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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.
| 1351. |
In the circuit shown in Fig. (i) find the phase difference between the currents through L and `R_(1)` (ii) find the phase difference between potential differece across C and `R_(2)` |
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Answer» (i) In Fig. as L and `R_(1)` are in parallel the potential difference acorss them are equal and in phase with eachother. Further, in resistance `R_(1)`, current and petentail difference are in phase with eachother, while in inductance, current lags behind the potential difference by a phase angle `pi//2` Hence, current through inductance L lags behind the current through `R_(1)` by a phase `pi//2` (ii) As C and `R_(2)` are in series, same current passes through them and is same phase, In `R_(2)`, current and potential difference are in phase with eachother. While n a capacitor potential difference lags behind the current by a phase angle `pi//2`. Hence, potential difference across the capacitor C lags behind the current in `R_(2)` by a phase anlge `pi//2`. |
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| 1352. |
How does self inductance of a solenoid change when number of turns is double keeping othe parameters same ? |
| Answer» As `L prop N^(2)`, `:.` when N is double, L becomes four times. | |
| 1353. |
Current is passed through a wire passing normally through the centre of a conducting loop, Fig. If currents is increased gradually, what will be the direction fo induced current in the loop ? |
| Answer» The magnetic lines of force due to current passed are parallel to the plane of the loop. Therefore magnetic flux linked with the loop is zero. Hence no current is induced in the loop. | |
| 1354. |
The plane of two circular conductors are perpendicular to each othe, as shown in Fig. If current in any one is changed, will there be induced current in the other ? |
| Answer» No. This is because magnetic flux due to changing current in any one conductor is parallel to the planeof other conductor. Therefore, magnetic flux linked with the other conductor is zero. Hence there will be no induced current. | |
| 1355. |
A radio can tune over the frequency range of a portion of MW broadcast band (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of `200 mu H`, what must be the range of its varialbe capacitor ? |
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Answer» Here, `v_(1) = 800 kHz = 8 xx 10^(5) Hz, v_(2) = 1200 k Hz = 12 xx 10^(5) Hz` `L = 200 mu H = 200 xx 10^(-6) H = 2 xx 10^(-4) H` `C_(1) = ? C_(2) = ?` From `v_(1) = (1)/(2 pi sqrt(LC_(1)))` `C_(1) = (1)/(4 pi^(2) L v_(1)^(2)) = (1)/(4 xx (22)/(7) xx (22)/(7) xx 2 xx 10^(-4) (8 xx 10^(5))^(2)) = 197.7 xx 10^(-12) F = 197.7 pF` Similarly, `C_(2) = (1)/(4 pi^(2) L v_(2)^(2)) = (1)/(4 xx (22)/(7) xx (22)/(7) xx 2 xx 10^(-4) (12 xx 10^(5))^(2)) = 87.8 xx 10^(-12) farad = 87.8 pF` |
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| 1356. |
A.d.c. motor acts as generator too. Is it ture ? |
| Answer» Yes. A.d.c. motor generates induced e.m.f. which oppose the supplied e.m.f. It is called back e.m.f. | |
| 1357. |
Why do we prefer a choke coil to a rhestat in controlling a.c.? |
| Answer» This is because no energy/power is dissipated in a choke coil. In a rhestat, some energy is dissipated in the form of heat. | |
| 1358. |
What is the basic difference in the design of an a.c. generator and d.c. generator ? |
| Answer» The slip ring arrangement in an a.c. generator is replaced by split ring arrangement or commutator arrangement in d.c. generator. | |
| 1359. |
Which generator do you prefer to install , a.c. or d.c. and why? |
| Answer» We perfer to install ac generator than dc generator as a.c. generator can produce three phase a.c. whereas the dc generator can produce single phase dc only, with the same amout of energy spent. | |
| 1360. |
You are given a fixed length of wire to design a generator. For a given magnetic field strength and given frequency of rotation, will you use one turn or two turn square coil to generate maximum peak e.m.f. ? |
