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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.
| 351. |
In the given LC circuit if initially capacitor C has change Q on it and `2C` has charge `2Q`. The polarities are as shown in the figure. Then after closing the switch s at `t = 0` A. energy wil get equally distributed in both the capacitor just after closing the switch.B. initial rate of growth of current in inductor will be 2Q/3CLC. maximum energy in the inductor will be `3Q^(2)//2C`D. none of these |
| Answer» Correct Answer - C | |
| 352. |
In the given LC circuit if initially capacitor C has change Q on it and `2C` has charge `2Q`. The polarities are as shown in the figure. Then after closing the switch s at `t = 0` A. energy will get equally distributed in both the capacitor just after closing the switch.B. initial rate of growth of current in inductor will be `2 Q//3CL`C. maximum energy in the inductor will be `3Q^(2)//2C`D. None of these |
| Answer» Correct Answer - C | |
| 353. |
In LC oscillations of a capacitor with an initial charge `q_(0)` is connected in parallel.A. Time period of oscillations is `(2pi)/sqrt(LC)`B. `Maximum current in circuit is `(q_(0))/sqrt(LC)`C. Maximum rate of change of current in circuit is `(q_(0))/(2C)`D. Maximum potential difference across the inductor is `(q_(0))/(2C)` |
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Answer» Correct Answer - A::B::C |
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| 354. |
A `100 Omega` iron is connected to 220 V, 50 wall plug. What is (i) peak pot. Diff. (ii) average pot.diff. over half cycle (iii) rms current ? |
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Answer» Here, `R = 100 Omega, E_(v) = 220 V`, `v = 50 c y c l e//s` (i) Peak pot.diff., `E_(0) = sqrt2 E_(v) = 1.414 xx 220 = 311 V` (ii) Average pot.diff. over half cycle, (iii) rms current,` I_(v) = (E_(v))/(R ) = (220)/(100) = 2.2 A` |
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| 355. |
The equation of a.c. in a circuit is `I = 50 sin 100 pi t`. Find (i) frequency of a.c. (ii) mean value of a.c. over positive half cycle (iii) rms value of current and (iv) value of current `1//600s` after it was zero. |
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Answer» Here, `I = 50 sin 100 pi t` Compare it with `I = I_(0) sin omega t = I_(0) sin 2 pi v t` `I_(0) 50 A, 2 pi v = 100 pi, v = 50 c//s` Mean value of a.c. over positive half cycle `= (2 I_(0))/(pi) = (2 xx 50)/(3.14) = 31.8 A` `I_(v) = (I_(0))/(sqrt 2) = (50)/(1.414) = 35.35 A` From `I = I_(0) sin omega t` `I = 50 sin 2 pi xx 50 xx (1)/(600)` `= 50 xx (1)/(2) = 25 A` |
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| 356. |
In the shown AC circuit phase different between current `I_(1)` and `I_(2)` is A. `(pi)/(2)-tan^(-1)(X_(L))/(R)`B. `tan^(-1)(X_(L)-X_(C))/(R)`C. `(pi)/(2)+tan^(-1)(X_(L))/(R)`D. `tan^(-1)(X_(L)-X_(C))/(R)+(pi)/(2)` |
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Answer» Correct Answer - A |
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| 357. |
In the given LC circuit if initially capacitor C has change Q on it and `2C` has charge `2Q`. The polarities are as shown in the figure. Then after closing the switch s at `t = 0` A. Energy will get equally distributed in both the capcitor just after closing the switch.B. Initial rate of growth of current in inductor is 2Q/3CLC. Maximumum energy in the inductor will be `4Q^(2)//3C`D. None of these |
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Answer» Correct Answer - D |
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| 358. |
In an RLC circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to :A. 4LB. 2LC. L/2D. L/4 |
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Answer» Correct Answer - B |
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| 359. |
A periodic voltage V varies with time t as shown in the figure 5.249. Tis the time period. The RMS value of the voltage for one cycle is : A. `(V_(0))/(g)`B. `(V_(0))/(2)`C. `V_(0)`D. `(V_(0))/(4)` |
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Answer» Correct Answer - D |
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| 360. |
