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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. |
Can the phenomenon of resonance be exhibited in RL or RC circuit. |
| Answer» Correct Answer - No, Resonance can be obtained in circuit having L and C. | |
| 2. |
The impedance of a sereis `RL` circuit is same as the series `RC` circuit when connected to the same `AC` source separately keeping the same resistance. The frequency of the source isA. `(1)/(sqrt(LC))`B. `(1)/(2pi sqrt(LC))`C. `(R )/(L)`D. `(1)/(RC)` |
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Answer» Correct Answer - B The impedance in `RL` circuit is same as the `RC` circuit: `:. sqrt(R^(2) + omega^(2) L^(2)) = sqrt(R^(2) + (1)/(omega^(2) L^(2)))` `implies omega^(2) = (1)/(LC)` or `f = (1)/(2 pi sqrt(LC))` |
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| 3. |
The equation of an alternating voltage is `E = 220 sin (omega t + pi // 6)` and the equation of the current in the circuit is `I = 10 sin (omega t - pi // 6)`. Then the impedance of the circuit isA. 10 ohmB. 22 ohmC. 11 ohmD. 17 ohm |
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Answer» Correct Answer - 2 `Z = (E_(0))/(l_(0))` |
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| 4. |
In an ac circuit, the current lags behind the voltage by `pi//3`. The components in the circuit areA. R and LB. R and CC. L and CD. Only R |
| Answer» Correct Answer - a | |
| 5. |
In an ac circuit, the current lags behind the voltage by `pi//3`. The components in the circuit areA. R and LB. L and CC. R and CD. only R |
| Answer» Correct Answer - A | |
| 6. |
A 200 km telephone wire has capacity of `0.014 muF km^(-1)`. If it carries an alternating current oif frequency 50kHz, what should be the value of an inductance required to be connected in series so that impedance is minimumn ? |
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Answer» Capacitance , `C = 0.014 xx 200muF = 2.8 xx 10^(-16)F` Frequency, `f = 50 k Hz = 50 xx 10^(3) cycl es s^(-1)` The impedance will be minimum at resonant frequency , when applied frequency will be same as resonant frequency `(f_(0) = f)`. The resonant frequency , `f_(0) = 1/(2pisqrt (LC)) implies L = 1/(4pi^(2) f_(0)^(2) C)` ` L = 1/(4 xx 22/7 xx 22/7 xx (50 xx 10^(3))^(2) xx (2.8 xx 10^(-6))) = 0.36 xx 10^(-5) H` |
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| 7. |
In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. `0V,2.0A`B. `0V,1.4A`C. `5.6V,1.4A`D. `8V,2.0A` |
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Answer» Correct Answer - C `X_L=omegaL=(5 xx10^(-3))(2000)=10 Omega` `X_C=1/(omegaC)=1/((2000)50xx10^(-6))=10Omega` Since, `X_L=X_C` circuit is in resonance. `Z=R=(6+4)=10Omega` `I_("rms")=V_("rms")/Z=((20//sqrt2))/10=1.414A` This is also the reading of ammeter. `V=4I_("rms")` `~~5.6` Volt. |
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| 8. |
A signal generator supplies a sine wave of `200V, 5kHz` to the circuit shown in the figureure. Then, choose the wrong statement. .A. The current in te resistive brance is `0.2 A`B. the current in the capacitive branch is `0.126 A`C. Total line current is `~~0.283A`D. Current in both the branches is same |
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Answer» Correct Answer - B `I_R=V_("rms")/R=20/100=0.2A` `X_C=1/(2pifC)=1/((2pi)(5xx10^3)(1/pixx10^-6))` `=100Omega` `:. I_C=V_("rms")/X_C+200/100=2A` `I_C is 90^@` ahead of the applied voltage and `I_R` is in phase with the applied voltage. Hence, there is as phase difference of `90^@` between `I_R` and `I_C` too. `:. I=sqrt(I_R^2+I_C^2)` `sqrt((2)^2(2)^2)` `=283A`. |
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| 9. |
In an `AC` circuit, the applied potential difference and the current flowing are given by `V=20sin100tvolt,I=5sin(100t-pi/2)` amp The power consumption is equal toA. `1000W`B. `40W`C. `20W`D. zero |
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Answer» Correct Answer - D `phi=90^@` between `V` and `I` functions. `:. P=V_("rms")I_("rms")cosphi=0` |
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| 10. |
In an `AC` circuit, the current is given by `i=5sin(100t-(pi)/2)` and the `AC` potential is `V=200sin(100t)`volt. Then the power consumption isA. `20` wattsB. `40` wattC. `1000` wattD. `0` watt |
