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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.
| 951. |
The resistance in the following circuit is increase at a particle instant. At this instant the value of resistanc eis `10Omega`. The current in the circuit will be now A. `i=0.5A`B. `igt0.5A`C. `ilt0.5A`D. `i=0` |
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Answer» Correct Answer - B If resistance is constant `(10Omega)` then steady current in the circuit `i=(5)/(10)=0.5A`. But resistance is increasing it means current through the circuit start decreasing. Hence inductance comes in picture which induces a current in the circuit in the same direction of main current. So `igt0.5A`. |
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| 952. |
Two identical conducting rings `A` and `B` of radius `R` are rolling over a horizontal conducting plane with same speed `v` but in opposite direction. A constant magnetic field `B` is present pointing into the plane of paper. Then the potential difference between the highest points of the two rings is A. zeroB. 2 BvrC. 4BvrD. none of these |
| Answer» Correct Answer - C | |
| 953. |
Two identical conducting rings `A` and `B` of radius `R` are rolling over a horizontal conducting plane with same speed `v` but in opposite direction. A constant magnetic field `B` is present pointing into the plane of paper. Then the potential difference between the highest points of the two rings is A. ZeroB. 2BvRC. 4BvRD. None of these |
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Answer» Correct Answer - C |
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| 954. |
A metal disc of radius a rotates with a constant angular velocity `omega` about its axis. The potential difference between the center and the rim of the disc is `("m = mass of electron, e = charge on electro")`A. (a) `(momega^(2)a^(2))/(e)`B. (b) `(1)/(2)(momega^(2)a^(2))/(e)`C. ( c) `(e omega^(2)a^(2))/(2m)`D. (b) `(eomega^(2)a^(2))/(m)` |
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Answer» Correct Answer - B ( b) Let `E` be the electric field at a distance `r` from the centre of the disc. Then `eE = m omega^(2)r` or `E = (m omega^(2)r)/(e)` ``:.` P.D. = int_(r=0)^(r=a)Edr` `=int_(0)^(a)(m omega^(2)a^(2))/(2e)` |
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| 955. |
A ring rotates with angular velocity `omega` about an axis perpendicula to the plane of the ring passing through the center of the ring (Fig. 3.77). A constant magnetic field `B` exists parallel to the axis. Find the emf induced in the ring. |
| Answer» Flux passing through the ring `phi=B.A` is a constant here, therefore emf induced in the coil is zero. Every point of this ring at the same potential, by symmetry. | |
| 956. |
A resistance `R` is connected between the two ends of the parallel smooth conducting rails.A conducting rod lies on these fixed horizontal rails and a uniform constant magnetic field `B` exists perpendicular to the plane of the rails as shown in the figure.If the rod is given a velocity `v` and released as shown in figure, it will stop after some time, which option are correct: A. The total work done by magnetic field is negative.B. The total work done by magnetic field is positive.C. The total work done by magnetic field is zero.D. loss in kinetic energy of conducting rod is equal to heat generate between `R`. |
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Answer» Correct Answer - C::D Magnetic field cannot do work |
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| 957. |
A copper wire ab of length l, resistance r and mass m start sliding at t=0 down a smooth, vertical, thick pair of connected conducting rails as shown in figure. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails which options are correct . A. The magnitude and direction of the induced current in the wire when speed of the wire v is `(vBl)/( r)`, a to bB. The downward acceleration of the wire at this instant `g-(B_(2)l^(2))/(mr)v`C. The velocity of the wire as a function of time `v_(m)(1-e^(-"gt"//v_(m))),("where" v_(m)=(mgr)/(B^(2)l^(2)))`D. The displacement of the wire as a function of time `v_(m)t-(v_(m)^(2))/(g)(1-e^(-"gt"//v_(m))),("where" v_(m)=(mgr)/(B^(2)l^(2)))` |
