed in a conductor by a given current in a given time is directly proportional to the resistance of the conductor. i.e. H R  , when i and t are constants. (iv) The amount of heat developed in a given conductor due to a given current is proportional to the time of flow of the current. i.e. H t  , when i and R are constants. ELECTRICAL ENERGY OR WORK If Q units of charge be carried between two points differing in potential by V, then electrical work done is W = Q × V Joule = Q × V × 107 ergs
• 2. POWER (i) The rate at which work is done is defined as power Q V Power Vi t     = i2R 2 V R  (ii) Thus electric power = potential difference × current i.e. 1 watt = 1 Volt × 1 Ampere = 1 Joule/sec = 107 ergs/sec (iii) The practical unit of power is kilowatt 1 kilowatt = 1000 watt (iv) Horse-power = 550 ft-lbs per sec. = 550 × 12 × 2.54 ×453.6 gm-cm/sec = 746 × 107 ergs/sec = 746 watt. Illustration 1: A coil of resistance 2 is immersed in 1 kgm of water and is connected to the terminals of the battery of internal resistance 4  and emf 6 volt for 3 minutes. The increase in temperature of water is : (A) 0.93C (B) 0.085C (C) 1.92C (D) 4.31C Solution: Heat developed in coil = Heat absorbed by water 2 Rt i ms t J = D 2 2 i Rt E Rt t Jms R r Jms æ ö D = = ç ÷ è ø + 2 3 6 2 180 0.085 C 4 2 4.2 1 10 ´ æ ö = ´ » ° ç ÷ è ø + ´ ´ Illustration 2: A Daniel cell has an emf of 1.08 volt and internal resistance 0.5. It is successively connected to two wires whose resistances are 2 and 3. The ratio of the amounts of heat developed in two wires will approximately be : (A) 4 : 3 (B) 3 : 4 (C) 2 : 3 (D) 3 : 2 Solution: Heat developed in a wire H = i2Rt or 2 2 E Rt (R r) + 2 R H (R r) µ + 2 1 1 2 2 2 1 H R R r H R R r æ ö æ ö + = ´ ç ÷ ç ÷ + è ø è ø 2 2 3.5 3 2.5 æ ö = ´ ç ÷ è ø 2 49 4 : 3 3 25 = ´ »
• 3. Illustration 3: For tungsten wire, the average temperature coefficient above 20C is 5.1  10–3C–1. The filament of an electric lamp has a cold resistance of 9.7 at 20C. When glowing it has a resistance of 121. Its temperature while glowing is : (A) 2250C (B) 2190C (C) 2270C (D) 2430C Solution: R1 = R20 (1 +  t) or 121 = 9.7  (1 + 5  10–3 t) t 2250 C = ° above 20C  the temperature of glowing = 2250 + 20 = 2270C Illustration 4: If a cell has an e.m.f. of 1.08 volt and internal resistance 0.5. Its terminals are connected to two wires of 1 and 2 in parallel. The current in each wire will be : (in ampere) (A) 0.617, 0.308 (B) 0.412, 0.206 (C) 0.824, 0.412 (D) 2.618, 1.309 Solution: Total current 1 2 1 2 E 1.08 1.08 6 i R R 1 2 7 r 2 3 R R ´ = = = + + + 1 i R µ  current in 1 resistance 2 2 1.08 6 i 0.617 amp. 3 3 7 ´ = = ´ = current in 2 resistance i 1.08 6 0.308 amp. 3 7 3 ´ = = = ´ Illustration 5: In the above problem the ratio heats generated in the two wires will be : (A) 1 : 2 (B) 2 : 1 (C) 1 : 3 (D) 3 : 1 Solution: Heat generated 2 V H t R = In parallel combination V = constant 1 2 2 1 H R 2 2:1 H R 1 = = = (v) Kilo-watt-hour or Board of Trade (B.O.T.) unit- (a) Energy consumed in a given time is the product of power and time (b) When power of one watt is consumed for an interval of one hour, then energy consumed = 1 watt × 1 hour = 1 watt – hour = 0.001 kilo-watt-hour (c) Board of Trade (B.O.T.) unit of electric power is kilo-watt-hour by which the consumption of electric energy is measured and charged by power supply authorities. 1 kilo watt-hour = 103 × 60 × 60 watt-sec = 36 × 105 Joule
• 4. (d) Rule for calculation of cost- (i) Number of B.O.T. units = Sum of wattages of all systems× time (in hour) 1000 (ii) Total cost = Number of B.O.T. units × rate of charge per unit (iii) Total cost = 2 2 V ( P)or( Vi) or( i R)or time in hour rate per unit R 1000             Illustration 6: A dwelling house is installed with 15 lamps, each of resistance 103 and 4ceiling fans each driven by 1/8th horse-power motor. If the lamps and fans are run on an average for 6 hours daily, then the number of B.O.T. units consumed by lamps in a month of 31 days will be: (A) 135 (B) 150 (C) 165 (D) 180 Solution: Number of B.O.T. units consumed by 15 lamps 2 V 15 6 31 R 1000 ´ ´ ´ = ´ 3 220 220 15 6 31 135 10 1000 ´ ´ ´ ´ = » ´ Illustration 7: In the above problem the number of B.O.T. units consumed by the fans in the month will be : (A) 193 (B) 173 (C) 143 (D) 113 Solution: No. of B.O.T. unit consumed by 4 fans 1 746 6 31 4 143 3 1000 ´ ´ ´ ´ = » ´ Illustration 8: In Illustration 6, the total cost of electric power consumed in the month, at the rate of Rs. 1.40 per B.O.T. unit, will be : (A) Rs. 279/- (B) Rs. 389/- (C) Rs. 421/- (D) Rs. 538/- Solution: Total number of B.O.T. units consumed = 135 + 143 = 278 cost of electric power consumed = 278  1.40 = Rs. 389.20 or  389/- CHEMICAL EFFECTS OF CURRENT Definitions of various terms (i) Electrolysis: The process of splitting up or decomposing a liquid by passing an electric current through it, is defined as electrolysis. (ii) Electrolyte: The compound, whether fused or in solution, which undergoes decomposition by an electric current, is defined as electrolyte. (iii) Anions and Kations: The decomposed substance appears in the form of ions. The ions appearing at the anode are known as anions and those at the cathode are known as cations.