A conducnting straight wire `PQ` of length `l` is fixed along as diameter of a non conducting ring as shown in the figure. The ring is given a pure rolling motion on as horizontal masgnetic field `B` in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire PQ at the position shown in figure will be A. BvlB. 2BvlC. 3Bvl2D. Zero |
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Answer» Correct Answer - C |
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| 361. |
A rod is rotating with a constant angular velocity `omega` about point `O` (its centre) in a magnetic field `B` as shown. Which of the folloiwng figure correctly shows the distribution of charge inside the rod? A. B. C. D. |
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Answer» Correct Answer - A |
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| 362. |
A straight conductig rod `PQ` is executing `SHM` in xy-plane from `x=-d` to `x=+d.` Its mean position is `x=0` and its length is along `y`-axis. There exists a uniform magnetic field B from `x=-d` to `x=0` pointing inward normal to the paper andfrom `x=0` to `x=+d` there exists another uniform magnetic field of same magnetic `B` but pointing outward normal to the plane of the paper. At the instant `t=0`, the rod is at `x=0` and moving to the right. The induced emf (epsilon) acorss the rod `PQ` `vs` time (`t`) graph will be A. B. C. D. |
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Answer» Correct Answer - A |
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| 363. |
In the figure shown `q` is in coulomb and t in second. At time `t=1 s` A. `V_(a)-V_(b)=4V`B. `V_(b)-V_(c)=1V`C. `V_(c)-V_(d)=16V`D. `V_(a)-V_(d)=20V` |
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Answer» Correct Answer - A::B::D |
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| 364. |
Assertion : A conducting equilateral loop `abc` is moved translationally with constant speed `v` in uniform inward magnetic field `B` as shown. Then : `V_a-V_b=V_b-V_c` Reason : Point a is at higher potential than point `b`.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Apply RHR, we can find that `V_agtV_b` |
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| 365. |
In the figure shown `q` is in coulomb and t in second. At time `t=1 s` A. `V_a-V_b=4V`B. `V_b-V_c=1V`C. `V_c-V_d=16V`D. `V_a-V_b=20V` |
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Answer» Correct Answer - A::B::C `q=2t^2` `i=(dq)/(dt)=4t` `(di)/(dt)=4A//s`s At `t=1s, q=2C,i=4A` and `(di)/(dt)=4A//s` `V_a-V_b=L(di)/(dt)=1xx4=4V` `V_b-V_c=q/C=2/2=1V` `V_C-V_d=ir=4xx4=16V` `V_a-V_d` is summation of above three i.e. `21V`. |
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| 366. |
Magnetic field in cylindrical region of radius `R` in inward direction is as shown in figure. A. an electron will expereince no force kept is `(2R,0,0)` if magnetic field increases with timeB. in the above situation, electron will experience the force in negative y-axisC. if a proton is kept at `(0, R/2,0)` and magnetic field is decreasing then it will experience the force in positive x-directionD. if a proton is kept at `(-R,0,0)` and magnetic field is increasing then it will experience force in negative y-axis |
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Answer» Correct Answer - B::C::D If `ox` magnetic field increases, then induced electric lines are anti clockwie. If `ox` magnetic field decreases, then induced electric lines are clockwise (both inside and outside the cylindrical region). On positive charge, force is the is in the direction of `E`. On negative chare, force is in the opposite. |
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| 367. |
The figure shows a non conducting ring of radius `R` carrying a charge `q`. In a circular region of radis `r`, a uniform magnetic field `B` perpendicular to the plane of the ring varies at a constant rate `(dB)/(dt)=beta`. The torque on the ring is A. `1/2qr^2beta`B. `1/2qR^2beta`C. `qr^2beta`D. zero |
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Answer» Correct Answer - A `El=(dphi)/(dt)=S(dB)/(dt)` `E(2pir)=(pir^2)(beta)` `:. E=r^2/(2R)betaimplies F=qE` and `tau=FR=qER =1/2qr^2beta` |