| Answer» Correct Answer - A::B::C::D | |
| 958. |
A conducting rod of length is moved at constant velocity `v_(0)` on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant magnetic field B perpendicular to the plane for the rails as shown in Fig. A resistance R is connected between the two ends of the rails. Then which of the following is/are correct: A. the power delivered by force will be constant with timeB. the power delivered by force will be increasing first and will decreaseC. the rate of power delivered by the external force will be increasing continuouslyD. the rate ow power delivered by external force will be decreasing continuously before becoming zero. |
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Answer» Correct Answer - D Force acting on the rod because of the induced current due to change in magnetic flux will try to oppose the motion of rod.Hence the acceleration of the rod will decrease with time `(dP)/(dt)=F(dv)/(dt)=Fxxa`.Thus rate of power delivered by external force will be decreasing continuously. |
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| 959. |
A conductor `ABOCD` moves along its bisector with a velocity of `1 m//s` through a perpendicular magnetic field of `1 wb//m^(2)`, as shown in fig. if all the four sides are of `1m` length each, then the induced emf between points `A` and `D` is |
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Answer» Correct Answer - b |
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| 960. |
In a series `RC` circuit with an AC source, `R = 300 Omega, C = 25 muF, epsilon_0= 50 V` and `v = 50/(pi) Hz`. Find the peak current and the average power dissipated in the circuit.A. 0.1 AB. 2 A/sC. 0.02 AD. 1A |
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Answer» Correct Answer - A `X_(C)=(1)/(2pifC)=(1)/(2pixx(50)/(pi)xx25xx10^(-6))` `=(1)/(25xx10^(-4))=4xx10^(2)=400Omega` `Z=sqrt(R^(2)+X_(C)^(2))=sqrt(9xx10^(4)+16xx10^(4))` `Z=5xx10^(2)=500Omega` `I_(0)=(e_(0))/(Z)=(50)/(500)=0.1A` |
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| 961. |
A coil of wire of a certain radius has `600` turns and a self-inductance of `108 mH`. The self-inductance of a `2^(nd)` similar coil of `500` turns will beA. 25 mHB. 50 mHC. 7.5 mHD. 75 mH |
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Answer» Correct Answer - D `L prop N^(2)" "therefore (L_(2))/(L_(1))=((N_(2))/(N_(1)))^(2)` `therefore L_(2)=75mH` |
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| 962. |
A circuit containing capacitors `C_1` and `C_2`, shown in the figure is in the steady state with key `K_1` and `K_2` opened. At the instant `t = 0`, `K_1` is opened and `K_2` is closed. (a) Find the angular frequency of oscillations of `L- C` circuit. (b) Determine the first instant `t`, when energy in the inductor becomes one third of that in the capacitor. (c) Calculate the charge on the plates of the capacitor at that instant. |
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Answer» Correct Answer - A::C With key `K_1` closed `C_1` and `C_2` are in series with the battery in steady state. `:. C_("net")=1muF` or `q_0=C_("net")V=20muC` a. With `K_1` opened and `K_2` closed, charge on `C_2` will remain as it is while charge on `C_1` will oscillate in `L-C_1` circuit. `omega=1/(sqrt(LC_1))` `=1/sqrt(0.2xx10^-3xx2xx10^-6)` `=5xx10^4rad//s` b. Since at `t=0`, charge is maximum `(=q_0)` Therefore, current will be zero. `1/2Li^2=1/3(1/2q^2/C)` or `i=q/sqrt(3CL)=(qomega)/sqrt3` From the expression `i=omegasqrt(q_0^2-q^2)` We have `(qomega)/sqrt3=omegasqrt(q_0^2-q^2)` or `q=sqrt3/2q_0` Since at `t=0` charge is maximum or `q_0,` so we can write `q=q_0cosomegat` or `(sqrt3q_0)/2=q_0cosomegat` or `omegat=pi/6` or `t=pi/(6omega)=pi/(6xx5xx10^4)` `=1.05xx10^-5s` c. `q=sqrt3/2q_0=sqrt3/2=xx20=10sqrt3muC` |
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| 963. |
An inductor `(L = 20 mH)`, a resistor `(R = 100 Omega)` and a battery`(varepsilon = 10 V) ` are connected in series. Find (a) the time constant, (b) the maximum current and (c ) the time elapsed before the current reaches 99% of the maximum value. |
