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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. |
Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance `1Omega//m`. Position of the conducting rod at `t=0` is shown. A time dependent magnetic field `B=2t` tesla is switched on at `t=0` At `t=0`, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed `5cm//s` by some external means. At `t=2s`, net induced emf has magnitudeA. (a) `0.12 V`B. (b) `0.08 V`C. ( c) `0.04 V`D. (d) `0.02 V` |
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Answer» Correct Answer - B (b) `(dB)/(dt) = 2 T//s` `E = -(AdB)/(dt) = -800 xx 10^(-4)m^(2) = 0.16 V` `I = (0.16)/(1 Omega) = 0.16 A, clockwise` At `t = 2 s, B = 4 T, (dB)/(dt) = 2 T//s ` `a = 20 xx 30 cm^(2)` `= 600 xx 10^(-4)m^(2), (dA)/(dt) = -(5 xx 20)cm^(2)//s` `= -100 xx 10^(-4)m^(2)//s` `E = -(dphi)/(dt) = -[(d(BA))/(dt)] = -[(BdA)/(dt) + (AdB)/(dt)]` `= -[4 xx(-100 xx 10^(-4)) + 600 xx 10^(-4) xx 2]` `= -[-0.04 + 0.120] = -0.08 V` Alternative: `phi = BA = 2t xx 0.2(0.4 - vt)` `E = -(dphi)/(dt) = 0.8 vt - 0.16` At `t = 2s` `E = 0.08 V` At `t =2s`, "length of the wire" `= (2 xx 30 cm) + 20cm = 0.8` Resistance of wire `= 0,8 Omega` Current through the rod `= (0.08)/(0.8) = (1)/(10)A` Force on the wire is `= ilB` `= (1)/(10) xx (0.2) xx 4 = 0.08 N` Same force is placed on the rod in opposite direction to make net force zero. |
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| 2. |
The unit of `L//R` is (where `L=` inductance and `R=` resistance)A. AmpereB. VoltC. per secD. sec |
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Answer» Correct Answer - D `e = L(dI)/(dt) :. IR = L(dI)/(dt) :. L/R = I(dt)/(dI)` `:. Unit of L/R = (A xx S)/A = sec`. |
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| 3. |
A series LCR circuit having `L = 10 mH`, `C = (400//pi^(2)) mu F` and R = 55 ohm is connected to 220 v variable frequency a.c. supply. (i) Find frequency of source, for which average power absorbed by the circuit is maximum (ii) Calculate the amplitude of current. |
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Answer» Here, `L = 10 mH = 10 xx 10^(-3) H = 10^(-2) H` `C = (400//pi^(2)) mu F = (4 xx 10^(-4))/(pi^(2))` `R = 55 ohm` `E_(v) = 220 V` (i) Average power absorbed by the circuit is maximum at resonance. The resonant frequency is `v_(0) = (1)/(2 pi sqrt(LC)) = (1)/(2 pi sqrt(10^(-2) xx (400)/(pi^(2)) xx 10^(-6))` `v_(0) = (1)/(2 xx 2 xx 10^(-3)) = (1000)/(4) = 25 Hz` (ii) Current is maximum at resonance `I_(0) = sqrt2 I_(0) = sqrt2 (E_(v))/(R ) = sqrt2 xx (220)/(55) = 4 sqrt2 A` |
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| 4. |
The household supply voltage as measured by an a.c. voltmeter is 200 meter is 220 volts. If the frequency of a.c. Supply is 50 Hz, then the equation of the line voltage, will beA. 1. `V=220 sin (100 pi t)`B. 2.`V=110 sin (50 pi t)`C. 3.`V=440 sin (100 pi t)`D. `4.V=311 sin (100 pi t)` |
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Answer» Correct Answer - D `omega = 2 pi f = 2 pi xx 50 = 10 pi` and `V_(0) = sqrt(2) V_(rms) = 1.414 xx 220 = 311 V` The equation is `V = V_(0)sin (omega t) = 311 sin (100 pi t)`. |
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| 5. |
The domestic power supply of 220 V, 50 Hz is connected to a resistor. What is the time taken by the alternating current flowing in the resistor, to change from its maximum value to turns value?A. `5 xx 10^(-3) s`B. `2.5 xx 10^(-3) S`C. `10 xx 10^(-3) S`D. `2.5 xx 10^(3)S` |
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Answer» Correct Answer - D Suppose that at time `t_(1) = 0, i=i_(max)` and `i_(rms) = (i_0)/(sqrt(2)) = i_(0) sin omega t_(2)` `:. Sin omega t_(2) = 1/(sqrt2) :. Omega t_(2) = (pi)/4 :. t_(2) = (pi)/(4 omega)` `:. t_(2) - t_(1) = (pi)/(4 omega)` `=(pi)/(4 xx 2 pi f) = 1/(8 xx 50 ) = 1/400` `:. t_(2)-t_(1) = 2.5 xx 10^(-3) s`. |
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| 6. |
A 200 ohm electric iron is connected to 220 V, 50 Hz a.c. supply. Calculate the average power delivered to iron. |
