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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.
| 901. |
A circuit of negligible resistance has an inductane of 10 mH and a capacitance of 0.1 `muF`. The resonant frequency of the circuit is nearlyA. 5 kHzB. 2.5 kHzC. 31.4 kHzD. 3.14 kHz |
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Answer» Correct Answer - A `f=(1)/(2pisqrt(LC))` `=(1)/(2xx3.14xxsqrt(10xx10^(-3)xx0.1xx10^(-6)))` `=5.036kHz` `~~5kHz` |
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| 902. |
In an AC circuit voltage applied is e=220 sin 100 t. if the impedance is `110Omega` and phase difference between the current and voltage is `60^(@)` the power consumption is equal toA. 55 WB. 110 WC. 220 WD. 330 W |
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Answer» Correct Answer - B `overline(P)=eIcosphi=(e^(2))/(Z)cosphi` `=(220xx220xxcos60)/(110)=110W` |
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| 903. |
An a.c. emf of e=220 sin 100`pit` is passed through the resistance of 1 `kOmega`. The average power of the a. c. circuit isA. 48.4 WB. 34.2 WC. 24.2 WD. 12.1 W |
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Answer» Correct Answer - C `overline(P)=(e_(0)I_(0))/(2)=(e^(2))/(2R)` `=(220xx220)/(2000)=24.2W` |
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| 904. |
A conductor is moving in the magnetic field B the induced current is I. if the magnetic field is doubled, the induced current willA. remain the sameB. be halfC. be doubleD. be four times |
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Answer» Correct Answer - C The induced emf is proportional to B. thus, induced current is also proportional to B. |
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| 905. |
A rectangular loop with a sliding connector of length `10 cm` is situated in uniform magnetic field perpendicular to plane of loop. The magnetic induction is `0.1` tesla and resistance of connector `(R )` is `1 ohm`. The sides `AB` and `CD` have resistances `2 ohm` and `3 ohm` respectively. Find the current in the connector during its motion with constant velocity one A. `(1)/(242)A`B. `(1)/(220)A`C. `(1)/(55)A`D. `(1)/(440)A` |
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Answer» Correct Answer - B `e=Bvl=(0.1)(1)(0.1)=(1)/(100)V` Net resistance `=1+(2xx3)/(2+3)=(11)/(5)Omega` `therefore` Current, `i=((1//100)/(11//5))=(1)/(220)A` |
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| 906. |
A glass rod of length l moves with velocity in a uniform magnetic field B. What is the e.m.f. induced in the rod ? |
| Answer» No induced e.m.f. is set up because glass rod is insulating. | |
| 907. |
Identify the incorrect statement. Induced electric fieldA. Is produced by varyig magnetic fieldB. is not conservative in natureC. cannot exist in a region not occuide by magnetic fieldD. None of the above |
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Answer» Correct Answer - B |
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| 908. |
When the current in the portion of the circuit shown in the figure is 2 A and increasing at the rate of `1As^(-1)`, then the measured potential difference `V_(ab)=8V`. However, when the current is 2 A and decreasing at the rate of `1As^(-1)`, then the measured potential difference `V_(ab)=4V`. The values of R and L are A. `3Omega` and 2 H respectivelyB. `3Omega` and 3 H respectivelyC. `2Omega` and 1 H respectivelyD. `3Omega` and 1 H respectively |
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Answer» Correct Answer - A From the figure, `V_(a)-iR-L(di)/(dt)=V_(b)` `therefore" "V_(a)-V_(b)=iR+L(di)/(dt)` According to given conditions, 8 = 2R + L ….(i) 4 = 2R - L …..(ii) Solving these two equation, we get `R=3Omega and L = 2H` |
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| 909. |
Is induced electric field conservative or non conservative ? |
| Answer» It is non convervative, as `oint vec E. vec dl = (d phi)/(dt) != 0` | |
