Saved Bookmarks
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.
| 401. |
Two toroidal solenoids are wound around the same pipe so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns and solenoid 2 has 400 turns. When the current in solenoid 1 is `6.52 A`, the average flux through each turn of solenoid 2 is `0.0320 Wb`. (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is `2.54 A`, what is the average flux through each solenoid 1? |
|
Answer» Correct Answer - A::B::C a. `M=(N_2phi_2)/i_2=((400)(0.032))/6.52=1.96H` b. `M=(N_1phi_1)/i_2` `phi_2=(Mi_2)/N_1=((1.96)(2.54))/700` `=7.12xx10^-3Wb` |
|
| 402. |
Two coils of self inductance 100 mH and 400 mH are placed very closed to each other. Find maximum mutual inductance between the two when 4 A current passes through them.A. `200 mH`B. `300 mH`C. `100sqrt2 mH`D. None of these |
|
Answer» Correct Answer - A `M_(max)=sqrt(L_(1)L_(2))=sqrt(100xx400)mH=200 mH` |
|
| 403. |
Two insulated wires are wound on the same hollow cylinder so as to form two solenoids sharing a common air-filled core. Let `l` be the length of the core, `A` the cross -sectional area of the core, `N_(1)` the number of times the first wire is wound around the core, and `N_(2)` the number of times the second wire is wound around the core. Find the mutual inductance of the two solenoids, neglecting the end effects. |
|
Answer» If a current `l_(1)` flows around the first wire then a unifrom axial magnetic field of strength `B_(1)=(mu_(0)N_(1)l_(1))/l` is generated in the core.The magnetic field in the region outside the core is of negligible magnitude.The flux linking a single turn of the second wire is `B_(1)A`.Thus, the flux linking all `N_(2)` turns of the second wire is `phi_(2)=N_(2)B_(1)A=(mu_(0)N_(1)N_(2)AI_(1))/l=MI_(1)` `therefore M=(mu_(0)N_(1)N_(2)A)/l` As described prevoiusly, `M` is a geometric quantity depending on the dimensions of the core and the manner in which the two wires are wound around the core, but not on the actual currents flowing through the wires |
|
| 404. |
Two coaxial coils are very close to each other and their mutual inductance is 5 mH. If a current (50 A) sin 500t is passed in one of the coils, then find the peak value of induced emf in the secondary coil. |
|
Answer» `i=50 sin 500t` `"As "epsilon=(-Mdi)/(dt)` `rArr" "epsilon=-(5xx10^(-3))50(cos 500t)xx500` `epsilon = -125cos 500t` `therefore" Peak value"=125V.` |
|
| 405. |
Find the mutual inductance of two concentric coils of radii `a_(1)` and `a_(2)(a_(1) lt lt a_(2))` if the planes of the coils are same. |
|
Answer» Let a current `i` flow in coil of radius `a_(2)` Magnetic field at the centre of coil `=(mu_(0)i)/(2a_(2))pia_(1)^(2)` or `Mi=(mu_(0)i)/(2a_(2))pia_(1)^(2)` or `M=(mu_(0)pia_(1)^(2))/(2a_(2))` |
|
| 406. |
A metalic ring connected to a rod oscillates freely like a pendulum. If now a magnetic field is applied in horizontal diection so that the pendulum now swings through the field, the pendulum will A. keep oscillating with the old time periodB. keep oscillating with a smaller time periodC. keep oscillating with a larger time periodD. come to rest very soon |
|
Answer» Correct Answer - D Emf induced in ring and it will opposes the motion.Hence due to the resistance of the ring all energy dissipates. |
|
| 407. |
Find the mutual inductance of two concentric coils of radii `a_(1)` and `a_(2)(a_(1) lt lt a_(2))` if the planes of the coils make an angle `theta` with each other. |