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Answer» Here, `L = 20 mH = 20 xx 10^(-3) H, R = 100 ohm`. `E = 10 V` Time constant, `tau = (L)/(R ) = (20 xx 10^(-3))/(100) = 2 xx 10^(-4) s` Maximum current, `I_(0) = (E)/(R ) = (10)/(100) = 10^(-1) A` Using `I = I_(0) (1 - e^(- t//tau))` `0.99 I_(0) = I_(0) (1 - e^(-t//tau))` `:. e^(-t//tau) = 0.01`. Solve to get `t = 0.92 xx 10^(-3) s` |
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| 964. |
An inductor (L =0.03 H) and a resistor `(R = 0.15k(Omega))` are connected in series to a battery of 15 V EMF in a circuit shown below. The key `K_(1)` is opened and Key `K_(2)` is closed simultaneously. At t =1 ms, the current in the circuit will be `(e^(5) = 150)` A. 6.7mAB. 0.67mAC. 100mAD. 67mA |
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Answer» Correct Answer - B (b) I_(0)=(15xx100)/(0.15xx10^(3))=0.1A` ltbr. `I_(oo)=0` I(t)=[I(0)-I(oo)]e^((-t)/(L//R))+i(oo)` I(t) 0.1 e^((-t)/(L//R))=0.1e^(R/L)` I(t)=0.1 e^((0.15xx1000)/(0.03)=0.67mA`. |
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| 965. |
Figure shows a circuit consisting of an ideal cell, an inductor `L`, and a resistor `R`, connected in series. Let switch `S` be closed at `t = 0`. Suppose at `t = 0`, the current in the inductor is `i_(o)`, then find out the equation of current as a function of time. |
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Answer» Correct Answer - `(1)/(R )[xi-(xi_(0)-I_(0)R)e^(Rt//L)]` |
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| 966. |
An ideal inductor, (having initial current zero) a resistor and an ideal battery are connected in series at time `t = 0`. At any time `t`, the battery supplies energy at the rate `P_(B)`, the resistor dissipates energy at the rate `P_(R )` and the inductor stores enegy at the rate `P_(L)`.A. `P_(B)=P_(g)+P_(L)` for all times tB. `P_(B)ltP_(L)` for all times tC. `P_(L)ltP_(g)` in steady stateD. `P_(B)gtP_(L)` only near the starting of the circuit. |
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Answer» Correct Answer - A |
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| 967. |
A bulb connected in series with a solenoid is lit by a.c. source, Fig. If the soft in core is introduced in the solenoid, will is bulb glow brighter ? |
| Answer» We know that the inductive reactance offered by the solenoid is `X_(L) = omega L = 2 pi v L`. On introducing soft iron core in the solenoid, its inductance L increases. Therefore, `X_(L)` increases As `I_(upsilon) = E_(upsilon) // X_(L)`, therefore, `I_(upsilon)` decreases. As glowing power of bulb `= I_(upsilon)^(2) R`, therefore, the bulb glows dimmer, and not brighter : | |
| 968. |
One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube maoves towards the other at a constant speed v. Then the emf induced in the circuit in terms of B, l and v where l is the width of each tube will be A. zeroB. 2 BlvC. BlvD. `-Blv` |
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Answer» Correct Answer - B Two emfs are induced in the closed circuit each of value Blv. Their two emfs are additive. So, `E_("net")=2Blv`. |
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| 969. |
Figure shows a conducting loop placed near a long straight wire carrying a current `i` as shown, if the current increases continuously, find the direction of the induced current in the loop. |
| Answer» `ox` magnetic field passing through loop is increasing. Hence induced current will produce magnetic field. So, induced current should be anti-clockwise. | |
| 970. |
Choose the correct optionsA. SI unit of magnetic flux is henry ampereB. SI unit of coefficient of self inductance is J/AC. SI unit of coefficient of self inductance is `("volt second")/("ampere")`D. SI unit of magnetic induction is weber |
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Answer» Correct Answer - A::C a.`L=(Nphi)/iimpliesphi=(Li)/N` So, `SI` unit of flux is Henry ampere. c. `L=(-e)/(/_i//_ )=-(-e/_ )/(/_ )` Hence, `SI` unit of L is `(V-s)/"ampere"` |
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| 971. |
