InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
How many calories of heat will be approximately developed in a `210 W` electric bulb in `5 min`?A. `15,000`B. `1050`C. `63,000`D. `80,000` |
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Answer» Correct Answer - A Heat in calories `= ( 210 xx 5 xx 60 )/( 4.2) = 15,000` |
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| 2. |
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 ifA. both the lengths and radius of the wire are halvedB. both the length and radius of the wire are doubledC. the radius of the wire is doubledD. the length of the wire is doubled |
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Answer» Correct Answer - B Heat developed is `H = (V^(2))/( R ) t` Heat developed will be doubled when `R` is halved. Further, `R = rho l//( pi r^(2)) and H = V^(2) pi r^(2) t // rho l`. So heat produced when both the length and radius of the wire are doubled. |
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| 3. |
Two electric bulbs , rated for the same voltage , have powers of `200 W and 100 W`, respectively. If their resistances are `r_(1) and r_(2)` , respectively , thenA. ` r_(1) = 2 r_(2)`B. ` r_(2) = 2 r_(1)`C. ` r_(2) = 4 r_(1)`D. ` r_(1) = 4 r_(2)` |
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Answer» Correct Answer - B `P = ( V^(2))/ ( R ) , (P_(1))/( P_(2)) = ( R_(2))/( R_(1)) or r_(2) = 2 r_(1)` |
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| 4. |
In figure circuit section `AB` absorbs energy at the rate of `5.0 W` when a current` i =1.0A` passes through it in the indicated direction.(a) What is the potential difference between points `A` and `B`? (b) Emf device `X` does not have internal resistance. What is its emf? (c) What is its polarity (the orientation of its positive and negative terminals)? A. `24 V`B. `32 V`C. 48 V`D. `12 V` |
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Answer» Correct Answer - C As section `AB` consumes `50 W at 1 A and P = VI`. S0 `V = V_(A) - V_(B) = (P)/(I) = (50)/(1) (W)/(A) = 50 V` As `R and C` are in series, `V = V_(R) + V_( C ) or V_( C ) = V - V_(R )` But here `V = 50 V and V_(R) = IR = 1 xx 2 = 2 V`. So `V_( C ) = 50 - 2 = 48 V`. Now as the element `C` absorbs energy and has no resistance , it is a source of emf with zero internal resistance ( i.e., ideal battery ). So emf is `E = V + Ir = 48 + 1 xx 0 = 48 V`. As in charging , positive and negative terminals of the charger are connected to the positive and negative terminals of the battery, respectively ,` B` is connected to the negative terminal of element `C`. |
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| 5. |
A heating coil is rated `100 W , 200 V`. The coil is cut in half and two pieces are joined in parallel to the same source . Now what is the energy `( "in" xx 10^(2) J)` liberated per second? |
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Answer» Correct Answer - 4 Let resistance of heating coil be `R` , then `100 = (200^(2)) /(R ) or R = ((200)^(2))/(100) Omega` Resistance of each cut part is `R//2`. Now , power dissipated is `P = 2(((220)^(2)) / ( R//2))` `= 4 ((220)^(2))/( R ) = 4 xx 100 = 400 W` |
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| 6. |
Three identical resistors are connected in series . When a certain potential difference is applied across the combination , the total power would be dissipated is `27 W`. How many times the power would be dissipated if the three resistors were connected in parallel across the same potential difference ? |
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Answer» Correct Answer - 9 For three identical resistors in series , `P_(s) = V^(2)// 3 R `. If they are now in parallel over the same voltage , `P_(p) = (V^(2))/(R_( eq)) = (V^(2))/(R//3) = ( 9 V^(2))/( 3 R) = 9P_(s) = 9 xx ( 27 W)` |
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| 7. |
The three resistance of equal value are arranged in the different combination shown below. Arrange them in increasing order of power dissipation.(I) (II) (IV) A. `III lt II lt IV lt I`B. `II lt III lt IV lt I`C. `I lt IV lt III lt II`D. `I lt III lt II lt IV` |
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Answer» Correct Answer - A `P_(I) = I^(2) ( 3 R ), P_(II) = I^(2) ((2 R ) /(3)) , P_(III) = I^(2)((R )/( 3)` `P_(IV) = I^(2) ((3R ) /(2))` `:. III lt II lt IV lt I` |