• 5. (iv) Ionisation: The phenomenon of separation of a molecule into oppositely charged ions is known as ionisation. (v) Equivalent weight- The ratio of the atomic weight of an element to its valency is defined as its equivalent weight. (vi) Atomic weight: The ratio of the average mass of an atom of an element of the mass of an atom of hydrogen, taken as 1.008, is defined as the atomic weight of that element. (vii) Valency: The valency of an element is the number of atoms of hydrogen or chlorine which combines with or is displaced by one atom of the element. (viii) Molecular weight: The molecular weight of a substance is the ratio of the mass of one molecule of the substance to the mass of an atom of oxygen which is taken as 16. (ix) Gram-equivalent: It is the weight in grams of an element which will combine with or replace 1 gm of hydrogen. Gram-equivalent = Gram- atom valency (x) Gram-molecule: The molecular weight of any substance. expressed in grams is defined as gram-molecule of that substance. (xi) Avogadro number: The number of molecules of a substance in its one-gram- molecule is known as Avogadro number. (xii) Normal solution: A solution containing one gram-equivalent of the solute per litre is called a normal solution. FARADAY'S LAWS OF ELECTROLYSIS (i) First law: The towal mass of ions liberated at an electrode, during electrolysis, is proportional to the quantity of electricity which passes through the electrolyte. i.e. m Q But Q =it  m it   Hence the first law may also be stated as follows The mass of ions liberated at an electrode during electrolysis is proportional to (a) the current following through the electrolyte, and (b) the time for which the current flows (ii) Second law: If same quantity of electricity is passed through different electrolytes, the masses of the substances (ions) deposited at the respective cathodes are directly proportional to their chemical equivalents (equivalent weights). i.e. m E  (chemical equivalent) (iii) Electro-chemical equivalent (E.C.E.): (a) The electro-chemical equivalent of an element is its mass in grams deposited on the electrode by the passage of 1 coulomb of charge through it i.e. by the passage of 1 ampere current for 1 second. (b) According to Faraday's first law- m i.t  or m = Z i.t Where Z is the constant of proportionality known as the electro- chemical equivalent of the substance. It is numerically equal to the mass in grams of the element deposited when a unit current flows in unit time. (c) According to Faraday's first law m = ZQ If same charge is passed two electrolytes, the 1 1 2 2 m Z m Z 
• 6. According to Faraday's second law 1 1 2 2 m E m E  1 1 2 2 m Z m Z   or Z2 = 2 1 1 E Z E  (d) E.C.E. of any substance = E.C.E. of hydrogen × chemical equivalent of the substance. (e) If silver is taken as the standard substance for which E.C.E. is 0.001118 gm per coulomb, then E.C.E. of any substance = Chemical equivalent of substance 0.001118 chemical equivalent of silver  (iv) Faraday: (a) The quantity of electricity (i.e. charge) required to liberate a gram equivalent of a substance during electrolysis. (b) As derived above 1 1 2 2 E Z E or constant F E Z Z    The constant F is known as one Faraday. For example for copper E = 31.5 gm and Z = 0.000329 gmC–1 31.5 F 96500 coulomb 0.000329    (c) According to Faraday's first law m = Z.i.t. = ZQ If p is the valency of the element, then electrons has to flow through the solution to deposit one atom.  charge required to deposit 1 mol of the substance = Npe, where N = Avogadro number and m = M  M = ZNpe M 1 E E Z p Ne Ne F     Where F = Ne = Faraday constant or F = 6.0229 × 1023 × 1.602 × 10–19 = 96487C  96500C Illustration 9: In producing chlorine through electrolysis 100 KW power at 125 volt is being consumed. If the E.C.E. of chlorine is 0.367  10–6 kg/coul, then the mass of chlorine liberated per minute will be: (A) 16.3  10–4 kg (B) 17.61  10–3 kg (C) 18.2  10–3 kg (D) 10–4 kg Solution: Mass of chlorine liberated P Zit Z t V = = 6 5 3 0.367 10 10 60 17.61 10 125 - - ´ ´ ´ = = ´ Illustration 10: The chemical equivalents of copper and silver are 32 and 108 respectively. When copper and silver voltammeters are connected in series and electric current is passed through them for sometime, 1.6 gm of copper is deposited. The mass of silver deposited will be :