`PQ` is an infinite current carrying conductor. `AB` and `CD` are smooth conducting rods on which a conductor `EF` moves with constant velocity `v` as shown. The force needed to maintain constant speed of `EF` is A. `1/(VR)[(mu_0IV)/(2pi) ln (b)/(a)]^2`B. `v/R[(mu_0IV)/(2pi)ln(a)/(b)]^2`C. `v/R[(mu_0IV)/(2pi)ln(b)/(a)]^2`D. None of these |
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Answer» Correct Answer - A `B_x=mu_0/(2pi)I/x` `de=B_x vdx=mu_0/(2pi) I/x v dx` `e=int_a^bde(mu_0Iv)/(2pi) ln (b/a)` `i=e/R=(mu_0Iv)/(2pir)ln(b/a)=`induced current `dF=(i)(dx)B_x` `[(mu_0Iv)/(2piR)ln(b/a)][mu_0/(2pi)I/x]dx` `F=int_a^bdF` |
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| 972. |
A metal rod of length 2 m is rotating with an angualr velocity of 100`"rads"^(-1)` in plane perpendicular to a uniform magnetic field of 0.3 T. The potential difference between the ends of the rod isA. 30 VB. 40 VC. 60 VD. 600 V |
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Answer» Correct Answer - C Induced emf, when rod rotates in a vertical plane perpendicular to magnetic field is given as `e=(1)/(2)Bl^(2)omega=(1)/(2)xx0.3xx2^(2)xx100=60V` |
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| 973. |
At an airport, a perosn is made to walk through the door wy of a metal detector, for security reasons. If `she//he` is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ? |
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Answer» The metal detector works on the principle of resonance in a.c. circuits. When we walk through a metal detectot, we are infact, walking through a coil of many turns connected to a capacitor turned suitably. If we carry some metal on our body, impedance of the circuit changes, bringing about significant changes in the current. This change in current is detected and the eletronic circuitry causes an alarm. |
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| 974. |
Calculate the inductance of a coil of 100 turns of wire would on an iron ring of radius 10 cm and `10 cm ^(2)` in cross-section, the relative permeability of iron being 700 |
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Answer» Correct Answer - 0.014 H |
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| 975. |
Two metallic rings ` A and B`, identical in shape and size but having different reistivities `rhoA and rhoB`, are kept on top of two identical solenoids as shown in the figure . When current `I` is switched on in both the solenoids in identical manner, the rings ` A and B` jump to heights `h_(A) and h_(B)`, repectively , with ` h_(A) gt h_(B)` . The possible relation(s) between their resistivities and their masses `m_(A) and m_(B)` is(are) A. (a) `rho_(A) gt rho_(B) and m_(A) = m_(B)`B. (b) `rho_(A) lt rho_(B) and m_(A) = m_(B)`C. ( c) `rho_(A) gt rho_(B) and m_(A) gt m_(B)`D. ( d) `rho_(A) lt rho_(B) and m_(A) lt m_(B)` |
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Answer» Correct Answer - B::D As `(dphi)/(dt) = emf is the same`, the current induced in the ring will depend upon the resistance of the ring. Larger the resistivity, smaller the current. |
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| 976. |
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value of zero as their temperature is lowered below a critical temperature `T_(C) (0)`. An interesting property of super conductors is that their critical temperature becomes smaller than `T_(C) (0)` if they are placed in a magnetic field, i.e., the critical temperature `T_(C) (B)` is a function of the magnetic field strength B. The dependence of `T_(C) (B)` on B is shown in the figure. . A superconductor has `T_(C) (0) = 100 K`. When a magnetic field of 7.5 Tesla is applied , its `T_(C)` decreases to 75 K. For this material one can difinitely say that whenA. `B=5` Tesla,`T_(C)(B)=80K`B. `B=5` Tesla,`75 K ltT_(C)(B) lt100K`C. `B=10` Tesla,`75 K ltT_(C)(B) lt100K`D. `B=10` Tesla,`T_(C)(B)=70K` |
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Answer» Correct Answer - B For `{:(B=0,T_(C)=100 k),(B=7.5T,T_(C)=75 k):}` |
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| 977. |