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Answer» As in known, the peak e.m.f. is given by `e_(0) = NAB omega`. As B and `omega`fixed, `e_(0)` will be maximum, when `N xx A =` maximum. Let L be length of given wire. If `N = 1, A = (L//4)^(2) = (L^(2))/(16) , NA = 1 xx (L^(2))/(16) = (L^(2))/(16)` If `N = 2, A = (L//8)^(2) = (L^(2))/(64) , NA = 2 xx (L^(2))/(64) = (L^(2))/(32)` Hence one turn square coil can be be used to generated maximum peak e.m.f. |
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| 1361. |
A generator develops an e.m.f. of 120 V and has a terminal potential difference of 115 V, when the amature current is 25 A. What is the resistance of the armature ? |
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Answer» Here, `E = 120 V, V = 115 V, I = 25 A` As `I = (E - V)/(R )` `:. R = (E - V)/(I) = (120 - 115)/(25) = (5)/(25) = 0.2 Omega` |
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| 1362. |
A conducting loop of cross-area `A` and resistance `R` is placed perpendicular to a magnetic field `B` . The loop is withdrawn completely from the field. The change, which flows through any cross-section of the wire in the processA. `(B)/(R)`B. `(BA)/(R)`C. `(B)/(2R)`D. `(2B)/(R)` |
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Answer» Correct Answer - B `Deltaq=(Deltaphi)/(R)=(BA)/(R)` |
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| 1363. |
A square- shaped copper coil has edges of length 50cm and contains 50 turns. It is placed perpendicular to 1.0 T magnetic field. It is removed form the magnetic field in 0.25 s and restored in its joriginal place in the next 0.25 s. Find the magnitude of the average emf induced in the loop during (a) its removal, (b) its restoration and (c) its motion.A. `50V`B. `55V`C. `60V`D. `45V` |
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Answer» Correct Answer - B `|bar(e)|=(Deltaphi)/(Deltat)=(NBA)/(Deltat)=(50xx1xx5050xx10^(-4))/(0.25)=50V` |
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| 1364. |
A coil of area 100` cm^(2)` having 50turns is perpendicular to a magnetic field of intensity 0.02T. The resistance of the coil is `2Omega`. If t is removed from magnetic field in is the charge flown through the coil is:A. 5CB. 0.5CC. 0.05CD. 0.005C |
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Answer» Correct Answer - B |
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| 1365. |
The frequency of ac mains in India isA. 30 HzB. 50HzC. 60HzD. 120Hz |
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Answer» Correct Answer - A |
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| 1366. |
A magnetic field of flux density 10 T act normal to a 50 turn coil of `100 cm^(2)` area. Find the emf induced in it if the coil is removed from the field in 1/20 s. |
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Answer» Correct Answer - 100 V |
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| 1367. |
Shows a square loop having 100 turns, an area of `2.5 xx10^(-3) m^2` and a resistance of `100 Omega`. The magnetic field has a magnitude B =0.40 T. Find the work done in pulling the loop out of the field, slowly and uniformly is 1.0 s. A. (a) `0`B. (b) `1mJ`C. ( c) `1muJ`D. (d) 0.1mJ` |
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Answer» Correct Answer - D (d) `W = Fl(NIBl)l = NBl^(2)((BVNl)/(R ))` `=(N^(2)B^(2)l^(2))/(R )V = (N^(2)B^(2)l^(3))/(R ) (l)/(t) = (N^(2)B^(2)l^(4))/(R t) = 0.1mJ` |
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| 1368. |
How does the sign of the phase angle `phi`, by which the supply voltage leads the current in an LCR serices circuit, change as the supply frequency is gradually increased from very low to very high values ? |
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Answer» The supply voltage leads the current in LCR series circuit by `/_ phi` , where `tan phi = (X_(L) - X_(C ))/(R ) = (2 pi v L - 1//2 pi v C)/(R )` When v is low, `tan phi` is negative ,`phi` is negative. When `v = v_(r) , 2 pi v L = 1//2 pi v C, tan phi = 0 , phi = 0^(@)`. When `v gt v_(r) , tan phi` is positive , `phi` is positive. |
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| 1369. |
A resistor of `100 Omega`, inductance of 1 H and a capacitor of capacitance `10.13 xx 10^(-6) F` are in series. This combination is conneceted to an A.C. source of 200 V, 50 Hz. Find the current in the circuit and the P.D. across the resistor. |