• 7. (A) 3.5 gm (B) 2.8 gm (C) 5.4 gm (D) none of these Solution: 1 1 2 2 E m E m = 2 2 1 1 E 108 m m 1.6 5.4 gm E 32 = ´ = ´ = Illustration 11: A silver and a copper voltmeters are connected across a 6 volt battery of negligible resistance. In half an hour, 1 gm of copper and 2 gm of silver are deposited. The rate at which energy is supplied by the battery will approximately be : (Given E.C.E. of copper = 3.294  10–4 g/C & E.C.E. of silver = 1.118  10–3 g/C) (A) 64 watt (B) 32 watt (C) 96 watt (D) 16 watt Solution: m m Zit or i Zt = = For silver voltmeter 1 1 3 1 m 2 i 0.994 amp Z t 1.118 10 1800 - = = = ´ ´ For copper voltmeter 2 2 4 2 m 1 i 1.687 amp. Z t 3.294 10 1800 - = = = ´ ´  power of circuit = V(i1 + i2) = 6  (0.994 + 1.687) = 6  2.681  16 W. SOME IMPORTANT POINTS (i) Capacity of a cell- It is expressed in kilowatt hours which involves both current and the voltage. (ii) Energy capacity- It is expressed in kilowatt hours which involves both current and the voltage. (iii) Efficiency- (a) It is the ratio of discharging capacity to the charging capacity. (b) If discharge takes place slowly, the maximum available energy is 80% of that spent in charging. (c) If the cell is short circuited, the discharge takes place suddenly and the cell is spoiled. APPLICATIONS OF ELECTROLYSIS The phenomenon of electrolysis has been used as a valuable tool in (i) identifying the components of certain liquids (ii) making accurate measurement of current (iii) calibrating an ammeter (iv) determining the electro-chemical equivalents of elements (v) electroplating (vi) producing pure metals (vii) electrolying THERMOELECTRICITY-SEEBECK EFFECT
• 8. (i) Thermocoule: If two wires of different metals are joined at their ends so as to from two junctions, then the resulting arrangement is called a thermocouple. (ii) Seebeck effect: (a) When the two junctions of a thermocouple are maintained at different temperatures, then a current starts flowing through the wires. This is called Seebeck effect and the e.m.f. developed in the circuit is called thermo-emf. (b) The Seebeck effect is perfectly reversible, i.e. if the hot and the cold junctions are interchanged, the direction of current is reversed. (c) The thermo-emf is of the order of a few micro-volts per degree temperature difference. THERMOELECTRIC SERIES (i) Seebeck arranged a number of metals in the form of a series called thermoelectic series. The arrangement of some of the metals forming the series are- Sb, Fe, Zn, Ag, Au, Mo, Cr, Sn, Pb, Hg, Mn, Cu, Co, Ni, Bi. (ii) In the above series, current flows through the cold junction from the metal which appear earlier to the metal which appears later. (iii) Greater the separation of the two metals in the series, greater is the thermo emf generated. VARIATION OF THERMO-EMF WITH TEMPERATURE- DIFFERENCE (i) If the cold junction is kept at 0o and the temperature of hot junction (t) is gradually increased, it is found that the thermo emf e first increases, attains a maximum value and then decreases to become zero again (Fig.) ti tn 0o C Thermo emf V Temperature of hot junction t (ii) If the temperature is further increased, the emf changes direction. (iii) Netural temperature (tn)- The temperature at which thermo-emf is maximum is defined as neutral temperature. tn is fixed for a given thermo-couple. (iv) Temperature of inversion (ti)- The temperature at which thermo-emf changes its direction is called the temperature of inversion. ti is an much above tn as the tenoeratyre if cikd hybctuib us bekiw tn. (v) Mathematically 2 1 e t t 2     where  and  are constants for a given thermo- couple. Illustration 12: The e.m.f. of a thermocouple, one junction of which is kept at 0C, is given by e = at + bt2 The neutral temperature will be: (A) a b (B) a b -