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| 368. |
In an oscillating `L-C` circuit in which `C = 4.00muF`, the maximum potential difference capacitor during the oscillations is `1.50 V` and the maximum current through `50.0 mA`. (a) What is the inductance `L`? (b) What is the frequency of the oscillations? (c) How much time does the charge on the capacitor take to rise from zero to its maximum value? |
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Answer» Correct Answer - A a. `1/2Li_0^2=1/2CV_0^2` `:. L=(CV_0^2)/i_0^2=((4xx10^-6)(0.5))/((50xx10^-3)^2)` `=3.6 xx 10^-3H` b. `f=1/(2pisqrtLC)=1/(2pisqrt((3.6xx10^-3)(4xx10^-6)))` `=0.133xx10^4Hz` `=1.33kHz` c.`t=T/4=1/(4f)=1/(4xx1.33xx10^3)s` `=0.188xx10^-3s` `=0.188ms` |
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| 369. |
In the given circuit, find the current through the `5 mH` inductor in steady state. |
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Answer» Correct Answer - A::C In steady state, main current from the battery is `i_0=E/R=20/5=4A` Now, this current distributes in inverse ratio of inductor. `:. i_5=(10/(10+5))(4A)=8/3A` |
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| 370. |
At `t = 0`, switch `S` is closed (shown in Fig.). After a long time, suddenly the inductance of the inductor is made `eta` times lesser `(L//eta)` than its initial value. Find out current just after the operation. |
| Answer» Using above result (note 4) `L_(1)i_(1)=L_(2)i_(2) rArr i_(2)=(etaepsilon)/R` | |
| 371. |
In the `L-C` circuit shown, `C =1muF`. With capacitor charged to `100V`, switch `S` is suddenly closed at time `t = 0`. The circuit then oscillates at `10^3Hz` (a) Calculate co and T (b) Express `q` as a function of time(c) Calculate ``L(d) Calculate the average current during the first quarter-cycle. |
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Answer» Correct Answer - A::D `a. omega=(2pi)/f,T1/f` b. At `t=0, q=q_0=CV_0=(100muC)` Now, `q=q_0cosomegat` c. `omega=1/sqrt(LC)impliesL=1/(omega^2C)` d. `|i|=|(dq)/(dt)|=q_0omegasinomegat` Average value of current in first quarter cycle `=(int_0^(T//4)dt)/(T//4)` |
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| 372. |
A metal bar is moving with a velocity of `v = 5 cm s^(-1)` over a U-shaped conductor. At `t = 0`, the external magnetic field is `0.1 T` direction out of the page and is increasing at a rate of `0.2 T s^(-1)`. Tale `l = 5cm`, and at `t = 0, x = 5 cm`. The current flowing in the circuit isA. (a) `2.5 A`B. (b) `5 A`C. ( c) `1 A`D. (d) `2 A` |
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Answer» Correct Answer - A (a) Let us take anticlockwise direction as positive, then area vector in upward direction will be positive `phi = BA = Blx` `e = -(dphi)/(dt) = -l[B(dx)/(dt) + x(dB)/(dt)]` `= -5 xx 10^(-2)[0.1 xx (-5 xx 10^(-2)) + 5 xx 10^(-2) xx 0.2]` `= -250 xx 10^(-6) V = -250 muV` emf is coming out to be negative, so it should be clockwise. Current: `I = (e)/(R ) = (250 xx 10^(-6))/(10^(-4)) = 2.5 A` |
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| 373. |
In the figure shown a `T-shaped` conductor moves with constant angular velocity `omega` in a plane perpendicular to uniform magnetic field B. The potential difference `V_A-V_B` is A. ZeroB. `(1)/(2)Bomegal^(2)`C. `2Bomega^(2)`D. `Bomega^(2)` |
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Answer» Correct Answer - A |
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| 374. |
A metal bar is moving with a velocity of `v = 5 cm s^(-1)` over a U-shaped conductor. At `t = 0`, the external magnetic field is `0.1 T` direction out of the page and is increasing at a rate of `0.2 T s^(-1)`. Tale `l = 5cm`, and at `t = 0, x = 5 cm`. The emf induced in the circuit isA. (a) `125mu V`B. (b) `250mu V`C. (c ) `100mu V`D. (d) `300mu V` |
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Answer» Correct Answer - B (b) Let us take anticlockwise direction as positive, then area vector in upward direction will be positive `phi = BA = Blx` `e = -(dphi)/(dt) = -l[B(dx)/(dt) + x(dB)/(dt)]` `= -5 xx 10^(-2)[0.1 xx (-5 xx 10^(-2)) + 5 xx 10^(-2) xx 0.2]` `= -250 xx 10^(-6) V = -250 muV` emf is coming out to be negative, so it should be clockwise. Current: `I = (e)/(R ) = (250 xx 10^(-6))/(10^(-4)) = 2.5 A` |
