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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.
| 651. |
In determination of frequency of a.c. mains using a sonometer, a.c. is supplied to electromagnetA. directlyB. through a step up transformerC. through a step down transformerD. none of the above |
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Answer» Correct Answer - C Electromagnet needs lower voltage. So, the supply is through a step down transformer. |
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| 652. |
The resonance length obatined by a student in determination of frequency by a student in determination of frequency of a.c. mains by sonometer is `26.7 cm`. If thickness of sonometer wire were doubled without changing any other parameter, the resonance length would beA. 26.7 cmB. 53.4 cmC. 13.4 cmD. none of the above |
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Answer» Correct Answer - C From `(1)/(lD) sqrt((T)/(pi d))` As n is `l xx D =` constant. When D is double become half. Therefore, new resonance length `= (26.7)/(2) = 13.35 cm = 13.4 cm` |
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| 653. |
A conductor of length 10 cm is moved parallel to itself with a speed of `10 m//s`, at right to a magentic induction of `10^(-4) Wb//m^(2)`. The e.m.f. Induced in the conductor isA. `10^(-4)V`B. `10^(-2)V`C. `0V`D. `10^(-6)V` |
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Answer» Correct Answer - A `e=Blv=10^(-4)xx0.1xx10=10^(-4)V` |
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| 654. |
If a conductor is moving in the north direction and magnetic field is applied vertically upwards, the change in flux is `2xx10^(-4)` Wb within 2s. Is the resistance of conductor is `5Omega`, then the magnitude of induced current will beA. 0.02 mAB. 0.2 mAC. 0.002 mAD. 2 mA |
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Answer» Correct Answer - A `e=(dphi)/(dt)=(2xx10^(-4))/(2)=1xx10^(-4)V` `I=(e)/(R)=(1xx10^(-4))/(5)=0.02mA` |
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| 655. |
A straight conductor of length 4 m moves at a speed of 10 m/s when the conductor makes an angle of `30^(@)` with the direction of magnetic induction 0.1 T. then the induced emf isA. 1 VB. 2 VC. 4 VD. 8 V |
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Answer» Correct Answer - B `e=Bvlsintheta` `=0.1xx10xx4xxsin30=2V` |
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| 656. |
An alternating e.m.f. is applied to a series combination of `L=2 H, C= 10 mu F and R=50 Omega`. For a particular value of the angular frequency of the applied e.m.f., resonance is produced. Then the impedance z of the combination isA. `20 Omega`B. `30 Omega`C. `40 Omega`D. `50 Omega` |
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Answer» Correct Answer - D In series resonance, `Z =R = 50 Omega`. |
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| 657. |
An alternating e.m.f. given by e=200 sin 50 t is applied to a circuit containing only a resistance of `50 Omega`. What is the value of r.m.s. current in the circuit?A. 2.828 AB. 28.28 AC. 0.2828 AD. 0.02828 A |
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Answer» Correct Answer - A `I_(0) = (e_0)/(R) = 200/50 = 4A` `:. I_(rms) = (I_0)/(sqrt2) = 4/(sqrt2) = 2sqrt(2) = 2 xx 1.414 = 2.828 A`. |
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| 658. |
A current of 10 A in primary of a circuit is reduced to zero at a unifrom rate in `10^(-3)`s. If coeff.of mutual inductance is 3 H, what is the induced e.m.f. in the secondary ? |
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Answer» Here, `I_(1) = 10 A, I_(2) = 0, dt = 10^(-3) s, M = 3 H, e = ?` As `e = - M (d I)/(dt) = (- M (I_(2) - I_(1)))/(dt) :.e = (-3(0 - 10))/(10^(-3)) = 3 xx 10^(4) V` |
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| 659. |
Three coils of self inductance, 5, 10 and 15 Mh are connected first in series an then in parallel. What is the ration of the net inductance in the two cases? |