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| 8. |
Three `10 Omega , 2 W` resistors are connected as in Fig. `7.34`. The maximum possible voltage between points `A and B` without exceeding the power dissipation limits of any of the resistors is A. `5 sqrt(3) V`B. `3 sqrt(5) V`C. `15 V`D. ` (5)/(3) V` |
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Answer» Correct Answer - B Power `= (V^(2))/( R )` or `V = sqrt(PR) = sqrt(2 xx 10) = sqrt(20) = 2sqrt(5) V` Clearly, voltage across single resistor of `10 Omega` cannnot exceed `2 sqrt(5) V`. Note that the resistance of parallel combination is half of `10 Omega` . Thus, the maximum possible voltage between `A and B` is `3 sqrt(5) V`. |
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| 9. |
An ideal gas is filled in a closed rigid and thermally insulated container. A coil of `100Omega` resistor carrying current 1 A for 5 minutes supplies heat to the gas. The change in internal energy of the gas isA. `10 kJ`B. `30 kJ`C. `20 kJ`D. `0 kJ` |
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Answer» Correct Answer - B The heat supplied under these conditions is the change in internal energy `Q = Delta U`. The heat supplied is `Q = i^(2) Rt xx 1 xx 1xx 100 xx 5 xx 60 = 30000 J = 30 kJ`. |
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| 10. |
If in the circuit shown in Fig.`7.55`, power dissipation is `150 W` , then find the value of `R ( "in" Omega)`. |
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Answer» Correct Answer - 6 `150 = (15 xx 15) /( R_(eq)) or R_(eq) = ( 15 xx 15)/( 150) Omega = (3) /(2) Omega` Now, `( 2R)/( 2+ R ) = (3) /( 2) or R = 6 Omega` |
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| 11. |
In the circuit shown in fig the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor isA. `1 Cals^(-1)`B. `2 Cals^(-1)`C. `3 Cals^(-1)`D. `4 Cals^(-1)` |
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Answer» Correct Answer - B Let current is `5 Omega` resistors be `I_(1)` and `in 4 Omega` resistors be `I_(2)`, then `I_(2)//I_(1) = (1//2)`. The heat generated in the ` 5 Omega` resistor is `10 cals^(-1)` Given `I_(1)^(2) xx 5 = 10 or (( 2 I ^(2))^(2) xx 5 = 10 or I_(2)^(2) = (1)/(2)`. Heat in the `4 Omega` resistor will be `(I_(2)^(2))^(2) xx 4 = 2 cal s^(-1)`. |
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| 12. |
The resistance in which the maximum heat is produced is given by `( Fig. 7.43 )` A. `2 omega`B. `6 Omega`C. `4 Omega`D. `12 Omega` |
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Answer» Correct Answer - A All resistances are in parallel. so for parallel combination `H = (V^(2))/( R ) t or H prop (1)/( R )` |
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| 13. |
Two electric bulbs have tungsten filament of same length. If one of them gives `60 W` and the other `100 W` , thenA. `100 W` bulb has thicker filamentB. `60 W` bulb has thicker filamentC. both filaments are of same thicknessD. it is not possible to get different wattages unless the lengths are different |
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Answer» Correct Answer - A `P = V^(2)//R` . If `P` is more , `R` is less . `R = rho l // a`. For less `R`, a is more . So the `100 W` bulb has thicker element . |
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| 14. |
The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp when not in use?A. `14,000 Omega`B. `400 Omega`C. `40 Omega`D. `4 Omega` |
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Answer» Correct Answer - C When hot , `R = (V^(2))/( P) = ( 200 xx 200)/( 100) = 400 omega` Hence, when cold, the resistance is `40 Omega`. |
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| 15. |
If resistance of the filament increases with temperature, what will be power dissipated in a `220 V - 100 W` lamp when connected to `110 V` power supplyA. 25WB. lt25WC. gt25WD. None of these |
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Answer» Correct Answer - C |
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| 16. |
Resistors `P , Q , and R` in the circuit have equal resistances. If the battery is supplying a total power of `12 W` , what is the power dissipated as heat in resistors `R`? A. `2 W`B. `6 W`C. `3 W`D. `8 W` |