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Answer» Correct Answer - D `P=V_(i)cos varphi` Phase difference `varphi=(pi)/2 implies P=zero` |
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| 11. |
In an `AC` circuit, the current is given by `i=5sin(100t-(pi)/2)` and the `AC` potential is `V=200sin(100)`volt. Then the power consumption isA. `20` wattB. `40` wattC. `1000` wattD. `0` watt |
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Answer» Correct Answer - D `P=Vi cos varphi` Phase difference `phi=(pi)/2 implies P=zero` |
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| 12. |
A 20volts `AC` is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is `12V`, the voltage across the coil isA. 16 voltsB. 10 voltsC. 8 voltsD. 6 volts |
| Answer» Correct Answer - a | |
| 13. |
A 20volts `AC` is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is `12V`, the voltage across the coil isA. 16 VB. 10 VC. 8 VD. 6 V |
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Answer» Correct Answer - A `V^(2)` = |
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| 14. |
A `12 ohm` resistor and a `0.21` henry inductor are connected in series to an `AC` source operating at 20volts,` 50` cycle/second. The phase angle between the current and the source voltage isA. `30^(@)`B. `40^(@)`C. `80^(@)`D. `40^(@)` |
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Answer» Correct Answer - C `tan varphi=(omegaL)/R=(2pixx50xx0.21)/12=5.5 implies varphi=80^(@)` |
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| 15. |
A `12 ohm` resistor and a `0.21` henry inductor are connected in series to an `AC` source operating at 20volts,` 50` cycle/second. The phase angle between the current and the source voltage isA. `30^(@)`B. `40^(@)`C. `80^(@)`D. `90^(@)` |
| Answer» Correct Answer - c | |
| 16. |
If an `8Omega` resistance and `6Omega` reactance are present in an AC series circuit then the impedence of the circuit will beA. 20 ohmB. 5 ohmC. 10 ohmD. `14sqrt(2)` ohm |
| Answer» Correct Answer - c | |
| 17. |
The impedence of a circuit consists of `3ohm` resistance and `4ohm` reactance. The power factor of the circuit isA. 0.4B. 0.6C. 0.8D. 1 |
| Answer» Correct Answer - b | |
| 18. |
If an `8Omega` resistance and `6Omega` reactance are present in an AC series circuit then the impedence of the circuit will beA. `2Omega`B. `14Omega`C. `4Omega`D. `10Omega` |
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Answer» Correct Answer - D Z = `sqrt R^(2) + underset(L)(X)^(2)` = 10 `Omega` |
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| 19. |
Ratio of impedence to capactive reactance hasA. no unitsB. ohmC. ampereD. tesla |
| Answer» Correct Answer - 1 | |
| 20. |
If an `8Omega` resistance and `6Omega` reactance are present in an AC series circuit then the impedence of the circuit will beA. `20 ohm`B. `5 ohm`C. `10 ohm`D. `14sqrt(2) ohm` |
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Answer» Correct Answer - C Impedance `Z=sqrt(Z^(2)+X^(2))=sqrt((8)^(2)+(6)^(2))=10 Omega` |
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| 21. |
In an AC circuit, a resistance of `Rohm` is connected is series with an inductance `L`. If phase angle between volage and current be `45^(@)`, the value of inductive reactance will beA. R/4B. R/2C. RD. cannot be found with the given data |
| Answer» Correct Answer - C | |
| 22. |
In an AC circuit, a resistance of `Rohm` is connected is series with an inductance `L`. If phase angle between volage and current be `45^(@)`, the value of inductive reactance will beA. `R//4`B. `R//2`C. `R`D. cannot be found with the given data |
| Answer» Correct Answer - C | |
| 23. |
In an AC circuit, a resistance of `Rohm` is connected is series with an inductance `L`. If phase angle between volage and current be `45^(@)`, the value of inductive reactance will beA. `R/4`B. `R/2`C. `R`D. Cannot be found with the given data |
| Answer» Correct Answer - c | |
| 24. |
In an AC circuit, a resistance of `Rohm` is connected is series with an inductance `L`. If phase angle between volage and current be `45^(@)`, the value of inductive reactance will beA. R / 4B. R / 2C. RD. cannot be found with the given data |
| Answer» Correct Answer - C | |
| 25. |