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Answer» Here, `L_(1) = 5 mH, L_(2) = 10 mH, L_(3) = 15mH` `L_(s) = L_(1)+ L_(2) + L_(3) = 5 + 10 + 15 = 30 mH` `(1)/(L_(P)) = (1)/(L_(1)) + (1)/(L_(2)) + (1)/(L_(3)) = (1)/(5) + (1)/(10) + (1)/(15) = (6 + 3 + 2)/(30) = (11)/(30)` `L_(p) = (30)/(11) mH (L_(s))/(L_(p)) = (32)/(30//11) = 11` |
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| 660. |
A metallic loop is placed in a non-uniform steady magnetic field. Will an emf be induced in the loop? |
| Answer» It is true that magnetic flux passing through the loop is calculated by integration but, it remains constant. | |
| 661. |
A coil is removed from a magnetic field (i) rapidly (ii) slowly. In which case, more work will be done ? |
| Answer» Work done will be more when the coil is removed rapidly. This is because in that case, opposing e.m.f induced in the coil will be more. | |
| 662. |
A current from A to B is increasing in magnitude. What is the direction of induced current. If any,in the loop as shown in the figure? |
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Answer» The current induced in the loop will be clockwise. According to right hand thumb rule, magnetic field passing through the loop due to current in AB is perpendicular to plane of paper and outward. The direction or current induced in the loop must be clock wise so as to opposethe increase in this field. |
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| 663. |
What is the magnitude of induced current in the circular loop ABCD or radius r, when a steady current of I ampere is passed through wire KL? |
| Answer» As current through wire KL is steady, magnetic flux linked with the loop ABCD is cosntant. As there is no charge in magnetic flux, induced current = 0. | |
| 664. |
A closed circular wire loop A lies in the plane of longer loop B, which is connected to the battery as shown in the figure. The direction of current induced in the loop A when the switch S is closed isA. clockwiseB. anticlockwiseC. no current is induced in A and loop A remains stationaryD. no current is induced in A, but A rotates clockwise |
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Answer» Correct Answer - B When switch S is closed then the current in loop is clockwise. Thus, emf induced in secondary coil is anticlockwise. |
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| 665. |
Find the current in the wire PQ for the configuration shown in Fig. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. `theta` is a fixed angle made by PQ travelling smoothly over two conducting parallel wires seperated by a distance d. |
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Answer» In Fig. PQ is a wire of negligible resistance travelling smoothly over two conducting parallel wires X and Y separated by distance (d). `theta` is fixed angle made by PQ and B is strength of magnetic field perperdicular to the plane of paper and outwards. Motional electric field `= upsion B`. Its direction is along 0 K, perpendicular to `vec upsilon` and `vec B`. `:.` e.m.f. induced in wire PQ e = length `PQ xx ` motional electric field along QP `= ((d)/(sin theta)) xx upsilon B sin theta = d upsilon B` current in the wire, `I = (e)/(R ) = (d upsilon B)/(R )` |
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| 666. |
A rod `PQ` is connected to the capacitor plates. The rod is placed in a magnetic field `(B)` directed downwards perpendicular to the plane of the paper. If the rod is pulled out of magnetic field with velocity `vec(v)` as shown in Figure. A. plate M will be positively chargeB. plate N will be positively chargedC. both plates will be similarly chargedD. no charge will be collected on plates |
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Answer» Correct Answer - A When rod is pulled out of the field emf is induced in the rod. The electrons will move from P to Q so plate M will be positively charged. |
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| 667. |
In an A.C. Circuit, the current is given by `I=6 sin(200 pi t+(pi)/6)A` The initial value of the current isA. 1AB. 2AC. 3AD. 4A |
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Answer» Correct Answer - C `I = 6 sin (200 pi t + (pi)/6))A` `:.` at time `t=0, I = 6 sin (pi)/6 = 6 xx 1/2 = 3A`. |