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Answer» Here, `R = 200 Omega, E_(v) = 220 V, v = 50 Hz` Av. Power/cycle, `P = E_(v) I_(v) = e_(v) (E_(v) // R) = (220 xx 220)/(200) = 242 W` |
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| 7. |
The primary winding of a transfomer has 50 turns while its secondary has 500 turns. If the primary is connected to an a.c. supply of 220 V, 50 Hz, then the output at the secondaray will beA. 220V, 50HzB. 2200V, 50HzC. 2200V, 500HzD. 22V, 5Hz |
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Answer» Correct Answer - B `(E_S)/(E_P) = (n_s)/(n_p) = 500/50 = 10` `:. E_(S) = 10 xx 220 = 2200 V` and frequency will be 50 Hz. |
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| 8. |
In India, domestic power suppy is at 220 V, 50 hz, while in U.S.A, its 110 V, 60 hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply. |
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Answer» For transfer of power `(= V xx I)` at higher voltage (220 V instead of 110 V), current carried by wires is just half. Therefore, such wires need not be very thick, saving lot of transmission material and reducing the cost of transmission. This is one advantage of 220 V supply. But to design a device of particular wattage, `P = (V^(2))/(R )` as `V^(2)` is 4 times, R must be four times. If not, the dissipation or power in the form of heat will be larger on 220 V supply. This is one disadvantange of this supply. |
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| 9. |
An e.m.f. `E =E_(0) cos omega t` is applied to a circuit containing L and R in series. If `X_(L)=R`, then the power dissipated in the circuit is given byA. `(E_(0)^(2))/(8R)`B. `(E_(0)^(2))/(4R)`C. `(E_(0)^(2))/(2R)`D. `(E_(0)^(2))/(R)` |
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Answer» Correct Answer - B `E= E_(0) cos omega t` Power = `(E_(0)I_(0)cos theta)/(2) = (E_(0))/2 (E_0)/z xx R/z` `= (E_(0)^(2)R)/(2 z^(2)) = (E_(0)^(2)R)/(2sqrt(R^(2)+R^(2))^(2))) = (E_(0)^(2)R)/(2 xx 2R^(2)) = (E_(0)^(2))/(4R)`. |
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| 10. |
Alternating current through an inductor lags behind the alternating voltage by `90^(@)`. What does it imply ? |
| Answer» It implies that at a given time, whatever is the value of alternating e.m.f. current will acquire that value after `t = T//4`. | |
| 11. |
The average power dissipated in A.C. circuit containing resistance, inductance and capacitance depends uponA. only on the effective value of currentB. only on the phase difference between e.m.f. and currentC. on the effecitive value of e.m.f. current and phase difference between themD. only on the effective value of e.m.f. |
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Answer» Correct Answer - C The average power dissipated in an L-C-R circuit depends upon the effecitive vlaues of e.m.f., current and phase difference between them. |
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| 12. |
In an ac circuit, the current lags behind the voltage by `pi//3`. The components in the circuit areA. only RB. L and CC. R and CD. R and L |
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Answer» Correct Answer - D In an L-R circuit, the current lags behind the voltage. For only R, I and E are in phase while in R-C circuit, leads the voltage. |
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| 13. |
A telephone wire of length `200 km` has a capacitance of `0.01 4 muF per km`. If it carries an `AC` of frequency `5kHz` what should be the value of an inductor required to be connected in series so that impedence of the circuit is minimum ?A. 0.35mHB. 35mHC. 3.5mHD. Zero |
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Answer» Correct Answer - B |
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| 14. |
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. 100 WB. 40 WC. 20 WD. 0 W |
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Answer» Correct Answer - D `overline(P)=(e_(0)I_(0))/(2)cosphi` `=(e_(0)I_(0))/(2)cos90` =0 W |
