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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.
| 1451. |
A metalic ring is attached with the wall of a room. When the north pole of a magnet is brought near to it, the induced current in the ring will be A. First clockwise, then anticlockwiseB. In clockwise directionC. In anticlockwise directionD. First anticlockwise, then clockwise |
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Answer» Correct Answer - C |
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| 1452. |
The figure shows certain wire segments joined together to form a coplaner loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time `I_(1) and I_(2)` are the currents in the segments ab and cd. Then, A. `I_(1) gt I_(2)`B. `I_(1) lt I_(2)`C. `I_(1)` is in the direction ba and `I_(2)` is in the direction cdD. `I_(1)` is in the direction ab and `I_(2)` is in the direction dc |
| Answer» Correct Answer - D | |
| 1453. |
Consider the situation shown in . The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separted by a distance l. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangualr region and a resistance R connects the rails outsided the field region At t= 0, the wire PQ is puched towards right with a speed `v_0`. find (a) the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instatn, (c ) the velocity v as a function of x and (d) the maximum distance the wire will move. A. `(2B^(2)l^(2)v)/(m(R + r))` towards leftB. `(B^(2)l^(2)v)/(m(R + r))` towards rightC. `(B^(2)l^(2)v)/(m(R + r))` towards leftD. none of these |
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Answer» Correct Answer - c |
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| 1454. |
Consider the situation shown in . The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separted by a distance l. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangualr region and a resistance R connects the rails outsided the field region At t= 0, the wire PQ is puched towards right with a speed `v_0`. find (a) the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instatn, (c ) the velocity v as a function of x and (d) the maximum distance the wire will move. A. `(Blv)/(r + R)`B. `(2Blv)/(r + R)`C. `(Blv)/(2 (r + R))`D. none of these |
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Answer» Correct Answer - a |
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| 1455. |
Consider the situation shown in the figure. The wire `PQ` has mass `m` , resistance `r` and can slide on the smooth, horizontal parallel rails separated by a distance `l` . The resistance of the rail is negligible. A uniform magnetic field `B` exists in the rectangular region and a resistance `R` connects the rail outside the field region. At `t=0` , the wire `PQ` is pushed toward right with a speed `v_(0)` . Find (a) the current in the loop at an instant when the speed of the wire `PQ` is `v` (b) the acceleration of the wire as this instant (c) the velocity `v` as a function of `x` (d) the maximum distance the wire will move (e) the velocity as a function of time A. `v = v_(0) - (2B^(2)l^(2)x)/(m(R + r))`B. `v = v_(0) - (B^(2)l^(2)x)/(m(R + r))`C. `v = v_(0) - (B^(2)l^(2)x)/(2m(R + r))`D. none of these |
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Answer» Correct Answer - b |
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| 1456. |
Shows a smooth pair of thick metallic rails connected across a battery of emf `epsilon` having a negligible internal resistance. A wire ab of length l and resistance r can slide smoothly on the rails. The entire system lie in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire is given a small velocity v towards right. (a) find the current in it at this instant. What is the direction of the current? (b) What is the force acting on the wire at this instant? (c) Show that after some time the wire ab will slide with a constant velocity. Find this velocity. A. `(1)/(r) (E - vBl)` , from b to aB. `(1)/(r) (E - vBl)` , from a to bC. `(1)/(2r)(E - vBl)` from b to aD. none of these |