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| 668. |
A vertical metal rod 1 m long performs linear SHM at right angles to `B_(H)` of `4xx10^(-5)Wb//m^(2)` with an amplitude of 5 cm and a period of `pis`. The values of maximum and minimum emfs induced in it areA. `0muV,4V`B. `4muV,0V`C. `0muV,0V`D. `4muV,4V` |
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Answer» Correct Answer - B `e_(max)=B_(H)lv_(max)` `=B_(H)laomega` `=B_(H)la(2pi)/(T)` `=(4xx10^(-5)xx1xx5xx10^(-5)xx2pi)/(pi)` `=4xx10^(-6)V` `=4muV` `e_("min")=B_(H)lv_("min")=B_(H)lxx0=0V` |
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| 669. |
A metal plate is getting heated . It can be becauseA. a direct current is passing through the plateB. it is placed in a time varying magetic fieldC. it is placed in a space varying magnetic field, but does not vary with timeD. a current (either direct or alternating) is passing through the plate |
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Answer» Correct Answer - A,B,D A metal plate is getting heated when a DC or AC current is passed through the plate, known as heating effect of current. Also, when metal plate is subjected to time varying magnetic field. The magnetic flux, linked with the plate changes and eddy currents comes into existance which make the plate hot. |
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| 670. |
Find the time required for a `50 Hz` alternating current to change its value from zero to the rms value. |
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Answer» Let to equation of aternating current be `I = I_(0) sin omega t = I_(0) sin 100 pi xx 0 = 0` At time t, `I = I_(rms) = (I_(0))/(sqrt 2)` `:. (I_(0))/(sqrt 2) = I_(0) sin omega t` `sin omega t = (1)/(sqrt 2) = sin pi // 4` `t = (pi)/(4 omega) = (pi)/(4 xx 2 pi xx 50) = (1)/(400) s` `t = (1)/(400) s = 2.5 xx 10^(-3) s` |
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| 671. |
An alternating current is given by `I=100 sin(5 pi t)`. How many times will be current become zero in one second?A. 50 timesB. 25 timesC. 40 timesD. 100 times |
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Answer» Correct Answer - A `omega t = 2 pi f t = 50 pi t :. 2 f = 50 :. f = 25 Hz` The current become zero, twice in one cycle. `:.` In one second it will be zero, 50 times . |
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| 672. |
A constant current of 1.5 A is maintained in a resistance of `5 Omega`. What is its r.m.s. value ?A. `(1.5)/(sqrt2)A`B. 1.5 AC. `1.5 sqrt(2)A`D. 0.75 A |
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Answer» Correct Answer - B For a constant current, there are no different values like r.m.s. value, peak value, etc. There is only value i.e. 1.5 A. |
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| 673. |
The current (I) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?A. B. C. D. |
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Answer» Correct Answer - D For inductor as we know indiced voltage For t = 0 to t = T/2 `V=L(dI)/(dt)=L(d)/(dt)((2I_(0)t)/(T))` = constant For t = T/2 to t = T `V=(LdI)/(dt)((-2I_(0)t)/(T))`= constant Therefore, anwer will represent by graph (d). |
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| 674. |
A coil resistance `20 W` and inductance `5 H` is connected with a `100 V` battery. Energy stored in the coil will beA. `41.5J`B. `62.50 J`C. `125J`D. `250J` |
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Answer» Correct Answer - B `U=(1)/(2)Li^(2)=(1)/(2)L((E)/(R ))^(2)=(1)/(2)xx5xx((100)/(20))^(2)=62.50J` |
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| 675. |
A coil resistance `20 Omega` and inductance `5 H` is connected with a `100 V` battery. Energy stored in the coil will beA. 6.25 JB. 62.5 JC. 65.2 JD. 26.5 J |
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Answer» Correct Answer - B `I=(V)/(R)=(100)/(20)=5A` `E=(1)/(2)LI^(2)=62.5J` |
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| 676. |