The value of current in tow series LCR circuits at resonance is same, thenA. both circuits must be having same value of capacitance and inductanceB. in both circuits ratio of L and C will be sameC. for both the circuits `(X_L)//(X_C)` must be same at that frequencyD. both circuits must have same impedance at all frequencies |
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Answer» Correct Answer - C `X_(L)=X_(C) at resonance (X_L)/(X_C) = 1` for both circuits. Impendance may be different if applied voltage is different. |
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| 26. |
The value of current in two series `LCR` circuits at resonance is same when connected across a sinusodial voltage source. Then:A. both circuits must be having same value of capacitance and inductorB. in both circuits ratio of `L` and `C` will be sameC. for both the circuits `X_(L)//X_(C )` must be same at that frequencyD. both circuits must have same impedance at all frequencies |
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Answer» Correct Answer - C `X_(L) = X_(C )` at resonance `:. (X_(L))/(X_(C )) = 1`. For both circuits |
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| 27. |
A physics lab is designed to study the transfer of electrial energy from one circuit to another by means of a magnetic field using simple transformers. Each transformer has two coils of wire electrically insulated from each other but wound a round a common core of ferromagnetic material. The two wires are close together but do not touch each other The primary `(1^(@))` coil is connected to a source of alternating `(AC)` current. The secondary `(2^(@))` coil is connected to resistor such as a light bulb. The `AC` source produces on oscillating voltage and current in the primary coil that produces an oscillating megnetic field in the core. material. This in turn induces an oscillating voltage and `AC` curent in the secondary coil Students collected the following data comparing the number of turns per coil `(N)`, the voltage `(V)` and the current `(I)` in the coils of three transformers. `|{:("Transformer","primary coil","Secondary coil"),(1,100 . 10V.10A,200.20V.5A),(2,100.10V.10A,50.5V.5A),(3,100.10V.10A,100.5V.20A):}|` Which of the following is a correct expression for `R`, the resistance of the load connected to the secondary coilA. `((V_(1^(@)))/(I_(1^(@))))((N_(2^(@)))/(N_(1^(@))))`B. `((V_(1^(@)))/(I_(1^(@))))((N_(2^(@)))/(N_(1^(@))))^(2)`C. `((V_(1^(@)))/(I_(1^(@))))((N_(1^(@)))/(N_(2^(@))))`D. `((V_(1^(@)))/(I_(1^(@))))((N_(1^(@)))/(N_(2^(@))))^(2)` |
| Answer» Correct Answer - D | |
| 28. |
A physics lab is designed to study the transfer of electrial energy from one circuit to another by means of a magnetic field using simple transformers. Each transformer has two coils of wire electrically insulated from each other but wound a round a common core of ferromagnetic material. The two wires are close together but do not touch each other The primary `(1^(@))` coil is connected to a source of alternating `(AC)` current. The secondary `(2^(@))` coil is connected to resistor such as a light bulb. The `AC` source produces on oscillating voltage and current in the primary coil that produces an oscillating megnetic field in the core. material. This in turn induces an oscillating voltage and `AC` curent in the secondary coil Students collected the following data comparing the number of turns per coil `(N)`, the voltage `(V)` and the current `(I)` in the coils of three transformers. `|{:("Transformer","primary coil","Secondary coil"),(1,100 . 10V.10A,200.20V.5A),(2,100.10V.10A,50.5V.5A),(3,100.10V.10A,100.5V.20A):}|` A trasnsformer with 40 turns in its primary coil is connected to a `120 V AC` source. If `20 W` of power if supplied to the primary coil, how much power is developed in the secondary coil ?A. `10 W`B. `20 W`C. `80 W`D. `160 W` |
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Answer» Correct Answer - B `V_(1) i_(1) = V_(2) i_(2)` (or) `P_(1) = P_(2)` |
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| 29. |
A conducting rod of length l is hinged at point O. It is a free to rotate in a verical plane. There exists a uniform magnetic field B in horizontal direction. The rod is released from the position shown. The potential difference between the two ends of the rod is proportional to A. `i^(3//2)`B. `i^(2)`C. `sin theta`D. `(sin theta)^(1//2)` |
