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. |
Find the dimension of the quantity L/(RCV), where symbols have usual meaning. |
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Answer» Correct Answer - `I^(-1)` `[L/(RCV)]=(sec)/(Amp//sec)=[I^(-1)]` `**L/R=` time constant |
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| 2. |
A wire forming one cycle of sine curve is moved in x-y plane with velocity `vec(V)=V_(x)hat(i)+V_(y)hat(j)`. There exist a magnetic field `vec(B)=-B_(0)hat(k)`. Find the motional emf decelop across the ends PQ of wire. |
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Answer» Correct Answer - `lambdaV_(y)B_(0)` `e=-(vec(B)xxvec(V)).vec(l)=-[B_(0)hat(k)xx(V_(x)hat(i)+V_(y)hat(j)].lambdahat(i)=B_(0)V_(y)lambda` |
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| 3. |
The phase diffenernce between the alternating current and emf is `(pi)/(2)`. Which of the following cannot be the constiuent of the circuit?A. `C` aloneB. `R, L`C. `L, C`D. `L` alone |
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Answer» Correct Answer - 2 `tan phi=X/R=oo=1/0 implies R=0` |
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| 4. |
A horizontal wire is free to slide on the verticle rails of a conducting frame as shown in figure. The wire has a mass `m` and length `l` and the resistance of the circuit is `R`. If a uniform magnetic field `B` is directed perpendicular to the frame, the terminal speed of the wire as it falls under the force of gravity is |
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Answer» Correct Answer - `(mgR)/(B^(2)l^(2))` `mg=IlB implies mg=e/RlB` `m mg=(Blv)/RlB implies v=(mgR)/(B^(2)l^(2))` |
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| 5. |
An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuit |
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Answer» Correct Answer - `2A, 400W` From given condition circuit is at resonance `tanphi=X_(L)/R implies sqrt(3)=X_(L)/R` `X_(L)=sqrt(3)R implies X_(L)=X_(C )=sqrt(3)R` `I_(rms)=V_(rms)/R=200/100=2 Amp` `P=I^(2)R=4xx100=400 Watt` |
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| 6. |
A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is L. A conducting massless rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table, A mass m, tied to the other end of the string hanges vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from rest, calculate. (i) the terminal velocity achieved by the rod, and the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity. |
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Answer» Correct Answer - `(i) V_(terminal)=(mgR)/(B^(2)Z^(2)); (ii) g/2` (i) `IlB=mg …(1)n` `I=e/R=(Blv)/R …(2)` From `(1)` and `(2)` `(B^(2)l^(2)v)/R=mg implies v=(mgR)/(B^(2)l^(2))…(3)` (ii) `ma=mg-IlB` `ma=mg-(Bl(v//2)lB)/R=mg-(B^(2)l^(2)v)/(2R)` `=mg-(mg)/2 …(4)` From (3) and (4) `a=g/2` |
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| 7. |
A uniform magnetic field of `0.08 T` is directed into the plane of the page and perpendicular to it as shown in the figure. A wire loop in the plane of the page has constant area `0.010 m^(2)`. The magnitude of magnetic field decrease at a constant rate of `3.0xx10^(-4)Ts^(-1)`. Find the magnitude and direction of the induced enf in the loop. |
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Answer» Correct Answer - `3 muV`, clockwise `phi=BA` `e=(dphi)/(dt)=A(dB)/(dt)=0.1xx3xx10^(-4)=3muV` since main flux is decreasing so induce flux will be in same direction so current is clockwise direction. |
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| 8. |
In series `LR` circuit, `X_(L) = 3 R`. Now a capacitor with `X_(C ) = R` is added in series. The ratio of new to old power factorA. `1`B. `2`C. `1/sqrt(2)`D. `sqrt(2)` |