• 9. (C) a 2b (D) a 2b - Solution: At neutral temperature e = maximum de 0 dt = or n a 2bt 0 + = n a t 2b = - . Illustration 13: In the above problem, the temperature of inversion will be : (A) a b (B) a b - (C) a 2b (D) a 2b - Solution: o i n t t t 2 + = i n o t 2t t = - or i a a t 2 0 2b b = ´ - - = - Illustration 14: In Illustration 12, the Peltier co-efficient will be : (A) (t – 273) (a + 2bt) (B) (t + 273) (a – 2bt) (C) (t – 273) (a – 2bt) (D) (t – 273) (a – 2bt) Solution: Peltier coeff. de T dT p = tC = T – 273 2 e a(T 273) b(T 273) = - + - d.w.r.t. T de a 2b(T 273) dT = + - de T T[a 2b(T 273)] dT p = = + - or (t 273) (a 2bt) p = + + . Illustration 15: In Illustration 12, the Thomson co-efficient will be : (A) 2a(t + 273) (B) 2a(t – 273) (C) 2b(t + 273) (D) 2b(t – 273) Solution: Thomson co-efficient 2 2 d e T dT s = de a 2b(T 273) dT = + - d.w.r.t. T 2 2 d e 2b dT = 2 2 d e T 2b(t 273) dT s = = +
• 10. SEEBECK COEFFICIENT The rate of change of thermo emf with temperature is called thermo-electric power or Seebeck coefficient (S). de X t dt       When t = tn, e is maximum de 0 dt   n n t 0 or t          PELTIER EFFECT (i) This effect is the converse of Seebeck effect. (ii) If a current is passed through a junction of two dissimilar metals, heat is eigher absorbed or evolved at the junction. (iii) On reversing the direction of current, the heating effect is also reversed. If the seebeck current is in a certain direction through the hot junction, then an external current sent in the same direction through this junction produces a cooling at this junction and a heating at the other junction. (iv) Peltier coefficient ()- It is the amount of heat absorbed or evolved at a junction per second when a current of one ampere flows through it. de T dT   THOMSON EFFECT (i) An emf is developed between two parts of a single metal if they are at different temperatures. This is called Thomson effect. (ii) Thomson coefficient- If de is the potential difference between two points in a metal which have a temperature difference dt, then the ratio 2 2 de d e T dT dT    is defined as the thomson coefficient. (iii) If one part of a metal is at a higher temperature than the other, the free electrons at hot part will have more kinetic energy and as a result these electrons will diffuse faster towards colder part than the electrons from cold to the hot part. This would result in the net transfer of electrons setting an electric current in the metal. APPLICATIONS OF THERMOELECTRIC EFFECTS (i) Measurement of temperature (ii) Detection of heat radiations (iii) Refrigeration (iv) Power generation.
• 11. OBJECTIVE ASSIGNMENT 1. Consider the two statements A and B (i) the neutral temperature does not depend on temperature of cold junction (ii) the inversion temperature does not depend on temperature of cold junction (A) both A and B are correct (B) A is correct but B is wrong (C) A is wrong but B is correct (D) both A and B are wrong. Sol: (B) 2. The neutral temperature of a thermocouple with cold junction at 20oC is 220oC. Its temperature of inversion is (A) 420oC (B) 120oC (C) 110oC (D) 440oC Sol: (A) We know n c i n t t t t - = - tc = neutral temperature ti = inversion temperature i n c t 2t t 440 20 420 C = - = ° - ° = ° 3. For a thermocouple if the cold junction is maintained at 0oC the inversion temperature is 680oC. Its Neutral temperature is (A) 1360oC (B) 680oC (C) 340oC (D) 170oC Sol: (C) We know inversion temperature = 2(neutral temperature) Neutral temperature 680 340 C 2 = = ° 4. Consider the following two statements A and B and identify the correct choice in the given answers. (i) Thermo e.m.f. is minimum at neutral temperature of a themocouple. (ii) When two junctions made of two different metallic wires are maintained at different temperatures, an electric current is generated in the circuit (A) A is false and B is true (B) A is true and B is false (C) Both A and B are false (D) Both A and B are true Sol: (A) At neutral temperature of a thermocouple, thermo emf should be maximum, but not minimum  A is false and B is true. 5. According to the Faraday's Law of electrolysis, the mass deposited or liberated at an electrode is proportional to : (A) m Q  (B) 2 m Q  (C) 2 m l  (D) m does not depend on Q Sol: (A) 6. Two resistors having equal resistances are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature changes. In a given time interval. (A) equal amounts of thermal energy must be produced in the resistors (B) unequal amounts of tgernak ebergt nat be oridyced (C) the temperature must rise equally in the resistors