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Answer» Correct Answer - a |
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| 1457. |
A variable frequency 230 V alternating voltage source is connected across a series combination of `L = 5.0 H, C = 80 mu F` and `R = 40 Omega` (i) Calculate the angurla frequency fo the source which drives the circuitin resonance. (ii) Obtain r.m.s. the impedance of the circuit and amplitude of current at resonance frequency (iii) Obtain r.m.s. potential drop across the three elements of the circuit at resonating frequency. (iv) How do you explain the observation that the algerbraic sum of voltages across the three elements obtained is greater than the supplied voltage. |
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Answer» (i) `omega_(r) = (1)/(sqrt(LC)) = (1)/(sqrt(5.0 xx 80 xx 10^(-6))) = 50 rad s^(-1)` (ii) At resonance, `Z = R = 40 ohm` `I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(Z) = (1.414 xx 230)/(40) = 8.13 A` (iii) `I_(v) = (E_(v))/(Z) = (230)/(40) = (23)/(4) A` `V_(R ) = I_(v) xx R = 23//4 xx 40 = 230` volt `V_(L) = I_(v) X_(L) = I_(v) (omega L) = 23//4 xx 50 xx 5.0 = 1437.5 V` `V_(C ) = I_(v) X_(C ) = (1_(v))/(omega C) = (23//4)/(50 xx 80 xx 40^(-6)) = 1437 .5 V` The three voltage, `vec V_(R)`, `vec V_(L)` and `vec V_(C )` are phasors/vector. Out of these, `vec V_(L)` and `vec V_(C )` are in opposite phase. Therefore, they cancel out. The vector sum of the three voltage `= vec V_(R ) = 230 V`, which is the source voltage. |
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| 1458. |
A circuit with `R=70 Omega` in series with a parallel combination of L=1.5H and `C=30 mu F` is a driven by a 230 V supply of angular frequency `300 rad//s`. (a) Find the impedance of the circuit. (b) What is the RMS value of total current ? (c ) What are the current amplitudes in the L and C arms of the circuit ? (d) How will the circuit behave at `omega=1//sqrt(LC)` ? |
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Answer» Correct Answer - `(a) 163.3 Omega (b) 1.414 A (c )0.653 A (d) Circuit current becomes zero |
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| 1459. |
In the adjoining `AC` circuit the voltmeter whose reading will be zero at resonance is A. `V_(1)`B. `V_(2)`C. `V_(3)`D. `V_(4)` |
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Answer» Correct Answer - B |
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| 1460. |
A coil of effective area `2 m^(2)` is rotated so as to cut a magnatic field of induction `7 x 10^(-5)` wb/m^2makes 100 revolution`//`sec, then the maximum e.m.f. Induced in the coil isA. 44 mVB. 88 mVC. 22 mVD. 200 mV |
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Answer» Correct Answer - B `e_(0)=nABomega = nAB cdot 2 pi n` `=2 xx 7 xx 10^(-5) xx 2 xx 22/7 xx 100 = 88 mV`. |
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| 1461. |
A straight conductor of length `0.4m` is moving with a velocity of `7 m//s`, in a magnetic field of induction `2 wb//m^2`. The value of the maximum induced e.m.f. In the conductor isA. 2VB. 3VC. 5.6VD. 2.8V |
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Answer» Correct Answer - C `e = Bvl = 2 xx 7 xx 0.4 = 5.6 V` |
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| 1462. |
A copper disc of radius `0.1 m` rotates about its centre with `10` revolutuion per second in a uniform magnetic field of `0.1` tesla with its plane perpendicular to the field. The emf induced across the radius of the disc isA. `(pi)/(10)V`B. `(2pi)/(10)V`C. `pixx10^(-2)V`D. `2pixx10^(-2)V` |
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Answer» Correct Answer - C `e=(1)/(2)Bomegar^(2)=(1)/(2)xx0.1xx2pixx10xx(0.1)^(2)=pixx10^(-2)V` |
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| 1463. |
In a region of uniform magnetic induction `B=10^(2)` tesla, a circular coil of radius `30 cm` and resistance `pi^(2)` ohm is rotated about an axis which is perpendicular to the directon of `B` and which form a diameter of the coil. If the coil rotates at `200rpm` the amplitude of the alternating current induced in the coil isA. `4pi^(2)mA`B. `30mA`C. `6mA`D. `200mA` |