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Answer» Correct Answer - A::D |
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| 30. |
A square loop is placed near a long straight current carrying wire as shown. Match the following table. |
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Answer» Correct Answer - (A) Q,S, (B) P,R, (C) P,R, (D) Q,S |
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| 31. |
In the circuit shown in the figure, the jockey J is being pulled towards right, so that the resistance in the circuit is increasing. It’s a value at some instant is `5Omega`. The current in the circuit at this instant will be A. 4AB. less than 4AC. more than 4AD. may be less than or more than 4A depending on the value of L |
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Answer» Correct Answer - C |
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| 32. |
A square loop of side `a` is placed in the same plane as a long straight wire carrying a current `i`. The centre of the loop is at a distance `r` from wire where `r gt gt a`. The loop is moved away from the wire with a constant velocity `v`. The induced `e.m.f` in the loop is A. `(mu_(0)iv)/(2pi)`B. `(mu_(0)iav)/(2pir)`C. `(mu_(0)ia^(2)v)/(2pir^(2))`D. `(mu_(0)ia^(3)v)/(2pir^(3))` |
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Answer» Correct Answer - C |
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| 33. |
A copper rod of mass `m` slides under gravity on two smooth parallel rails `l` distance apart set at an angle `theta` to the horizontal. At the bottom, the rails are joined by a resistance `R`. There is a uniform magnetic field perpendicular to the plane of the rails. the terminal valocity of the rod isA. `(mgR cos theta)/(B^(2)I^(2))`B. `(mgR sin theta)/(B^(2)I^(2))`C. `(mgR tan theta)/(B^(2)I^(2))`D. `(mgR cos theta)/(B^(2)I^(2))` |
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Answer» Correct Answer - B |
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| 34. |
An `L-C` circuit consists of an inductor coil withh `L=50.0mH` and `20.0muF` capacitor. There is negligible resistance in the circuit. The circuit is driven by a voltage source with `V=V_0cosomegat`. If `V_0=5.00mV` and the frequency is twice the resonance frequency, determine. a. the maximum charge on the capacitor. b. the maximum current in the circuit. c. the phase relationship between the voltages across the inductor, the capacitor and the sourse. |
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Answer» Correct Answer - A::B::C::D `omega=2/(sqrtLC)=2/(sqrt(5xx10^-3xx20xx10^-6))` `632.4.5rad//s` `X_L=omegaL=(6324.5)(5xx10^-3)=31.62Omega` `X_C=1/(omegaL)=1/(6324.5xx20xx10^-6)=7.9Omega` `:. Z=X_L-X_C=23.72Omega` (a) Maximum voltage across capacitor `=i_0X_C=(0.211)(7.9)=1.67mV` `:.` Maximum charge `q_0=(20xx10^-6)(1.67xx10^-3)=33.4nC` (b) `i_0=V_0/Z=5/23.72mA=0.211mA` (c) Since `X_LgtX_C`, current in the circuit will lag behind the applied voltage by `pi//2`. Further voltage across the inductor will lead this current by `pi//2`. Therefore, applied voltage and voltage across inductor are in phase. Voltage across the capacitor will lag the circuit current by `pi//2`. Therefore, phase difference between `V_L` and `V_C` will be `180^@`. |
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| 35. |
A `40Omega` electric heater is connected to a `200V, 50Hz` main supply. The peak value of electric current flowing in the circuit is approx.A. `2.5A`B. `5A`C. `7A`D. `10A` |
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Answer» Correct Answer - C `i_(r.m.s)=(V_(r.m.s))/(R)=(200)/(40)=5A implies i_(0) =i_(r.m.s)sqrt(2)=7.07A` |
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| 36. |
A Choke coil is needed to operate an arc lamp at `160 V` (rms) and `50 Hz`. The lamp has an effective resistnce of `5 Omega` when running at 1`0 A("rms")`. Calculate the inductance of the choke coil. If the same arc lamp is to be operated on `160 V (DC)`, what additional resistance is required ? Compare the power loses in both cases. |