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Answer» Correct Answer - D `cos phi_(1)=R/sqrt(R^(2)+X_(L)^(2))=R/sqrt(R^(2)+(3R)^(2))=1/sqrt(10)` `cos phi_(2)=R/sqrt(R^(2)+(X_(L)-X_(C ))^(2))=R/(R^(2)+(2R)^(2))=1/sqrt(5)` `(cos phi_(2))/(cos phi_(1))=sqrt(2)/1` |
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| 9. |
A vertical bar magnet is dropped from from position on the axis of a fixed metallic coil as shown in fig-I. In fig - II the magnet is fixed and horizontal coil is dropped. The accelaration of the magnet and coil are `a_(1)` and `a_(2)` respectively then A. `a_(1)gt g, a_(2) gt g`B. `a_(1)gt g, a_(2) lt g`C. `a_(1)lt g, a_(2) lt g`D. `a_(1)lt g, a_(2) gt g` |
| Answer» Correct Answer - C | |
| 10. |
Statement 1: In the purely resistive element of a series LCR, ac circuit the maximum value of rms current increase with increase in the angular frequency of the applied emf. Statement 2: `I_(max)=(epsilon_(max))/(Z), Z=sqrt(R^(2)+(omega L-(1)/(omega C)^(2)))`, where `(I_(max))` is the peak current in a cycle.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D The maximum value of ems current `=epsilon_(rms)/z =epsilon_(rms)/R`. It does not upon `omega`. |
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| 11. |
An AC current is given by `I=I_(0)+I_(1)` sin wt then its rms value will beA. `sqrt(I_(0)^(2)+0.5I_(1)^(2))`B. `sqrt(I_(0)^(2)+0.5I_(0)^(2))`C. `0`D. `I_(0)//sqrt(2)` |
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Answer» Correct Answer - A `I_(rms)=sqrt(1/Tint_(0)^(T)(I_(0)+I_(1) sin omegat)^(2)dt)` `sqrt(1/T int_(0)^(T)I_(0)^(2)dt+int_(0)^(T)I_(1)^(2) sin^(2) omegatdt+int_(0)^(T)(2I_(0)I_(1))/T sin 2omegatdt)` `=sqrt(I_(0)^(2)+(I_(1)^(2)//2))=sqrt(I_(0)^(2)+0.5I_(1)^(2))` |
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| 12. |
In (figure9a)a solenoid produce a magnetic field whose stength increases into the plane of the page. An induced emf is established in a conduction loop surrounding the solenoid, and this emf lights bulbs A and B. In figure (b) P and Q are shorted. After the short is inserted A. Bulb A goes out bulb B gets brighterB. Bulb B goes out bulb A gets brighterC. Bulb A goes out bulb B gets dimmerD. Bulb B goes out bulb A gets dimmer |
| Answer» Correct Answer - A | |
| 13. |
A long solenoid of `N` turns has a self-induced `L` and area of cross-section A. When a current `i` flows through the solenoid, magnetic field inside it has magnitude `B` . The current `i` is equal toA. `BAN//L`B. `BANL`C. `BN//AL`D. `B//ANL` |
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Answer» Correct Answer - A `phi=Li implies NBA=LIimpliesI=(NBA)/L` |
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| 14. |
L, C and R represent the physical quantities, inductance, capacitance and resistance respectively. The combination(s) which have the dimensions of frequency areA. `1//RC`B. `R//L`C. `1/sqrt(LC)`D. `C//L` |
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Answer» Correct Answer - A, B, C `RC` = time constant `L/R=` time constant. |
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| 15. |
In an ac circuit the potential differences across an inductance and resistance joined in series are, respectively, 16 V and 20 V. The total potential difference across the circuit is |
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Answer» Correct Answer - `20 V` `V=sqrt(V_(R)^(2)+V_(L)^(2))=sqrt((12)^(2)+(16)^(2))=sqrt(144+256)` =20 volt |
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| 16. |
Suppose the emf of the battery, the circuit shown varies with time `t` so the current is given by `i(t)=3+5t`, where `i` is in amperes & `t` is in seconds. Take `R=4Omega, L=6H` & find an expression for the battery emf as function of time. |