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Answer» Correct Answer - C Amplitude of the current `I_(0)=(e_(0))/(R )=(omegaNBA)/(R )=(2pivNB(pir^(2)))/(R )` `i_(0)=(2pixx1xx10^(-2)xxpi(0.3)^(2))/(pi^(2))=6xx10^(-3)A=6mA` |
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| 1464. |
The length of each side of a square coil of 10 turns is 10 cm. It rotates in a magnetic field of flux density `25 xx 10^(-3)T`. If the maximum induced e.m.f. is 20 mV, then the angulare velocity of the coil will beA. 2 rad`//`secB. 4 rad`//`secC. 6 rad`//`secD. 8 rad`//`sec |
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Answer» Correct Answer - D `e = nAB omega :. Omega = e/(nAB)` `(20 xx 10^(-3))/(10 xx 100 xx 10^(-4) xx 25 xx 10^(-3)) = 8 rad//s`. |
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| 1465. |
A copper disc of radius `0.1 m` rotates about its centre with `10` revolutuion per second in a uniform magnetic field of `0.1` tesla. The emf induced across the radius of the disc isA. `(pi)/(10)V`B. `(2pi)/(10)V`C. `10pimV`D. `20pimV` |
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Answer» Correct Answer - C The induced emf between centre and rim of the rotating disc is `E=(1)/(2)BomegaR^(2)` `=(1)*(2)xx0.1xx2pi10xx(0.1)^(2)=10pixx10^(-3)"volt"` |
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| 1466. |
A coil having `500` square loops each of side `10 cm` is placed normal to a magnetic flux which increase at the rate of `1.0 "tesla"/"second"`. The induced r.m.f. in volts is |
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Answer» Here, `N = 500, A = 10 xx 10 cm^(2) = 10^(2) xx 10^(4) m^(2)=10^(-2) m^(2)` `(dB)/(dt) = 1.0 T//s, e = ?` `e = N (d phi)/(dt) = N A(dB)/(dt) = 500 xx 10^(-2) xx 1.0 = 5 ` volt |
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| 1467. |
A coil of area `0.05 m^(2)` and 500 turns is placed in a magnetic field of strength `4 xx 10^(-5)` tesla. If it is rotated through `90^(@)` in 0.1 sec, then the magnitude of e.m.f. Induced in the coil will beA. 10 mVB. 20 mVC. 5 mVD. 15 mV |
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Answer» Correct Answer - A `e = (d phi)/(dt) = (nAB)/t = (500 xx 5 xx 10^(-2) xx 4 xx 10^(-5))/(10^(-1))` `:. E = 10^(-2) V = 10 mV`. |
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| 1468. |
A metal rod of length 1m, rotates about its one end in a plane at right angles to a horizontal magnetic field of induction `7/22 xx 10^(-4)T`. If its frequency of rotation is 10 Hz, then the magnitude of induced e.m.f. IsA. 5mVB. 1mVC. 0.5VD. 1V |
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Answer» Correct Answer - B Flux `phi=B.A. Area swept//sec= (dA)/(dt) = f cdot pi r^(2)` `:. E = (d phi)/(dt)=B(dA)/(dt)=Bf cdot pi r^(2)` `:. E = 7/22xx10^(-4)xx10xx22/7xx1^(2)=1 mV`. |
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| 1469. |
The length of each side of a square coil of 10 turns is 10 cm. This coil rotates in a magnetic field of induction 0.02 Tesla. If the maximum induced e.m.f. in the coil is 20 mV, then the angular velocity of the coil will beA. `5 rad//s`B. `10 rad//s`C. `2.5 rad//s`D. `15 rad//s` |
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Answer» Correct Answer - B `omega = e/(nAB) = 10 rad//s`. |
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| 1470. |
A 800 turn coil of effective area `0.05 m^(2)` is kept perpendicular to a magnetic field `5xx10^(-5)` T. When the plane of the coil is rotated by `90^(@)` around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:A. `0.012 V`B. `0.05 V`C. ``0.1 V`D. `0.2 V` |
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Answer» Correct Answer - D `e=(-NBA(costheta_(2)-costheta_(1))/(Deltat)` `=-(500x4xx10^(-4)xx0.1(cos90-cos0))/(0.1)=0.2V` |
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| 1471. |
When a rod of length l is rotated with angular velocity of `omega` in a perpendicular field of induction B , about one end , the emf across its ends isA. `Bl^(2)omega`B. `0.5 Bl^(2)omega`C. `Bl omega`D. `0.5 Blomega` |