• 12. (D) the temperature must rise unequally in the resistors. Sol: (A) 7. A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if (A) both the length and radius of the wire are halved (B) both the length and radius of the wire are doubled (C) the radius of the wire is doubled (D) the length of the wire is doubled Where the time remains the same in both the cases. Sol: (B) 2 2 2 2 V V r r H R L L p = = µ r 2 1 2 2 1 r L H H' r L H 1 2 1 H' 4 1 2 æ ö æ ö = ´ ç ÷ ç ÷ è ø è ø Þ = ´ = H' 2H Þ = 8. Which of the following plots may represent the thermal energy produced in a resistor in a given time as a function of the electric current? (A) a (B) b (C) c (D) d d c b a i Sol: (A) 2 H Rt U = I = 2 U Þ µ I It represents a parabola. 9. Two heater wires, made of the same material and having the same length and the same radius, are first connected in series and then in parallel to a constant potential difference. If the rates of heat produced in the two cases are Hs and Hp respectively, then s p H H will be : (A) 1 2 (B) 2 (C) 1 4 (D) 4 Sol: (C) Since 1 2 1 2 1 2 l l , r r & = = r = r So resistance 1 2 R R R = = 2 2 s eq V V H R 2R = = 2 2 p eq V V H R R 2 = =
• 13. 2 s p 2 V 2R H 1 H 4 V R 2 æ ö ç ÷ è ø = = æ ö ç ÷ ç ÷ ç ÷ è ø 10. A hot electric iron has a resistance of 80 and is used on a 200 V source. The electrical energy spent, if it is used for 2hr. will be (A) 800 Wh (B) 1000 Wh (C) 2000 Wh (D) 8000 Wh Sol: (B) 2 2 V (200) (P) 500W R 80 = = = Therefore energy spent = power  time = 500  2 = 1000 Wh. 11. Antimony and Bismuth are usually used in a thermocouple, because (A) higher thermo e.m.f. is produced (B) lower thermo e.m.f. is produced (C) constant thermo e.m.f. is produced (D) negative thermo e.m.f. is produced Sol: Since antimony and bismuth form the pair of metals producing higher thermo emf in comparison with other pairs of thermocouple. Therefore pair is usually used in thermocouple. 12. At 40C resistance of platinum is 3.14 and at 100C its resistance is 3.76. Its temperature coefficient is (A) 3.75  10–1/C (B) 3.75 × 10–2/C (C) 0.00329/C (D) None of these Sol: (C) 2 1 1 2 1 R R R (t t ) - a = ´ - 3.76 3.14 0.00329 3.14 (100 40) - Þ a = = ´ - 13. Two resistance filaments of same length are connected first in series and then in parallel. Fine the ratio of power dissipated in both cases assuming that equal current flows in the main circuit. (A) 1 : 4 (B) 4 : 1 (C) 1 : 2 (D) 2 : 1 Sol: (D) Required ratio Series Combination R R R R i/2 i/2 i Parallel Combination
• 14. 2 2 2 2 i R i R 2 :1 i i R R 2 2 + = æ ö æ ö + ç ÷ ç ÷ è ø è ø 14. Thermocouple thermometer is based on (A) Seeback effect (B) Crompton effect (C) Peltier effect (D) Photoelectric effect. Sol: (A) 15. A wire has a resistance of 3.1 at 30C and a resistance 4.5 at 100oC. The temperature coefficient of resistance of the wire (A) 0.0064oC–1 (B) 0.0034oC–1 (C) 0.0025oC–1 (D) 0.0012oC–1 Sol: (A) 100 30 30 R R 4.5 3.1 R t 3.1 (100 30) - - a = = ´ D ´ - 1 1.4 1.4 0.0064 C 3.1 70 217 - = = = ° ´ 16. Faraday constant (F), chemical equivalent (E) and electrochemical equivalent (Z) are related as : (A) F = EZ (B) F = Z E (C) E F Z  (D) 2 E F Z  Sol: (C) We know frmo the Faraday's second law of electrolysis that electrochemical equivalent 1 M 1 (Z) E F P F = ´ = ´ or E F Z = 17. Two heater coils separately take 10 minutes to boil a certain amount of water. If both coils are connected in series, time taken to boil the same amount of water will be : (A) 15 min. (B) 20 min. (C) 7.5 min. (D) 2.5 min. Sol: (B) 2 V Q t t R R = Þ µ 1 1 2 2 t R t R = 2 2 10 R t 20 min t 2R Þ = Þ = 18. Resistance of a copper coil is 4.64 at 40oC and 5.6 at 100oC. Then its resistance at 0oC is (in ): (A) 4 (B) 0.96 (C) 5.12 (D) 4.2 Sol: (A) t 0 0 0 0 0 0 R R 5.6 R 4.64 R R 4t R 100 R 40 - - - a = = = ´ ´ ´ 0 R 4 Þ = W 19. Resistance of a coil at 100oC is 4.2 and the temperature coefficient of resistance of its material is 0.004oC–1. Then its resistance at 0oC is () : (A) 5 (B) 3 (C) 4 (D) 3.5