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| 375. |
A rectangular loop of sides `a` and `b` is placed in `xy`-placed. A uniform but time varying magnetic field of strength `B=20thati+10t^2hatj+50hatk` is present in the region. The magnitude of induced emf in the loop at time isA. 20+20rB. 20C. 20rD. Zero |
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Answer» Correct Answer - C |
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| 376. |
As the bar shown in figure moves in a direction perpendicular to the field, is an external force required to keep it moving with constant speed. |
| Answer» Circuit is not closed. So, current is zero or magnetic force is zero. | |
| 377. |
A rectangular loop of sides `a` and `b` is placed in `xy`-placed. A uniform but time varying magnetic field of strength `B=20thati+10t^2hatj+50hatk` is present in the region. The magnitude of induced emf in the loop at time isA. `20+20t`B. `20`C. `20t`D. zero |
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Answer» Correct Answer - D `S=(ab)hatkrarr `perpendicular to xy-plane `phi=B.S=(50)(ab)=`constant `(dphi)/(dt)=0` `:. e=0` |
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| 378. |
Two different coils have self inductances `L_1=9mH` and `L_2=2mH`. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current the induced voltage and the energy stored in the first coil are `i_1,V_1` and `W_1` respectively. Corresponding values for the second coil at the same instant are `i_2,V-2` and `W_2` respectively. Then,A. `i_1/i_2=1/4`B. `i_1/i_2=4`C. `W_1/W_2=1/4`D. `(V_(1))/(V_(2)) =4` |
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Answer» Potential difference across an inductor: `VpropL`, if rate of change of current is constant `(V=-L(di)/(dt))` `:. V_2/V_1=L_2/L_1=2/8=1/4` or `V_1/V_2=4` Power given to the two coils is same i.e. `V_1 i_1 =V_2i_2` or `i_1/i_2=V_2/V_1=1/4` Energy stored `W=1/2Li^2` `:. W_2/W_1=(L_2/L_1)(i_2/i_1)^2=(1/4)(4)^2` or `W_1/W_2=1/4` `:.` The correct options are a, c and d. |
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| 379. |
The network shown in the figure is a part of complete circuit. What is the potential difference `V_B-V_A` when the current `I` is `5 A` and is decreasing at a rate of `10^3A//s`? A. `5V`B. `10V`C. `15V`D. `20V` |
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Answer» Correct Answer - C `V_A-1xx5+15+(5xx10^-3)(10^3)=V_B` `:. V_B_V_A=15V` |
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| 380. |
Assertion: Current flowing in the circuit is `i=2t-8`. At `t=1s, V_a-V_b=+4V` Reason: `V_a-V_b` is `+4 V` all the time.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A::B `(di)/(dt)=2A//s` `V_a-V_b=L (di)/(dt)` `=(2)(2)=4V` |
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| 381. |
Two identical conductors `P` and `Q` are placed on two friction less rails `R` and `S` in a uniform magnetic field directed into the plane. If `P` is moved in the direction shown in figure with a constant speed, then rod `Q` A. will be attracted towards PB. will be repelled away from PC. will remain stationaryD. may be repelled or attracted towards P |
| Answer» Correct Answer - A | |
| 382. |
Two identical coaxial circular loops carry a current `i` each circulating int the same direction. If the loops approch each other the current inA. the current in each loop will decreaseB. the current in each loop will increaseC. the current in each loop will remain the sameD. the current in one loop will increases and in the other loop will decrease |
| Answer» Correct Answer - A | |
| 383. |
Two circular coils `A` and `B` are facing each other as shown in figure. The current `i` through `A` can be altered A. there will be repulsion between A and B if `i` is increasedB. there will be attraction between A and B if `i` is increasedC. there will be neither attraction nor repulsion when `i` is changedD. attraction or repulsion between A and B depends on the direction of current. It does not depend whether the current is increased or decreased. |