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Answer» Here, `R = 100 Omega, L = 1 H`, `C = 200 V, v = 50 Hz, I_(v) = ?, V = ?` `E_(v) = 200 V, v = 50 Hz, I_(v) = ?, V = ?` `X_(L) = omega L = 2 pi v L = 2 xx 3.14 xx 50 xx 1 = 314 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (1)/(2 xx 3.14 xx 50 xx 10.13 xx 10^(-6))` `= (1000)/(3.14 xx 10.13) = 314.38 Omega ~= 314 Omega` As `X_(L) = X_(C )`, therefore, circuit is in reasonace. `I_(v) = (E_(v))/(R ) = (200)/(100) 2 A` `V = I_(v) xx R = 2 xx 100 = 200 V` |
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| 1370. |
The variation of induced emf `(E)` with time `(t)` in a coil if a short bar magnet is moved along its axis with a constant velocity is best represent as A. B. C. D. |
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Answer» Correct Answer - B The ploarity of emf will be opposite in the two cases while the magnet enters the coil and while the magnetic leaves the coil. Only in option (b) polarity is changing . |
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| 1371. |
The variation of induced emf `(F)` with time `(t)` in a coil if a short bar magnet is moved along its axis with a constant velocity is best represent as B. C. D. |
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Answer» Correct Answer - B As the magnet moves towards the coil, the magnetic flux increase (nonlinearly). Also there is a change in polarity of induced emf when the magnet passes on to the other side of the coil. |
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| 1372. |
The variation of induced emf `(E)` with time `(t)` in a coil if a short bar magnet is moved along its axis with a constant velocity is best represent as A. B. C. D. |
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Answer» Correct Answer - B As the magnet move towards the coil, the magnetic flux increases (non-linearly). Also, there is a change in polarity of induced emf when the magnet passes on the other side of the coil. |
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| 1373. |
The current `i` in a coil varies with time as shown in the figure. The variation of induced emf with time would be A. B. C. D. |
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Answer» Correct Answer - D We know induced induced emf `e=-L(di)/(dt)` During `0` to `(T)/(4),(di)/(dt)`=constant so, induced emf `e=-ve` for `(T)/(4)` to `(T)/(2),(di)/(dt)=0` induced emf `e=0` For `(T)/(2)` to `(3T)/(4),(di)/(dt)`=constant sol induced emf `e=+ve` |
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| 1374. |
The induced emf in a coil of 10 H inductance in which current varies from 9 A to 4 A in 0.2 s isA. 200 VB. 250 VC. 300 VD. 350 V |
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Answer» Correct Answer - B `because` Induced emf, `e=-(di)/(dt)` Given, L = 10 H, `Deltai = 9-4=5A, dt =0.2s` emf, `e=10xx(5)/(0.2)=250V` |
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| 1375. |
(a) the magnetic field in a region varies as shown in . Calculate the average induced emf in a conducting loop of area `2.0xx10^(-3) m^2` placed perpendicular to the field in each of the 10ms intervals shown. (b) In which intervals is the emf not constant? Neglect the behaviour near the ends of 10 ms intervals. |
| Answer» Correct Answer - A::B::C::D | |
| 1376. |
Assertion : A square loop is placed in x-y plane as shown in figure. Magnetic field in the region is` B =- B_0x hatk`. The induced current in the loop is anti-clockwise. Reason : If inward magnetic field from such a loop increases, then current should be anti-clockwise.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Due to non uniform magnetic field (a function of `x`) magnetic flux passing through the loop obtained by integration. But that remains constant with time Hence,` (dphi)/(dt)=0` `or e=0` Magnetic field is along `-hatk` direction or in `ox` magnetic is increasing. Hence, induced current should produce `o.` magnetic field. Or induced curent shoud be anti cloclwise. |
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| 1377. |
A magnetic flux through a stationary loop with a resistance `R` varies during the time interval `tau` as `phi=at(tau-t)`. Find the amount of the generated in the loop during that timeA. `(atau^2)/(2R)`B. `(a^2tau^2)/(3R)`C. `(2a^2tau^3)/(3R)`D. `(atau)/(3R)` |