|
Answer» If `i` current flow in the larger coil, magnetic field produced at the centre will be perpendicular to the plane of larger coil. Now the area vector of smaller coil which is perpendicular to the plane of smaller coil will make an angle `theta` with the magnetic field. Thus flux `=vecB.vecA=(mu_(0)i)/(2a_(2))*pia_(1)^(2)*cos theta or M=(mu_(0)pia_(1)^(2) cos theta_(1))/(2a_(2))` |
|
| 408. |
Find the mutual inductance of two concentric coils of radii `a_(1)` and `a_(2)(a_(1) lt lt a_(2))` if the planes of the coils are perpendicular. |
| Answer» Let a current `i` flow in the coil of radius `a_(1)`. The magnetic field at the centre of this coil will now be parallel to the plane of smaller coil and hence no flux will passs through it, hence M=0 | |
| 409. |
Find the mutual inductance of two concentric coils of radii `a_(1)` and `a_(2)(a_(1) lt lt a_(2))` if the planes of the coils are same. |
|
Answer» Let a current I flow in coil of radius `a_(2)`. Magnetic field at the centre of coil `=(mu_(0)i)/(2a_(2))pia_(1)^(2)` `or Mi=(mu_(0)i)/(2a_(2))pia_(1)^(2) or M=(mu_(0)pia_(1)^(2))/(2a_(2))` |
|
| 410. |
Two circular loops `P` and `Q` are concentric and coplanar as shown in figure. The loop `Q` is smaller than. `P`. If the current `I_1` flowing `i` loop `P` is decreasing with time, then the curren `I_2` in the loop `Q` A. ClockwiseB. zeroC. counter clockwiseD. in a direction that depends on the ratio of the loop radii |
|
Answer» Correct Answer - C The induced current will be in such a direction so that it opposes the change due to which it is produced. |
|
| 411. |
Figure shows tow concentric coplanar coils with radii a and b `(a lt lt b)`. A current i=2t flows in the smaller loop. Neglecting self inductance of larger loop (a) Find the mutual inductance of two coils. (b) Find the emf induced in the larger coil (c) If the resistance of the larger loop is R find the current in it as a function of time. |
|
Answer» (a) To find mutual inductance , it does not matter in which coil we consider current and in which flux is calculated (Reciprocity theorem) Let current `i` be flowing in the larger coil. Magnetic field at the centre `=(mu_(0)i)/(2b)`. flux through the smaller coil `=(mu_(0)i)/(2b)pia^(2) " " :. M=(mu_(0))/(2b)pia^(2)` (b) |emf induced in larger coil|`=m[((di)/(dt))" in smaller coil"]=(mu_(0))/(2b)pia^(2) (2)=(mu_(0)pia^(2))/(b)` (c) current in the larger coil `=(mu_(0)pia^(2))/(b R)`. |
|
| 412. |
A thin non conducting ring of mass `m`, radius a carrying a charge `q` can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant `t=0`, a uniform magnetic field is switched on which is vertically downward and increases with time according to the law `B=B_0t`. Neglecting magnetism induced due to rotational motion of ring. The magnitude of an electric field on the circumference of the ring isA. `aB_0`B. `2aB_0`C. `1/2aB_0`D. zero |
|
Answer» Correct Answer - C `El=(dphi)/(dt)` `:. E(2pia)=(pia^2)B_0` or `E=1/2aB_0` |
|
| 413. |
Two parallel long straight conductors lie on a smooth plane surface. Two other parallel conductors rest on them at right angle so as to form a square of side a. A uniform magnetic field B exists at right angle to the plane containing the conductors. Now, conductors starts moving outward witha constant velocity `v_0` at `t=0`. Then, induced current in the loop at any time is `(lamda` is resistance per unit length of the conductors)A. `(aBv_0)/(lamda(a+v_0t))`B. `(aBv_0)/(2lamda)`C. `(Bv_0)/lamda`D. `(Bv_0)/(2lamda)` |