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| 15. |
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. 1000 WB. 40 WC. 20 WD. 0 W |
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Answer» Correct Answer - D In the given problem emf leads the current by `90^(@)`. Thus, `overline(P)=e_(rms)I_(rms)cosphi=e_(rms)I_(rms)cos90=0` |
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| 16. |
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. 16VB. 10VC. 8VD. 6V |
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Answer» Correct Answer - A |
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| 17. |
The magnetic field of a cyclindrical magnet that has a pole face radius `2.8 cm` can be varied sinusoidally between the minimum value `16.8 T` and the maximum value `17.2 T` at a frequency of `50//piHz`. Cross section of the magneic field created by the nagnet is shown in Fig. 3.211. At a radial distance of `2cm` from the axis, find the amplitude of the electric field `(in units of xx 10^(2) mNC^(-1))` induced by the magnetic field variation. |
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Answer» Correct Answer - 2 `int vec(E)dvec(l) = -A(dB)/(dt)` As `B = 17 + (0.2) sin(omegat + phi)`, `E(2pir) = -pir^(2)(0.2)omega cos (omehat + phi)` `E = -(r )/(2)(0.2)omega cos (omegat + phi)` Magnitude of the amplitude `= (R )/(2)(0.2)omega = 2 xx 10^(2)mN//C` |
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| 18. |
A uniform magnetic field of induction `B` is confined to a cyclindrical region of radius `R`. The magnetic field is increasing at a constant rate of `dB//dt` (tesla`//` second). A charge `e` of mass `m`, placed at the point `P` on the periphery of the fixed experiences an acceleration : A. (a) `(1)/(2)(eR)/(m)(dB)/(dt)` toward leftB. (b) `(1)/(2)(eR)/(m)(dB)/(dt)` toward rightC. ( c) `(eR)/(m)(dB)/(dt)` toward leftD. (d) zero |
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Answer» Correct Answer - A (a) Induced electric field at point `P`: `E = (R )/(2)(dB)/(dt)` towards right acceleration of electron: `a = (eE)/(m) = (eR)/(2m)(dB)/(dt)` towards left |
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| 19. |
A uniform magnetic field of induction `B` is confined to a cyclindrical region of radius `R`. The magnetic field is increasing at a constant rate of `dB//dt` (tesla`//` second). A charge `e` of mass `m`, placed at the point `P` on the periphery of the fixed experiences an acceleration : A. `(1)/(2)(eR)/(m)(dB)/(dt)` towards leftB. `(1)/(2)(eR)/(m)(dB)/(dt)` towards rightC. `(eR)/(m)(dB)/(dt)` toward leftD. `(1)/(2)(eR)/(m)(dB)/(dt)` zero |
| Answer» Correct Answer - A | |
| 20. |
A neutral metallic ring is placed in a circular symmetrical uniform magnetic field with its plane perpendicular to the field.If the magnitude of field starts increasing with time, then:A. the ring starts translatingB. the ring starts rotating about its axisC. the ring slightly constractsD. the ring starts rotating about a diameter. |
| Answer» Correct Answer - C | |
| 21. |
A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time the magnet.A. will move with an acceleration gB. will move with almost constant speedC. will stop in the tubeD. will oscillate |
| Answer» Correct Answer - B | |
| 22. |
A wire of fixed length is wound on a solenoid of length `l` and radius `r`. Its self-inductance is found to be `L`. Now, if the same wire is wound on a solenoid of length `l//2` and radius `r//2` then the self-inductance will beA. 2LB. LC. 4LD. 8L |
| Answer» Correct Answer - A | |
| 23. |
The number of turns, cross-sectional area and length for four solenoids are given in the following table. The solenoid with maximum self inductance is :A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - D | |
| 24. |
The inductance of a solenoid `0.5m` long of cross-sectional area `20cm^(2)` and with `500` turns isA. `12.5 mH`B. `1.25mH`C. `15.0mH`D. `0.12mH` |
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Answer» Correct Answer - B `L=L=(mu_(0)N^(2)A)/(l)` `=(4pixx10^(-7)xx(500)^(2)xx20xx10^(-4))/(0.5)=1.25mH` |
