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. |
Choose the quantity whose SI is not ohm.A. ResistanceB. ReactanceC. CapacitanceD. Impedance |
| Answer» Correct Answer - C | |
| 2. |
Identify the graph which correctly reperesents the variation of capacitive reactance `X_C` with frequencyA. B. C. D. |
| Answer» Correct Answer - D | |
| 3. |
Mutual inductance of two coils is 0.005H, ac in primary coil, `I=I_(0)sinomegat`, where `I_(0)=10A` and `omega=100pi` rad/s. What is the maximum emf in the secondary coil? |
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Answer» ac in primary coil, `I=I_(0)sinomegat` `therefore (dI)/(dt)=omegaI_(0)cosomegat` So, `e_(2)=-M(dI)/(dt)` [here M = mutual inductance] `=-MomegaI_(0)cosomegat` So the maximum emf in the secondary coil `=MomegaI_(0)=0.005xx100pixx10=15.7V` |
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| 4. |
A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction `(1)/(pi)Wb//m^(2)` in such a way that its axis makes an angle `60^(@)` with the field. The magnetic flux linked with the disc isA. 0.01 WbB. 0.02 WbC. 0.06 WbD. 0.08 Wb |
| Answer» Correct Answer - B | |
| 5. |
A semicircular wire of radius r is rotating with angular velocity `omega` in a uniform magnetic field B with its radius as axis. If the resistance of the circuit be R and if the axis of rotation remains perpendicular to B, what will be the average power produced in each period? |
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Answer» Magnetic flux, `phi=BAcosomegat=B.(pir^(2))/(2).cosomegat` `therefore` emf induced, `e=-(dphi)/(dt)=(Bpir^(2))/(2)omegasinomegat` Power, `P=(e^(2))/(R)=((Bpir^(2)omega)^(2))/(4R)sin^(2)omegat` In a complete period, the average of `sin^(2)omegat=(1)/(2)` So, the average power, `barP=((Bpir^(2)omega)^(2))/(8R)` |
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| 6. |
Write down the dimensional formula for induced emf? |
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Answer» Induced emf e has the dimension of potential different V. Potential difference `(V)=("work (W)")/("charge (Q)")` `therefore [e]=[V]=([W])/([Q])=(ML^(2)T^(-2))/(lT)=ML^(2)T^(-3)l^(-1)` |
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| 7. |
A circular conducting coil of radius a and resistance R is placed with its plane perpendicular to a magnetic field. The magnetic field varies with time according to the equation `B=B_(0)sinomegat`. Obtain the expression for the induced current in the coil. |
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Answer» As we know, induced current, `I=("induced emf(e)")/("resistance (R)")` Again, `e=-(dphi)/(dt)` = rate of change of magnetic flux with time `therefore I=((-dphi)/(dt))/(R)=-(1)/(R).(d)/(dt)(BAcos0^(@))` or, `I=-(A)/(R).(d)/(dt)(B_(0)sinomegat)[because B=B_(0)sinomegat]` `=-(AomegaB_(0))/(R)cosomegat=-(pia^(2)omegaB_(0))/(R)cosomegat` |
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| 8. |
A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is 15 m/s and is directed perpendicular to the field. The emf induced between the two ends of the conductor isA. 0.10 VB. 0.15 VC. 1.50 VD. 15.00 V |
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Answer» Correct Answer - B The emf induced between the two ends of the conductor `=Blv=0.1xx0.1xx15=0.15V` |
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| 9. |
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the mef developed across the cut if the velocity of the loop is 1 m. `s^(-1)` in a direction normal to the Suppose in this case the loop is stationary but the current feeding the electromagnet produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T. `s^(-1)`. If the cut is joined and the loop has a resistance of `1.6 Omega`, how much power is dissipated by the loop as heat? What is the source of this power? |
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Answer» Induced emf, `e=(dphi)/(dt)=(dB)/(dt).A` `=0.02xx8xx2xx10^(-4)=3.2xx10^(-5)V` Induced current, `I=(e)/(R)=(3.2)/(1.6)xx10^(-5)A=2xx10^(-5)A` Power dissipated `=exxI=6.4xx10^(-10)W` The source of this power is the external agency which brings change in magnetic field. |