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value of zero as their temperature is lowered below a critical temperature `T_(C) (0)`. An interesting property of super conductors is that their critical temperature becomes smaller than `T_(C) (0)` if they are placed in a magnetic field, i.e., the critical temperature `T_(C) (B)` is a function of the magnetic field strength B. The dependence of `T_(C) (B)` on B is shown in the figure. . In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields `B_1`(solid line) and `B_2` (dashed line). If `B_2` is larget than `B_1` which of the following graphs shows the correct variation of R with T in these fields?A. B. C. D. |
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Answer» Correct Answer - A As the magnetic field is greater, the critical temperature is lower and as `B_(2)` is larger than `B_(1)`.Graph `A` is correct. |
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| 978. |
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as there is lowered below a critical temperature `T_(C)(0)`. An interseting property of supercoductors is that their critical temperature becomes smaller than `T_(C)(0)` if they are placed in a magnetic field , i.e., the critical temperature `T_(C) (B)` on B si shown in the figure. A superconductor has `T_(C) (0)=100K`. When a magnetic field of 7.5 Tesla is applied , its `T_(C)` decreases to 75 . For this material one can definitely say that when :A. B=5 Tesla, `T_(C)(B)=80K`B. B=5 Tesla,`75K lt T_(C)(B) lt 100K`C. B=1 Tesla, `75K lt T_(C)(B) lt 100 K`D. B=10 Tesla, `T_(C)(B)=70K` |
| Answer» Correct Answer - B | |
| 979. |
Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value of zero as their temperature is lowered below a critical temperature `T_(C) (0)`. An interesting property of super conductors is that their critical temperature becomes smaller than `T_(C) (0)` if they are placed in a magnetic field, i.e., the critical temperature `T_(C) (B)` is a function of the magnetic field strength B. The dependence of `T_(C) (B)` on B is shown in the figure. . In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields `B_1`(solid line) and `B_2` (dashed line). If `B_2` is larget than `B_1` which of the following graphs shows the correct variation of R with T in these fields?A. B. C. D. |
| Answer» Correct Answer - A | |
| 980. |
A rectangular frame ABCD, made of a uniform metal wire, has a straight connection between E and F made of the samae wire, as shown in fig. AEFD is a square of side 1m, and EB=FC=0.5m. The entire circuit is placed in steadily increasing, uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of the magnetic field is `1T//s`. The resistance per unit length of the wire is `1omega//m`. Find the magnitude and directions of the currents in the segments AE, BE and EF. |
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Answer» Correct Answer - `(3)/(11)A,(3)/(11)A,(1)/(22)A` |
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| 981. |
A coil of induction 50 H is connected to a battery of emf 2 V through a resistance of 10 ohm. What is the time constant of the circuit and maximum value of current in the circuit ? |
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Answer» Here, `L = 50 H, E = 2 V, R = 10 ohm, tau = ? I_(0) = ?` `tau = (L)/(R ) = (50)/(10) = 5 s` `I_(0) = (E)/(R ) = (2)/(10) = 0.2 A` |
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| 982. |
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) value of current `(1)/(300) s` after it was zero. |
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Answer» Here, `I = 50 sin 100 pi t`. Compare with the stantard form : `I = I_(0) sin 2 pi v t` (i) Mean value of a.c. over pos. half cycle ` = (2 I_(0))/(pi) = (2 xx 50)/(3.14) = 31.8 A` (iii) rms value of current, `I_(v) = (I_(0))/(sqrt2) = (50)/(1.414) = 35.35 A` (iv) `I = I_(0) sin omega t` `= 50 sin 100 pi xx (1)/(300) = (50 sqrt3)/(2) = 43.3 A` |
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| 983. |
A capacitor of `1 muF` is placed in series with a resistor of 2 mega-ohm and a battery of emf 2 V. Calculate the time after which the charge will grow to 86.74% of its max value. |