In figure,`xi=100 V,R_(1)=10 Omega,R_(2)=20 Omega, R_(3)=30 Omega` and `L= 2H`. Find `i_(1) & i_(2)` (a)imeediately after switch `S_(w)` is closed (b)a long time after ( c)immediately after `S_(w)` is opended again (d)a long time later. |
| Answer» Correct Answer - A::B::C::D | |
| 677. |
An e.m.f. of `15` volt is applied in a circuit containing `5` henry inductance and `10` ohm resistance. The ratio of the current at time `t=oo` and at `t=1` second isA. `(E^(1//2))`B. `(e^(2))/(e^(2)-1)`C. `(1-e^(-1))`D. `e^(-1)` |
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Answer» Correct Answer - B `i=i_(0)(1-e^(-Rt//L))` `i_(0)=(E)/(R )` (steady current) when `t=oo` `i_(oo)=(E)/(R )(1-e^(-oo))=(5)/(10)=1.5` `i_(1)=1.5(1-e^(-R//L))=1.5(1-e^(2))` `implies(i_(oo))/(i_(1))(1)/(1-e^(-2))=(e^(2))/(e^(2)-1)` |
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| 678. |
The circuit shown in the figure is used to transfer energy from one capacitor to another. Initially capacitor of capacitance `C_(1) = C_(0)` is charged to a potential difference of `V_(0)`. Switch `S_(1)` is closed at time t = 0. After some time `S_(1)` is opened and `S_(2)` is closed simultaneously. At time `t = T, S_(2)` was opened and it was found that the potential difference across capacitor of capacitance `C_(2) = (C_(0))/(9)` was `3V_(0)`. Find the smallest possible value of time T. The coil has inductance L. Assume no resistance. |
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Answer» Correct Answer - `T_(min) = (2pi)/(3) sqrt(LC_(0))` |
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| 679. |
The circuit shown in figure has two identical capacitors one of which carries a charge Q and the other is having no charge. Switch S is closed. Find the maximum value of current in the circuit if self inductance of the loop is L and each capacitance C. Neglect resistance of the loop. |
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Answer» Correct Answer - `i_(max) = (Q)/(sqrt(2LC))` |
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| 680. |
The dimensions of magnetic flux areA. `[MLT^(-2)A^(-2)]`B. `[ML^(2)T^(-2)A^(-2)]`C. `[ML^(2)T^(-1)A^(-2)]`D. `[ML^(2)T^(-2)A^(-1)]` |
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Answer» Correct Answer - D Dimensional formula of formula flux = `[ML^(2)T^(-2)A^(-1)]` |
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| 681. |
A certain volume of copper is drawn into a wire of radius a and is wrapped in shape of a helix having radius `r ( gt gt a)`. The windings are as close as possible without overlapping. Self inductance of the inductor so obtained is `L_(1)`. Another wire of radius 2 a is drawn using same volume of copper and wound in the fashion as described above. This time the inductance is `L_(2)`. Find `(L_(1))/(L_(2))` |
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Answer» Correct Answer - `8:1` |
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| 682. |
In the circuit shown in figure, the current through each resistor is I. Find the currents through the resistors immediately after the switch ‘S’ is opened. How much heat will get dissipated in the circuit after thw switch is opened ? |
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Answer» Correct Answer - `2I,I,I,3LI^(2)` |
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| 683. |
“The current through an inductor cannot change instantaneously. However, the potential difference across an inductor can change abruptly”. Do you agree with this statement. Give reason. |
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Answer» Correct Answer - Statement is true |
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| 684. |
The circuit shown in the Figure is in steady state. Find the rate of change of current through L immediately after the switch S is closed. |
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Answer» Correct Answer - `(2epsilon)/(3L)` |
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| 685. |
A coil has 600 turns which produces `5 xx 10^-3 Wb//turn` of flux when `3 A` current flows in the wire. This produced `6 xx 10^-3` Wb/turn in 1000 turns secondary coil. When the opened, the current drops to zero in `0.2 s` in primary. Find (a) mutual inductance,(b) the induced emf in the secondary,(c) the self-inductance of the primary coil. |