• 15. Sol: (B) t 0 R R (1 t) = + a 0 0 4.2 R (1 0.004 100) 4.2 R 3 1.4 Þ = + ´ Þ = = W 20. An electric kettle has two heating elements. One brings it to boil in ten minutes and the other in fifteen minutes. If the two heating filaments are connected in parallel, the water in the kettle will boil in (A) 5 minutes (B) 25 minutes (C) 8 minutes (D) 6 minutes. Sol: (D) 2 2 1 2 V V Q 10 15 R R = ´ = ´ 1 2 R 2 R 3 Þ = 2 2 1 1 2 1 2 V V Q 10 t R R R R R = ´ = æ ö ç ÷ + è ø 1 2 R 5 10 1 t t R 3 t 6 min æ ö Þ = + = ç ÷ è ø Þ = 21. The resistance of a conductor is 5 ohm at 50oC and 6 ohm at 100oC. Its resistance at 0oC is (A) 2 ohm (B) 1 ohm (C) 4 ohm (D) 3 ohm. Sol: (C) 1 0 0 R R R t - a = 0 0 0 0 0 0 5 R 6 R 10 2R 6 R 50R 100R - - Þ a = = Þ - = - 0 R 4 = W 22. The mass liberated at any electrode when electricity is passed through it is directly proportional to (A) q2 (B) q (C) 1 q (D) q3 Sol: (B) m Z t Zq m q = I = Þ µ 23. Which one of the following causes production of heat when current is set up in a wire? (A) fall of electrons from higher orbits to lower orbits (B) interatomic collisions (C) interelectron collisions (D) collisions of conduction electrons with atoms. Sol: (D) 24. In the Seeback series bismuth occurs first followed by Cu and Fe among other. The Sb is the last in the series. If V1 be the thermo-emf at the given temperature difference for Bi-Sb themocouple and V2 be hat for Cu-Fe thermocouple, then
• 16. (A) V1 = V2 (B) V1 < V2 (C) V1 > V2 (D) data insufficient. Sol: (D) 25. Which of the following is a characteristic temperature for the themocouple ? (A) temperature of cold junction (B) temperature of hot junction (C) temperature of inversion (D) neutral temperature. Sol: (D) 26. When a copper voltameter is connected with a battery of electromagnetic field 12V, 2gm of copper is deposited in 30 minutes. If the same voltmeter is connected across a 6V battery, the mass of copper deposited in 45 minutes would be (A) 1 gm (B) 1.5 gm (C) 2 gm (D) 2.5 gm Sol: (B) m z t, m zvt µ I µ 2 2 2 1 1 1 m v t m v t = 2 2 2 1 1 1 v t 6 45 m m 2 1.5 gm v t 12 30 Þ = ´ = ´ ´ = 27. The electro-magnetic field of a Cu-Fe thermo couple varies with temperature  of the hot junction (cold junction at 0oC) as E = 14 – 0.022. The neutral temperature is given by (A) 350oC (B) 400oC (C) 450oC (D) 500oC. Sol: (A) 2 dE E 14 0.02 , 0 d = q - q = q 350 C Þ q = ° 28. Same current is being passed through a copper voltmeter and a silver voltmeter. The rate of increase in weights of the cathodes in the two voltameters will proportional to: (A) atomic masses (B) atomic numbers (C) relative densities (D) none of these. Sol: (A) 29. Which of the following acts a depolariser in Leclanche cell? (A) ZnCl2 (B) Mn2O3 (C) MnO2 (D) NH4Cl Sol: (C) 30. In the circuit shown the 5 resistor develops 20W power due to current flowing through it, then power dissipated in 4 resistance is (A) 4 W (B) 6 W (C) 10 W (D) 20 W Sol: (A) If the current in 5  branch is i, then current in 4  branch will be i/2. 2 2 2 2 P i R i 5 20 i 2A P' i R' (1) 4 4W = = ´ = = = = ´ = 1. t 0 4R R (1 t) = + µ 3 0 0 3R R (1 4 10 t) - = + ´ 3 1 t 500 C 2 10- Þ = = ° ´ (B) 2. H Pt 210 300 J = = ´ 210 300 cal 15000 cal 4.2 ´ = =
• 17. 3. Q t 1.6 60 96C = I = ´ = 20 19 96 No. of Cu ions 3 10 3.2 10 ++ - = = ´ ´ For Test 1. At what temperature will the resistance of a copper wire become three times its value at 0oC? [Temperature coefficient of resistance for copper = 4 × 10–3 per oC] (A) 400oC (B) 450oC (C) 500oC (D) 550oC. Sol: (C) 2. How much heat is developed in a 210 watt electric bulb in 5 minutes ? [Mechanical equivalent of heat = 4.2 joule/calorie] (A) 7500 calories (B) 15000 calories (B) 22500 calories (D) 30000 calories. Sol: (B) 3. A current of 1.6 A is passed through a solution of CuSO4. How many Cu++ ions are liberated in one minute ? [Electronic charge = 1.6 × 10–19 coulomb]. (A) 3 × 1020 (B) 3 × 1019 (C) 6 × 1020 (D) 6 × 1019 Sol: (A) 4. The smallest temperature difference that can be measurred with a combination of a thermocouple of thermo e.f.f. 30V per degree and a galvanometer of 50 ohm resistance, capable of measuring a minimum current of 3 × 10–7 amp, is : (A) 0.5 degree (B) 1.0 degree (C) 1.5 degree (D) 2.0 degree. Sol: (A) Minimum voltage that