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Answer» As for lamp `V_(R)=IR=10xx5=50 V`,so when it is connected to `160 V` `ac` source through a choke in series. `V^(2)=V_(R)^(2)+V_(L)^(2) , V_(L)=sqrt(160^(2)-50^(2))=152 V` and as, `V_(L)=IX_(L)=IomegaL=2pifLI` So, `L=V/(2pifI)=152/(2xxpixx50xx10)=4.84xx10^(-2)H` Now the lamp is to be operated at `160 V dc`, instead of choke if additional resistance `r` is put in series with it `V=I(R+r)`,i.e.`160=10(5+r)` i.e.,`r=11 Omega` In case of `ac` ,as choke has no resistance , power loss in the choke will be zero while the bulb will consume, `P=I^(2)R=10^(2)xx5=500 W` However, in case of `dc` as resistance `r` is to be used instead of choke the power loss in the resistance `r` will be `PL=10^(2)xx11=1100 W` while the bulb will still consume `500 W`, i.e. when the lamp is run on resistance `r` instead of choke more than double the power consumed by the lamp is wasted by the resistance `r`. |
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| 37. |
A bulb is rated at `100V`, `100W`. It can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb at its rated power with the help of an ac source of `200V` and `50Hz`.A. `(pi)/(sqrt(3)) H`B. `100 H`C. `(sqrt(2))/(pi) H`D. `(sqrt(3))/(pi) H` |
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Answer» Correct Answer - 4 Resistance of bulb is `R = (100 xx 100)/(100) = Omega` Rated current is `(100)/(100) = 1A` In ac, `I_("rms") = (V_("rms"))/(Z) , Z = 200 Omega` `sqrt(100^(2) + (omega L)^(2)) = 200 implies omega^(2) L^(2) = 30000` and `L = sqrt((30000)/((100 pi)^(2))) = (sqrt(3))/(pi)` henry. |
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| 38. |
A `120 V, 60 Hz` a.c. power is connected `800 Omega` non-inductive resistance and unknown capacitance in series. The voltage drop across the resistance is found to be `102 V`, then voltage drop across capacitor isA. `8 V`B. `102 V`C. `63 V`D. `55 V` |
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Answer» Correct Answer - 3 `V^(2) = V_(R )^(2) + V_(C )^(2)` `V_(C )^(2) = V^(2) - V_(R )^(2)` `V_(C )^(2) = (120)^(2) - (102)^(2)` `V_(C ) = 63 V` |
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| 39. |
Assertion: The division are equaly marked on the scale of `AC` ammeter. Reason: Heat produced is directely proportional to the currentA. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - D An `AC` ammeter is consturcted on the basics of heating effect of the electric current. Since heat produced varies as square of current `(H=I^(2)R)`. Therefore the division marked on the scale of `AC` ammeter are not equally spaced. |
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| 40. |
A conducting rod `PQ` of length `l=1.0m` is moving with a uniform speed `v2.0m//s` in a uniform magnetic field `B=4.0T` directed into the paper. A capacitor of capacity `C=10muF` is connected as shown in figure. Then A. `q_(A)=+80muC and q_(B)=-80muC`B. `q_(A)=-80muC and q_(B)=+80muC`C. `q_(A)=0=q_(B)`D. charge stored in the capacitor increases expontially with time |
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Answer» Correct Answer - A |
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| 41. |
A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant, uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statements(s) from the followingA. The entire rod is at same electric potentialB. There is an electric field in the rodC. The electric potential is highest at the center of the rod and decreases towards its endsD. The electric potential is lowest at the center of the rod and increases towards its ends |
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Answer» Correct Answer - B |
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| 42. |
In a cylindrical region uniform magnetic field which is perpendicular to the plane of the figure is in increasing with time and a conducting rod `PQ` is placed in the region.If `C` is the centre of the circle then A. P wil be at thigher potential than QB. Q will be at higher potential than PC. Both P and Q will be at zero potentialD. No potential difference will be developed across the rod |
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Answer» Correct Answer - A |
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| 43. |
`AB` and `CD` are fixed conducting smooth rails placed in a vertical plane and joined by a constant current source at its upper end. `PQ` is a conducting rod which is free to slide on the rails. A horizontal uniform magnetic field exists in space as shown in figure. If the rod `PQ` is released from rest then, A. The rod PQ may move downward with ocnstant accelerationB. The rod PQ may move upward with constant accelerationC. The rod wil move downward with decreasing acceleration and finally acquire a constant velocityD. either a or b |