| 910. |
As a result of change in the magnetic flux linked to the closed loop shown in the fig, an e.m.f. V volt is induced in the loop. The work done (joule) in taking a charge Q coulomb once along the loop is A. qVB. zeroC. 2qVD. `(qV)/(2)` |
| Answer» Correct Answer - A | |
| 911. |
When the current in the portion of the circuit shown in the figure is `2A` and increasing at the rate of `1A//s`,the measured potential difference `V_(a)-V_(b)=8V`.However when the current is `2A` and decreasing at the rate of `1A//s`, the measured potential difference `V_(a)-V_(b)=4V`.The values of `R` and `L` are: A. 3 ohm and 2 Henry respectivelyB. 2 ohm and 3 Henry respectivelyC. 10 ohm and 6 Henry respectivelyD. 6 ohm and 1 Henry respectively |
| Answer» Correct Answer - A | |
| 912. |
Asseraton: The work done by a charge in a closed (induced) current carrying loop is non-zero. Reason: Induced electric field is non-conservative in nature.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `DeltaW=q(DeltaV)` Here `DeltaV=` non-zero in a closed loop. |
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| 913. |
A super conducing loop having an inductance `L` is kept in a magnetic field which is varying with respect to time. If `phi` is the total flux, `epsilon`=total induced `emf`, then:A. `phi`=constantB. `l=0`C. `epsilon=0`D. `epsilon ne 0` |
| Answer» Correct Answer - A::C | |
| 914. |
Two parallel vertical metallic rails `AB` and `CD` are separated by `1m`. They are connected at the two ends by resistances `R_1` and `R_2` as shown in the figure. A horizontal metallic bar `l` of mass `0.2 kg` slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in `R_1` and `R_2` are `0.76W` and `1.2W` respectively `(g=9.8m//s^2)` The value of `R_2`isA. `0.6Omega`B. `0.5Omega`C. `0.4Omega`D. `0.3Omega` |
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Answer» Correct Answer - D At terminal velocity `iLB=mg` `:. i=(mb)/(LB)=(0.2xx98)/(1xx0.6)` `i=3.27A` ...(i) `e=BvL(v=`terminal velocity) `=(0.6)(v)(1)` `e=0.6v` `P_(R_1)=e^2/R_1` `:. 0.76=(0.36v^2)/R_1` …..(ii) `P_(R_2)=e^2/R_2` `:. 1.2=(0.36v^2)/R_2` .........(iii) `R_1` and `R_2` are in parallel `:. R_("net")=(R_1R_2)/(R_1+R_2)`..........iv `ii=e/(R_("net")` ............(v) Solving these five equation we can get the results. |
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| 915. |
Two parallel vertical metallic rails `AB` and `CD` are separated by `1m`. They are connected at the two ends by resistances `R_1` and `R_2` as shown in the figure. A horizontal metallic bar `l` of mass `0.2 kg` slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in `R_1` and `R_2` are `0.76W` and `1.2W` respectively `(g=9.8m//s^2)` The terminal velocity fo the bar L will beA. `2m//s`B. `3m//s`C. `1m//s`D. None of these |
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Answer» Correct Answer - C At terminal velocity `iLB=mg` `:. i=(mb)/(LB)=(0.2xx98)/(1xx0.6)` `i=3.27A` ...(i) `e=BvL(v=`terminal velocity) `=(0.6)(v)(1)` `e=0.6v` `P_(R_1)=e^2/R_1` `:. 0.76=(0.36v^2)/R_1` …..(ii) `P_(R_2)=e^2/R_2` `:. 1.2=(0.36v^2)/R_2` .........(iii) `R_1` and `R_2` are in parallel `:. R_("net")=(R_1R_2)/(R_1+R_2)`..........iv `ii=e/(R_("net")` ............(v) Solving these five equation we can get the results. |
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| 916. |
Two parallel vertical metallic rails `AB` and `CD` are separated by `1m`. They are connected at the two ends by resistances `R_1` and `R_2` as shown in the figure. A horizontal metallic bar `l` of mass `0.2 kg` slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in `R_1` and `R_2` are `0.76W` and `1.2W` respectively `(g=9.8m//s^2)` The value of `R_1` isA. `0.47Omega`B. `0.82Omega`C. `0.12Omega`D. None of these |