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Answer» Here, `C = 1 mu F = 10^(-6) F`, `R = 2 mega-ohm = 2 xx 10^(6) Omega, V = 2`volt `t = ? (q)/(q_(0)) = (86.47)/(100)` `tau = RC = 2 xx 10^(6) xx 10^(-6) = 2 s` From `q = q_(0) (1 - e^(-t//tau))` `(q)/(q_(0)) = 1 - e^(-t//tau)` or `e^(-t//tau) = 1 - (q)/(q_(0)) = 1 = (86.74)/(100) = (13.53)/(100)` `e^(-t//2) = (13.53)/(100)` or `e^(t//2) = (100)/(13.53)` `(t)/(2) = (log_(e) 100 - log_(e) 13.53)` `= 2.303 (2 - 1.1310)` `t = 2 xx 2.303 (0.8690) = 4.0 s` |
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| 984. |
A 1.5 mH inductor in an LC circuit stores a maximum energy of `30 mu J`. What is the maximum current in the circuit ? |
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Answer» Here, `L = 1.5 mH = 1.5 xx 10^(-3) H`, `E = 30 mu J = 30 xx 10^(-6) J, I = ?` From `E = (1)/(2) LI^(2)` `I = sqrt((2E)/(L)) = sqrt((2 xx 30 xx 10^(-6))/(1.5 xx 10^(-3))) = 0.2 A` |
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| 985. |
A 1.5 mH inductor in LC circuit stores a maximum energy of `17 mu` J. What is the peak current ? |
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Answer» Here, `L = 1.5 mH = 1.5 xx 10^(-3) H, E_(0) = 17 mu J = 17 xx 10^(-6) J, I_(0) = ?` From `E_(0) = (1)/(2) LI_(0)^(2) , I_(0) = sqrt((2 E_(0))/(L)) = sqrt((2 xx 17 xx 10^(-6))/(15 xx 10^(-3))) = sqrt((34)/(15) xx 10^(-2)) = 0.15 A` |
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| 986. |
The frequency of voltage source used in ac circuit is 5 kHz. The circuit contains a coil of inductance of 0.5 mH, a capacitor of capacitance of `10muF` and resistor of resistance 8 `Omega` are connected in series with source. The impedance of circuit isA. `14.9Omega`B. `1.49Omega`C. `1.49Omega`D. `18.9Omega` |
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Answer» Correct Answer - A `X_(L)=2pifL` `=2xx3.14xx5xx10^(3)xx0.5xx10^(-3)` `=15.7Omega` `X_(C)=(1)/(2pifC)=(1)/(2xx3.14xx5xx10^(3)xx10xx10^(-6))` `=3.185Omega` `Z=sqrt(R^(2)(X_(L)-X_(C))^(2))` `=sqrt(64+(15.7-3.2)^(2))=14.84Omega` |
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| 987. |
Inductance of a coil is 5 mH is connected to AC source of 220 V, 50 Hz. The ratio of AC to DC resistance of the coil isA. `5Omega`B. `0Omega`C. infinityD. data is incomplete |
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Answer» Correct Answer - C `(X_(ac))/(X_(dc))=(2pifc)/(0)=oo` |
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| 988. |
An alternating voltage `E=200sqrt(2)sin(100t)` is connected to a `1` microfarad capacitor through an AC ammeter. The reading of the ammeter shall beA. `10 mA`B. `100 mA`C. `200 mA`D. `20 mA` |
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Answer» Correct Answer - D In a circuit, at an instant `t`, alternating voltage is given by `e = 200 sqrt2 sin 100 t` volts Comparing it, with `e = e_(0) sin omega t` We have, `e_(0) = 200 sqrt2 V, omega = 100 rad s^(-1)` `X_(C) = (1)/(omega C) = (1)/(100 xx (1 xx 10^(-6))) =(200 sqrt2 // sqrt2)/(10^(4))` `= 2 xx 10^(-2) A = 20 mA` |
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| 989. |
An alternating voltage `E = 200sqrt(2) sin (100 t)V` is applied to a `2 mu F` capacitor through an A.C. ammeter. The reading of the ammeter isA. 4 mAB. 40 mAC. 2 mAD. 3 mA |
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Answer» Correct Answer - B `I_(0)=(e_(0))/(X_(C))` `I_(0)=(e_(0))/((1)/(WC))=wce_(0)` `I=100xx2xx10^(-6)xx200sqrt(2)` `I_(rms)=(I_(0))/(sqrt(2))=(100xx2xx10^(-6)xx200sqrt(2))/(sqrt(2))=40mA`. |
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| 990. |
Calculate the self-inductance of a coil of `100` turns, if a current of `2A` gives rise to magnetic flux of `50muWb` through the coil. Also calculate the magentic energy stored in the medium surrounding the coil for the above value of current. |
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Answer» `Nphi=Li` `100xx50xx10^(-6)=Lxx2` `L=2.5xx10^(-3)H=2.5mh` `U=(1)/(2)Li^(2)=(1)/(2)xx2.5xx10^(-3)xx(2)^(2)` `=5xx10^(-3)J` |
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| 991. |