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| 10. |
What will be the self-inductance of a solendoid of length 1m, diameter 12cm and number of turns 4000? How much energy will be stored in its magnetic field due to a current of 2A through it? `(mu_(0)=4pixx10^(-7)H.m^(-1))`. |
| Answer» Correct Answer - 0.227H, 0.454J | |
| 11. |
The motion of copper plates is damped when it is allowed to oscillate between the two poles of a magnet. If slots are cut in the plate, how will the damping be affected? |
| Answer» The damping is due to eddy currents induced in the copper plate. If slots are cut in the plate, the eddy current loops become much shorter-as a result, the net value of the eddy current decreases. Then the damping becomes less. | |
| 12. |
A coil of resistance 100 `Omega` having 100 turns is placed in a magnetic field. A galvanometer of resistance `400 Omega` is connected in series with it. If the coil is brought from the present magnetic field to another magnetic field in `(1)/(10)s`, determine the average emf and the current. Given, the initial and final magnetic flux linked with each turn of the coil are 1 mWb and 0.2 mWb respectively. |
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Answer» Change in magnetic flux for each turn `=0.2-1=-0.8mWb` So, change in magnetic flux for 100 turns `=100xx(-0.8)mWb=-0.08Wb` Hence, the magnetic of average emf induced = the negative of the rate of change of magnetic flux `=-(-(0.08Wb)/((1)/(10)s))=0.8V` The equivalent resistance of the circuit = 100 + 400 = 500 `Omega`. Hence, the average induced current = `(0.8V)/(500Omega)=0.0016A=1.6mA`. |
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| 13. |
A transformer is an electrical device that is used for …………………. It ………………………..work on …………………… |
| Answer» Correct Answer - changing a.c. voltages ; does not ; d.c. | |
| 14. |
A d.c. motor converts ………………………into ………………………. |
| Answer» Correct Answer - direct current energy ; mechanical energy of rotation | |
| 15. |
Inside a parallel plate capacitor the electric field E varies with time as `t^(2)`. The variation of induced magnetic field with time is given byA. `t^(2)`B. no variationC. `t^(3)`D. t |
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Answer» Correct Answer - D Varying electric field induces magnetic field and this induced magnetic field is proportional to the rate of change in electric field. According to the question, electric field is proportional to `t^(2)and(d)/(dt)(t^(2))=2t`, so induced magnetic field is proportional to t. |
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| 16. |
In hydroelectric power station,………………of falling water is converted into ………………………… |
| Answer» Correct Answer - kinetic energy ; electrical energy | |
| 17. |
In d.c. generator,…………………….generator of a.c. generator is replaced by……………….. |
| Answer» Correct Answer - slip ring ; split ring arrangement | |
| 18. |
A d.c. generator produces…………………from………………… |
| Answer» Correct Answer - direct current energy ; mechanical energy | |
| 19. |
An a.c generator is based on the phenomenon of…………………………… |
| Answer» Correct Answer - electromagnetic induction | |
| 20. |
An a.c. generator is a machine that produces…………….from……………… |
| Answer» Correct Answer - alternating current energy ; mechanical energy | |
| 21. |
The frequency of a.c. generated depend onA. speed of rotation of coilB. amplitude of a.cC. size of coilD. all the above |
| Answer» Correct Answer - A | |
| 22. |
What is the role of the split-ring in an electric motor?A. a.c. generatorB. d.c. generatorC. choke coilD. Transformers |
| Answer» Correct Answer - B | |
| 23. |
Ohmic resistance R can reduce ………………..but indcutor L can reduce………………..only. |
| Answer» Correct Answer - both a.c. and d.c ; a.c. | |
| 24. |
The from factor of an a.c. generator is given byA. `(I_(av))/(I_(0))`B. `(I_(0))/(I_(av))`C. `(I_(av))/(I_(v))`D. `(I_(v))/(I_(av))` |