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Answer» Correct Answer - A At terminal velocity `iLB=mg` `:. i=(mb)/(LB)=(0.2xx98)/(1xx0.6)` `i=3.27A` ...(i) `e=BvL(v=`terminal velocity) `=(0.6)(v)(1)` `e=0.6v` `P_(R_1)=e^2/R_1` `:. 0.76=(0.36v^2)/R_1` …..(ii) `P_(R_2)=e^2/R_2` `:. 1.2=(0.36v^2)/R_2` .........(iii) `R_1` and `R_2` are in parallel `:. R_("net")=(R_1R_2)/(R_1+R_2)`..........iv `ii=e/(R_("net")` ............(v) Solving these five equation we can get the results. |
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| 917. |
Asseration: The self-inductance `(L)` is given by `phi` (magnetic flux) =`L` `i` (current). Reason: When current is increased, self-inductance increases.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C `L` is dependent only upon geometrical parameter. |
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| 918. |
Two parallel vertical metellic rails `AB` and `CD` are separated by `1m`. They are connecting at two ends by resistances `R_(1)` and `R_(2)` as shown in Fig. 3.96. A horizontal metallic bar `L` mas `0.2 kg` slides without friction vertically down the rails under the action of gravity. There is a uniform horozontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, thwe power dissipated in `R_(1) and R_(2)` are `0.76 and 1.2W`, respectively. Find the terminal velocity of the bar `L` and the values of `R_(1) and R_(2)`. |
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Answer» Correct Answer - A For terminal velocity `Mg=ILB` here `I=epsilon/(R_(eq))=(BV_(0)L)/(R_(eq))` `Mg=(B^(2)L^(2)V_(0))/(R_(1)R_(2)//R_(1)+R_(2))` `V_(0)=Mg.(R_(1)R_(2))/(R_(1)+R_(2)) 1/(B^(2)L^(2))`..(i) Given that `I_(1)^(2)R_(1)=0.76`..(ii) & `I_(2)^(2)R_(2)^(2)=1.2`..(iii) where `I_(1)=epsilon/R_(1)` and `I_(2)=epsilon/R_(2)` Solve (i), (ii) and (iii) method II (Better sol.) power of gravitational force `mg V_(T)=0.76+1.20` So,`V_(T)=1 m//s` `epsilon=BV_(T)l=0.6 "volt" , P_(1)=epsilon^(2)/R_(1) , thereforeR_(1)=epsilon^(2)/P_(1)=((0.6)^(2))/0.76` `R_(1)=36/76Omega` similarly `R_(2)=0.36/1.20=3/10Omega` `R_(1)/R_(2)=30/19` |
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| 919. |
Statement I: An eletric lamp is connected in series with a long solenoid of copper with air core and then connected to an ac source. If an iron rod is inserted in the solenoid, the lamp will become dim. Satement II: If an iron rod is inserted in the solenoid, the inductance of the solenoid increases.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A If inductance of solenoid increases, than reachtance of circuit too increase, than obviously current will decrease and light becomes dim. |
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| 920. |
Assertion: Only a charge in magnetic flux will maintain an induced current in the coil. Reason: The presence of large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.A. AB. BC. CD. D |
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Answer» Correct Answer - C Induced current is due to emf induced which is directly proportional to the rate of changeof magnetic flux lined with the coil. Mere pressure of magnetic flux does not maintain current in the coil. Assertion is ture, but reason is false. |
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| 921. |
In the circuit shown switch `S` is connected to position `2` for a long time and then joined to position `1`.The total heat produced in resistance `R_(2)` is: A. `(LE^(2))/(2R_(2)^(2))`B. `(LE^(2))/(2R_(1)^(2))`C. `(LE^(2))/(2R_(1)R_(2))`D. `(LE^(2)(R_(1)+R_(2))^(2))/(2R_(1)^(2)R_(2)^(2))` |
| Answer» Correct Answer - A | |
| 922. |
Assertion: Only a charge in magnetic flux will maintain an induced current in the coil. Reason: The presence of large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.A. If both asseration and reason are true and reason is the correct explanation of assertion.B. If both asseration and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C Presence of magnetic flux cannot produce current. |
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| 923. |
