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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.
| 551. |
A coil is rotated with constant angular speed in a uniform magnetic field about an axis that is perpendicular to the field. Tell, with reason, if the following statement is true– “The emf induced in the coil is maximum at the instant the magnetic flux through the coil is zero”. |
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Answer» Correct Answer - 1 |
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| 552. |
A stretchable conducting ring is in the shape of a circle. It is kept in a uniform magnetic field (B) that is perpendicular to the plane of the ring. The ring is pulled out uniformly from all sides so as to increase its radius at a constant rate `(dr)/(dt) = V` while maintaining its circular shape. Calculate the rate of work done by the external agent against the magnetic force when the radius of the ring is `r_(0)`. Resistance of the ring remains constant at R. |
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Answer» Correct Answer - `(4pi^(2)B^(2)r^(2)v^(2))/(R)` |
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| 553. |
A rod `PQ` of length `l` is rotating about end `P`, with an angular velocity `omega`. Due to electrifugal forceed the free electrons in the rod move toward the end `Q` and an emf is created. Find the induced emf. |
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Answer» The accumulation of free electrons will create an electric field which will finally belance the centrifugal forces and a steady state will be reached. In the steady state `m_(e)omega(2)x=eE`. `V_(P)-V_(0)=int_(x-0)^(x=l) barE.d barx=int_(0)^(l)(m_(e)omega^(2)x)/(e )dx=(m_(e)omega^(2)l^(2))/(2e)` |
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| 554. |
Two identical small balls, made of insulting material are attached to the ends of a light insulating rod of length d. The rod is suspended from the ceiling by a thin torsion free fibre as shown in figure. Each ball is given a charge q. There is a uniform magnetic field `B_(0)`, pointing vertically down, in a cylindrical region of radius R. The fibre is along the axis of the cylindrical region. The system is initially at rest. Now the magnetic field is suddenly switched off. Calculate the angular velocity acquired by the system. Each ball has mass m. |
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Answer» Correct Answer - `omega = (2qR^(2)B_(0))/(md^(2))` |
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| 555. |
A uniform circular ring of radius R, mass m has uniformly distributed charge q. The ring is force to rotate about its own axis (which is vertical) without friction. In the space a uniform magnetic field B, directed vertically down ward exists in a cylindrical region. The cylindrical region is co-axial with B and has a radius greater than R. If B increasesa at a constant rate `(dB)/(dt)=alpha`. angular acceleration of the ring will.A. Directly proportional to RB. Directly proportional to qC. Directly proportional to mD. Independent of R and m |
| Answer» Correct Answer - B | |
| 556. |
A current of 1 A flows through a coil, when a 100 V d.c. is applied to it. But a current of 0.5 A flows through the same coil, when a 100V a.c. of frequency 50 Hz is applied. The resistance and inductance of the coil are given by (take `pi^(2)=10`)A. `100 Omega , sqrt(0.2) H`B. `50 Omega , sqrt(0.3) H`C. `100 Omega , sqrt(0.3) H`D. `100 Omega , sqrt(0.2) H` |
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Answer» Correct Answer - C `R = 100/1 = 100 Omega for d.c.` for a.c., `I = V/(sqrt(R^(2)+L^(2)omega^(2)))` `:. R^(2)+L^(2)omega^(2) = (V^2)/(I^2) = (100 xx 100)/((1)/(2) xx (1)/(2)) = 4 xx 10^(4)` `:. 4 xx 10^(4) = (100)^(2) + 4pi^(2)L^(2)f^(2) = 10^(4)(1+10L^(2))` `3 = 10 L^(2) :. L = sqrt(0.3)H` (Thus `R = 100 Omega and L = sqrt(0.3)H)`. |
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| 557. |
An alternating voltage is applied to a series L-C-R circiut. If the current leads the voltage by `45^(@)`, thenA. `X_(L)=X_(C)-R`B. `X_(L)=X_(C)+R`C. `X_(C)=X_(L)+R`D. `R=X_(L)+X_(C)` |
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Answer» Correct Answer - B The phase diff. Between I and V is given by `tan theta = (X_L-X_C)/(R)` `:. tan 45^(@) = 1 = (X_L-X_C)/R` `:. X_L-X_C = R or X_L=X_C+R`. |