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| 25. |
If in a coil rate of change of area is `5m^(2)//millisecond` and current become `1 amp` from `2 amp` in `2xx10^(-3)` sec. magnitude of field id `1 teslsa` then self-inductance of the coil isA. `2H`B. `5H`C. `20H`D. `10H` |
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Answer» Correct Answer - D `Nvarphi=Liimplies(Ndvarphi)/(dt)=(Ldi)/(dt)impliesNB(dA)/(dt)=(Ldi)/(dt)` `implies(1xx1xx5)/(10^(-3))=Lxx((2-1)/(2xx10^(-3)))impliesL=10H` |
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| 26. |
L,C and R represent the physical quantities inductance, capacitance and resistance respectively. Which of the following combinations have dimensions of frequency?A. `(1)/(RC)`B. `(R )/(L)`C. `(1)/(sqrt(LC))`D. `(C )/(L)` |
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Answer» Correct Answer - A::B::C (a) `(1)/(RC) = (1)/(R (Q//V)) = (V//R)/(Q) = (I)/(Q) = (I)/(It) = s^(-1)` (b) `(R )/(L) = (Omega)/(Omega s) = s^(-1)` (c ) `(1)/(sqrt(LC)) = (1)/(sqrt((L//R)(RC))) = (1)/(sqrt(s.s)) = s^(-1)` |
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| 27. |
A conducting ring of radius `1` meter is placed in an uniform magnetic field `B` of `0.01` tesla coscilliating with frequency `100Hz` with its plane at right angles to `B`. What will be the induced electric field.A. `pivolt//m`B. `2volt//m`C. `10volt//m`D. `20volt//m` |
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Answer» Correct Answer - B In a constant magnetic field conducting ring oscillates with a frequaency of `100Hz`. i.e., `T=1//100s`, in time `T//2` flux links with coil changes from `BA` to zero. :. Induced emf `=("change in flux")/("time")` `=(BA)/(T//2)=(2BA)/(T)` `=2Bxxpir^(2))/(T)` `(2xx0.01xxpixx1^(2))/(1//100)=4piV` induced electric field along the circle, using maxwell equatiion `ointE.dl=-(dphi)/(dt)=A(dB)/(dt)=e` :. `E=(1)/(2pir)xx(pir^(2)xx(dB)/(dt))=(e)/(2pir)=(4pi)/(2pir)=2V//m` |
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| 28. |
A metallic metre stick translates in a direction making an angle of `60^(@)` with its length.The plane of motion is perpendicular to a uniform magnetic field of `0.1 T` that exists in the space.Find the `emf` induced between the ends of the rod if the speed of translation is `0.2 m//s`. |
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Answer» Correct Answer - A::B::C `epsilon=BV(L sin theta)=0.1xx0.2 xx 1 sin 60^(@)=sqrt3xx10^(-2) V` |
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| 29. |
A square frame with side a and a long straight wire carrying a current `i` are located in the same plane as m shown in figure. The fram translates to the right with a constant velocity `v`. Find the emf induced in the frame as a function of distance `x`. |
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Answer» Correct Answer - `[(mu_(0)ia^(2)xv)/(2pix(x+a))]` |
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| 30. |
A wire ring of radius R is fixed in a horizontal plane. The wire of the ring has a resistance of `lambda Omega m^(–1)`. There is a uniform vertical magnetic field B in entire space. A perfectly conducting rod (l) is kept along the diameter of the ring. The rod is made to move with a constant acceleration a in a direction perpendicular to its own length. Find the current through the rod at the instant it has travelled through a distance `x = (R)/(2)`. |
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Answer» Correct Answer - `(18B)/(5pi) (sqrt(3aR))/(lambda)` |
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| 31. |
There is a uniform magnetic field (B) perpendicular to the plane of the figure in a circular region of radius a centred at O. ABCD is a conducing loop in the plane of the figure with its arms BC and DA along two radial lines from O having an angle `theta` between them. AB is circular arc of radius OB = d centred at O. CD is also a circular arc of radius `OC = b` centred at O. The magnetic field is changed at a rate `(dB)/(dt)`. (a) The emf induced in the loop ABCD. (b) Find emf induced in arc AB. |