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Answer» Correct Answer - A Let resistors `P, Q , and R` have resistance `r`. The effective resistance across the source is `R_(eff) = r || r = r + (( r ) ( r ))/( r + r) = r + ( r ) /(2) = ( 3 r) /(2)` Current drawn from source is `I_(s)^(2) R_(eff) = 12` or `I_(s) = sqrt((12)/(R_(eff))) = sqrt((8)/(r )) A` Since `Q and R` have equal resistance `r`, each draws a current of `I` , which is given by `I = (1)/(2) I, sqrt((2)/( r ))r = 2 W` |
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| 17. |
Figure `7.37` shows a network of three resistances. When some potential difference is applied across the network , thermal powers dissipated by `A, B and C` are in the ratio A. `2:3:4`B. `2:4:3`C. `4:2:3`D. `3:2:4` |
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Answer» Correct Answer - C |
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| 18. |
A room `AC` run for `5` hour at a voltage of `220V` The wiring of the room constant of `Cu` of `1mm` ratio and a length of `10 m` consumption per day is `10` commerclal unit What fraction of it goes in the joule heated in wire? What would happen if the wiring is made of aluminum of the same distances? `[rho_(cu) = 1.7 xx 10^(-8) Omega,rho_(A1) = 2.7 xx 10^(-8) Omega m]` |
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Answer» Correct Answer - A |
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| 19. |
Which of the following plots may represent the thermal energy produced in a resistor for a given current as a function of time? A. aB. bC. cD. d |
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Answer» Correct Answer - D |
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| 20. |
Two heating coils, one of fine wire and the other of thick wire of the same material and of the same length are connected in series and in parallel. Which of the following statement is correct ?A. In series fine wire liberates more energy while in parallel thich wire will liberate more energy.B. In series fine wire liberates less energy while in parallel thick wire will liberate more energyC. Both will liberate quicklyD. In series the thick wire will liberate more while in paralel it will liberate less energy |
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Answer» Correct Answer - A |
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| 21. |
Statement I: Internal resistance of a battery is drawn parallel to a battery in electical circuit. Statement II: Heat generated in a battery is due to internal resistance.A. Statement I is True , Statement II is True , Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True , Statement II is NOT a correct explanation for statement I.C. Statement I is True , Statement is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - D Internal resistance is drawn in series with battery. |
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| 22. |
In the previous problem , the power lost in the cable during transmission isA. `3.15 kW`B. `12.5 kW`C. `6.25 kW`D. `25 kW` |
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Answer» Correct Answer - C power lost `= 25 xx 25 xx 10 W = 6250 W = 6.25 kW` |
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| 23. |
A ` 2k W` heater used for `1 h` every day consumes the following electrical energy in `30 days`A. `60 units`B. `120 units`C. `15 units`D. none of the above |
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Answer» Correct Answer - A `2 k W xx 30 h = 60 kWh = 60 units` |
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| 24. |
The heat generated through `4 Omega and 9 Omega` resistances separately , when a capacitor pf `100 my F` capacity charged to `200 V` is discharged one by one , will beA. `2 J and 8 J, respectivelyB. `8 J and 2 J, respectivelyC. `2 J and 4 J, respectivelyD. `2 J and 2 J, respectively |
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Answer» Correct Answer - D `U = (1)/(2) CV^(2) = (1)/(2) xx 100 xx 10^(-6) xx 200 xx 200 J = 2 J` |
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| 25. |
If the length of the filament of a heater is reduced by `10%`, the power of the heater willA. increase by about 9%B. increase by about 11%C. increase by about 19%D. decrease by about 10% |