can be measured by the galvanometer 7 5 V ig G 3 10 50 1.5 10 volt - - = = ´ ´ = ´ Smallest temperature difference that can be mesured 5 6 1.5 10 t 0.5 C 30 10 - - ´ D = = ° ´ 5. If 2.2 kilowatt power is transmitted through a 10 ohm line at 22000 volt, the power loss in the form of heat will be- (A) 0.1 watt (B) 1 watt (C) 10 watt (D) 100 watt. Sol: (A) 2 2 2 P R 2200 2200 10 H R 0.1W V 22000 22000 ´ ´ = I = = = ´ 6. For electroplating a spoon, it is placed in the voltameter at (A) the position of anode (B) the position of cathode (C) exactly in the middle of anode and the cathode (D) anywhere in the clectrolyte Sol: (B) Positive ios get deposited on cathode. 7. The quantity of electricity needed to liberate one gram equivalent of an element is- (A) 1 ampere (B) 96500 amperes
• 18. (C) 96500 coulomb (D) 96500 faradays. Sol: (C) The amount of electricity required to liberate 1 gm equivalent is 1 Faraday or 96500 coulomb. 8. When current is passed through a junction of two dissimilar metals, heat is evolved or absorbed at the junction. This process is called- (A) Seebeck effect (B) Joule effect (C) Peltier effect (D) Thomson effect. Sol: (C) Peltier effect 9. A 5oC rese in temperature is observed in a conductor by passing a current. When the current is doubled the rise in temperature will be approximately- (A) 10oC (B) 16oC (C) 20oC (D) 12oC. Sol: (C) As 2 H µ I , on doubling the current, heat produced and hence rise in temperature becomes fuor times.  Rise in temperature 4  5 = 20C 10. Three equal resistors connected across a source of e.m.f. together dissipate 10 watt of power. What will be the power dissipated in watts if the same resistors are connected in parallel across the same source of e.m.f.? (A) 10 (B) 10 3 (C) 30 (D) 90. Sol: (D) In series 2 2 V Power 10W V 30R 3R = = = In parallel 2 V 30R 3 P' 90W R R 3 ´ = = = æ ö ç ÷ è ø 11. If nearly 105 coulombs liberte 1 gm-equivalent of aluminium, hen the amount of aluminum, (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 amp. will be- (A) 0.6 gm (B) 0.09 gm (C) 5.4 gm (D) 10.8 gm. Sol: (C) m zit or m it = µ 1 1 1 2 2 2 m i t m i t = or 5 2 9 10 m 50 20 60 = ´ ´ or 2 m 5.4 gm = 12. Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing, the ratio of the heat produced is : (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4 Sol: (C) 2 Q R or Q R = I µ s s p p Q R 2R 4 :1 R Q R 2 = = = æ ö ç ÷ è ø 13. Constantan wire is used in making standard resistances, because its (A) specific resistance is low (B) density is high (C) temperature coeff. of resistance is negligible
• 19. (D) melting point is high. Sol: (C) The temperature coefficient of resistance of constantan is quite negligible, hence its resistance almost remains constant. 14. As the temperature of hot junction increases, the thermo cmf : (A) always increases (B) always decreases (C) may increase of decrease (D) always remains constant Sol: (C) Thermo e.m.f. Temp. of hot junction 0o C t The variation of thermo-e.m.f. with temperature of hot junction is as shown in the figure. 15. A certain charge liberates 0.8 gm of Oxygen. The same charge will liberate how many gms of Silver? (A) 108 gm (B) 10.8 gm (C) 0.8 gm (D) 108 gm 0.8 . Sol: (B) g g A A 0 0 M E M E = or g A 108 0.8 m 10.8 g 8 ´ = = 16. An electric kettle has two heating coils. When one coil is used, water in the kettle boils in 5 minutes, while when second coil is used, same water boils in 10 minutes. If the two coils, connected in parallel are used simultaenously, the same water will boil in time. (A) 3 min. 20 sec. (B) 5 min. (C) 7 min. 30 sec. (D) 2 min. 30 sec. Sol: (A) 2 V t Q t R R = Þ µ 1 1 2 1 2 2 t R 5 R 2R t 10 R = = Þ = In second case 1 1 1 2 1 1 2 R R 5 R R 2 t R 3 R R = = + 10 t min 3 min 20 sec. 3 Þ = = 17. An external resistance R is connected to a battery of e.m.f. V and internal resistance r. The joule heat produced in resistor R is maximum when R is equal to