| Answer» Correct Answer - D | |
| 25. |
A series resonance circuit is called an ……………..and a……………………is called ……………… |
| Answer» Correct Answer - acceptor circuit ; parallel resonance circuit ; rejector/filter circuit | |
| 26. |
A consdenser ………………….a.c. to pass through but…………………….d.c. |
| Answer» Correct Answer - allows ; blocks | |
| 27. |
Q factor of resonance is given byA. `(1)/(R ) sqrt((L)/(C ))`B. `(1)/(R ) sqrt((C )/(L))`C. `(1)/(L) sqrt((R )/(C ))`D. `(1)/(C ) sqrt((L)/(R ))` |
| Answer» Correct Answer - A | |
| 28. |
The…………… of alternating current varies …………………….. With time and its………………….is reversed |
| Answer» Correct Answer - magnitude ; continuosly ; direction ; periodically | |
| 29. |
The alternating current from a source is represented by `I = 0.5 sin 314 t`. The frequecy of a.c. isA. 314 HzB. 100 HzC. 50 HzD. zero |
| Answer» Correct Answer - C | |
| 30. |
The dimensions of inductvie………………………and ……………….are the same as those of ……………….. |
| Answer» Correct Answer - reactance ; capacitative reactance ; resistance. | |
| 31. |
In an a.c. circuit containing L only, alternating current………………….alternating voltage by a phase angle of……………….. |
| Answer» Correct Answer - lags behind ; `90^(@)` | |
| 32. |
Phase difference between voltages across L and C in series isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `360^(@)` |
| Answer» Correct Answer - C | |
| 33. |
In an a.c. circuit containing R only…………………and…………………are in ………………..phase. |
| Answer» Correct Answer - alternating current ; alternating voltage ; same | |
| 34. |
The peak value of 220 a.c. isA. 220 VB. `(220)/(sqrt2) V`C. 440 VD. `220 sqrt2 V` |
| Answer» Correct Answer - D | |
| 35. |
220 V a.c. means…………………… and an a.c. of 1 A means……………………. |
| Answer» Correct Answer - `E_(v) = 220 V ; I_(V) = 1 A`. | |
| 36. |
The r.m.s. value or………………………value or ………………….value of a.c. is ……………………..the peak value of a.c. |
| Answer» Correct Answer - virtual ; effective ; 0.707 times | |
| 37. |
Ordinary d.c. ammerter and d.c. votmeter, when used in ………………….record……………………reading |
| Answer» Correct Answer - a.c. circuit ; zero | |
| 38. |
The efficiency of d.c. motor is given by `eta =`A. `("back e.m.f.")/("applied e.m.f.")`B. `("applied e.m.f.")/("back e.m.f.")`C. back e.m.f. `xx` applied e.m.f.D. none of the above |
| Answer» Correct Answer - A | |
| 39. |
Out of the following , choose the wrong statement :A. A transformer cannot work on d.c.B. A transformer connot change the frequency of a.c.C. A tranformer can produce a.c. powerD. In a transformer, when a.c. voltage is raised n times, the alternating current reduces to `1//n` time. |
| Answer» Correct Answer - C | |
| 40. |
The magnetic flux `(phi)` linked with a coil varies with time (t) as `phi=at^(n)`, where a and n are constants. The induced emf in the coil is e. Which of the following are correct?A. if `0ltnlt1,e=0`B. if `0ltnlt1,ene0` and |e| decreases with timeC. if n = 1, e is constantD. if `ngt1,|e|` increases with time |
| Answer» Correct Answer - A::C | |
| 41. |
The magnetic flux linked with a coil varies with time t as `phi=at^(2)+bt+c`, where a, b and c are constants. The emf induced in the coil will be zero at a time ofA. `(b)/(a)`B. `-(b)/(a)`C. `(b)/(2a)`D. `-(b)/(2a)` |
| Answer» Correct Answer - D | |
| 42. |
The mutual inductance between two adjacent coils is 1.5 H. If the current in the primary coil changes from 0 to 20 A in 0.05 s, determine the average emf induced in the secondary coil. If the number of turns in the secondary coil is 800, what change in flux will be observed in it? |
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Answer» The average value of induced emf, `e=-M(DeltaI)/(Deltat)=-1.5xx((20-0))/(0.05)=-(1.5xx20)/(0.05)=-600V` Hence, the magnitude of the induced emf = 600 V. Again, from the relation `phi=MI` we get, change in flux, `Deltaphi=M.DeltaI=1.5xx(20-0)=30Wb`. |
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| 43. |