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Answer» Correct Answer - `42+20t volt` `E=iR+L(di)/(dt)=(3+5t)4+6xx5=42+20t` |
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| 17. |
A cionducing rod moves with constant velocity `v` perpendicular to the straight wire carrying a current I as shown compute that the emf generated between the ends of the rod A. `(mu_(0)vIl)/(pir)`B. `(mu_(0)vIl)/(2pir)`C. `(2mu_(0)vIl)/(pir)`D. `(mu_(0)vIl)/(4pir)` |
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Answer» Correct Answer - B `e=Blv=mu_(0)/(4pi)(2I)/rlv` `e=(mu_(0)Ilv)/(2pir)` |
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| 18. |
The power in ac circuit is given by `P=E_(rms)I_(rms)cosphi`. The vale of cos phi in series LCR circuit at resonance is:A. zeroB. `1`C. `1/2`D. `1/sqrt(2)` |
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Answer» Correct Answer - B At resonance `cosphi=1` |
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| 19. |
An inductor of inductance `2.0 mH` s connected across a charged capacitor of capacitance `5.0 mu F` and the resulting `LC` circuit is set oscillating at its natural frequency. Let `Q` denote the instantaneous charge on the capacitor and `I` the current in the circuit. It is found that the maximum value of charge `Q` is `200 mu C`. a. When `Q = 100 mu C`, what is the value of `|(dI)/(dt)|`? b. When `Q = 200 mu C`, what is the value of `I`? c. Find the maximum value of `I`. d. When `I` is equal to one-half its maximum value, what is the value of `|Q|`? |
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Answer» Correct Answer - `(a) 10^(4) A//s (b) 0 (c ) 2A (d) 100sqrt(3)muC` Given that `Q=Q_(0) cosomegat` `omega=1/sqrt(LC)=1/sqrt(2xx10^(-3)xx5xx10^(-6))=10^(4) rad//sec` `(a) Q=Q_(0) cosomegat implies 100=200 cosomegat` `cosomegat=1/2` `since Q=Q_(0) cosomegat implies I=(dQ)/(dt)=Q_(0) omega sinomegat` `(dI)/(dt)=-Q_(0)omega^(2) cosomegat` `|(dI)/(dt)|=omega^(2)(Q_(0)cosomegat)=omega^(2)Q=10^(8)xx100xx10^(-6)` `10^(4) Amp//sec` `(b) Q=Q_(0)cosomegat` `200=200 cosomegat implies cosomegat=1` `sinomegat=0` `I=(dQ)/(dt)=-Q_(0)omega sin omegat=0` `(c ) l_(max)=Q_(0)omega=200xx10^(-6)xx10^(4)=2Amp` `(d) I=-I_(0) sinomegat` `I_(0)/2=-I_(0)sinomegat` `sinomegat=sin (7pi)/6 implies omegat=(7pi)/6` `cosomegat=cos(7pi)/6=-cospi/6=-(-sqrt(3))/2` `Q=Q_(0)cosomegat=200muCxx(-sqrt(3)/2)` `=-100sqrt(3)muC` `|Q|=100sqrt(3) muC` |
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| 20. |
Asseration: An emf is induced in a along solenoid by a bar magnet that moves while totally inside it along the solenoid axis. Reason: As the magnet moves inside the solenoid the flux through individual turns of the solenoid changes.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D Even though flux through individual lines changes, it remains unchanged for the solenoid as a whole. Therefore no emf is induced in the long solenoid. |
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| 21. |
There is a series `LCR` curcuit (as shown). An alternating source of emf having voltage `V=V_(0) sin omega t` is applied between `M` & `N`. Here `V_(M)-V_(N)=V_(0) sin omega t` and `1/(omegaC)-omegaL=R` The potential difference across `R` has got a phase difference of `theta` of alternating source of enf. whereA. `theta=pi/2`B. `theta=0`C. `theta=pi/4`D. `theta=pi` |
| Answer» Correct Answer - C | |
| 22. |
There is a series `LCR` curcuit (as shown). An alternating source of emf having voltage `V=V_(0) sin omega t` is applied between `M` & `N`. Here `V_(M)-V_(N)=V_(0) sin omega t and 1/(omegaC)-omegaL=R` R.M.S value of potential difference across capacitor will beA. `V_(0)/sqrt(2)`B. `V_(0)RomegaC`C. ZeroD. `V_(0)/(2RomegaC)` |