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Answer» Correct Answer - B `e=|(dphi)/(dt)|=(atau=2at)` `i=e/R=(atau-2at)/R` `H=int_0^taui^2Rdt` |
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| 1378. |
A magnetic field `B = B_(0) sin (omega t) hat k` covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d, Fig. The wires are in the x-y plane. The wire AB (of length d) has resistance R and parallel wires have negligible resistance. If AB is moving with velocity `upsilon`, what is the current in the circuit ? What is the force needed to keep the wire moving at constant velocity ? |
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Answer» Let us assume that the parallel wires at are y=0 i.e., along x-axis and `y=d`. At `t=0`, AB has `x=0`, i.e, along y-axis and moves with a velocity v. Let at time t, wire is at `x(t)=vt`. Now, the motional emf across AB is `=(B_0sinomegat)vd(-hatj)` emf due to change in field (along) OBAC) `=-B_0omegacosomegatx(t)d` Total emf in the circuit =emf due to change in field (along OBAC)+the motional emf across AB `=-B_0d[omegaxcos(omegat)+vsin(omegat)]` Electric current in clockwise direction is given by `=(B_0d)/(R)(omegaxcosomegat+vsinomegat)` The force acting on the conductor is given by `F=ilBsin90^@=ilB` Substituting the values, we have Force needed along `i=(B_0d)/(R)(omegaxcosomegat+vsinomegat)xxdxxB_0sinomegat` `=(B_0^2d^2)/(R)(omegaxcosomegat+vsinomegat)sinomegat` This is the required expression for force. |
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| 1379. |
The loop shown moves with a velocity `v` in a uniform magnetic field of magnitude `B`, directed into the paper. The potential differene between point `P` and `Q` is e. Then A. `e=(1)/(2)BLv`B. e=BLvC. P is positive with respect to QD. Q is positive with respect to P |
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Answer» Correct Answer - A::D |
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| 1380. |
A circular brass loop of radius a and resistance R is placed with it plane perpendicular to a magnetic field, which varies with time as `B = B_(0) sin omega t`. Obtain the expression for the induced current in the loop. |
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Answer» Induced current, `I = (I n d u c e d e.m.f.)/(resistance) (e) /(R )` ` = - (d phi // dt)/(R ) = (-1)/(R ) (d)/(dt) (BA cos 0^(@))` `I = - (A)/(R ) (d)/(dt) (B_(0) sin omega t) = - (AB_(0))/(R ) cos omega t (omega)` `= -(A omega B_(0))/(R )` |
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| 1381. |
A square of side L meters lies in the x-y plane in a region, where the magnetic field is give by `B = B_(0) (2 hati + 3 hat j + 4 hatk)`T, where `B_(0)` is constant. The magnitude of flux passing through the square isA. `2B_(0)L^(2)Wb`B. `3B_(0)L^(2)Wb`C. `4B_(0)L^(2)Wb`D. `sqrt29B_(0)L^(2)Wb` |
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Answer» Correct Answer - C The magnetic flux linked with uniform surface fo area of A in uniform magnetic field is given by `phi=B.A` Here, `A=L^(2)hatkand B=B_(0)(2hat"i"+3hat"j"+4hatk)T` `therefore" "phi=B.A=B_(0)(2hat"i"+3hat"j"+4hatk).L^(2)hatk=4B_(0)L^(2)Wb` |
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| 1382. |
Now assume that the straight wire carries a current of 50A and the loop is moved to the right with a constant velocity, `upsilon=10m//s`. Calculate the induced emf in the loop at the instant when x=0.2m. Take `a=0.1m` and assume that the loop has a large resistance. |
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Answer» Given current `I=50A, "Velocity" V=10m//s, x=0.2m and a=0.1m` Induce emf e `=(-d theta)/(dt)=(-d)/(dt)[I(mu_(0)a)/(2pi)log_(e)((a+x)/(x))]` `e=(-mu_(0)aI)/(2pi).d/(dt)(log(x+a)-logx)` `=(-mu_(0)aI)/(2pi)((1)/((x+a))d/dt(x+a)-1/x.d/dt(x))=(-mu_(0)aI)/(2pi)((V)/(x+a)-(V)/(x))` `=(-mu_(0)aI)/(2pi).(a^(2).IV)/(x(a+x))Rightarrow e=(4pixx10^(-7))/(2pi)xx((0.1)^(2)xx(50)xx(10))/(0.2xx(0.1+0.2))=(2xx10^(-7)xx50xx10^(-2)xx10)/(0.2xx0.3)` `e=1.67xx10^(-5)V` |
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| 1383. |
In the circuit shown in figure `L=10H, R=5Omega, E=15V.` The switch `S` is closed at `t=0`. At `t=2s` the current in the circuit is A. `3(1-(1)/(e))A`B. `3(1-(1)/(e^(2)))A`C. `3((1)/(e))A`D. `3((1)/(e^(2)))A` |
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Answer» Correct Answer - C |