Two identical coils each of self-inductance L , are connected in series and are placed so closed to each other that all the flux from one coil links with the other. The total self-inductance of the system isA. LB. 2LC. 3LD. 4L |
| Answer» Correct Answer - D | |
| 924. |
A neutral metal bare moves at a constant velocity v to the right through a region of uniform magnetic field directed out the page, as shown. Therefore, A. positive charges accumulate to the left side and negative charges to the right side of the rodB. negative charges accumulate to the left side and positive charges to the right side of the rodC. positive charges accumulate to the top end and negative charges to the bottom end of the rodD. negative charges accumulate to the top end and positive charges to the bottom end of the rod |
| Answer» Correct Answer - D | |
| 925. |
The magnetic field `B` at all points within `a` circular region of the radius `R` is uniform space and directed into the plane of the page in figure. If the magnetic field is increasing at a rate `dB//dt` what are the magnitude and direction of the force on as stationary positive point charge `q` located at points `a,b, c`? (Point a is a distance `r` above the centre of the region, point `b` is a distance `r` to the right to the centre and point `c` is at the centre of the region). |
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Answer» Inside the circular region at distance `r` `El=(dphi)/(dt)=S((dB)/(dt))` `:. E(2pir)=(pir^2).(dB)/(dt)` `E=r/2 (dB)/(dt)` `F=qE=(qr)/2(dB)/(dt)` At points `a` and `b`, distance from centre is `r` `:. F=(qr)/2 (dB)/(dt)` At point `C` distance `r=0` `:. F=0` `ox` magnetic field is increasing. Hence, induced current in an imaginary loop passing through `a` and `b` should produce `o`. magnetic field. Hence induced current through an imaginary circular loop passing through a and `b` shoud be anti clockwise. Force on positive charge is in the direction of induced current. Hence, force at a is towards left and force at `b` is upwards. |
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| 926. |
In a closed loop, which has some inductance but negligible resistance, uniform but time varying magentic field is applied directed into the plane of the loop. Variation of field with time is shown in Fig. Initially current in loop was zero. Then . A. emf induced in the loop is zero at t = 2 sB. current in the loop will be maximum at r = 2 sC. direction of emf in the loop will change at t = 2 sD. None of the above |
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Answer» Correct Answer - A::C `e=-(dphi)/(dt)=-A.((dB)/(dt))=-A("Slope of B-t graph")` At t = 2, slope is zero and it changes its sign. |
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| 927. |
In a closed loop, which has some inductance but negligible resistance, uniform but time varying magentic field is applied directed into the plane of the loop. Variation of field with time is shown in Fig. Initially current in loop was zero. Then . A. emf induced in the loop is zero at `t = 2 s`B. current in the loop will be maximum at `t = 2 s`C. direciton of emf in the loop will not change at `t = 2 s`D. none of above |
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Answer» Correct Answer - A As `phi = BA` and `|e| = (d phi)/(dt)`, therefore, from Fig. we find that at `t = 2 s, B = max, phi = max , e = 0`. The flux increases from 0 to 2 sec and it decreases from `t = 2 s` to `t = 4s`. Therefore, the direction of e.m.f. in the loop is reversed at `t = 2 s`. Choice (a) is correct. |
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| 928. |
A rod lies across frictionless rails in a uniform magnetic field `vec(B)` as shown in figure. The rod moves to the right with speed `V`. In order to make the induced emf in the circuit to be zero, the magnitude of the magnetic field should A. (a) not changeB. (b) increase linearly with timeC. ( c) decrease linearly with timeD. (d) decrease nonlinearly with time |
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Answer» Correct Answer - D (d) `Bl Vt = constant` `B = ( C)/(lVt)` |