|
Answer» Correct Answer - C At `t=t` side of square, `l=(a+2v_0t)` Area, `S=l^2=(a+2v_0t)^2` `phi=BS=B(a+2v_0t)^2` `e=(dphi)/(dt)=4Bv_0(a+2v_0t)` `R=lamda[4l]=4lamda(a+2v_0t)` `:. i=e/r=(Bv_0)/R` |
|
| 414. |
Two circular loops `P` and `Q` are concentric and coplanar as shown in figure. The loop `Q` is smaller than. `P`. If the current `I_1` flowing `i` loop `P` is decreasing with time, then the curren `I_2` in the loop `Q` A. flows in the same direction as that of `P`B. flows in the opposite direction as that of `Q`C. is zeroD. None of the above |
|
Answer» Correct Answer - A Magnetic field through `Q` (by `I_2)` is downwards. By decreasing `I_1`, downward magnetic field through `Q` will decrease. Hence, induced current in `Q` should produce magnetic field in same |
|
| 415. |
A uniform but time varying magnetic field `B=(2t^3+24t)T` is present in a cylindrical region of radius `R =2.5 cm` as shown in figure. The variation of electric field at any instant as a function of distance measured from the centre of cylinder in first problem isB. C. D. |
|
Answer» Correct Answer - C From Eq. (i) of above problem we can se that `Epropr` i.e. `E-r` graph is a straight line passing through origin. |
|
| 416. |
A uniform but time varying magnetic field `B=(2t^3+24t)T` is present in a cylindrical region of radius `R =2.5` cm as shown in figure. The force on an electron at `P` at `t=2.0` s isA. `96xx10^-21N`B. `48x10^-21N`C. `24xx10^-21N`D. zero |
|
Answer» Correct Answer - B `(dB)/(dt)=(6t^2+24)T//s` At `t=2s, (dB)/(dt)=48T//s` `El=(dphi)/(dt)=S((dB)/(dt))` or `E(2pir)=pir^2((dB)/(dt))` ` :. E=r/2.(dB)/(dt)` ...(i) `F=qE=(qr)/2 (dB)/(dt)` `=((1.6xx10^-19)(1.25xx10^-2))/2(48)` `=48xx10^-21N` |
|
| 417. |
The magnetic flux through a coil perpendicluar to its plane and directed into paper is varying according to the relation `phi=(2t^(2)+4t+6)mWb`. The emf induced in the loop at t=4 s isA. 0.12 VB. 2.4 VC. 0.02 VD. 1.2 V |
|
Answer» Correct Answer - C Given, `phi=(2t^(2)+4t+6)mWb` AS, `epsi=(dphi)/(dt)=(d)/(dt)(2t^(2)+4t+6)xx10^(-3)Wbs^(-1)` `=(4t+4)xx10^(-3)V` At `t=4s` `epsi=(4xx4+4)xx10^(-3)V=20xx10^(-3)=0.02V` |
|
| 418. |
A uniform but time varying magnetic field `B=(2t^3+24t)T` is present in a cylindrical region of radius `R =2.5 cm` as shown in figure. In the previous problem, the direction of circular electric lines at `t=1 s` isA. clockwiseB. anti clociwiseC. no current is inducedD. cannot be predicted |
|
Answer» Correct Answer - B `ox` Magnetic field is increasing. Hence `o.` magnetic field is produced b a conducting circular loop placed there. For producing magnetic field induced current should be anti clockwise. Direction of induced circular electric lines are also anti clockwise. |
|
| 419. |
A uniform but time varying magnetic field exists in a cylindrical region as shown in the figure. The direction of magnetic field is into the plane of the paper and its magnitude is decreasing at a constant rate of `2xx10^-3T//s`. A particle of charge `1muC` is moved slowly along circle of radius `1m` by an external force as shown in figure. The plane of the circle of radius `1m` by an external force as shown in figure. The plane of he circle lies in the plane of the paper and it is concentric with the cylindrical region. The work done by the external force in moving this charge along the circle will be A. zeroB. `2pixx10^-9J`C. `pixx10^-9J`D. `4pixx10^-6J` |