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Answer» Correct Answer - B |
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| 26. |
Consider the circuit in the figure. (a) how much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity? (b) Electrons give up energy at the rate of `Rl^(2)` per second to the thermal energy. What time scale would the number associate with energy in problem (a)? n = number of electron/volume `=10^(29)//m^(3)`. Length of circuit = 10cm cross-section = `A = (1mm)^(2)`. |
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Answer» Correct Answer - A |
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| 27. |
A constant current `i` is passed through a resistor. Taking the temperature coefficient of resistance into account, indicate which of the plots shown in Figure best represents the rate of production of thermal energy in the resistor A. aB. bC. cD. d |
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Answer» Correct Answer - D |
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| 28. |
Two identical heaters rated `220 V , 1000 W` are paced in series with each other across `220 V` line , then the combined power isA. ` 1000 W`B. `2000 W`C. `500 W`D. `4000 W` |
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Answer» Correct Answer - C Let the resistance of the two heaters be denoted by `R_(1) and R_(2)`, respectively. Then ` R_(1) = (V^(2))/(P_(1)) and R_(2) = (V^(2))/(P_(2)))` If the resistance of the series combination is denoted by `R_(s)` and the corresponding power by `R_(s) = R_(1) + R_(2)`. So `(V^(2))/( P_(s)) = (V^(2))/(P_(1)) + (V^(2))/(P_(2))` or `P_(s) = (P_(1)P_(2))/(P_(1) + P_(2)) = ( 1000 xx 1000 )/(2000) = 500 W` |
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| 29. |
A electric kettle ( rated accurately at `2.5 kW`) is used to heat `3 kg` of water from `15^(@)C` to boiling point . It takes `9.5 min`. Then the amount of heat that has been lost isA. `3.5 xx 10^(5) J`B. `7 xx 10^(8) J`C. `3.5 xx 10^(4) J`D. `7 xx 10^(8) J` |
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Answer» Correct Answer - A Energy consumed in `9.5 min ` is `2.5 xx 1000 xx 9.5 xx 60 J = 1425000 J` Heat usefully consumed is ` 3 xx 4.2 xx 1000 xx (100 - 15 ) J = 1071000 J` Loss `= 3.54 xx 10^(5) J` |
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| 30. |
If a given volume of water in a `220 V` heater is boiled in `5 min`, then how much time will it take for the same volume of water in a `110 V` heater to be boiled?A. `20 min`B. `30 min `C. `25 min`D. `40 min` |
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Answer» Correct Answer - A `H = V^(2) t // R`. When voltage is halved , the heat becomes one - fourth. Hence , time taken to heat the water becomes four times. |
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| 31. |
It takes `16 min` to boil some water in an electric kettle. Due to some defect it becomes necessary to remove `10 %` turns of the heating coil of the kettle . After repairs , how much time will it take to boil the same mass of water ?A. `17.7 min`B. `14.4 min`C. `20.9 min`D. `13.7 min` |
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Answer» Correct Answer - B If `N` is the initial number of turns in the coil and `r` is the radius of coil, then its resistance is `R = rho(L)/(A) = rho (N 2 pi r )/(A) or (V^(2) t A ) /( 4.2 rhoN 2 pi r ) = Q = msd theta`or `(t)/(n) = ( msd theta xx 4.2 rho 2 r )/( V^(2) A) = constant` or `(t_(1) // N_(1))/( t_(2)//N_(2)) = 1 or t_(2) = (N_(2))/(N_(1)) t_(1) = ((( 9)/( 10) N_(1))/(N_(1))) t_(1) = ( 9) /( 10) xx 16 min` |
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| 32. |
How many `60 W` lamps may be safely run on a `230 V` circuit fitted with a ` 5 A` fuse?A. `2`B. `19`C. `20`D. `4` |
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Answer» Correct Answer - B Watt ` = "volt xx ampere"` `rArr 60 = 230 xx I or = (6//23) A` If `n` lamps are used in parallel , each allowing `6// 23 A` , then total current , `n xx ( 6 // 23 ) lt = 5` or `n le 19.1 or n = 19 ` |
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| 33. |
How many `60 W` lamps may be safely run on a `230 V` circuit fitted with a ` 5 A` fuse?A. `11`B. `22`C. `33`D. `66` |