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| 929. |
The current through the coil in figure (i) varies as shown in figure (ii). Which graph best shows the ammeter `A` reading as a function of time? A. (a) B. (b) C. ( c) D. (d) |
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Answer» Correct Answer - A (a) Let `M` is mutual inductance of coil, then the flux in second coil is `phi_(2) = Mi_(1)` ltbr. `e_(2) = (dphi_(2))/(dt) = M(di_(1))/(dt)` `i_(2) = (e_(2))/(R ) = (M)/(R )(di)/(dt)` `rarr` `i_(2) prop (di)/(dt)` |
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| 930. |
An a.c source of frequency `1000 Hz` is connected to a coil of `(200)/(pi) mH` and negligible resistance. If effective current through the coil is `7.5 mA`, what is the voltage (in volt) across the coil ? |
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Answer» Correct Answer - 3 Here, `v = 1000 Hz, L = (200)/(pi) = (0.2)/(pi) H` `I_(upsilon) = 7.5 mA = 7.5 xx 10^(-3) A, E_(upsilon) = ?` As resistance of coil is negilgible, `:. E_(upsilon) = I_(upsilon) (X_(L)) = I_(upsilon) (2 pi v L)` `= 7.5 xx 10^(-3) (2 pi xx 1000 xx (0.2)/(pi)) = 3 V` |
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| 931. |
An a.c. voltage is represented by `E = 220 sqrt2 cos (8 pi) t`. How many times will the current become zero in 1 sec ? |
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Answer» Correct Answer - 8 Compare `E = 220 sqrt2 cos (8 pi) t` with the standard form: `E = E_(0) cos omega t` `omega = 8 pi = (2 pi)/(T) :. T = (2 pi)/(8 pi) = (1)/(4) sec`. In `(1)/(4) sec`., current becomes zero, twice `:.` In 1 sec., current will become zero 8 times |
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| 932. |
A rectangular loop of sides 10 cm and 5 cm with a cut is stationary between the pole pieces of an electromagnet. The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced, so that the field decreased from its initial value of 0.2 T at the rate of `0.02Omega`. If the cut is joined and the loop has a resistance of `2.0Omega`, then the power dissipated by the loop as heat isA. 5 nWB. 4 nWC. 3 nWD. 2 nW |
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Answer» Correct Answer - A Here, area, `A=10xx5=50cm^(2)=50xx10^(-4)m^(2)` `(dB)/(dt)=0.2Ts^(-1)rArrR=2Omega` `therefore" emf, e"=(dphi)/(dt)=A.(dB)/(dt)=50xx10^(-4)xx0.02=10^(-4)V` Power dissipated in the form of heat `=(e^(2))/(R)=(10^(-4)xx10^(-4))/(2)=0.5xx10^(-8)W=5xx10^(-9)W=5nW` |
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| 933. |
A conducting wire is moving towards right in a magnetic field `B`. The direction of induced current in the wire is shown in the figure. The direction of magenetic field will be A. In the plane of paper pointing towards rightB. In the plane of paper pointing towards leftC. perpendicular to the plane of paper and downwardsD. perpendicular to the plane of paper and upwards |
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Answer» Correct Answer - C According to fleming right hand rule, the direction of `B` will be perpendicular to the plane of paper and act downward. |
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| 934. |
Suppose the loop in Textual Exercise 4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of `0.02Ts^(-1)` .If the cut is joined and the loop has a resistance of `1.6Omega` how much power is dissipated by the loop as heat ? What is the source of this power ? |
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Answer» Area of loop `=8xx 2= 16cm^(2)= 16 xx 10^(-4)m^(2)` Rate of change of magnetic field `(dB)/(dt)=0.02T/S` Resistance of loop `R= 1.62Omega` Induced emf of loop `e=(dphi)/(dt)=(d)/(dt) (BA) (therefore phi=BA)` `e=A (dB)/(dt) Rightarrow e=16xx10^(-4) xx 0.02 Rightarro e=0.32 xx 10^(-4)V` Induced current in the loop `I=e/R=(0.32xx10^(-4))/(1.6)=0.2 xx 10^(-4)A Power of source as heat `P=I^(2)R=(0.2xx10^(-4))x1.6=6.4xx10^(-10)W Rightarrow P=6.4xx10^(-10)W` The agency which changing the magnetic field with time is the source of this power. |