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| 558. |
In a circuit containing an inductance of zero resistance, the current leads the applied a.c. voltage by a phase angle atA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `(-90^(@))` |
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Answer» Correct Answer - D Lags behind V by `90^(@)` or leads by `(-90)^(@)`. |
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| 559. |
Two solenoids have identical geometrical construction but one is made of thick wire and the other of thin wire. Which of the following quantities are different for the two solenoids?A. `(i), (ii)`B. `(iii), (iv)`C. `(ii), (iv)`D. all |
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Answer» Correct Answer - D The number of turns, area of cross-section and total length of wires will be different. |
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| 560. |
In the simple `L-R` circuit, can the emf induced across the inductor ever be greater than the emf of the battery used to produce the current? |
| Answer» Correct Answer - `E=V_R+V_L` | |
| 561. |
A circular disc of radius `0.2 m` is placed in a uniform magnetic fied of induction `(1)/(pi) ((Wb)/(m^(2)))` in such a way that its axis makes an angle of `60^(@)` with The magnetic flux linked with the disc isA. 0.02WbB. 0.06WbC. 0.08WbD. 0.01Wb |
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Answer» Correct Answer - A Here, `r=0.2m,B=(1)/(pi)Wbm^(-2),theta=60^(@)` `:.` Magnetic flux, `phi=BAcostheta=B(pir^(2))costheta` `(1)/(pi)pi(0.2)^(2)cos60^(@)=0.02Wb` |
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| 562. |
An a.c. generator is connected to a sealed box through a pair of terminals. The box may cantain R,L,C or series combination of any two of the three elements. Measurements made outside the box show that `E = (75 sin omega t)` volt and `I = 1.2 sin (omega t + pi//5) A` (i) Name the circuit elements (ii) What is the power factor of the circuit (iii) What is the rat at which energy is delivered by the generator to the circuit ? |
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Answer» Here, `E = (75 sin omega t)` volt and `I = 1.2 sin (omega t + pi//5)` As current leads the e.m.f by a phase angle `pi//5`, therefore, the box must contain a series combination of resistor (R ) and capacitor (C ). (ii) Power factor of the circuit is `cos phi = cos pi//5 = 0.81` (iii) Rate at which energy is delivered by the generator to the circuit is `P = E_(v) I_(v) cos phi` `P = (75)/(sqrt2)xx(1.2)/(sqrt2)xx0.81 = 36.45` watt |
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| 563. |
A conducting loop rotates with constant angular velocity about its fixed diameter in a uniform magnetic field, whose direction is perpendicular to that fixed diameter.A. The `emf` will be maximum at the moment flux is zero.B. The `emf` will be `0` at the moment when flux is zero.C. The `emf` will be maximum at the moment when plane of the loop is parallel to the magnetic fieldD. The phase difference between the flux and the `emf` is `pi//2` |
| Answer» Correct Answer - A::B::C::D | |
| 564. |
The potential difference between the plates of a parallel plate capacitor of capacitance `2 mu F` is changing at the rate of `10^(5)V//s`. What is the displacement current in the dielectric of the capacitor?A. 1AB. 0.5AC. 0.2 AD. 0.75A |
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Answer» Correct Answer - C `I_(d)= C(dV)/(dt) = (xx10^(-6))xx10^(5)` `:. i_(d) = 2 xx 10^(-1)= 0.2 A`. |
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| 565. |
A variable frequency a.c. source is connected only to a parallel plate capacitor. What is the effect on the displacment current `(I_D)`, if the frequency is decreased?A. `I_(d)` increaseB. `I_(d)` decreaseC. `I_(d)` does not changeD. `I_(d)` varies between `0 to oo` |
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Answer» Correct Answer - B Capacitive reaction `X_(C) = 1/(2pifC)`. When f si decreased, `X_(C)` is increase and `1 prop 1/(X_C)` `:.` Conduction current `(I_C)` decreases `:. i_(d)` is decreased (as `I_(d)-I_(C))`. |
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| 566. |
A variable frequency a.c. source is connected only to a parallel plate capacitor. What is the effect on the displacment current `(I_D)`, if the frequency is decreased?A. `I_(D)` will decreaseB. `I_(D)` will increaseC. `I_(D)` will not changeD. `I_(D)` may increase or decrease, depending upon the values of capacitance and frequency |