|
Answer» Correct Answer - B `El=(dphi)/(dt)=S(dB)/(dt)` `:. E(2pir)=(pir^2)(dB)/(dt)` or `E=r/2 (dB)/(dt)` `F=qE=(qr)/2(dB)/2` `W=Fd=(2pir)` `=pir^2q((dB)/(dt))` `(22/7)(1)^2(10^-6)(2xx10^-3)` `=2pixx10^-9J` |
|
| 420. |
A unifrom magnetic field B exists in a cylindrical region of radius 1 cm as shown in figure. A uniform wire of length 16 cm and resistance 4.0`Omega` is bent into a square frame and is placed with one side along a diameters of the cylindrical region. If the magnetic field increaes at a constant rate of 1 T/s find the current induced in the frame. |
| Answer» Correct Answer - A::D | |
| 421. |
A uniform magnetic field `B` exists in a cylindrical region of radius `I0cm` as shown in figure. A uniform wire or length `80cm` and resistance `44.0Omega` is bent into a square frame and is placed with one side along a diameter of the cylindrical region. if the magnetic field increases at a constant rate of `0.0I0T//sec` , the current induced in the frameA. `3.9xx10^(-5)A`B. `4.0xx10^(5)A`C. `4.1xx10^(-5)A`D. `3.9xx10^(-4)A` |
|
Answer» Correct Answer - A Only half area of ring area of ring will be considered. `therefore phi_(B)=BS=B.(pir^(2))/(2)` `thereforee=(dphi_(B))/(dt)=(pir^(2))/(2).(dB)/(dt)=(pi(0.1)^(2))/(2)xx0.01=(pi)/(2)xx10^(-4)V` `thereforei=(e)/(R)=((pi)/(2)xx10^(-4))/(4)=(pi)/(8)xx10^(-4)=3.9xx10^(-5)A` |
|
| 422. |
The value of impedance in parallel LC circuit at resonance is (assuming inductor and capacitor to be ideal)A. minimumB. maximum currentC. infiniteD. zero |
| Answer» Correct Answer - C | |
| 423. |
In a region of space, magnetic field exists in a cylindrical region of radius a centred at origin with magnetic field along negative z-direction. The field is given by `vecB=-B_(0)thatk`. The force experienced by a stationary charge q placed at (r, 0, 0), where `r gt a` isA. `qB_(0)`B. `(qB_(0)a^(2))/(2r)`C. `(qB_(0)r)/(2)`D. Zero |
| Answer» Correct Answer - B | |
| 424. |
A small square loop of wire of side `l` is placed inside a large square loop of wire of side `L (L gt gt l)`. The loops are coplanar and their centre coincide. What is the mutual inductance of the system ? |
|
Answer» Magnetic field produced by a current `i` in a large square loop at its centre is `B=(mu_(0)i2sqrt2)/(piL)` Therefore`phi=Mi==(mu_(0)i2sqrt2)/(piL)xxl^(2) rArr M=(2sqrt2mu_(0)l^(2))/(piL)` |
|
| 425. |
A capacitor in LC oscillatory circuit has a maximum potential of V volts and maximum energy E. When the capacitor has a potential `V_(1)` volts and energy `E_(1)` joules, what is the emf across the inductor and energy stered in the magnetic fieldA. `V-V_(1),E-E_(1)`B. `V-V_(1), E_(1)`C. `V_(1),E_(1)`D. `V_(1),E-E_(1)` |
| Answer» Correct Answer - D | |
| 426. |
A small square loop of `n_(1)` turns and side l is placed at the centre of a large coplanar, circular loop of radius r( r gt gt l) and `n_(2)` tuns. The mutual inductacne of the combination isA. `(mu_(0)n_(1)n_(2)pir^(2))/(l)`B. `(mu_(0)n_(1)n_(2)l^(2))/(2r)`C. `(mu_(0)n_(2)^(2)l^(2))/(2r)`D. `(mu_(0)n_(1)^(2)l)/(2r)` |
| Answer» Correct Answer - B | |
| 427. |
A short solenoid of length 4 cm, radius 2 cm and 100 turns is placed inside and on the axis of a long solenoid of length 80 cm and 1500 turns. A current of 3 A flows through the short solenoid. The mutual inductance of two solenoid isA. 0.12 HB. `5.3xx10^(5)H`C. `3.52xx10^(-3)H`D. `8.3xx10^(-5)H` |