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Answer» a. `M=(N_2phi_2)/i_1=((1000)(6xx10^-23))/3=2H` b. `|e_2|=M((/_i_1)/(/_ ))=((2)(3))/0.2=30V` c. `L_1=(N_1phi_1)/i_1=((600)(5xx10^(-3)))/(3)=1H` |
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| 686. |
The magnetic flux linked with a circuit of resistance `100 ohm` increase from `10` to `60` webers. The amount of induced charge that flows in the circuit is (in coulomb)`A. 0.5B. 5C. 50D. 100 |
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Answer» Correct Answer - A The amount of induced charge is given by `q=(1)/(R)Deltaphi=(10)/(100)(60-10)` `["where," phi=10Wb, phi_(2)=60Wb, R=100Omega]` `therefore` Charge, `q=(50)/(100)=0.5C` |
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| 687. |
A circular loop of radius r is made of a wire of circular cross section of diameter ‘a’. When current I flows through the loop the magnetic flux linked with the loop due to self induced magnetic field is given by `phi = mu_(0)r [ln((16r)/(a)) - (7)/(4)]l`. The resistivity of the material of teh wire is `rho` and `r gt gt a`.Switch S is closed at time `t = 0` so as to connect the loop to a cell of emf `V`. Find the current in the loop at time t. |
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Answer» Correct Answer - `i = (Va^(2))/(8 rho r) [1- e^(-t//tau)]` Where `tau = (mu_(0)a^(2))/(8rho) [In ((16r)/(a)) - (7)/(4)]` |
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| 688. |
The unit of magnetic flux isA. `Wbm^(-2)`B. WbC. HD. `Am^(-1)` |
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Answer» Correct Answer - B Magnetic flux is a scalar quantity. Its SI unit is weber. |
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| 689. |
In the circuit shown, find the value of resistance R in terms of inductance L and capacitance C such that the current through the cell remains constant forever after the switch is closed. |
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Answer» Correct Answer - `R = sqrt((L)/(C))` |
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| 690. |
In the circuit arrangement shown in Figure, the three way switch S is kept in position 1 for a long time. (a) Find the potential difference across the inductor immediately after the switch is thrown from posi- tion 1 to position 2. (b) After being left in position 2 for a long time, the switch is moved to position 3. Find the potential difference across R immediately after the switch is moved to position 3. Assume no time lag in moving the switch from one position to another. |
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Answer» Correct Answer - (a) E (b) zero |
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| 691. |
The magnetic flux linked with a vector area `vec(A)` in a uniform magnetic field `vec(B)` isA. `BxxA`B. ABC. `B.A`D. `(B)/(A)` |
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Answer» Correct Answer - C Magnetic flux linked with a vector area A in a uniform magnetic field B is given by `phi=B.A` |
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| 692. |
In the circuit shown in the figure, `E = 50.0 V, R = 250 Omega` and `C = 0.500 muF.` The switch `S` is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of `150 V`. What is the inductance `L`? |
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Answer» Correct Answer - B In steady state when switch was closed, `i_0=E//R=(1//5)A=0.2A` After switch is opened it becomes `L-C` circuit in which peak value current is `0.2 A`. `:. 1/2 Li_0^2=1/2CV_0^2` or `L=V_0^2/i_0^2.C` `=((150)^2)/((0.2)^2)xx0.5xx10^-6` `=0.28H` |
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| 693. |
In the circuit shown in figure, the switch `S` is closed at `t=0`. If `V_L` is the voltge induced across the inductor and `i` is the instantaneous current, the correct variation of `V_L` versus is given by B. C. D. |