| Answer» Correct Answer - D | |
| 23. |
STATEMENT-1: In a series LCR circuit at resonance, the voltage across the capacitor or inductor may be more than the applied voltage. STATEMENT-2: At resonance in a series LCR circuit, the voltages across inductor and capacitor are out of phase.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - B | |
| 24. |
There is a series `LCR` curcuit (as shown). An alternating source of emf having voltage `V=V_(0) sin omega t` is applied between `M` & `N`. Here `V_(M)-V_(N)=V_(0) sin omega t and 1/(omegaC)-omegaL=R` Potential difference across inductor i.e. `V_(Q)-V_(N)` isA. `V_(0)/(sqrt(2)R) omegaL sin (omegat+(3pi)/4)`B. `(V_(0)omegaL)/(sqrt(2)R) sin omegat`C. `(V_(0)omegaL)/(sqrt(2)R) sin (omegat+pi/2)`D. `(V_(0)omegaL)/(sqrt(2)R) cos omegat` |
| Answer» Correct Answer - A | |
| 25. |
STATEMENT-1: If a changed perticle is released from rest in a time varying magnetic field, it moves in a circle. STATEMENT-2: In time varying magnetic field electric field is induced.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - D | |
| 26. |
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating, It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :A. development of air current when the plate is placed.B. induction of electrical charge on the plateC. shielding of magnetic lines of force as aluminimum is a paramagnetic material.D. electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. |
| Answer» Correct Answer - 4 | |
| 27. |
For L-R circuit, the time constant is equal toA. twice the ratio of the energy strored in the magnetic field to the rate of dissipation of energy in the resistanceB. ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistanceC. half the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistanceD. square of the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance |
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Answer» Correct Answer - A `tau=L/R=(2xx1/2LI^(2))/(I^(2)R)` |
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| 28. |
A ring of resistance `10Omega`, radius `10 cm` and `100` turns is rotated at a rate revolutions per second about its diameter is perpendicular to a uniform magnetic field of induction `10 mT`. The amplitude of the current in the loop will be nearly `("Take": pi^(2)=10)`A. `200A`B. `2A`C. `0.002A`D. none of these |
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Answer» Correct Answer - B `phi=NAB cos omegat` `e=NBAomega sinomegat` `e_(0)=NBAomega` `I_(0)=(NBAomega)/R` `=(100xx10^(-2)xxpixx(.1)^(2)2pixx100)/10=2A` |
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| 29. |
Current growth in two L-R circuits (b) and (c ) as shown in figure (a). Let `L_(1), L_(2), R_(1)` and `R_(1)` be the corresponding values in two circuits. Then A. `R_(1)gt R_(2)`B. `R_(1)=R_(2)`C. `RL_(1)gt L_(2)`D. `L_(1)lt L_(2)` |
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Answer» Correct Answer - B, D `i=E/R(1-e^(-t//tau)) =I_(0)(1-e^(-t//tau))` `I_(0)=E/R_(1)=E/R_(2) implies R_(1)=R_(2)` `tau_(2) gt tau_(1) implies L_(2)/R_(1) gt L_(2)/R_(2)` `L_(2) gt L_(1)` |
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| 30. |
Two metallic rings ` A and B`, identical in shape and size but having different reistivities `rhoA and rhoB`, are kept on top of two identical solenoids as shown in the figure . When current `I` is switched on in both the solenoids in identical manner, the rings ` A and B` jump to heights `h_(A) and h_(B)`, repectively , with ` h_(A) gt h_(B)` . The possible relation(s) between their resistivities and their masses `m_(A) and m_(B)` is(are) A. `rho_(A) gt rho_(B) and m_(A)=m_(B)`B. `rho_(A) lt rho_(B) and m_(A)=m_(B)`C. `rho_(A) gt rho_(B) and m_(A)gt m_(B)`D. `rho_(A) lt rho_(B) and m_(A)lt m_(B)` |