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Answer» Correct Answer - A For a capacitor `X_(C) = 1/(omega C) = 1/(2 pi fC)` `:.` If the frequecy is decreased, the impedance will increase and the conduction current will decrease. As `I_(C) = I_(D)`, hence the displacement current will decrease. |
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| 567. |
A variable frequency a.c. source is connected to a capacitor. What is the effect on the displacement current `(I_d)` when the frequency of the a.c. source is increased from 500 Hz to 1000 Hz?A. `I_(d)` will remain constantB. `I_(d)` will decreaseC. `I_(d)` will be doubledD. `I_(d)` will become half |
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Answer» Correct Answer - C The reactance of a capacitor `X_(C) = 1/(omega C) = 1/(2 pi f C)` If the frequency is increased, the reactance is decreased and hence the current (I) is increased. As displacement current `(I_d)`= coduction current `I_(d)` will be doubled if the frequency is doubled. |
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| 568. |
An AC is given by the equation `i=i_(1)cos omegat+i_(2)sin omegat`. The r.m.s. current is given byA. `(1)/(sqrt(2))(I_(1)^(2)+I_(2)^(2))^(1//2)`B. `(1)/(sqrt(3))(I_(1)^(2)+I_(2)^(2))^(1//2)`C. `(1)/(sqrt(3))(I_(1)^(2)+I_(2)^(2))^(3//2)`D. `(1)/(sqrt(3))(I_(1)^(2)+I_(2)^(2))^(3//2)` |
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Answer» Correct Answer - A `I_(rms)=sqrt(overline(I^(2)))` `=sqrt(overline((I_(1)cosomegat+I_(2)sinomegat)^(2)))` `=sqrt(I_(1)^(2)overline(cos^(2)omegat)+I_(2)^(2)overline(sin^(2)omegat)+2overline(I_(1)I_(2)sinomegatcosomegat))` `=sqrt(I_(1)^(2)xx(1)/(2)xxI_(2)^(2)xx(1)/(2)+0)` `=sqrt((1)/(2)(I_(1)^(2)+I_(2)^(2)))=(1)/(sqrt(2))(I_(1)^(2)+I_(2)^(2))^(1//2)` |
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| 569. |
In the part of a circuit branch shown in figure-5.304 the potential difference `V_(ab)` at t=1s is: A. 30VB. `-30V`C. 20VD. `-20V` |
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Answer» Correct Answer - D |
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| 570. |
When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid areA. `100Omega,5.05Omega`B. `210Omega,5.50H`C. `200Omega,5.55H`D. `200Omega,0.55H` |
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Answer» Correct Answer - D `R=(V_(dc))/(I_(dc))` `=(100)/(1)=100Omega` `Z=(e_(rns))/(I_(rns))` `=(100)/(0.5)=200Omega` `therefore X_(L)=sqrt(Z^(2)-R^(2))=173Omega` `L=(X_(L))/(2pif)=0.55H` |
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| 571. |
A 100 turn coil of area `0.1 m^(2)` rotates at half a revolution per second. It is placed in a magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil. |
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Answer» Here, `N = 100, A = 0.1 m^(2), v = 0.5 rps` `B = 0.01 T, e_(0) = ?` `e_(0) = NAB omega = NAB (2 pi v)` `= 100 xx 0.1 xx 2 xx 3.14 xx 0.5` `e_(0) = 0.314 V` |
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| 572. |
A 110 V d.c source replace an a.c source such that heat produced is same in the two case. What is the rms value of alternting voltage source. |
| Answer» As heat produced is the same, rms voltage of a.c. source = d.c. voltage `= 100 V` | |
| 573. |
Calculate the time taken by 60 Hz a.c source to reach its negative peak value from zero value ? |
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Answer» Here, `v = 60 Hz, T = (1)/(v) = (1)/(60) s` From negative peak value to zero value, time taken `t = (T)/(4) = (1)/(50 xx 4) s = (1)/(240)` second. |
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| 574. |
A charged 30 `mu F` capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit ? |
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Answer» Here, `C = 30 mu F = 30 xx 10^(-6) F, L = 27 mH = 27 xx 10^(-3) H, omega_(r) = ?` `omega_(r) = (1)/(sqrt(LC)) = (1)/(sqrt(27 xx 10^(-3) xx 30 xx 10^(-6))) = (10^(4))/(9) = 1.1 xx 10^(3) rad//s` |
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| 575. |
The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases asA. l and increaseB. l decreases and A increasesC. A increases and A decreasesD. both l and A decrease |