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| 1384. |
In the circuit shown in figure `L=10H, R=5Omega, E=15V.` The switch `S` is closed at `t=0`. At `t=2s` the current in the circuit is A. `3(1-1/e)A`B. `3(1-1/e^2)A`C. `3(1/e)A`D. `3(1/e^2)A` |
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Answer» Correct Answer - A `tau_L=L/R=2s` `i_0=E/R=3A, t=2s` `i=i_0(1-e^(-t//tau_L))` Substituting the given values, we can find i. |
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| 1385. |
A 10 V battery connected to `5omega` resistance coil having inductance 10 H through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is 2ms. The average emf induced across the coil isA. `4xx10^(4)V`B. `2xx10^(4)V`C. `2xx10^(2)V`D. `1xx10^(4)V` |
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Answer» Correct Answer - D Here, curernt `I=(V)/(R)=(10)/(5)=2A` Final current, when the swithc is opened is zero. `:. (dI)/(dt)=(0-2)/(2xx10^(-3))=-1xx10^(-3) A s^(-1)` As `epsi=-L(dI)/(dt)=-10(-1xx10^(-3))=10^(4) V` |
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| 1386. |
On passing a current of 2A in a coil inductance 5H, the energy stored in it will be-A. 100 jouleB. 10 jouleC. 20 jouleD. 5 joule |
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Answer» Correct Answer - B `1/2Li^(2)=1/2 xx50xx10^(-3) xx2^(2)=10 J` |
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| 1387. |
An average induced e.m.f. of `1V` appears in a coil when the current in it is changed from `10` `A` in one direction to `10A` in opposite direction in `0.5` sec. self-inductance of the coil isA. `25mH`B. `50mH`C. `75mH`D. `100mH` |
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Answer» Correct Answer - A `|e|=L(di)/(dt)impliesI=(Lxx{10-(-10)})/(0.5)` `impliesL=25mH` |
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| 1388. |
The magnetic flux linked with a coil, in webers, is given by the equation `f=3T^(2)+4t+9`. Then the magnitude of induced e.m.f. at `t=2` second will beA. `2` voltB. `4` votlC. `8` voltD. `16` volt |
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Answer» Correct Answer - D `e=-(dvarphi)/(dt)=-(d)/(dt)3(t^(2)+4t+9)=-(6t+4)` `=-[6(2)+4]=-16implies|e|=16` volt |
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| 1389. |
A coil of inductance `L=50muH` and resistance = `0.5Omega` is connected to a battery of emf = 5V. A resistance of `10Omega` is connected parallel to the coil. Now, at the same instant the connection of the battery is switched off. Then, the amount of heat generated in the coil after switching off the battery isA. 1.25 mJB. 2.5 mJC. 0.65 mJD. 0.12 mJ |
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Answer» Correct Answer - D Energy, `U=(1)/(2)Li_(0)^(2)=(1)/(2)L((V)/(R))^(2)` `=(1)/(2)xx50xx10^(-6)((5)/(0.5))^(2)J=2.5mJ` Now, after switching off the battery this energy is dissipated in coil (resistance = `0.5Omega`) and resistnace `(10Omega)` in the ratio of their resistances (`H=i^(2)Rt` or `HpropR`). Therefore, heat generated in the coil. `((5)/(0.5))^(2)((0.5xx10)/(0.5+10))2.5mJ=0.12mJ` |
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| 1390. |
A coil of resistance `10Omega` and an inductance `5H` is connected to a `100` volt battery. Then energy stored in the coil isA. `12.5J`B. `125J`C. `25.0J`D. `250J` |
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Answer» Correct Answer - D `U=(1)/(2)Li^(2)=U=(1)/(2)xx5xx(100)/(10^(2))=250J` |
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| 1391. |
Two inductor coils of self inductance `3H` and `6H` respectively are connected with a resistance `10Omega` and battery `10 V` as shown in figure.The ratio of total energy stored at steady state in the inductors to that of heat developed in resistance in `10` seconds at the steady state is (neglect mutual inductance between `L_(1)` and `L_(2)`): A. `1/10`B. `1/100`C. `1/1000`D. `1` |
| Answer» Correct Answer - B | |
| 1392. |
Some magnetic flux is changed from a coil of resitance `10 Omega`. As a result, an induced current is developed it, which varies with time as shown in Fig. 3.213. Find the magnitude of the change in flux through ythe coil in weber. A. `8`B. `2`C. `6`D. `4` |
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Answer» Correct Answer - B `I=|(1)/(2)(dphi)/(dt)|` `|sphi|=|Irdt|` `dphi=("area of triangle")xxR` `=(1)/(20xx4xx0.1)xx10=2Wb` |