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Answer» Correct Answer - B Current required by each bulb is `I = (P)/(v) = (100)/(220)A` If `n` bulbs are joined in parallel , then `nI = I_(fuse) or n xx (100)/(220) = 10 or n = 22` |
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| 34. |
A 220 volt, 1000 watt bulb is connected across a 110 volt mains supply. The power consumed will beA. `1000 W`B. `750 W`C. `500 W`D. `250 W` |
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Answer» Correct Answer - D `P = V^(2)// R`. If `V` is halved ,`P` is reduced by a factor of `4` . So new power is `(1000 //4) W , i.e, 250 W`. |
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| 35. |
A 100 watt bulb working on 200 volt and a 200 watt bulb working on 100 volt haveA. resistance in the ratio of `4:1`B. maximum current ratings in the ratio of `1:4`C. resistance in the ratio of `2:1`D. maximum current ratings in the ratio of `1:2` |
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Answer» Correct Answer - B |
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| 36. |
The power rating of an electric motor that draws a current of `3.75 A`, when operated at `200 V`, is nearlyA. ` 54 W`B. `1 hp`C. `500 W`D. `750 hp` |
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Answer» Correct Answer - B `P = 200 xx 3.75 W = 750 W ~~ 1 hp` |
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| 37. |
A cable of resistance `10 Omega` carries electric power from a generator producing `250 kW at 10,000 V`. The current in the cable isA. `1000 A`B. `250 A`C. `100 A`D. `25 A` |
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Answer» Correct Answer - D `P = VI or I = (P)/( V) = ( 250 xx 1000)/( 10000) A = 25 A` |
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| 38. |
An element with `emf epsilon ` and interval resistance `r` is connected across an external resistance `R`. The maximum power in external circuit is `9 W`. The current flowing through the circuit in these conditions is `3 A`. Then which of the following is // are correct ?A. ` epsilon = 6 V`B. ` r = R `C. `r = 1 Omega`D. `r = 3 Omega` |
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Answer» Correct Answer - A::B::C Current through the circuit is `I = epsilon //( r + R )`. Power dissipated in external resistance is `P = I^(2) R = (epsilon^(2))/(( r + R )^(2)) R` We know that power dissipated is maximum when `r = R`. `P_(max) = (epsilon^(2) r) /( 2r)^(2) = (epsilon^(2))/( 4 r ) = 9 W , i = ( epsilon ) /( 2r) = 3 A` Solve to get , ` epsilon = 6 V and r = 1 Omega`. |
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| 39. |
Two bulbs consume same energy when operated at `200 V and 300 V` , respectively . When these bulbs are connected in series across a dc source of `500 V`, thenA. ratio of potential difference across them is `3/2`B. ratio of potential difference across them is `9/4`C. ratio of power concumed across them is `4/9`D. ratio of power consumed across them is `2/3` |
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Answer» Correct Answer - C |
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| 40. |
Two bulbs consume same energy when operated at `200 V and 300 V` , respectively . When these bulbs are connected in series across a dc source of `500 V`, thenA. ratio of potential difference across them is `3//2`B. ratio of potential difference across them is `4//9`C. ratio of power produced in them is `2 // 3`D. ratio of power produced in them is `2//3` |
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Answer» Correct Answer - B `P = (V^(2))/( R) or R = (V^(2))/( p) or R prop V^(2)`, i.e., `(R_(1))/(R_(2)) = ((200)/(300))^(2) = (4)/(9)` when connected in series , potential drops is in the ratio of their resistances . So, `(V_(1))/(V_(2)) = (R_(1))/(R_(2)) = (4)/(9)` Now, `P = I^(2) R` or `P prop R ( "in series I is the same" )` or `(P_(1))/(P_(2)) = (R_(1))/(R_(2)) = ( 4) /( 9)` |
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| 41. |
Three identical cells , each having an `emf 1.5 V` and a constant internal resistance `2.0 omega`, are connected in series with a `4.0 omega resistor R`, first as in circuit (i) , and second as in circuit (ii) . Then ( power in R in circuit (i))/( Power in R in circuit (ii)) = A. `9.0`B. `7.2`C. `1.8`D. `3.0` |