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Answer» Correct Answer - B The self-induction L of a solenoid of length l and area of cross section A with fixed number of turns is `L = (mu_(0) N^(2) A)/(l)` Obviously, L increases when l decreases and A increases. Choice (b) is correct. |
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| 576. |
A solenoid of radius 3 cm and length 1 m has 600 turns per metre. What is its self inductance? Will the value of self inductance change if is would on an iron piece? |
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Answer» Here, `r = 3 cm = 3 xx 10^(-2) m, l = 1 m,` `N = 600, L = ?` `A = pi r^(2) = pi (3 xx 10^(-2))^(2) = 9 pi xx 10^(-4) m^(2)` `L = (mu_(0) N^(2) A)/(l) = (4 pi xx 10^(-7) (600)^(2) 9 pi xx 10^(-4))/(1)` `= 1.28 xx 10^(-3) H` If the solenoid is wound on an iron core, value of its self inductance will change. |
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| 577. |
A rectangular loop of sides 8 cm and 2 cm with a small cut is stationary in a uniform magnetic field directed normal to the loop. The magneic field is reduced form its initial value of 0.3 T at the rate of 0.2 T `s^(-1)`. If the cut is joined and the loop has a resistance of `1.6 Omega`, how much power is dissipated by the loop as heat ? |
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Answer» At instant t, let the radius of the loop be r. Therefore, magnetic flux linked with the loop at this instant, `phi = pi r^(2) B` `(d phi)/(dt) = 2 pi r B (dr)/(dt)` When `r = 2 xx 10^(-2) m` and `(dr)/(dt) = 10^(-3) m//s`, `e = (d phi)/(dt) = 2 xx (22)/(7) (2 xx 10^(-2)) (0.02) xx 10^(-3)` `= 2.5 xx 10^(-6) V = 2.5 mu V` |
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| 578. |
Predict the direction of induced current in the situation described by follow (e) |
| Answer» As the key is just released, the current which is flowing anticlockwise goes on decreasing. Thus, the induced current developed in such a sense the magnetic field due to left coil increases (which is towards right). So, the magnetic field due to the right coil should also towards right and hence the induced current is in anticlockwise, i.e., x to yx direction. | |
| 579. |
Predict the direction of indcued current in the situation described by the following Fig. |
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Answer» (a) South pole developes at q, current induced must be clockwise at q. `:.` in the coil, induced current is from p to q (b) Coil pq in this case would develop S-pole at q and coil XY would also develop S pole at X Therefore induced current in coil pq will be from q to p and induced current in coil XY will be from X to Y. (c ) Induced current in the right loop will be along XYZ. (d) Induced current in the left loop will be along XYZ as seen from front. (e) Induced current in the right coil is from X to Y. (f) No current is induced because magnetic lines of force lie in the plane of the loop. |
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| 580. |
The current is a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of `0.8 A s^(-)`. Find the emf indcued in it. |
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Answer» Here, `N = 240, l = 12 cm = 12 xx 10^(2) m`, `r = 2 cm = 2 xx 10^(-2) m, A = pi r^(2)` `dI//dt = 0.8 As^(-1), e = ?` `L = (mu_(0) N^(2) A)/(l) = (4 pi xx 10^(-7) (240)^(2) xx pi (0.02)^(2))/(0.12)` `|e| = (L dI)/(dt) = (4 pi xx 10^(-7) (240)^(2) xx 3.14 xx 4 xx 10^(-4) xx 0.8)/(0.12)` `= 6 xx 10^(-4) V` |
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| 581. |
(a) Suppose the windings og the armature in the d.c. motor or Example 4 connot tolerate a current of more than 20 amp. WhaT do you think will happen if the armature gets jammed and cannot rotate when the motor is connected to the supply ? (b) If the supply connection of the d.c. motor in Ex. are removed and the motor is used as a generated by connecting the shaft of its armature to an external mechanical rotor of speed 3000 r.p.m., how much e.m.f. will be generated ? |
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Answer» (a) When the armature gets jammed and cannot rotate, back e.m.f. `E = 0`. `:.` When connected to the supply. current `I = (V)/(R ) = (200)/(8.5) = 23.53` amp. Which exceeds the safe limit of 20 amp. Therefore. Armature will burn out. (b) The e.m.f. generated is obviously equal to the back e.m.f. `E = 157.5` volt, as calculate in Example 8. |