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Answer» Correct Answer - B Induced e.m.f. `(e) = (d phi)/(dt) = B cdot (dA)/(dt)` But `dA = pil^(2) and dt=T and T = (2 pi)/(omega)` `:. e = (B cdot pi l^(2))/((2pi//omega))=1/2 Bl^(2)omega`. |
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| 1472. |
A metal rod `1/(sqrt(pi))` m long rotates about one of its ends in a plane perpendicular to a magnetic field of induction `4 xx 10^(-3)T`. If the emf induced between the ends of the rod is 16 mv, then the number of revolutions made by the rod per second isA. 3 rpsB. 4 rpsC. 5 rpsD. 6 rps |
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Answer» Correct Answer - B The induced emf `e = (d phi)/(dt) = d/(dt)(AB) = B(dA)/(dt)` In one rotation the rod traces out a circle of radius l or an area `A = pi l^(2)`. `:.` In n rotations it will trace out an area of `n pi l^(2)` `:. e = B(dA)/(dt) = B cdot (n pi l^(2))/(dt)` But `n/(dt) = ("No. of rev. performed")/("time") = "frequency" f` `:. e = B f pi l^(2) = Bf cdot A` `:. F = (e)/(B pi l^(2)) = (16 xx 10^(-3))/(4 xx 10^(-3) xx [pi((1)/(sqrt(pi)))^(2)])` `= 4/(pi xx 1/(pi)) = 4 r.p.s.` |
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| 1473. |
A metal rod of length 1m is rotated about one of its ends in a place at right angle to a uniform magnetic field of induction `5xx10^(-3)` T. if it makes 1800 rotations per minute, then the emf iniduced between its ends isA. 0.471 VB. 4.71 VC. 4.17 VD. 1.47 V |
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Answer» Correct Answer - A `e=BA f=Bpir^(2)f=Bpil^(2)f` `=5xx10^(-3)xx3.14xx1xx30=0.471V` |
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| 1474. |
When a rod of length l is rotated with angular velocity of `omega` in a perpendicular field of induction B , about one end , the emf across its ends isA. `Bomegal^(2)`B. `2Bomegal^(2)`C. `(1)/(2)Bomegal^(2)`D. `(3)/(2)Bomegal^(2)` |
| Answer» Correct Answer - C | |
| 1475. |
A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity `omega` . The induced e.m.f between the two ends isA. `(1)/(2) B omegal^(2)`B. `(3)/(4) B omega l^(2)`C. `B omega l^(2)`D. `2 B omega l^(2)` |
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Answer» Correct Answer - a |
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| 1476. |
When a rod of length l is rotated with angular velocity of `omega` in a perpendicular field of induction B , about one end , the emf across its ends isA. `Bl^(2)omega`B. `0.5Bl^(2)omega`C. `Blomega`D. `0.5Blomega` |
| Answer» Correct Answer - B | |
| 1477. |
A half metre rod is rotating abou tone fixed end perpendicular to uniform magnetic field `4xx10^(-5)T` with angular velocity 720 rpm. The emf induced across its ends isA. 0.24 VB. 0.36 VC. 0.12 VD. 0.36 mV |
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Answer» Correct Answer - D `e=BA f` `=B pir^(2)f` `=(4xx10^(-5)xx3.14xx0.25xx720)/(60)` |
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| 1478. |
Induced emf produced in a coil rotating in a magnetic field will be maximum when the angle between the axis of coil and direction of magnetic field isA. `0^(@)`B. `90^(@)`C. `45^(@)`D. `180^(@)` |
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Answer» Correct Answer - B If the angle between the axis of coil and direction of magnetic field is `90^(@)` then magnetic flux through the coil is maximum. Hence, induced emf produced in a coil is maximum. |
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| 1479. |
A power transformer is used to step up an alternating e.m.f. of `220 V` to `11kv` to trannsmit `4.4 kW` of power. If the primary coil has `1000` turns, what is the current rating of the secondary?(Assume `100%` efficiency for the transformer)A. `4A`B. `0.4A`C. `0.04A`D. `0.2A` |
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Answer» Correct Answer - B `P_(P)=I_(P)*e_(P)` `therefore I_(P)=(P_(P))/(e_(P))=(4.4xx10^(3))/(220)=20A` `I_(S)=(e_(P))/(e_(S))xxI_(P)=(4.4xx10^(3))/(11xx10^(3))=0.4A` |
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| 1480. |