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Answer» Correct Answer - (a) Zero (b) `(a^(2) theta)/(2) (dB)/(dt)` |
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| 32. |
A conducing circular loop is placed in a uniform magnetic field of indution `B` tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate `(dr//dt)`. Then the induced e.m.f. at the instant when the radius is `r` is:A. `pi r B ((dr)/(dt))`B. `2 pi r B ((dr)/(dt))`C. `pi r^(2) ((dr)/(dt))`D. `((pi r^(2))/(2))^(2) ((dt)/(dt))` |
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Answer» Correct Answer - B Here: `A = pi r^(2)`, and `phi = BA = B pi r^(2)` As `e = (d phi)/(dt) = (d)/(dt) (B pi r^(2)) = pi B (2 r) (dr)/(dt)` `= 2 pi r B (dr)/(dt)` |
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| 33. |
A conducing circular loop is placed in a uniform magnetic field of indution `B` tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate `(dr//dt)`. Then the induced e.m.f. at the instant when the radius is `r` is:A. `pirB(dr//dt)`B. `2pirB(dr//dt)`C. `pir^(2)(dB//dt)`D. `(pir^(2)//2)B(dr//dt)` |
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Answer» Correct Answer - B Induced emf, `e=B.(dS)/(dt)=2pirB.(dr)/(dt)` where, `(dr)/(dt)` is the rate at which radius of loop is shrinking. |
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| 34. |
A conducing circular loop is placed in a uniform magnetic field of indution `B` tesla with its plane normal to the field. Now, radius of the loop starts shrinking at the rate `(dr//dt)`. Then the induced e.m.f. at the instant when the radius is `r` is:A. `pi r B((dr)/(dt))`B. `pi r^(2)((dB)/(dt))`C. `2pi rB((dr)/(dt))`D. `((pir^2)/(2))^(2)B cdot(dr)/(dt)` |
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Answer» Correct Answer - C Area of the coil = `pi r^(2)` `|e| = (d phi)/(dt)=d/(dt)[AB]` `=d/(dt)[pir^(2)B] = pi B d/(dt)[r^2]` `pi B 2r (dr)/(dt) = 2 pi r B((dr)/(dt))`. |
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| 35. |
A magnet is allowed to fall towards a metal ring. During the fall Its accelertion isA. Equal to gB. Greater than gC. Less than gD. Equal to the product of g and the radius of the ring |
| Answer» Correct Answer - C | |
| 36. |
A region of width L contains a uniform magnetic field B directed into the plane of the figure. A square conducing loop of side length `l ( lt L)` is kept with its side AB at the boundary of the field region (see Figure). The loop is pushed into the field region with a speed such that it just manages to exit the field region. Calculate the time needed for the entire loop to enter the field region after it is pushed. Mass and resistance of the loop is M and R respectively. |
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Answer» Correct Answer - `(MRl n2)/(B^(2)l^(2))` |
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| 37. |
A short magnet is allowed to fall from rest along the axis of a horizontal conducting ring. The distance fallen by the magnet in one second may beA. `5m`B. `6m`C. `4m`D. none of these |
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Answer» Correct Answer - C In case of free fall, `d=1/2g t^2` `=1/2(10)(1)^2=5m` Here due to repulsion from inducd effects `a lt g` `:. dlt5m` |
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| 38. |
A loop is kept so that its center lies at the origin of the coordinate system. A magnetic field has the induction `B` pointing along `Z` axis as shown in the figure A. No EMF and current will be induced in the loop if it rotates about Z-axisB. EMF is induced but no current flow sif the loop is a fiber when it rotates about Y-axis.C. EMF is indued and induced current flows in the loop if the loop is made of copper `&` is rotated about Y-axis.D. If the loop moves along Z-axis with constant velocity, no current flows in it. |
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Answer» Correct Answer - A::B::C |
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| 39. |