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| 582. |
Given the input current 15 A and input voltage of 100 V for a step up transformer having 90% efficiency. Find the output power and voltage in the secondary if output current is 3 A. |
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Answer» Here, `I_(p) = 15 A, E_(p) = 100 V`, `eta = 90%, P_(0) = ? E_(s) = ? I_(s) = 3 A` As `eta = ("Output power")/("Input power") = (E_(s) I_(s))/(E_(p) I_(p))` `:. (90)/(100) = (E_(s) xx 3)/(100 xx 15) , E_(s) = (90 xx 100 xx 15)/(3 xx 100)` `= 450 V` Output power ` = P_(o) = E_(s) I_(s)` `= 450 xx 3 = 1350 W` |
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| 583. |
Ajit had a high tension tower erected on his farm land. He kept complaninng to the authorities to remove it as it was occupying a large portion of his land. His uncle who was a teacher, explained to him the need for erecting these toward for efficient transmission of power. As Ajit realized its significance, he stopped complaining. Read the above passage and answer the following questions : (i) Why is necessary to transport power at high voltage ? (ii) A low power factor implies larger power loss. Explain. (iii) Write two values each, displayed by Ajit and his uncle. |
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Answer» (i) Power has to be transported at high voltage to reduce the eneryg losses and also the cost of transmission. (ii) As `P = E_(v) I_(v) cos phi`. To transmit a given power P at a given voltage `E_(V)` if `cos phi` is small ,`I_(v)` has to be increased. Therefore, power loss `= I_(v)^(2) R` will increase. (iii) Ajit understood the necessary of high tension tower which did not affect his land adversely. His uncle made practial use of his knowledge by convincing Ajit that high tensions towers are essential and they cause no harm to the farm land. |
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| 584. |
A metallic wire bent in the form of a semi-circle of radius 0.1 m is moved in a direction parallel to its plane, but perperdicular to a magnetic field `B = 20 mT with a vel. Of 10 `m//s`. Find the e.m.f. induced in the wire. |
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Answer» Here, `vec A = (30 hat I + 16 hat j + 23 hat k) cm^(2)` Magnitude of area vector `A = sqrt(30^(2) + 16^(2) + 23^(2))` `= sqrt(900 + 256 + 529)` `= sqrt(1685) = 41 cm^(2)` `14 xx 10^(-4) m^(2)` `phi = 5 mu Wb = 5 xx 10^(-6) Wb, B = ? theta = 0^(@)` From `phi = BA cos theta` `B = (phi)/(A cos 0^(@)) = (phi)/(A) = (5 xx 10^(-6))/(41 xx 10^(-4))` `= 1.22 xx 10^(-3) T` |
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| 585. |
The magnetic field in a region is given by `vecB=B_(0)/Lxhatk` where `L` is a fixed length.A conducting rod of length lies along the `X`-axis between the origin and the point `(L,0,0)`.If the rod moves with a velocity `vecv=v_(0) hatj` the `emf` induced between the ends of the rod. |
| Answer» Correct Answer - B | |
| 586. |
A solenoid has an inductance of 10 H and a resistance `R= 5 Omega` . It is connected ta a 10 V battery . How long will be it take for the magnetic energy to reach `(1)/(4)` of its maximum value ? |
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Answer» Correct Answer - `1.386 s` |
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| 587. |
In the circuits show `epsilon =15V, R_(1)= 1 Omega , R_(2) = 1 Omega, R_(3) = 2 Omega and L= 15 H` . Find the current `i_(1),i_(2),i_(3)` (i) immediately after the switch is closed (ii) immediately after the opening from the closed position (iii) sufficiently long after , the switch is opened f rom the colsed position, |
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Answer» Correct Answer - `(a) 7.5 A, 7.5A, 0 (b) 9 A, 6A, 3A (c) 0 A, 3 A , 3A` |
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| 588. |
A wire is bent in the form of a `V` shape and placed in a horizontal plane.There exists a uniform magnetic field `B` perpendicular to the plane of the wire. A uniform conducting rod starts sliding over the `V` shaped wire with a constant speed v as shown in the figure. If the wire no resistance, the current in rod wil A. Increase with timeB. Decrease with timeC. Remain constantD. Always be zero |
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Answer» Correct Answer - A |