A conducting rod is rotated about one end in a plane perpendicular to a uniform magnetic field with constant angular velocity. The correct graph between the induced emf (e) across the rod and time (t) isA. B. C. D. |
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Answer» Correct Answer - C `e=(Bomegal^2)/2=`constant |
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| 1481. |
An emf induced in a coil rotating in a uniform magnetic field is given byA. `e=e_(0)sin omegat`B. `e_(0)=esinomegat`C. `e=sinomegat`D. `e=e_(0)sin omegat` |
| Answer» Correct Answer - A | |
| 1482. |
A rectangular coil ABCD which is rotated at a constant angular veolcity about an horizontal as shown in the figure. The axis of rotation of the coil as well as the magnetic field B are horizontal. Maximum current will flow in the circuit when the plane of the coil is A. inclined at `30^(@)` to the magnetic fieldB. perpendicular to the magnetic fieldC. inclined at `45^(@)` to the magnetic fieldD. parallel to the magnetic field |
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Answer» Correct Answer - B As the coil is rotated, angle `theta` (angle which is normal to the coil makes with B at any instant t) changes, therefore magnetic flux `phi` linked with the coil changes and hence an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then `e=-(dphi)/(dt)=-(d)/(dt)(NBS cos omegat)` where, N in number of turns in the coil. or `e=-NSB(d)/(dt)(cosomegat)=-NSB(-sinomegat)omega` or `e=NBSomega sinomegat` ....(i) The induced emf will be maximum when, sin `omega`t i.e., maximum `therefore" "e_("max")=e_(0)=NSBomegaxx1or e=e_(0)sin omegat` Therefore, e would be maximum, hence current is maximum (as `i_(0)=e_(0)//R`), when `theta=90^(@)` i.e., normal to plane of coil is perpendicular to the field or plane of coil is parallel to magnetic field. |
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| 1483. |
A rod of length `10 cm` made up of conducting and non-conducting material (shaded part is non-conducting). The rod is rotated with constant angular velocity `10 rad//s` about point `O`, in constant magnetic field of `2 T` as shown in the figure. The induced `emf` between the point `A` and `B` of rod will be: A. `0.029 v`B. `0.1 v`C. `0.051 v`D. `0.064 v` |
| Answer» Correct Answer - C | |
| 1484. |
Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform. With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain. |
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Answer» Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T` Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega` Velocity of rod `V=12" cm/s "12 xx 106(-2) m//s.` When key is open, there is no net force on the electrons because the presence of excess charge at P and Q sets up an electric field and magnetic forcé on the electrons is balanced by force on them due to force by the electric field. So, there is no net force on the rod. |
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| 1485. |
Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform. What is the retarding force on the rod when K is closed? |
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Answer» Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T` Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega` Velocity of rod `V=12" cm/s "12 xx 106(-2) m//s.` When the key Kis closed, current flow in the loop and the current carrying wise experience a retarding force in the magnetic field which is given by `"Force"=BIl=B.e/R.l =(0.5xx9xx10^(-3)xx15xx10^(-12))/(9xx10^(-3))=7.5xx10^(-2)N` |
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| 1486. |
Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform. How much power is required (by an external agent) to keep the rod moving at the same speed `(=12" cm s"^(-1))` when K is closed ? How much power is required when K is open? |
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Answer» Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T` Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega` Velocity of rod `V=12" cm/s "12 xx 10^(-2) m//s.` To keep the rod moving at the same speed the required power `="retarding force "xx" velocity "=7.5xx10^(-2)xx12xx10^(-2)=9xx10^(-3)W` |
