"TH CLASS :-10 Unit :- Electricity SUBJECT:-PHYSICS1. Question: Two bulbs have ratings 100 W, 220 V and 60 W, 220 V respectively. Whichone has a greater resistance?2 Answer: P=VI=V /R For the same V, R is inversely proportional to P. Therefore, the bulb 60 W, 220 V has a greater resistance.2. Question: A torch bulb has a resistance of 1 O when cold. It draws a current of 0.2 Afrom a source of 2 V and glows. Calculate(i) the resistance of the bulb when glowing and(ii) explain the reason for the difference in resistance.Answer:(i) When the bulb glows:V = I R ---- Ohm's law R = V/I = 2/.2 =10 O(ii) Resistance of the filament of the bulb increases with increase in temperature. Hence when itglows its resistances is greater than when it is cold.3. Question: Calculate the resistance of 1 km long copper wire of radius 1 mm. (Resistivity-8 of copper = 1.72 x 1 0 Answer: L = 1 km = 1000 m-3 R = 1 mm = 1 x 1 0p = 1.72 x 1 0-8?? m-3 x -3 -6 Area of cross section = ? r2= 3.14 x 1 0 1 0 = 3.14 x 1 0-6 R = pl/A = (1.72 x 1 0-8x 1000 ) / 3.14 x 1 0 = 5.5 ?4. Question: When a potential difference of 2 V is applied across the ends of a wire of 5 mlength, a current of 1 A is found to flow through it. Calculate: (i) the resistance per unit length of the wire (ii) the resistance of 2 m length of this wire (iii) the resistance across the ends of the wire if it is doubled on itself.Answer: (i) V = I R ----- Ohm's law R=V/I=2/1= 2 OhmResistance per unit length: 2/5= 0.4 Ohm/m(ii) Resistance of 2 m length of the wire = 0.4 x 2=0.8 ohm(iii) When the wire is doubled on itself:(a) the area of cross-section is doubled. If A is the original C.S. area, now it is 2 A. (b) The length becomes half i.e.L/2 Resistance of this wire =R' = p (l/2)/(2A) = 1/4(p(L/A)But p(L/A) = 2 ohmR' = 1/4 x 2=0.5 Ohm5. How much work is done in moving 4 C across two point having pd. 10 vSolution : W = VQ = 10 x 4 = 40J6. How much energy is given to each coulomb of charge passing through a 9 v battery?Solution:Potential difference = Work done = Potential difference × chargeWhere, Charge = 1 C and Potential difference = 6 VWork done = 9×1 = 9 Joule. 7. 100j of work is done in moving a charge of 5 C from one terminal of battery to another .What is the potential difference of battery?Solution: V = W/Q =100j/5C = 20 V-3 - 68. If 4 x 10J of work is done in movinga particles carrying a charge of 16 x 10 Cfrom infinity to point P .What will be the potential at a point?Solution: the potential at a point is work done to carry unit from one point to another-3 - 6 = (4 x 10) /(16 x 10 C) = 250 V9. Calculate the current and resistance of a 100 W ,200V electric bulb.Solution:Power,P = 100W and Voltage,V = 200VPowerP= VISo, Current I = P/v = 100/200 = 0.5AResistance R = V/I = 200/0.5 = 400W. 10.Calculate the power rating of the heater coil when used on 220V supply taking 5 Amps.Solution:Voltage ,V = 220V andCurrent ,I = 5A,Power,P = VI = 220 × 5 = 1100W = 1.1 KW.11. A lamp can work on a 50 volt mains taking 2 amps.What value of the resistance mustbe connected in series with it so that it can be operated from 200 volt mains giving the samepower.Solution: Lamp voltage ,V = 50V andCurrent ,I = 2 amps.Resistance of the lamp = V/I = 50 / 2 = 25 ?Resistance connected in series with lamp = r.Supply voltage = 200 volt. andCircuit current I = 2 ATotal resistance Rt= V/I= 200/2= 100?Rt = R + r=>100 = 25 + r=>r = 75?12. Calculate the work done in moving a charge of 5 coulombs from a point at a potential of210 volts to another point at 240 voltsSolution: Potential diffrence= 210- 240 = 30 V So, W.= V xQ = 30Vx5C = 150 Joules13. How many electrons pass through a lamp in one minute if the current be 220 mA?Solution:I= 220 mA = 0.22 A I = Q/T 0.22 = Q/60 Q= 0.22 x 60 = 13.2 C 18 No of electron carry 1 C charge = 6 x 1018 18 No of electron carry 13.2 C charge = 6 x 10x 13.2 C = 79.2 x 1014.Calculate the current supplied by a cell if the amount of charge passing through the cellin 4 seconds is 12 C ?Solution:I = Q/t = 12/4 = 3A15. A 2 Volt cell is connected to a 1 O resistor. How many electrons come out of thenegative terminal of the cell in 2 minutes?Solution: V = IR =>I = V/R = 2/1 = 2 AI = Q/t=>Q = It = 2 x 2 x 20 = 80 C18 No of electron carry 1 C charge = 6 x 1018 1820 No of electron carry 80 Ccharge = 6 x 10x 80 C = 108x 10= 1. 08x 10"