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Answer» Correct Answer - B, D Since the rings are indentical and the solenoids are identical, magnetic field `B=mu_(0) n I` and hence magnetic flux phi=BxxA (here A = area of the ring) is the same for both the rings. Hence the magnetitude of induced enf `(e)` is the same. Since `h_(A) gt h_(B)`, the force exerted on ring `A` greater than that on `B`. Hence current induced in A greater than that in `B`, i.e. `I_(A) gt I_(B) implies e/R_(A) gt e/R_(B) implies R_(B) gt R_(A)` Now `R=(rl)/A`. Hence `rho_(A) lt rho_(B)`. Mass `m_(A)` can be less than, equal to or greater than `m_(B)`. Hence the current choices are `(B)` and `(D)` |
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| 31. |
A current of `2 A` is increasing at a rate of `4 A s^(-1)` through a coil of inductance `1 H`. Find the energy stored in the inductor per unit time in the units of `J s^(-1)`.A. `2 J//s`B. `1 J//s`C. `16 J//s`D. `4 J//s` |
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Answer» Correct Answer - C `P=ei=L(di)/(dt)xxi=2xx4xx2` `P=16 J//sec` |
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| 32. |
A conducting rod `PQ` of length `5 m` oriented as shown in figure is moving with velocity `(2 m//s)hat(i)` without any rotation in a uniform magnetic field `(3hat(j)+4hat(k))` Tesla. Emf induced in the rod is A. 32 VoltsB. 40 VoltC. 50 VoltD. none |
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Answer» Correct Answer - A `vec(l)=5cos53^(@)hat(i)+5 sin 53^(@)hat(j)=3hat(i)+4hat(j)` `e=|(vec(V)xxvec(B)).vec(l)|=|2hat(i)xx(3hat(j)+4hat(k)).(3hat(i)+4hat(j))|` `=|(6hat(k)-8hat(j)).(3hat(i)+4hat(j))|=32` volt |
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| 33. |
In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L). At the start, the capicitor was uncharged. when switch `(S_1)` is closed and `(S_2)` is kept open, the time constant of this circuit is `tau`. which of the following is correctA. after time interval `tau`, change on the capacitor is `CV//2`B. after time interval `2tau`, charge on the capacitor is `CV(1-e^(-2))`C. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.D. after time interval `2tau`, charge on the capacitor is `CV(1-e^(-1))` |
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Answer» Correct Answer - B `Q=Q_(0)(1-e^(-t//tau))=CV(1-e^(-2))` |
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| 34. |
For L-R circuit, the time constant is equal toA. twice the ratio of the energy strored in the magnetic field to the rate of dissipation of energy in the resistanceB. the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.C. half of the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistanceD. square of the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance |
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Answer» Correct Answer - A `tau=(2xx1/2LI^(2))/(I^(2)R)=L/R` |
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| 35. |
The adjoining figure shows two bulbs `B_(1)` and `B_(2)` resistor `R` and an inductor and `L`. When the switch `S` is turned off A. The bulb `B_(2)` lights up earlier than `B_(1)` and finally both the bulbs shine equally bright.B. `B_(1)` light up earlier and finally both the bulbs acquire equal brightness.C. `C_(2)` lights up earlier and finally `B_(1)` shines brighter than `B_(2)`.D. `B_(1)` and `B_(2)` light up togather with equal brightness all the time. |
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Answer» Correct Answer - A Due to induction effect inductor delay the growth of current. |