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| 1393. |
Some magnetic flux is changed from a coil of resitance `10 Omega`. As a result, an induced current is developed it, which varies with time as shown in Fig. 3.213. Find the magnitude of the change in flux through ythe coil in weber. |
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Answer» Correct Answer - 2 `Deltaphi = R(Deltaq) = Rint idt` =R[area under i-t graph] `= (1)/(2)(4)(0.1)(10) =2 Wb` |
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| 1394. |
An inductor of 2 H and a resitance of `10Omega` are conncts in series with a bttery of 5 V. The intial rate of change of current isA. `0.5As^(-1)`B. `2.0As^(-1)`C. `2.5As^(-1)`D. `0.25As^(-1)` |
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Answer» Correct Answer - C Growth of current in the circuit is `i=i_(0)(1-e^(-Rt//L))` `rArr" "(di)/(dt)=(d)/(dt)i_(0)-(d)/(dt)(i_(0)e^(-Rt//L))` `therefore" "(di)/(dt)=0-i_(0)(-(R)/(L))e^(-Rt//L)=(i_(0)R)/(L)e^(-Rt//L)` Initially, t = 0, `therefore" "(di)/(dt)=(i_(0)R)/(L)=(E)/(L)=(5)/(2)=2.5As^(-1)` |
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| 1395. |
Some magnetic flux is changed from a coil of resitance `10 Omega`. As a result, an induced current is developed it, which varies with time as shown in Fig. 3.213. Find the magnitude of the change in flux through ythe coil in weber. A. 2B. 4C. 6D. none of these |
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Answer» Correct Answer - A From `dq = I dt = (dphi)/(R )` `d phi = (I dt) R` `= ("area under I = t graph") xx R` `= ((1)/(2) xx 4 xx 0.1) xx 10 = 2 web` |
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| 1396. |
Some magnetic flux is changed from a coil of resitance `10 Omega`. As a result, an induced current is developed it, which varies with time as shown in Fig. 3.213. Find the magnitude of the change in flux through ythe coil in weber. A. `2`B. `4`C. `6`D. none of these |
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Answer» Correct Answer - A `|dq|=(dphi)/(R )=I dt=`area under `i-t` graph :. `dphi=`(Area under `i-t` graph )`R` `=(1)/(2)xx4xx0.1xx(10)=(2)Wb`. |
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| 1397. |
Two long parallel horizonal rails , a distance d apart and each having a resistance `lambda` per unit length are jointed at one end by a resitance R. A perfectly conducting rod MN of mass m is free to slide along the rails . A variable force F is applied to the rod MN , such that ,as the rd moves, a constant current i flows through R. (a) Find the velocity v and the force F as function of the function of the distance x of the rod from R. (b) What fraction `eta` of the work done by F per second in converted into heat ? |
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Answer» Correct Answer - `v=((R+ 2xlambda)i)/(Bd) F=B+(2mlambdar^(2)(R + 2 x lambda))/(B^(2)d^(2)),eta =(R)/((R+2 x lambda)+(2 lambdax (R+2x lambda )^(2))/(B^(3)d^(3) |
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| 1398. |
A rod of length l and mass m rests on two smooth parallel conductors shorted at one end by an inductor L and open at the end . The circuits is in a unifrom field B perpendcular in to the plance . The conductor is suddenly imparted an initial velocity `V_(0)` direction to the right . Show that the motion is simple harmonic . Find its angular frequency and amplitude . |
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Answer» Correct Answer - `omega= Bl//sqrt(mL),A= v_(0) sqrt(mL)//Bl` |
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| 1399. |
In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit isA. 1.5 WB. 2.5 WC. 2 WD. 3 W |
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Answer» Correct Answer - B `overline(P)=(e_(0)I_(0))/(2)"cos"(pi)/(3)=2.5W` |
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| 1400. |
In an `AC` circuit, `V` and `I` are given by `V=100sin(100t)volts, I=100sin(100t+(pi)/3)mA`. The power dissipated in circuit isA. 250 WB. 25 WC. 2.5 WD. 5 W |
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Answer» Correct Answer - C Power (P) = `E_(V)I_(V) cos phi` `=100/(sqrt2) xx 100/(sqrt2) xx 10^(-3) xx 1/2` `=2500 xx 10^(-3) = 2.5 W`. |
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