A short bar magnet having magnetic dipole moment M is moving along the axis of a fixed conducting (non magnetic) ring of radius R. The axis of the ring is along z direction. (a) Write the z component of magnetic field due to the magnetic dipole at a point P in the plane of the ring, at the instant the magnet is at a distance z from the centre of the ring. Position of point P can be defined in terms of angle `theta` as shown. (b) Write the magnetic flux through the ring due to the magnetic field produced by the magnet as a function of z. (c) Write the magnitude of emf induced in the ring at the instant shown if speed of the magnet at the moment is v. |
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Answer» Correct Answer - (a) `b_(z)=(mu_(0))/(4pir^(3))[2 cos^(2) theta -sin^(2)theta]` (b) `phi=(mu_(0)MR^(2))/(2(R^(2)+z^(2))^(3//2))` (c) `E_("in")=(3mu_(0)MR^(2)vz)/(2(R^(2)+z^(2))^(5//2))` |
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| 40. |
A bar magnet is kept along the axis of a conducting loop. When the magnet is moved along the axis, a current is seen in the loop. Which force is responsible for driving electrons in the loop if the observer is in (a) Reference frame attached to the loop. (b) Reference frame attached to the magnet. |
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Answer» Correct Answer - (a) Inudced electric field applies the force (b) Magnetic force. |
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| 41. |
Two coaxial circular loops of raadii `r_(1)` and `r_(2)` are separated by a distance `x` and carry currents `i_(1)` and `i_(2)` respectively. Calculate the mutual inductance. What is the force between the loops ? |
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Answer» Magnetic field due to loop `(1)` at `(2)` `B_(1)=(mu_(0)i_(1)r_(1)^(2))/(2(r_(1)^(2)+x^(2))^(3//2))` , along the axis Flux passing through `(2)` `phi=B_(1)A_(2)=(mu_(0)i_(1)r_(1)^(2))/(2(r_(1)^(2)+x^(2))^(3//2))pir_(2)^(2)` Mutual inductance `M=(phi_(2))/(i_(1))=(mu_(0)pir_(1)^(2)r_(2)^(2))/(2(r_(1)^(2)+x^(2))^(3//2))` Magnetic moment of loop `(2)` `M_(2)=i_(2).pir_(2)^(2)` , along the axis `P.E.` of loop `(2)` ltbr. `U=-vec(M)_(2).vec(M)_(1)=-M_(2)B_(1)=-(i_(2)pir_(2)^(2)mu_(0)i_(1)r_(1)^(2))/(2(r^(2)+x^(2))^(3//2))` `F=-(dU)/(dx)=(mu_(0)i_(1)i_(2)pir_(1)^(2)r_(2)^(2))/(2)(-(3)/(2))(r^(2)+x^(2))^(-5//2)(2x)` `=-(3mu_(0)i_(1)i_(2)pir_(1)^(2)r_(2)^(2)x)/(2(r^(2)+x^(2))^(5//2))` `-ve` sign shows that foorce is attractive. |
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| 42. |
A very small circular loop of area `5xx10^(-4)m^(2)` , resistance `2Omega` and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius `0.1m` . A constant current of `1A` is passed in the bigger loop and the smaller loop is rotated with angular velocity `omega rad//sec` about a diameter. Calculate (a) the flux linked with the smaller loop, (b) induced emf (c) induced current in the smaller loop, as a function of time. |
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Answer» Correct Answer - `[(a) 10pixx10^(-10) Wb; (b) 10^(-9) omega; (c ) 2.5xx10^(-10) omega sin omega t]` |
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| 43. |
Two inductors `L_(1)` and `L_(2)` are connected in parallel and a time varying current flows as shown. the ratio of current `i-(1)//i_(2)` A. `L_(1)//L_(2)`B. `L_(2)//L_(1)`C. `(L_(1)^(2))/((L_(1)+L_(2)^(2))`D. `(L_(2)^(2))/((L_(1)+L_(2)^(2))` |
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Answer» Correct Answer - B The inductors are in parallel. Therefore, potential difference across them is same, hence, `V_(1)=V_(2)` or `L_(1)((di_(1))/(dt))=L_(2)((di_(2))/(dt))` or `L_(1)(di_(1))=L_(2)(di_(2))` integrating, we get `L_(1)i_(1)=L_(2)i_(2)` or `(i_(1))/(i_(2)=(L_(2))/(L_(1))` |
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| 44. |
A coil has an inductance of `2.5 H` and a resistance of `0.5r`. If the coil is suddenly connected across a `6.0` volt battery, then the time required for the current to rise `0.63` of its final value isA. `3.5` secB. `4.0` secC. `4.5` secD. `5.0` sec |