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| 1487. |
Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform. Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? |
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Answer» Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T` Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega` Velocity of rod `V=12" cm/s "12 xx 106(-2) m//s.` Yes, an excess positive charge developes at P and the same amount of negative charge developes at Q as the key is open. When the key K is closed, the induced current flows and maintains the excess charge. |
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| 1488. |
Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform. What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular? |
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Answer» Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T` Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega` Velocity of rod `V=12" cm/s "12 xx 10^(-2) m//s.` When the field is parallel to length of rails `theta-0^(@)`, induced emf `=e=BVl sin theta=0` `(therefore sin theta^(@)=0)`. In this situation, the moving rod will not cut the field lines so that flux change is zero and hence induced emf is zero. |
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| 1489. |
An air cored solenoid is of length 0.3 m, area of cross section `1.2 xx 10^(-3) m^(2)` and has 2500 turns. Around. The solenoid and the coil are electrically insulated from eachother. Calculate the e.m.f. induced in the coil if the initial current of 3 A in the solenoid is reversed in 0.25 s. |
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Answer» Here, `l = 0.3 m, A = 1.2 xx 10^(-3) m^(2), N_(1) = 2500` `N_(2) = 350, e = ?, dI = - 3 - (3) = - 6 A` `dt = 0.25 s`. `e = - M (dI)/(dt) = - (mu_(0) N_(1) N_(2) A)/(l) ((dI)/(dt))` ` = (- 4pi xx 10^(-7) xx 2500 xx 350 xx 1.2 xx 10^(-3) (-6))/(0.3 xx 0.25)` `e = 0.106 V` |
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| 1490. |
An air-cored solenoid with length 30 cm, area of cross-section `25 cm^(2)` and number of turus 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of `10^(3)s`. How much is the average back emf induced across the ends of the open switch in the circuit ? Ignore the variation in magnetie field near the ends of the solenoid. |
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Answer» Given, length of solenoid `l=30" cm "30 xx 10^(-2)m` Area of cross-section `A=25"cm "25 xx 10^(-4)m^(2)` Number of turns N=500, Current `I_1=2.5A, I_(2)=0`, Brief time `dt=10^(-3)s`. Induced emf in the solenoid `e=(dphi)/(dt)=(d(BA))/(dt) (therefore phi=BA)` Magnetic field induction B at a point well inside the long solenoid carrying cuurent I is `B=mu_(0)nI("where, n"=N/l)` `e=NA (dB)/(dt)=A(d)/(dt)(mu_(0)N/l.I)` `e=A(m_(0)N)/(l).(dI)/(dt)` `e=500xx25xx10^(-4)xx4xx3.14xx10^(-7) xx (500)/(30xx10^(-2))xx(2.5)/(10^(-3)) Rightarrow e=6.5V` |
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| 1491. |
Calcualte the self-inducatance of the air cored solined of length 80 cm and has 500 turns and its cicular cross- section has diamter of 2 cm.A. `150.6 muH`B. `162.2 muH`C. `123.3 muH`D. `102.5 mu H` |
| Answer» Correct Answer - C | |
| 1492. |
A series LCR circuit is connected to an a.c. source 200 V, 50 Hz. The voltages acorss the resistor, indcutor and capacitor are respectively 200 V, 250 V and 250 V. (i) The algerbriac sum of the three voltages is greater than the source voltage. How is the paradox resolved ? (ii) If `R = 40 Omega`, calculate the current in the circuit. |
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Answer» Here, `E_(v) = 200 V, v = 52 Hz,` `V_(R ) = 200 V, V_(C ) = 250 V, V_(L) = 250 V` Now, `V_(R ) + V_(C ) + V_(L) = 200 + 250 + 250` `= 700 V`, which is greater than the source voltage 200 V. This is because voltages across C and L are in oppsoite phase and canel out. The sum continues to be 200 V, the source voltage. `I_(v) = (E_(v))/(Z) = (E_(v))/(R ) = (200)/(40) = A` |
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| 1493. |