| Answer» Correct Answer - A | |
| 428. |
Two identical circular loops of metal wire are lying on a table without touching each other. Loop-A carries a current which increases with time. In response, the loop-BA. Is attracted by loop AB. Is repelled by loop AC. Remains stationaryD. None of these |
| Answer» Correct Answer - B | |
| 429. |
In a lossless transformer an alternating current of `2` amp is flowing in the primary coil. The number of turns in the primary and secondary coils are `100` and `20` respectively. The value of the current in the secondary coil isA. `0.08A`B. `0.4A`C. `5A`D. `10A` |
|
Answer» Correct Answer - D `(N_(s))/(N_(p))=(i_(p))/(i_(s))impliesi_(s)=i_(p)xx(N_(p))/(N_(s))=2xx(100)/(20)=10A` |
|
| 430. |
In a lossless transformer an alternating current of `2` amp is flowing in the primary coil. The number of turns in the primary and secondary coils are `100` and `20` respectively. The value of the current in the secondary coil isA. 0.08 AB. 0.4 AC. `5A`D. `10A` |
|
Answer» Correct Answer - D `I_(S)=(n_(P))/(n_(S))xxI_(P)=(25xx4)/(20)xx2=10A` |
|
| 431. |
In a step up transformer, the turn ratio is 1:10. A resistance of `200 Omega` connected across the secondary draws a current of 0.5 A. What is the primary voltage?A. 5 VB. 10 VC. 20 VD. 2.5 V |
|
Answer» Correct Answer - B `V_(s) = R xx I_(s) = 200 xx 0.5 = 100 V` `:. (V_p)/(V_s) = (n_p)/(n_s) :. V_(p) = 100 xx 1/10 = 10 V`. |
|
| 432. |
An ideal transformer has `500` and `5000` turn in primary and secondary windings respectively. If the primary voltage is connected to a `6V` battery then the secondary voltage isA. `0`B. `60V`C. `0.6V`D. `6.0V` |
|
Answer» Correct Answer - A Transformer works on ac only. |
|
| 433. |
A power tansmission line feeds input input power at 2200 V to a step down transformer with its primary windings having 3000 turns. Find the number of turns in secondary to get the power output at 220 V. |
|
Answer» Here, `E_(p) = 2200 V, n_(p) = 3000, n_(s) = ? E_(s) = 220 V` For a step down transformer, `(n_(s))/(n_(p)) = (E_(s))/(E_(p)) = (220)/(2200) = (1)/(10)` `n_(s) = (n_(p))/(10) = (3000)/(10) = 300` |
|
| 434. |
The number of turns in the primary and secondary coils of an ideal transformer are 2000 and 50 respectively. The primary coil is conneted to a main supply of 120 V and socondary to a night bulb of `0.6 Omega`. Calculate (i) Voltage acorss the secondary. (ii) Current in the bulb, (iii) Current in primarya coil, (iv) Power in primary and secondary coils, (iv) Power in primary and secondary coils. |
|
Answer» Here, `n_(p) = 2000, n_(s) = 50`, `E_(p) = 120 V , R_(s) = 0.6 Omega` `E_(s) = ?, I_(s) = ?, I_(p) = ?, P_(p) = ?, P_(s) = ?` (i) As `(E_(s))/(E_(p)) = (n_(s))/(n_(p))` `:. E_(s) = E_(p) . (n_(s))/(n_(p)) = 120 xx (50)/(2000) = 3 V` (ii) As `I_(s) = (E_(s))/(R ) :. I_(s) = (3)/(0.6) = 5 A` (iii) As `(I_(p))/(I_(s)) = (E_(s))/(E_(p))` `:. I_(p) = (E_(s))/( E_(p)) xx I_(s) = (3)/(120) xx 5 = 0.125 A` (iv) Power in primary, `P_(p) = E_(p) xx I_(p) = 120 xx 0.125 = 15 W` Power in secondary, `P_(s) = E_(s) xx I_(s) = 3 xx 5 = 15 W` |
|
| 435. |
An ideal transformer has primary and secondary coils of 200 turn and 40 turns respectively. If the current in the primary coil is 3A. Then the value of current in the secondary coil wil beA. a.`10A`B. b.15AC. c.1.5 AD. d.`5A` |