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| 935. |
Two straight and parallel wires A and B are being brought towards each other. If current in A be i. what will be the direction of induced current in B? If A and B are being taken away from each other, then ? |
| Answer» In the first case, induced current in B will be opposite to i. so that A and B repel eachother and the operation of bringing them closer is opposed. Similarly, in the second case, induced current in B will be in the direction of i. | |
| 936. |
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm `s^(-1)` in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of `10^(-3) T cm^(-1)` along the negative x-direction (that is it increases by `10^(-3) T cm^(-1 )`as one moves in the negative x-direction), and it is decreasing in time at the rate of `10^(-3) T s^(-1)`. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 `mOmega`. |
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Answer» Side of the square loop, s = 12 cm = 0.12 m Area of the square loop, A =` 0.12 xx 0.12 = 0.0144 m^(2)` Velocity of the loop, v = 8 cm/s = 0.08 m/s Gradient of the magnetic field along negative x-direction, `(dB)/(dx)=10^(-3)Tcm^(-1)=10^(-1)Tm^(-1)` And, rate of decrease of the magnetic field, `(dB)/(dt)=10^(-3)Ts^(-1)` Resistance of the loop, `R=4.5 Omega=4.5xx10^(-3)Omega` Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as: `(dphi)/(dt)=xx(dB)/(dx)xxv` `=144xx10^(-4)m^(2)xx10^(-1)xx0.08` `=11.52xx10^(-5)Tm^(2)s^(-1)` Rate of change of the flux due to explicit time variation in field B is given as: `(dphi)/(dt)=Axx(dB)/(dt)` `=144xx10^(-4)xx10^(-3)` `=1.44xx10^(-5)Tm^(2)s^(-1)` Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as: `e=1.44xx10^(-5)+11.52xx10^(-5)` `=12.96xx10^(-5)` V Induced current, `i=e/R` `(12.96xx10^(-5))/(4.5xx10^(-3)` `i=2.88xx10^(-2)`A Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction. |
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| 937. |
The flux linked with a circuit is given by `phi=t^(3)+3t-7` . The graph between time (`x`-axis) and induced emf (`y`-axis) will beA. straight line through originB. straight line with positive interceptC. stairght line with negative interceptD. parabola not through the origin |
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Answer» Correct Answer - D `phi=t^(3)+3t-7` `e=-(dphi)/(dt)=-(3t^(2)+3)` `ev//st` graph: `y=-3x^(2)-3` , this is parabola, not through origin. |
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| 938. |
A magnet N-S is suspended from a spring and while at oscillates, the magnet moves in and out of the coil C. The coil is connected to a galvanometer G. Then, as the magnet oscillates,A. G shows deflection to the left and right with constant amplitudeB. G shows deflection on one sideC. G shows no deflectionD. G shows deflection to the left and right but the amplitude steadily decreases |
| Answer» Correct Answer - D | |
| 939. |
A current is induced in coil `C_1` due to the motion of current carrying coil `C_2`. (a) Write any two ways by which a large deflection can be obtained in the galvanometer G. (b) Suggest an alternative device to demonstrate the induced current in place of galvanometer. |
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Answer» (a)When coil `C_(1)` and `C_(2)` held stationary.Coil `C` is connected to a sensitive galvanometer `G` and coil `C_(2)` is connected to a battery through a trapping ket `K`. When the key is released the galvanometer shows again a momentary deflection.but in the opposite direction The pointer in the galvometer returns to zero almost instantly.The galvanometer direction increases dramatically when an iron rod is inserted into the coils along their axis. (b) Ammeter can be used in place Galvanometer to read the current. |
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| 940. |