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Answer» Correct Answer - D `t=tau=(L)/(R )=(2.5)/(0.5)=5`sec |
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| 45. |
An ideal coil of `10` henry is joined in series with a resistance of `5` ohm and a battery of `5` volt. `2` second after joining, the current flowing in ampere in the circuit will beA. `e^(-1)`B. `(1-e^(-1))`C. `(1-e)`D. `e` |
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Answer» Correct Answer - B `i=(E)/(R)(1-e^(-(Rt)/(L)))=(5)/(5)(1-e^(-(5xx2)/(10)))=(1-e^(-1))` |
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| 46. |
An inductor of `2` henry and a resistance of `10` ohms are connected in series with a battery of `5` volts. The initial rate of change of current isA. `0.5amp//sec`B. `2.0amp//sec`C. `2.5amp//sec`D. `0.25amp//sec` |
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Answer» Correct Answer - C `i=(E)/(R)(1-e(-Rt//L)` `(di)/(dt)=(E)/(R)e^(-(Rt)/(L)).(R)/(L)=(E)/(L)e^(-(Rt)/(L))` `t=0, (di)/(dt)=(E)/(L)=(5)/(2)=2.5A//sec` |
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| 47. |
A conducting rod AB of mass M and length L is hinged at its end A. It can rotate freely in the vertical plane (in the plane of the Figure). A long straight wire is vertical and carrying a current I. The wire passes very close to A. The rod is released from its vertical position of unstable equilibrium. Calculate the emf between the ends of the rod when it has rotated through an angle `theta` (see Figure). |
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Answer» Correct Answer - `(mu_(0)I)/(2pi sin theta) sqrt(3gL(1-cos theta))` |
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| 48. |
A metallie rod of 1 m length is rotated with a frequency of 50 revis, with one end hinged at the centre and the other end at the eireumference of a circular metallic ring of radius 1m, about an axis passing through the centre and perpendicular to the plane of the ring as in figure. A constant and uniform magnetic field of the parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring? |
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Answer» Method 1: As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distribuled over the ring. Thus, the resulting separalion of charges produces an emi acros the ends of the rod: At a certain value of emi, there is no more flow of electrons and a steady state is reached. Using equation `(epsi=-Bl (dx)(dt)=BlV)`, the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by `depsi=Bvdr"Hence"` `epsi=intdepsi=underset(0)overset(R)intBvdr=underset(0)overset(R)intBomegardr=(BomegaR^(2))/(2)` Note that we have used `upsilon= omegar`. This gives `epsi=1/2 xx 1.0 xx2pixx 50 xx (1)^(2)=157V` Method II : To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistance is then equal to the induced emf and equals `B xx` (rate of change of area of loop). If `theta` is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by `piR^(2)xx (theta)/(2pi)=1/2R^(2)theta` where R is the radius of the circle. Hence, the induced emf is `piR^(2)xx (theta)/(2pi)=1/2 R^(2)theta` `epsi=B xx (d)/(dt)[1/2 R^(2)theta]=1/2BR^(2) (d theta)/(dt)=(B omegaR^(2))/(2)` This expression is identical to the expression obtained by Method I and we get the same value of `epsi`. |
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| 49. |
Could a current be induced in a coil by rotating a magnet inside the coil ? If so, how ? |
| Answer» Yes. By holding the magnet along the axis of the coil and turning the magnet about the diameter of the coil. | |
| 50. |
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. `4A`B. less than 4 AC. more thanD. may be less than or more than 4 A depending on the value of L |
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Answer» Correct Answer - C Since, yor are decreasing current in the circuit by increasing resistance of the circuit. Induced emf across inductor will support 20 V battery. Hence, net emf of the circuit is greater than 20 V or current in the circuit is more than 4 A. |
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