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| 36. |
A `DC` ammeter cannot measure alternating current becauseA. A.C. current pass through d.C. ammeterB. A.C change directionC. average value of current for complete cycle is zeroD. D.C ammeter will get damaged |
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Answer» Correct Answer - 3 The full cycle of alternating current consists of two half cycles. For one half, current is positive and for second half, current is negative. There-fore, for an a.c. cycle, the net value of current average out zero. While for the half cycle, the value of current is different at different points. Hence, the alternating current cannot be measured by D.C. ammeter |
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| 37. |
An AC voltage source of variable angular frequency `(omega)` and fixed amplitude `V_(0)` is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When `(omega)` is increasedA. The bulb glows dimmerB. The bulb glows brighterC. Total impedence of the circuit is unchangedD. Total impendance of the circuit increases |
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Answer» Correct Answer - B Impendance `Z=[R^(2)+1/((omegaC)^(2))]^(1//2), I_(0)=V/Z` As `omega` increases, `Z` decreases. Hence the current `I_(0)` increass. Therefore, the blub glows brighter, which is choice `(B)`. |
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| 38. |
A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of side a(altltR) having two turns is placed with its centre at `=sqrt(3)R` along the axis of hte circular wire loop, as shown in figure. The plane of the square loop makes an angle of `45^(@)` with respect to the z-axis. If the mutual inductance between the loops is given bu `(mu_(0)a^(2))/(2^(p//2)R)`, then the value of p is |
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Answer» Correct Answer - 7 `B=(mu_(0)IR^(2))/(2(4R^(2))^(3//2))=(mu_(0)I)/(16R)` `M=(BA cos 45^(@))/l= 1/sqrt(2)=1/2^(7//2)(m_(0)a^(2))/R` |
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| 39. |
If the total charge stored in the `LC` circuit is `Q_(0)`, then for `tgt= 0`A. the charge on the capacitor is `Q=Q_(0) cos(pi/2+t/sqrt(LC))`B. the charge on the capacitor is `Q=Q_(0) cos(pi/2-t/sqrt(LC))`C. the charge on the capacitor is `Q=-LC(d^(2)Q)/(dt^(2))`D. the charge on the capacitor is `Q=-1/sqrt(LC)(d^(2)Q)/(dt^(2)` |
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Answer» Correct Answer - C `Q-Q_(0) cosomegat …(1)` `(dQ)/(dt)=-Q_(0)omegasinomegat implies(d^(2)Q)/(dt^(2))=-Q_(0)omega^(2) cosomegat…(2)` From `(1)` and `(2)` `(d^(2)Q)/(dt^(2))=-omega^(2)Q` `implies Q=-1/omega^(2)(d^(2)Q)/(dt^(2))=-LC(d^(2)Q)/(dt^(2))` |
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| 40. |
A series `L-C-R` circuit containing a resistance of `120Omega` has resonance frequency `4xx10^5rad//s`. At resonance the voltages across resistance and inductance are `60V` and `40V`, respectively. Find the values of `L` and `C`.At what angular frequency the current in the circuit lags the voltage by `pi//4`? |
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Answer» Correct Answer - `0.2 mH, 1/32 muF, 8xx10^(5) rad//s` At resonance `V_(L)=V_(C )=40V` `V=V_(R)=60V` and `I_(max)=60/120=0.5 Amp`. `I_(max)X_(L)=40 implies X_(L)=80 Omega` `omegaL=80 implies L=0.2 mH` `X_(C )=80 implies 1/(4xx10^(5))C=80` `C=1/32xx10^(-6)=1/32 muF` Let at frequency `omega` current lags the voltage by `45^(@)` `tanphi=x/R implies X=R implies X_(L)-X_(C )=R` `omegaL=1/(omegaC)=R` Putting the value of `X_(L), X_(C )` and `R` we get `omega=8xx10^(5) rad//s` |
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