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Answer» Correct Answer - D |
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| 44. |
In `L-C-R` series `AC` circuit,A. If `R` is increased, then current will decreaseB. If `L` is increased, then current will decreaseC. If `C` is increased, then current will increaseD. If `C` is increased, then current will decrease |
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Answer» Correct Answer - A `I=V/Z` `=V/sqrt(R^2+(omegaL~1/(omegaC))^(2))` By increasing `R` current will definitely decrease by change in `L` or `C`, current may increase or decrease. |
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| 45. |
In the above problem further choose the correct options.A. The given values are at frequency less than the resonance frequencyB. The given values are at frequency more than the resonance frequencyC. If frequency is increased from the given value, impedance of the circuit will increaseD. If frequency is decreased from the given value, current in the circuit may increase or decrease |
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Answer» Correct Answer - B::C::D (b) At resonance frequency `(omega_r)` `X_LgtX_C` In the given values `X_LgtX_C`. Hence `omegaltomega_r` As, `X_L=omegaLimpliesX_Lpropomega` As `X_L=omegaLimpliesX_Lpropomega` and `X_C=1/(omegaC)impliesX_Cprop1/omega` (c) If frequency is increased from the given value, `X_L` will further increase. So, `X_L-X_C` will increase. Hence, net impedance will decrease. (d) If frequency is decreased from the given value., then `X_C` will increase and `X_L` will decrease. So, `X_L-X_C` may be less than the previous value of `X_C-X_L` may be greater than the previous. So `Z` may either increase or decrease Hence, current may decrease or increase. |
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| 46. |
In a circuit `L, C` and `R` are connected in series with an alternating voltage source of frequency `f`. The current lead the voltages by `45^(@)`. The value of `C` is :A. `(1)/(pi upsilon(2pi upsilon L-R))`B. `(1)/(2pi upsilon(2pi upsilon L-R))`C. `(1)/(pi upsilon(2pi upsilon L+R))`D. `(1)/(2pi upsilon(2pi upsilon L+R))` |
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Answer» Correct Answer - D As the current leads the voltage by `45^(@)`, therefore, `X_(C )gt X_(L), tan phi = (X_(C )-X_(L))/(R )=tan 45^(@)=1` `rArr X_(C )-X_(L)=R` or `X_(C )=X_(L)+R=omega L+R` `(1)/(omega C)=omega L+R, C=(1)/(omega(omega L+R))=(1)/(2pi upsilon(2pi upsilon L+R))` |
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| 47. |
In a circuit `L, C` and `R` are connected in series with an alternating voltage source of frequency `f`. The current lead the voltages by `45^(@)`. The value of `C` is : |
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Answer» As current leads the voltage by `45^(@)` `:.tan theta = (X_(C ) - X_(L))/(R ) = tan 45^(@) = 1` `:. X_(C ) - X_(L) = R` or `X_(C ) = X_(L) + R` or `(1)/(omega C) = omega L + R implies C = (1)/(omega (omega L + R))` `C = (1)/(2 pi f (2 pi fL + R))` |
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| 48. |
Identify the graph which correctly reperesents the variation of capacitive reactance `X_C` with frequencyA. B. C. D. |
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Answer» Correct Answer - B `X_(C)=1/(omegaC)=1/(2pifC)implies X_(C) prop1/f` |
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| 49. |
Identify the graph which correctly reperesents the variation of capacitive reactance `X_C` with frequencyA. B. C. D. |
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Answer» Correct Answer - B `X_C=1/(omegaC)=1/(2pifC)` or `X_Cprop1/f` i.e. `X_C` versus `f` graph is a rectangular hyperbola. |
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| 50. |
Which one of the follwing represents the variation of capacitive reactance `(X_C)` with the frequency (v) of the voltage source ? .A. B. C. D. |
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Answer» Correct Answer - D `X_(C)=1/(omegac)=1/(2pivC)` `X_(C)V=1/(2piC)="Constant"` |
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