Fig. shows series LCR circuit with `L = 5.0 H, C = 80 mu F, R = 40 Omega` connected to a variable frequency 240 V source. Calculate (i) the angular freuquency of the source which drives the circuit at resonance. (ii) Current at the resonanting frequency. (iii) the rms pot. drop across the capacitor at resonance. |
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Answer» Here, `L = 5.0 H`, `C = 80 mu F = 80 xx 10^(-6) F, R = 40 ohm` `E_(v) = 240 V, omega = ?, I = ?, V_(C ) = ?` At resonance, `X_(L) = X_(C )` `omega = (1)/(omega C)` or `omega = (1)/(sqrt(LC))` `omega = (1)/(sqrt(5.0 xx 80 xx 10^(-6))) = (1000)/(20) = 50 rad//s` (ii) At resonance. `Z = R =40 Omega` `I_(v) = (E_(v))/(Z) = (240)/(40) = 6 A` (iii) `V_(C ) = I_(v) X_(C )` `= (I_(v))/(omega C) = (6)/(50 xx 80 xx 10^(-6)) = (6000)/(4) = 1500 V` |
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| 941. |
Answer the following questions : Fig shows an inductor L and a resistance R connected in parallel to battery through a switch. The resistance of R is the same as that of the coil that makes L. Two identical bulbs are put in each arm of the circuit. (i) Which of the bulb be equally bright after some time ? |
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Answer» (i) The bulb `B_(2)` in arm R lights up earlier because induced e.m.f. across L opposes growth of current in `B_(1)` (ii) Yes, both the bulbs will be equally bright after some time. This is because once the current has reached its maximum value, self inductance plays no role. Resistance of L and R is same, and both the bulbs are identical and are connected to same source of potential difference. |
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| 942. |
The natural frequency of the circuit shows in Fig. is A. `(1)/(2 pi sqrt(LC))`B. `(1)/(2 pi sqrt(2LC))`C. `(2)/(2 pi sqrt(LC))`D. none of these |
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Answer» Correct Answer - A In Fig., the two condensers are in series and so are the two inductors. Therefore, `L_(s) = L + L = 2 L` `(1)/(C_(s)) = (1)/(C ) + (1)/(C ) = (2)/(C ) , C_(s) = C//2` `:.` Natural frequency of the circuit, `v = (1)/(2 pi sqrt(L_(s) C_(s))) = (1)/(2 pi sqrt(2 L xx C //2)) = (1)/(2 pi sqrt(LC))` |
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| 943. |
Figure shows the to views of a rod that can slide without friction. The resistor is `6.0Omega` and a `2.5T` magnetic field is directed perpendicularly downward into the paper. Let `l=1.20m`. a. Calculate the force `F` required to move the rod to the right at a constant speed of `2.0 m/s`, b. At what rate is energy deliered to the resistor? c. Show that this rate is equal to the rate of work done by the applied force. |
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Answer» Given, resistance, R =`6.0Omega` Magnetic field, `B=2.57T` Length of the rod, `l=1.20m` The motional emf in the rod is given as, e = Bvl `or" "e=(2.5)(2.0)(1.2)V=6.0V` The current in the circuit, `i=(e)/(R)=(6.0)/(6.0)=1.0A` (i)The magnitude of force F required with be equal to the magnetic force acting on the rod, which oppses the motion. `therefore" "F=F_(m)=lvB=ilB` `or" "F=(1.0)(1.2)(2.5)N=3N` (ii) Rate by which energy is delivered to the resistor is `P_(1)=i^(2)R=(1)^(2)(6.0)=6W` (iii) The rate by which work is done by the applied force is, `P_(2)=Fv=(3)(2.0)=6WrArrP_(1)=P_(2)` |
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| 944. |
A with of length 50 cm moves with a velocity of 300 m/min, perpendicular to a magnetic fiel. If the emf induced in the wire is 2 V, then the magnitude of the field in tesla isA. 2B. 5C. `0.8`D. `2.5` |
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Answer» Correct Answer - C When a wire of length l moves with velocity v, perpendicular to a magnetic field B, then the induced emf is produced. The magnitude of induced emf is given by `|epsilon|=Blv`. Given l = 50 cm = 0.5 m, v = 300 m/min = 5 m/s `|epsilon|=2Vor B=(|epsilon|)/(lv)=(2)/(0.5xx5)=0.8T` |