| Answer» Correct Answer - A | |
| 384. |
Two coils are at fixed location: When coil 1 has no corrent and the current in coil 2 increase at the rate of `15.0 A s^(-1)`, the emf in coil 1 is `25 mV`, when coil 2 has no current and coil 1 has a current of `3.6 A`, the flux linkange in coil 2 isA. 16 mWbB. 10mWbC. 4.00 mWbD. 6.00 mWb |
| Answer» Correct Answer - D | |
| 385. |
An inductor coil stores U energy when i` current is passed through it and dissipates energy at the rate of P. The time constant of the circuit, when this coil is connected across a battery of zero internal resistance isA. `(4U)/P`B. `U/P`C. `(2U)/P`D. `(2P)/U` |
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Answer» Correct Answer - C `U=1/2Ll^(2) rArr P=I^(2)R` or `(2U)/P=L/R=tau`. |
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| 386. |
When induced `emf` in inductor coil is `50%` of its maximum value then stored energy in inductor coil in the given circuit at that instant will be:- A. 2.5mJB. 5mJC. 15mJD. 20mJ |
| Answer» Correct Answer - A | |
| 387. |
When induced `emf` in inductor coil is `50%` of its maximum value then stored energy in inductor coil in the given circuit at that instant will be:- A. `2.5 mJ`B. `5 mJ`C. `15 mJ`D. `20 mJ` |
| Answer» Correct Answer - A | |
| 388. |
The current (in Ampere) in an inductor is given by `l=5+16t`, where `t` in seconds. The selft - induced emf in it is `10mV`. Find (a) the self-inductance, and (b) the energy stored in the inductor and the power supplied to it at `t =1s` |
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Answer» a. `dI//dt=16A//s` `:.L=|e/(dI//dt)|=(10xx10^-3)/16` `=0.625xx10^-3H=0.625mH` b. At `t=1s, I=21A` `U=1/2LI^2=1/2xx(0.625xx10^-3)(21)^2` `=0.137J` `P=Ei=(10xx10^-3)(21)` `=0.21J//s` |
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| 389. |
At the instant when the current in an inductor magnitude of the self-induced emf is `0.0640 A//s`, the magnitude fo the self- induced emf is `0.0160 V`. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is `0.720A`? |
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Answer» Correct Answer - A::B::D `a. L=|e/(di//dt)|=0.016/0.064=0.25H` `b L=(Nphi)/i` `:. =(Li)/N=((0.25)(0.72))/400` `4.5xx10^-4Wb` |
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| 390. |
A rod of length l rotates with a uniform angular velocity omega about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the lrod isA. `2Bomegal^(2)`B. `(1)/(2)omegaBl^(2)`C. `Bomegal^(2)`D. zero |
| Answer» Correct Answer - D | |
| 391. |
Two circular coils can be arranged in any of the three situation shown in the figure. Their mutual inductance will be A. Maximum in situation `(A)`B. Maximum in situation `(b)`C. Maximum in situation `(C)`D. The same in all situations |
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Answer» Correct Answer - A The mutual inductance between two coils depends on their degree of flux linkage, i.e., the fraction of flux linked with one coil which is also linked to the other coil. Here, the two coils in arrangement (a) are placed with their planed parallel. This will allow maximum flux linkage. |
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| 392. |
When an ac source of emf`e=E_(0) sin (100 t)` is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be `(pi)//(4)` as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements. A. `R=1kOmega,C=10muF`B. `R=1kOmega,C=1muF`C. `R=1kOmega,C=10H`D. `R=1kOmega,L=1H` |
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Answer» Correct Answer - A |
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| 393. |
A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced A. in `AD`, but not in `BC`B. In `BC`, but not in `AD`C. neither in `AD` nor in `BC`D. in both `AD` and `BC` |
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Answer» Correct Answer - D Both `AD` and `BC` are straight conductors moving in a uniform magnetic field and emf will be induced in both. This will cause elecric fields in both, but no net current flows in the circuit. |
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| 394. |