A small d.c. motor operates at 110 V d.c. what is back e.m.f. when its effieincy is maximum ? |
| Answer» `E = V//2 = 110//2 = 55 V` | |
| 992. |
A (current vs time) graph of the current passing through a solenoid is shown in Fig. For which time is the back electromotive force (u) a maximum? If the back emf t = 3 s is e, find the back emf at t = 7 s, 15 s and 40 s OA, AB and BC are straight line segments. |
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Answer» As back e.m.f. `e = L (dI)/(dt)`, e will be maximum when `dI//dt` = max. The graph in Fig. shows that `dI//dt` is maximum for 5 x lt 6 lt 10 s. For `0 s lt t lt 5 s, (dI)/(dt) = (1)/(5)` `:.` At `t = 3 sec , e = L (dI)/(dt) = (L)/(5)` For `5 s lt t lt 10 s, (dI)/(dt) = (-2-1)/(5) = (-3)/(5) :.` At `7 s, e = L(dI)/(dt) = L (-(3)/(5)) = -3 e` For `10 s lt t lt 30 s, (dI)/(dt) = (2)/(20)=(1)/(10) :.` At `t = 15 s, e = L (dI)/(dt) = (L)/(10) = (e)/(2)` For `t gt 30 s, (dI)/(dt) = 0 :.` At `t = 40 s, e = 0` |
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| 993. |
Air cored chokes are used for reducing high frequency a.c., why? |
| Answer» Inductive reactance of a choke is `X_(L) = omega L = 2 pi v L`. When v is high, L need not be made high. Therefore, air cored choke will serve the purpose. | |
| 994. |
Iron cored chokes are used for reducing low frequency a.c., why ? |
| Answer» With iron core, coefficient of self inductance (L) increase. Therefore, `X_(L) = omega L = 2 pi v L` becomes large, even when v is low. Hence the current `I_(v) = E_(v)//X_(L)` reduces. | |
| 995. |
When an AC voltage of 220 V is applied to the capacitor CA. the maximum voltage between plates is 220 VB. the current is in phase with the applied voltageC. the charge on the plates is in phase with the applied voltageD. power delivered to the capacitor is zero |
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Answer» Correct Answer - C::D When an a.c. voltages of 220 V is applied to a capacitor C, the charge on the plates is in phase with the applied voltage. As current developed leads the applied voltage by a phase angle of `90^(@)`, therefore, power delivered to the capacitor per cycle is `P = E_(upsilon) I_(upsilon) cos 90^(@)` = zero. Choices (c) and (d) are correct. |
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| 996. |
An a.c. emf of e=220 sin `omega` t is applied across the capacitor, the power consumption is |
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Answer» Correct Answer - A `overline(P)=e_(rms)I_(rms)cosphi` `=e_(rms)I_(rms)cos90` `=0W` |
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| 997. |
An e.m.f of 40 mV is induced in a solenoid, when the current in it changes at the rate of `2 A//s`. The self inductance of the solenoid isA. 5 mHB. 10 mHC. 20 mHD. 40 mH |
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Answer» Correct Answer - C `e = L(dI)/(dt) :. 40 xx 10^(-3) = L xx 2` `:. L = 20 xx 10^(-3) H = 20 mH` |
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| 998. |
A wire is moving with velocity `12xx10^(-2)`m/s in a magnetic field of 0.5 T. if the induced emf is 9 mV, the length of wire isA. 150 cmB. 15 cmC. 1.5 cmD. 0.15 cm |
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Answer» Correct Answer - B `l=(e)/(Bv)` `=(9xx10^(-3))/(0.5xx12xx10^(-2))=15cm` |
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| 999. |
A conductor is moving with the velocity v in the magnetic field and induced current is I. If the velocity of conductor becomes double, the induced current will beA. 0.5 IB. 1.5 IC. 2 ID. 2.5 I |
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Answer» Correct Answer - B When the velocity of conductor becomes double, area intercepted becomes twice. Therfore induced current becomes twice |
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| 1000. |
Aconductor of lengfht 0.4 m is moving with a speed of 7 m/s perpendicular to a magnetic field of intensity` 0.9 Wb//m^(2)` .The induced emf across the coduct isA. 1.26 VB. 2.52 VC. 5.04 VD. 25.2 V |
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Answer» Correct Answer - B `e=Blv=0.9xx0.4xx7=2.52V` |
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