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| 589. |
Which of the following material is ferromagneticA. BismuthB. NickelC. QuartzD. Aluminium |
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Answer» Correct Answer - C |
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| 590. |
Statement-1 : When we cousider mutual induction of two coils, their self induction is taken as zero. Statement-2 : This is because mutual induction of two coils is over and above self induction of each coil.A. AB. BC. CD. D |
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Answer» Correct Answer - D Statement-1 is false, mutual induction of two coils is in addition to their self induction. Statement-2 is however true. |
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| 591. |
Where is the power dissipaiton in an alternating current circuit ? In resistance ? In inductance ? In capacitance ? |
| Answer» Power is dissipated an a.c. circuit in resistance only. | |
| 592. |
Why are parallel resonance circuits called rejector circuit/filter circuits/antiresonance circuits ? |
| Answer» This is because parallel resonance circuits reject the current corresponding to parallel resonance frequencies. There circuits are used in the transmitting circuits. | |
| 593. |
Statement-1 : A parallel resonance circuit is called a rejector circuit. statement-2 : At resonance frequency, current is completely cut off.A. AB. BC. CD. D |
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Answer» Correct Answer - A Both the statements are ture, and statement-2 is correct explanaiton of statement-1. |
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| 594. |
In an `LR` circuit, current at `t = 0` is `20 A`. After `2 s` it reduces to `18 A`. The time constant of the circuit is :A. In`(10/9)`B. `2`C. `2/("In"(10/9))`D. `2 "In"(10/9)` |
| Answer» Correct Answer - C | |
| 595. |
In an `LR` circuit, current at `t = 0` is `20 A`. After `2 s` it reduces to `18 A`. The time constant of the circuit is :A. In `((10)/(9))`B. 2C. `(2)/(In((10)/(9))`D. `2 In ((10)/(9))` |
| Answer» Correct Answer - C | |
| 596. |
In the given circuit, let `i_(1)` be the current drawn battery at time `t=0` and `i_(2)` be steady current at `t=oo` then the ratio `(i_(1))/(i_(2))` is A. `1.0`B. `0.8`C. `1.2`D. `1.5` |
| Answer» Correct Answer - B | |
| 597. |
In the given circuit, let `i_(1)` be the current drawn battery at time `t=0` and `i_(2)` be steady current at `t=oo` then the ratio `(i_(1))/(i_(2))` is A. `1.0`B. `0.8`C. `1.2`D. `1.5` |
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Answer» Correct Answer - B Initially the inductor offers infinite resistance hence `i_(1)` is `1A`.Finally at steady state inductor offers zero resistance and current `i_(2)` is `1.25 A` in the battery. |
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| 598. |
At parallel resonance frequency, is current zero or maximum ? |
| Answer» At parallel resonance frequency, current is zero. | |
| 599. |
Switch `S` of the circuit shows in Fig. is closed at `t = 0`. If `e` denotes the induced emf in `L` and `i` the current flowing through the circuit at time `t`, then which of the following graphs correctly represents the variation of `e` with `i`? A. B. C. D. |
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Answer» Correct Answer - C At the time `t=0`, `e` is max and is equal to `E`, but current `i` is zero . As the time passes, current through the circuit increase but induced elf decreases. |
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| 600. |
In the circuit shown in Fig. Sliding contact is moving with uniform velocity towards right. Its value at some instance is `12 (Omega)`. The current in the circuit at this instant of time will be A. `0.5A`B. More than `0.5A`C. less than `0.5A`D. may be less or more than `0.5A` depending on the value of `L`. |
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Answer» Correct Answer - B Since resistance of the circuit is increasing and hence current in the circuit is decreasing so `(di)/(dt)` is negative. Current in the circuit is given by `i=(6-L(di)/(dt))/(12)` Since `(di)/(dt)` is negative so the value of numerator will be more than `6V` and hence current in the circuit at that instant willl be more than `0.5A`. |
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