A long solenoid having 450 turns per m carries a current of 1.6 A. At the centre of the solenoid a coil of 180 turns with cross sectional area `3.5 cm^(2)` is placed having its axis parallel to the field produced by the solenoid. What will be the amount of induced emf when the direction of current in the solenoid is reversed within 0.03 s? |
| Answer» Correct Answer - 3.78 mV | |
| 44. |
Cross sectional area of a solenoid is `10cm^(2)`. Half of its cross section is filled with iron `(mu_(r)=450)` and the remaining half with air `(mu_(r)=1)`. Calculate the self-inductance of the solenoid if its length is 2 m and number of turns is 3000. |
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Answer» If `alpha_(1)` part of cross section of the solenoid is filled with a substance of relative permeability `mu_(r_(1))` and the remaining part of cross section `alpha_(2)` with another substance of relative permeablity `mu_(r_(2))` then self-inductance of the solenoid is, `L=(mu_(0)n^(2)A)/(l)(mu_(r_(1))alpha_(1)+mu_(r_(2))alpha_(2))` Here `mu_(0)=4pixx10^(-7)` H/m, number of turns, n = 3000, length of solenoid, l = 2 m, area of cross section, `A=10cm^(2)=0.001m^(2),alpha_(1)=0.5andalpha_(2)=0.5`. `therefore` Self-inductance of the solenoid, `L=(4pixx10^(-7)xx(3000)^(2)xx0.001)/(2)(1xx0.5+450xx0.5)H` `=2pixx9xx10^(-4)xx(0.5+225)H` = 1.27 H (approx). |
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| 45. |
A square loop of wire of side 10 cm is placed at angle of `45^(@)` with a magnetic field that changes uniformly from 0.2 T to zero in 1 second. Find the current induced in the loop of resistance `1 Omega`. |
| Answer» Correct Answer - 1.4 mA | |
| 46. |
A small square loop of wire of side y is placed inside a large square loop of side `x(xgtgty)`. The loops are coplanar and their centres coincide. Find the mutual inductance of the system. |
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Answer» Let I be the current flowing through a square loop of side L. Magnetic field at the centre of the loop, `B=4B_(1)` [where `B_(1)` is magnetic field at the centre of the loop due to one of its sides] Now, for the large square loop, L = x `therefore B_(1)=(mu_(0))/(4pi).(I)/((x)/(2))(sin45^(@)+sin45^(@))` [`because` distance of the centre from each side of the large loop = `(x)/(2)`] `=(mu_(0))/(2pi).(I)/(x)((1)/(sqrt(2))+(1)/(sqrt(2)))=(mu_(0)I)/(sqrt(2)pix)` `therefore B=(4mu_(0)I)/(sqrt(2)pix)=(2sqrt(2)mu_(0)I)/(pix)` Now, magnetic flux linked with the small square loop, `phi=Bxx` area of the small square loop `=Bxxy^(2)=(2sqrt(2)mu_(0)Iy^(2))/(pix)" ... (1)"` If M be the mutual inductance between the two loops,then `phi=MI" ... (2)"` From equations (1) and (2), `M=(2sqrt(2)mu_(0)y^(2))/(pix)` |
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| 47. |
Induced emf is directly proportional to the rate of change with time of magnetic_______linked with a coil. |
| Answer» Correct Answer - flux | |
| 48. |
How does the mutual inductance of a pair of coils change when distance between the coils is increased and |
| Answer» The flux linkage decreases in this case, so the mutual inductance decreases. | |
| 49. |
Statement I: Induced emf in a conductor is proportional to the time rate of change of associated magnetic flux. Statement II: In case of electromagnetic induction transfer of energy takes place in a manner so that total energy is conserved.A. Statement I is true, statement II is true, statement II is a correct explanation for statement IB. Statement I true, statement II is true, statement II is not a correct explanation for statement IC. Statement I is true, statement II is falseD. Statement I is false, statement II is true |
| Answer» Correct Answer - B | |
| 50. |
If the emf induced in an electrical circuit be e and the current induced be I then,A. both e and I depend on the resistance of the circuitB. none of e and I depends on the resistance of the circuitC. e depends on the resistance of the circuit but not iD. I depends on the resistance of the circuit but not e |
| Answer» Correct Answer - D | |