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Answer» Correct Answer - D `i=0(1-e^(-t//tau_L))=E/R (1-e^(-t//tau_L))` `=E/R -(Ee^(-t//tau-L))/R=i_0-(V_C/R) (as V_L =Ee^(-t//tau_L))` `:. V_L=(i_0R)-(R)i` `V_L` versus i graph is a straight line with positive intercept and negastive slope. |
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| 694. |
Initally switch `S_(1)` is closed, after a long time, `S_(1)` is opened and `S_(2)` is closed. Find the heat produced in coil after long time. |
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Answer» When `S_(1)` is closed, in steady-state current in the coil `i_(0)=E//R` . The energy stored in inductor `=(1)/(2)Li_(0)^(2)` . After closing off `S_(2)` , after long time, heat generated in the coil i.e. in `R` `=((R)/(R+R_(0)))((1)/(2)Li_(0)^(2))` `=((R)/(R+R_(0)))((1)/(2)L(E^(2))/(R^(2)))=(LE^(2))/(2R(R+R_(0))` OR After closing of `S_(2)` , current in the coil i.e. in `R` `i=i_(0)e^-((R+R_(0))/(L))t` Heat produced in coil i.e. in `R` `H=int_(0)^(oo)i^(2)Rdt=i_(0)^(2)int_(0)^(00)e^(2(R+R_(0)t)/(L)dt` `=(E^(2))/(R^(2))R(|e^(-(2(R+R_(0)))/(L)t)|_(0)^(oo))/((2R+R_(0))/(L))` `=-(LE^(2))/(2R(R+R_(0)))|e^(-e^(oo))-e^(0)=(1)/(e^(oo))-1=(1)/(oo-1)=0-1|` `=(LE)^(2)/(2R(R+R_(0)))` |
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| 695. |
During current growth in an L-R circuit the time constant is the time in which the magnitude of current becomesA. `l_(0)`B. `l_(0)//2`C. `0.63l_(0)`D. `0.37l_(0)` |
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Answer» Correct Answer - C Time interval, during which the current in an inductive circuit rises to `63%` of its maximum value is defined as time constant. |
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| 696. |
In the circuit shown, switch S is kept closed and the circuit is in steady state. (a) Find reading of the ideal voltmeter (b) Now the switch is opened. Find the reading of the voltmeter immedi- ately after the switch is opened. (c) Fid the heat dissipated in resistance R after the switch is opened. |
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Answer» Correct Answer - (a) E (b) 2E with polarity reversed (c) `(1)/(2)E^(2)((L)/(R^(2))+C)` |
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| 697. |
A mental rod AB of length l is rotated with a constant angular velocity `omega` about an axis passing through O and normal to its length . Potenial difference between ends of rod in absence of external magnetic field ( Where , `e ` = electric change) A. zeroB. `(m omega^(2)l^(2))/(4e)`C. `(m omega^(2)l^(2))/(2e)`D. `(m omega^(2)l^(2))/(8e)` |
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Answer» Correct Answer - B `eV= m omega^(2)x " " rArrV= (momega^(2)X)/(e)` `dV= (momega^(2))/(e) dx rArr V=(momega^(2))/(e).^(3//4)int_(1//4)" x dx "= (momega^(2)l^(2))/(4e)` |
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| 698. |
When is the magnetic crossing a given surface area held in a magnetic field maximum ? |
| Answer» The magnetic flux is maximum when area is held perpendicular to the direction magnetic field. | |
| 699. |
A coil is held in a magnetic field perpendcular to its plane. When will the average e.m.f. induced in the coil be maximum ? (i) magnetic field is strong but it does not change (ii) magnetic field is small, but its rate of change is large (iii) magnetic field is large, but its rate of change is small. |
| Answer» As `e = (d phi)/(dt) = - (d B)/(dt)`, therefore, induced e.m.f. will be maximum, when rate of change of magnetic field is large, even through the field is small. Choice (ii) is correct. | |
| 700. |
A coil Q is connected to low voltage bulb B and placed near another coil P as showns in Fig. Give reasons to explain the following observation : (a) The bulb B lights (b) Bulb gets dimmer if coil Q is moved toward left. |
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Answer» (i) The bulb B light on account of emf induced in coil Q due to mutual induction between P and Q (ii) When coil Q is moved towards left, magnetic flux linked with Q decreases, and may even reduce to zero at some distance. The emf induced may decrease and the bulb B gets dimmer. |
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