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| 945. |
A conducing rod of length `l` is falling with a velocity `v` perpendicular to a unifrorm horizontal magnetic field `B` . The potential difference between its two ends will beA. 2 B/vB. B/vC. `(1)/(2)B//v`D. `B^(2)l^(2)v^(2)` |
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Answer» Correct Answer - B The potential difference between two ends of conducting rods will be Bvl. |
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| 946. |
A rectangular coil of 25 turns, area of `25cm^(2)` and resistance of `4 ohm//turn` is placed perpendicular to a varying magnetic field, which changes at the rate of `500 T//s`. The induced current in the coil isA. 0.3125 AB. 0.3225 AC. 31.25 AD. 3.225 A |
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Answer» Correct Answer - A `e = NA(dB)/(dt) = 25 xx 25 xx 10^(-4) xx 500 = 31.25 V` and `R = 25 xx 4 = 100 Omega` `:. I = e/R = 31.25/100 = 0.3125 A`. |
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| 947. |
Suppose a coil of area `5m^(2)`, resistance `10Omega` and number of turns 200 held perpendicular to a uniform magnetic field of strengh `0.4`T. The coil is now turned through `180^(@)` in time 1 s. What is (i) average induced emf (ii) average induced current (iii) total charge that flows through a given cross-setion of the coil? |
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Answer» Given, area of coil, `S=5m^(2)`, resistance, `R=10Omega`, number of turns, N = 200, magnetic field, B = `0.4T,Deltat=1s`. When the plane of coil is perpendicular to the magnetic field i.e.,`theta_(i)=0^(@)`. And after it is rotated through `180^(@)`, then `theta_(f)=180^(@)`. `rArr"Initial flux,"phi_(i)=NBS cos 0^(@)=NBS` `=200xx0.4xxs=400Wb` and final flux =NBS `cos 180^(@)=-NBS=-400Wb` `rArr"Change in flux",|Deltaphi_(B)|` `=NBS-(-NBS)=2NBS=800Wb` (i) Average induced emf `(epsilon)=(|Deltaphi|)/(Deltat)=(2NBA)/(Deltat)=800V` (ii) Average induced emf `=(epsilon)/(R)=(2NBA)/(Deltat)=80A` (iii) Average change `=(DeltaQ)/(Deltat)=(2NBA)/(RDeltat)` `rArr" Total charge "DeltaQ=(2NBA)/(R)=80C` |
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| 948. |
A stsnding wave `y = 2A sin kx cos omegat` is set up in the wire `AB` fixed at both ends by two vertical walls (see Fig. 3.197). The region between the walls contains a constant magnetic field `B`. Now answer the following question: In the above qestion, the time when the emf becomes maximum for the time isA. (a) `(2pi)/(omega)`B. (b) `(pi)/(omega)`C. ( c) `(pi)/(2omega)`D. (d) `(pi)/(4omega)` |
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Answer» Correct Answer - C ( c) `y = 2A sin kx cos omegat` `v = (dy)/(dt) = -2A sin kx omega sin omegat` `v_(max) = -2A omega sin kx, k = (3pi)/(AB)` `e = int_(0)^(l = AB) bv_(max)dx = -2aomegaBint_(0)^(AB) sin kx dx = + (2omegaAB)/(k)[cos(3pi)/(AB)AB - cos theta] = (-4(AB)omega)/(k)` `omegat = (pi)/(2)` `t = (pi)/(2omega)` For second harmonic `k = (2pi)/(AB)` |
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| 949. |
`A and B` two metallic rings placed at opposite sides of an infinitely long straight conducting wire as shown in Fig. 3.157. If current in the wire is slowly decreased, the direction of the induced current will be A. clockwise in `A` and anticlockwise in `B`B. anticlockwise in `A` and clockwise in `B`C. clockwise in both `A` and `B`D. anticlockwise in both `A` & `B` |
| Answer» Correct Answer - B | |
| 950. |
`A and B` two metallic rings placed at opposite sides of an infinitely long straight conducting wire as shown in Fig. 3.157. If current in the wire is slowly decreased, the direction of the induced current will be A. Clockwise in A and anticlockwise in BB. Anticlockwise in A and clockwise in BC. Clockwise in both A and BD. Anticlockwise in both A `&` B |
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Answer» Correct Answer - C |
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