|
Answer» Correct Answer - B `(I_s)/(I_p) = (n_p)/(n_s) :. I_(s) = 200/40 :. I_(s) = 15 A` |
|
| 436. |
A step down transformer has a turn ratio of `20 : 1`. If 8 volt are developed across `0.4 Omega` secondary, then the primary current will be:A. `1 A`B. `2 A`C. `4 A`D. `0.5 A` |
|
Answer» Correct Answer - A `(n_p)/(n_s) = 20/1 = (V_p)/(V_s) = (I_s)/(I_p)` `:. I_(s) = 8/(4//10) = 20 A :. I_(p)=I_(s) xx 1/20 = 1A`. |
|
| 437. |
in a step-up transformer, the turn ratio is `1:2` leclanche cell (e.m.f. 1.5V) is connected across the primary. The voltage devloped in the secondary would beA. 20VB. 30VC. 40VD. zero |
|
Answer» Correct Answer - D Zero, A transformer does not work on D.C. |
|
| 438. |
In a step up transformer, the turn ratio is 3 : 2. A battery of EMF 4.5V is connected across the primary windings of transformer. The voltage developed in the secondary would be:A. 4.5VB. 30VC. 1.5VD. Zero |
|
Answer» Correct Answer - A |
|
| 439. |
A coil when connected across a 10 V d.c. supply draws a current of 2 A. When it is connected across 10 V - 50 hz a.c. supply the same coil draws a current of 1 A. Explain why? Hence determine self inductance of the coil. |
|
Answer» The coil draws lesser current in 2nd case because of inductive reactance of the coil. In 1st case, `R = (V)/(I) = (10)/(2) = 5 Omega` In second case, `Z = (V)/(I) = (10)/(1) = 10 Omega` Inductive reactance, `X_(L) = sqrt(Z^(2) - R^(2)) = sqrt(10^(2) - 5^(2)) = 5 sqrt 3` `2 pi v L = 5 sqrt 3` `L = (5 sqrt 3)/(2 pi v) = (5 sqrt 3)/(2 xx 3.14 xx 50) = 0.0288 H` |
|
| 440. |
A coil of self-inductance `L` is connected in series with a bulb `B` and an `AC` source. Brightness of the bulb decreases whenA. number of turns in the coil is reducedB. a capacitance of reactance `X_(C)=X_(L)` is included in the same circuitC. an iron rod is inserted in the coilD. frequency of the AC source is decreased |
|
Answer» Correct Answer - D When an iron rod is inserted in the coil, then value of `X_(L)` increases (`X_(L)=omegaL=omega(mu_(0)N^(2)A)/(l)`) (`thereforeu` (Peameability of soft iron) gt absolute permeability `(mu_(0))`) In LR series ckt, `I_(V)=(E_(V))/(sqrt(R^(2)+x_(L)^(2)))` |
|
| 441. |
Two coils A and B have mutual inductance `2xx10^(-2)` Henry if the current in he primary coil is i=5 sin `(10pit)` then the maximum value of emf induced in coil B isA. `pi` voltB. `pi//2` voltC. `pi//3` voltD. `pi//4` volt |
|
Answer» Correct Answer - A `e_(0)=omegaMI_(0)` `=10pixx2xx10^(-2)xx5=pi`. |
|
| 442. |
In an oscillator, for sustained oscillations, Barkhausen criterion is `Abeta` equal to (A = voltage gain without feedback and `beta` = feedback factor)A. zeroB. `1/2`C. 1D. 2 |
|
Answer» Correct Answer - C Barkhausen criterion is AB = 1 (theory). |
|
| 443. |
The primary coil of an ideal stepup transformer has 100 turns and the transformer ratio is also 100. The input voltage and power are 220 V and 1100 W. Calculate (i) number of turns in secondary (ii) current in primary (iii) voltage across secondary (iv) current in secondary (v) power in secondary. |
|
Answer» Here, `n_(P) = 100, K = 100, E_(P) = 220 V`, `P_(i) = 1100 W` (i) From `K = (n_(s))/(n_(P))` (ii) From `P_(i) = E_(P) I_(P)` `I_(P) = (P_(i))/(E_(P)) = (1100)/(250) = 5 A` `(E_(s))/(E_(p)) = (n_(s))/(n_(P)) = K` (iv) `(I_(s))/(I_(P)) = (n_(P))/(n_(s)) = (1)/(K)` `I_(s) = (I_(P))/(K) = (5)/(100) = 0.05 A` (v) Power in secondary `= P_(0) = P_(i) = 1100 W` |