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Answer» Correct Answer - A Power `P = I^(2)R = ((V)/( R + 3 r ))^(2) R or P prop V^(2)` Power ratio `= ((V_(1))/(V_(2)))^(2) = (( 4.5)/( 1.5))^(2) = 3^(2) = 9 ` |
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| 42. |
An electric bulb rated for `500 W` at `100 V` is used in a circuit having a `200 V` supply. The reistance `R` that must be put in series with bulb, so that the bulb delivers `500 `W is ……….`Omega`.A. `10Omega`B. `20 Omega`C. `50 Omega`D. `100 Omega` |
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Answer» Correct Answer - B |
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| 43. |
In the circuit in Fig. `7.27`, bulb `B` does not glow although ammeter `A` indicates that the current is flowing . Why does the bulb not glow? A. The bulb is fused .B. tThere is a break in the circuit between bulb and ammeter.C. The variable resistor has too large resistance.D. There is a break in the circuit between the bulb and the variable resistance. |
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Answer» Correct Answer - C It is a case of weak current. |
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| 44. |
Statement I : In the circuit in `Fig. 7.46`, both cells are ideal and of fixed emf, the resistor `R_(1)` has fixed resistance and the resistance of resistor `R_(2)` can be varied ( but `R_(2)` is always non zero). Then the electric power delivered to the resistor of resistance `R_(1)` is independent of the value of resistance `R_(2)`. Statement II: If potential difference across a fixed resistance is unchanged , the poweer delivered to the resistor remains constant. A. Statement I is True , Statement II is True , Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True , Statement II is NOT a correct explanation for statement I.C. Statement I is True , Statement is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - A The potential difference across resistance `R_(1)` is always `|E_(1) - E_(2)|`. Hence , assertion and reason are true and the reason is the correct explanation of the assertion. |
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| 45. |
A voltmeter and an ammeter are connected in series to an ideal cell of `emf E`. The voltmeter reading is `V`, and the ammeter readings is `I`. Then (i) `V lt E` (ii) the voltmeter resistance is `V// I` (iii) the potential difference across the ammeter is `E - V` (iv) Voltmeter resistance + ammeter resistance = E//I` Correct statements areA. `I and ii`B. `ii and iii`C. `iii and iv`D. all |
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Answer» Correct Answer - A::B::C::D Treat all voltmeters and ammeters as resistances. Draw the circuit and find the currents and potential differences for each section. |
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| 46. |
Battery shown in figure has e.m.f. `E` and internal resistance `r`. Current in the circuit can be varied by sliding the contact `J`. If at any instant current flowing through the circuit is `I`, potential difference between terminals of the cells is `V`, thermal power generated in the cell is equal to `eta` fraction of total electrical power generated in it, then which of the following graphs is correct ? A. B. C. D. Both a and b are correct |
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Answer» Correct Answer - D |
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| 47. |
In the following circuit, `18Omega` resistor develops 2J/sec due to current flowing through it. The power developed across `10Omega` resistance is A. 125 WB. 10 WC. 4/5 WD. 25 W |
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Answer» Correct Answer - B |
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| 48. |
A variable current flows through a `1Omega` resistor for 2 s. Time dependence of the current is shown in the graph. A. Total charge flows through the resistor is `10^(@)`C.B. Average current through the resistor is 5A.C. Total heat produced in the resistor is 50 J.``D. Maximum power during the flow of current is 100 W. |
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Answer» Correct Answer - A::B::D |
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