• 20. (A) r (B) r 2 (C) 2r (D) infinitely large. Sol: (A) According to maximum power transfer theroem, transfer of power from the surce to the load will be maximum when load resistance is equal to the internal resistance of the source.  R = r 18. If Tc, Tn and Ti denote the temperatures of cold junction, neutral temperature and inversion temperature of a thermo-couple respectively, then : (A) Tc + Ti = Tn (B) Ti + Tc = 2Tn (C) Tc + Ti = 2Tn (D) Tc – Ti = 2Tn Sol: (A) c i n T T T 2 + = c i n T T 2T Þ + = 19. The electrochemical equivalent of a material in an electrolyte depends on (A) The nature of the material (B) The current through the electrolyte (C) The amount of charge passes through electrolyte (D) The amount of material present in electrolyte. Sol: (A) m Zit ZQ = = m Z Q = It depend on the nature of material. 20. A wire of length I and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number of N similar cells is now connected in series with a wire of the same material and cross section but the length 2L. The temperature of the wire is raised by the same amount T in the same time t. The value of N is- (A) 4 (B) 6 (C) 8 (D) 9. Sol: (B) The heat generated in the wire is givein by 2 V t H R = In the first case, we have 2 p (3E) t mC T R D = . . . (1) in the second case, we have 2 p (NE) t (2m) C t ( R L) 2R D = µ . . . (2) Dividing Equation (2) by Equation, (1) we get 2 2 N / 2 2 N 36 N 6 9 = Þ = Þ = 21. Two wires having resistance of 2 and 4 are connected to same voltage. Ratio of heat dissipated in the resistances is- (A) 1 : 2 (B) 4 : 3 (C) 2 : 1 (D) 5 : 2. Sol: (C) As heat dissipated 2 V /R = , so it is inversely proportional to 1 2 2 1 H R 2 1/R H R 1 = = 22. If thermo e.m.f. is given by E = at + bt2, then neutral temperature is-
• 21. (A) a 2b  (B) 2b a (C) a b  (D) b a  Sol: (A) At neutral temperature dE/dt = 0 or a + 2bTN = 0 N a T 2b = - 23. Two parallel and straight conductors are at a distance d apart, each have current i following through them. The force per unit length is (A) 0i 4 d m p (B) 2 0i 2 d m p (C) 2 0i 4 d m p (D) 0 2 i d m Sol: (B) Force due to magnetic field induction = B12, where B1 is the magnetic field due to first 0 1 2 4 d m I = × p 2 0 i F 2 d m = × p 24. Thomson coefficient of a conductor is 10V/K. The two ends of it are kept at 50C and 60C respectively. Amout of heat abosrbed by the conductor when a charge of 10 C flows through it is : (A) 1000 J (B) 100 J (C) 100 mJ (D) 1 mJ Sol: (D) Amount of heat absorbed by the conductor H =  Q dt H = (10  10–6) (10) (10) = 10–3 J = 1 mJ 25_57. In an electroplating experiment 4A current for 2 minutes deposits m gram of silver. When 6A current flown for 40 seconds then the deposited silver will be (A) 4m (B) m/2 (C) m/4 (D) 2m Sol: (B) m  i t 1 1 1 2 2 2 2 m i t m 4 120 or m i t m 6 40 ´ = = ´ or 2 m m 2 = . 26_58. The temperature coefficient of a resistance wire is 0.00125 per degree. The resistance at 300K is 1 ohm. At what temperature its resistance will be 2 ohm. (A) at 1154 K (B) at 1100 K (C) at 600 K (D) at 1400 K Sol: (B) 2 1 t t 2 1 R R [1 (t t )] = + a - 2 2 1[1 0.00125(t 300)] = + - or 2 1 t 300 800 0.00125 - = = or 2 t 1100 K =
• 22. 27_59. Two electric heater wires are made from the same material and have the same dimensions. First these are connectted in series and then these are connected in parallel (to a 220 V AC source) then the ratio of heat generated will be (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 Sol: (D) 2 V 1 H R R = µ 1 2 2 1 R H R 2 1: 4 H R 2R = = = 28_60. The mass of a product liberated on anode in an electrochemical cell depends on (A) ( t)1/2 (B)  t (C)  / t (D) 2 t (Where t is the time period for which the current is passed). Sol: (B) m  Q But Q =  t  m   t or m = Z t 29_61. If i is the inversion temperature, n is the neutral temperature, c is the temperature of the cold junction, then (A) i + c = n (B) i – c = 2n (C) i c n 2 q + q = q (D) c – i = 2n Sol: (C) Neutral temperature is the mean of temperature of inversion and temperature of cold junction, i.e., i c n 2 q + q q = . 30_75. The thermo e.m.f. of a thermo-couple is 25V/C at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as 10–5 A, is connected with the thermocouple. The smallest temperature difference that can be detected by this system is (A) 16C (B) 12C (C) 8C (D) 20C Sol: (A) Minimum p.d. to be measured 5 4 V iG 10 40 4 10 V - - = = ´ = ´ Minimum temp. diff. detected 4 6 4 10 16 C 25 10 - - ´ = = ° ´ .