When an ac source of emf`e=E_(0) sin (100 t)` is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be `(pi)//(4)` as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements. A. `R = 1 k Omega, C = 10 mu F`B. `R = 1 k Omega, C = 1 mu F`C. `R = 1 k Omega, L = 10 mH`D. `R = 10 k Omega, L = 10 mu H` |
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Answer» Correct Answer - A Fig. shows that current `i` leads the emf `e` by a phase angle `pi//4`. Therefore, the circuit can be `R - C` circuit alone. `tan phi = (X_(C ))/(R ) = (1)/(omega CR) = "tan" (pi)/(4) = 1` or `CR = (1)/(omega)` From `e = E_(0) sin 100 t, omega = 100 rad//s` `:. CR = (1)/(omega) = (1)/(100)` when `R = 1 K Omega = 10^(3) Omega` `C = (1)/(omega R) = (1)/(10^(5)) = 10^(-5) F = 10 mu F` |
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| 395. |
When an AC source of `emf e=E_(0) sin(100t)` is connected across a circuit i in the circuit, the phase difference between the emf e and the current i in the circuit is observed to be `(pi//4)`, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements A. `R=1k Omega, C=10 mu F`B. `R=1k Omega, C=1 mu F`C. `R=1k Omega, L=10 mu F`D. `R=1k Omega, L=1 mu F` |
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Answer» Correct Answer - A Since,current leads emf (as seen from the graph), therefore, this is an R-c circuit. `tan(phi)=(X_(C)_X_(L))/(R)` Here, `(phi)=(45^@)` `:. (X_C)=R` [`(X_L)=0` as there is no inductor] `1/(omega C)=R implies RC(omega)=1` `:. RC=(1)/(100)(s^(-1))`. |
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| 396. |
A 20 mH coil is connected in series with a `2 k Omega` resistor and a 12 V battery. Calcualte (i) time constant of the circuit, (ii) time during which current decrys to 10% of its maximum value. |
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Answer» Here, `L = 20 mH = 20 xx 10^(-3) H`, `R = 2 k Omega = 2 xx 10^(3) ohm, V = 12 V` (i) Time constant `= (L)/(R ) = (20 xx 10^(-3))/(2 xx 10^(3)) = 10^(-5) s` (ii) For decay of current, `I = I_(0) e^(-(R )/(L) t)` `(I_(0))/(10) = I_(0) e^(-10^(5) t)` `:. e^(10^(5) t) = 10` `10^(5) t = 2.303 log_(10) 10` `t = (2.303)/(10^(5)) = 2.303 xx 10^(-5) s` |
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| 397. |
Electric charge `q` is distributed uniformly over a rod of length `l`. The rod is placed parallel to a ling wire carrying a current `i`. The separation between the rod and the wire is . The force needed to move the rod along its length with a uniform velocity `v` isA. `(mu_(0)//qv)/(2nd)`B. `(mu_(0)iqv)/(4pia)`C. `(mu_(0)iqvl)/(2pia)`D. `(mu_(1)iqvl)/(4pia)` |
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Answer» Correct Answer - C |
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| 398. |
Assertion Mutual inductance of two coils depends on the distance between the coils and their orientation. Reason It does not depend on the magnetic material filled between the coils. |
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Answer» Correct Answer - c The mutual inducrtance in case of a medium of ralative permeability `mu_(r)` present is `M=mu_(r)mu_(0)n_(1)n_(2)pi r_(1)^(2)l` |
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| 399. |
Two solenoids `A` and `B` spaced close to each other and sharing the same cylindrical axis have `400` and `700` turns, respectively. A current of `3.50A` in coil `A` produced an average flux of `300muT-m^2` through each turn of `A` and `a` flux of `90.0mT-m^2` through each turn of `B`. a. Calculate the mutual inductance of the two solenoids. b.What is the self inductance of `A`? c. What emf is induced in `B` when the current in `A` increases at the rate of `0.5A//s`? |
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Answer» Here, `M = ? dI = 6 - 2 = 4 A, dt = 0.1 s, e = 1 V` From `e = M dI//dt` `M = (e. dt)/(dI) = (1 xx 0.1)/(4) = 0.025 H` |
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STATEMENT - 1 : Mutual induction depends on geometry and orientation of loops w.r.t. each other STATEMENT - 2 : A capacitor acts as an infinite resistance for AC. STATEMENT - 3 : The A. C. voltage across a resistance can be measured using a hot-sire volt meter.A. TFTB. FFTC. TTFD. TFF |
| Answer» Correct Answer - A | |