|
| 444. |
A transmitter transmits at a wavelength of `300 m`. A condenser of capacitance `2.4 muF` is being used. The value of the inductance for the resonant circuit is approximatelyA. `10^(-8)H`B. `10^(-6)H`C. `10^(-9)H`D. `10^(-7)H` |
|
Answer» Correct Answer - A `v=flamda` `f=(v)/(lamda)` `therefore (1)/(2pisqrt(LC))=(v)/(lamda)` `(1)/(4pi^(2)LC)=(v^(2))/(lamda^(2))` `therefore L=(lamda^(2))/(4pi^(2)v^(2)C)=(9xx10^(4))/(4xx9.87xx9xx10^(16)xx2.4xx10^(-6))` `=10^(-8)H` |
|
| 445. |
The magnetic field in a certain region is given by `B = (4.0veci -1.8 veck) xx 10^-3 T`. How much flu passes through a `5.0 cm^2` area loop in this region if the loop lies flat on the xy-plane?A. `-800` nWbB. `-600` nWbC. `-900` nWbD. `-450` nWb |
|
Answer» Correct Answer - C `phi=-AB costheta` `=-5xx10^(-4)xx18xx10^(-4)` `=-90xx10^(-8)Wb` |
|
| 446. |
An 220V AC voltage at a frequency of 40 cycles/s is applied to a circuit containing a pure inductance of 0.01H and a pure resistance of `6 Omega` in series. Calculate (a) The current supplied by source (b) The potential difference across the resistance (c ) The potential difference across the inductance (d) The time lag between maxima of current and EMF in circuit |
|
Answer» Correct Answer - `(a) 33.83 mA (b) 202.98 V (c ) 96.83 V (d) 0.01579 s` |
|
| 447. |
When a voltage measuring device is connected to a.c. mains the meter shows the steady input voltage of `220 V`. This meansA. input voltage can not be AC voltage, but a DC voltageB. maximum input voltage is 220 VC. the meter reads no `upsilon` but `lt upsilon^(2)gt` and is calibrated to read `sqrt(lt upsilon^(2)gt)`D. the pointer of meter is stuck by some mechanical defect |
|
Answer» Correct Answer - C The volmeter connected to A.C. mains is calibrated to read root mean square value or virtual value of a.c. voltage, i.e., `sqrt(lt upsilon^(2) gt)`. Choice ( c) is correct. |
|
| 448. |
A Source of 200 V - 50 Hz a.c. is connected to a resistance of 10 ohm, a capacitor C and an ammeter in series. If reading of ammeter is 2 A, what is the capacity of condenser ? |
|
Answer» Here, `E_(v) = 200 V, v = 500 Hz, R = ohm, C = ? I = 2 A` From `Z = (E_(v))/(I_(v)) = (200)/(2) = 100 ohm` Again `R^(2) + X_(C)^(2) = Z^(2)` `X_(C) = sqrt(Z^(2) - R^(2)) = Z^(2)` `X_(C) = sqrt (Z^(2) - R^(2)) = sqrt(100^(2) - 10^(2)) = 99.5` As `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `:. C = (1)/(2 pi v X_(C)) = (1)/(2 xx 3.14 xx 50 xx 99.5) = (1)/(314 xx 99.5) = 3.2 xx 10^(-6) F = 3.2 mu F` |
|
| 449. |
In LCR series circuit an ac emf of 2 volt and frequency 50 Hz is applied across the combination. If resistance is `4Omega`, capacitance is `8muF` and inductance is `10^(-2)H` then the voltage across inductor will beA. (3/5 V)B. (5/3 V)C. (2/3 V)D. (0.02 V) |
|
Answer» Correct Answer - D `I=(e)/(Z)` `X_(L)=2pifL` `=2xx3.14xx50xx10^(-2)=3.14Omega` `X_(C)=(1)/(2pifC)=(1)/(2xx3.14xx50xx8xx10^(-6))` `=(0.3185xx10^(4))/(8)=(3185)/(8)=397Omega` `Z=sqrt(R^(2)(X_(L)-X_(C))^(2))` `Z=397Omega` `e_(R)=I*R=(e)/(Z)*R` `=(2)/(397)xx4=0.02V`. |
|
| 450. |
In series LCR circuit at resonance,A. current is maximum and impedance is maximumB. current is maximum and impedance is minimumC. current is minimum and impedance is maximumD. current is minimum and pmpedance is minimum |
| Answer» Correct Answer - B | |