A current of 30 mA is taken by a `4 mu F` capacitor connected acorss an alternating current line having frequency of 500 Hz. Calculate reactance of the capacitor and voltage across the capacitor. |
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Answer» Here, `I_(v) = 30 mA = 30 xx 10^(-3) A`, `C = 4 mu F = 4 xx 10^(-6) F` `v = 500 Hz, X_(C ) = ? V _(C ) = ?` `X_(C ) = (1)/(omega C) = (2)/(2 pi v xx C)` `(1)/(2 xx 3.14 xx 500 xx 4 xx 10^(-6))` `= (1000)/(12.56) = 79.6 Omega` `V_(C ) = I_(v) X_(C ) = 30 xx 10^(-3) xx 79. 6 = 2.4 V` |
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| 1494. |
In a transformer with transformation ratio 0.1, 220 volt a.c. is fed to primary. What voltage is obtained acorss the secondary ? |
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Answer» `K = (n_(s))/(n_(p)) = (e_(s))/(e_(p))` `0.1 = (e_(s))/(220) :.e_(s) = 22` volt |
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| 1495. |
A step down transformer is used at 220 V to provide a current of 0.5 A to a 15 W bulb. If the secondary has 20 turns, find the number of turns in primary coil has current that flows in primary coil. |
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Answer» Here, `E_(P) = 220 V, I_(s) = 0.5 A, P_(s) = 15 W`, `n_(s) = 20, n_(P) = ? I_(P) = ?` `E_(s) = (P_(s))/(I_(s)) = (15)/(0.5) = 30 V` As `(E_(P))/(E_(s)) = (n_(P))/(n_(s)) :.n_(P) = (E_(P))/(E_(s)) xx n_(s) = (220)/(30) xx 20` = 147 Also, from `E_(P) I_(P) = E_(s) I_(s)` `I_(P) = (E_(s) I_(s))/(E_(P)) = (P_(s))/(E_(P)) = (15)/(220) = 0.068 A` |
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| 1496. |
An electric motor operator on `15 V` supply, draws a current of `5 A` and yields mechanical power of `60 W`. The energy lost as heat in one hour (in kJ) is.A. `0.54`B. `5.4`C. 54D. 540 |
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Answer» Correct Answer - C `V = 15 V, I = 5 A, P = 60 W, t = 1 h = 3600 sec`. Power supplied `= VI = 15 xx 5 = 75 W` Mechaical power yielded `P = 60 W` Power lost = power lost `= 75 - 60 = 15 W` Energy lost = Power lost `xx` time `= 15 xx 3600 = 54000 J = 54 kJ` |
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| 1497. |
A step down transformer has a trum ratio of 5:1. it is connected to 220 V, 50 Hz a.c. Mains supply. The secondary voltage and the frequency of secondary voltage are given byA. 44V, 10 HzB. 110V, 50 HzC. 44V, 50 HzD. 1100V, 10 Hz |
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Answer» Correct Answer - C `5/1=(V_p)/(V_s) = 220/(V_s) :. V_(s) = 220/5 = 44 V` There is no change in frequency. |
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| 1498. |
An electric motor operating on a `50 V dc` supply draws a current of `12 A`. If the efficiency of the motor is `30 %`,estimate the resistance of the windings of the motor. |
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Answer» Here, `V = 50` volt, `I = 12` ampere., `eta = 30%, R = ?` Input electric power ` = VI = 50 xx 12 = 600` watt As efficiency is 30%, therefore, power dissipated as heat is 70% of this power `P = (70)/(100) (VI) = (70)/(100) xx 600 = 420 W` As `P = I^(2) :. R = (P)/(I^(2)) = (420)/(12 xx 12) = 2.9 Omega` |
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| 1499. |
In a step-up transformer the voltage in the primary is `220 V` and the currrent is `5A`. The secondary voltage is found to be `22000V`. The current in the secondary (neglect losses)isA. `5A`B. `50A`C. `500A`D. `0.05A` |
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Answer» Correct Answer - D `(V_(p))/(V_(s))=(i_(s))/(i-(p))implies(220)/(22000)=(i_(s))/(5)implies(i_(s)=0.05` amp |
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| 1500. |
An electric motor operating on a `60 V` dc supply draws a currrent of `10A`. If the effeciency of the motor is `50%`, the resistance of its winding isA. `6Omega`B. `4Omega`C. `2.9Omega`D. `3.1Omega` |
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Answer» Correct Answer - C `eta=(e)/(E)xx100impliese=0.3E` Now, `i=(E-e)/(R )=(220-210)/(2)=(10)/(2)=5A` |
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