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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.
| 151. |
In figure, there is conducting ring having resistance `R` placed in the plane of paper in a uniform magnetic field `B_(0)`. If the rings is rotating in the plane of paper about an axis passing through point `O` and perpendicular to the plane of paper with constant angular speed `omega` in clockwise direction, then A. Point `A` will be at higher potential than `O`.B. The potential of point `B` and `C` will be same.C. The current in ring will be zero.D. The current in the ring will be `(2B_(0) omega. r^(2))/(R)`. |
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Answer» Correct Answer - A::B::C The change in magnetic flux is zero, hence the current in the ring will be zero. |
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| 152. |
In figure, there is conducting ring having resistance `R` placed in the plane of paper in a uniform magnetic field `B_(0)`. If the rings is rotating in the plane of paper about an axis passing through point `O` and perpendicular to the plane of paper with constant angular speed `omega` in clockwise direction, then A. point `O` will be at higher potential than `A`B. the potential of point `B` and `C` will be differentC. the current in the ring will be zeroD. the current in the ring will be `2B_(0) omega r^(2)//R` |
| Answer» Correct Answer - C | |
| 153. |
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. `74 mH`B. `75 mH`C. `76 mH`D. `77 mH` |
| Answer» Correct Answer - B | |
| 154. |
When the number of turns in a coil is doubled without any change in the length of the coil, its self-inductance becomesA. Four timesB. DoubledC. HalvedD. Unchanged |
| Answer» Correct Answer - A | |
| 155. |
A hundred turns of insulated copper wire are wrapped around an iron cylinder of area `1xx10^(-3) m^(2)` and are connected to a resistor. The total resistance in the circuit is `10` ohms. If the longitudinal magnetic induction in the iron changes from `1` weber `m^(-2)`, in one direction to `1` weber `m^(-2)` in the opposite direction, how much charge flows through the circuitA. `2xx10^(-2) C`B. `2xx10^(-3) C`C. `2xx10^(-4) C`D. `2xx10^(-5) C` |
| Answer» Correct Answer - A | |
| 156. |
The figure shows four wire loops, with edge length of either `L` or `2L`. All four loops will move through a region of uniform magnetic field `vecB` (directed out of the page) at the same constant velocity . Rank the four loops according to the maximum magnitude of the e.m.f. induced as they move through the field, greatest first A. `(e_(c)=e_(d)) lt (e_(a)=e_(b))`B. `(e_(c)=e_(d)) gt (e_(a)=e_(b))`C. `e_(c) gt e_(d) gt e_(b) gt e_(a)`D. `e_(c) lt e_(d) lt e_(b) lt e_(a)` |
| Answer» Correct Answer - B | |
| 157. |
The magnetic field in the cylindrical region shown in figure increase at a constant rate of `20mT//sec`. Each side of the square loop `ABCD` has a length of `1 cm` and resistance of `4Omega`. Find the current in the wire `AB` if the switch `S` is closed A. `1.25 xx10^(-7) A`, (anti-clockwise)B. `1.25xx10^(-7) A` (clockwise)C. `2.5xx10^(-7) A` (anti clockwise)D. `2.5xx10^(-7) A` (clockwise) |
| Answer» Correct Answer - A | |
| 158. |
The magnetic flux linked with coil, in weber is given by the equation, `phi = 5t^(2)+3t+16`. The induced emf in the coil in the fourth second isA. `10 V`B. `30 V`C. `45 V`D. `90 V` |
| Answer» Correct Answer - A | |
| 159. |
A step-down transformer is connected to main supply `200V` to operate a `6V`, `30W` bulb. The current in primary isA. 3AB. `1.5 A`C. `0.3A`D. `0.15 A` |
| Answer» Correct Answer - D | |
| 160. |
An ideal transformer has `100` turns in the primary and `250` turns in the secondary. The peak value of the ac is `28 V`. The r.m.s. secondary voltage is nearest toA. 50 VB. 70 VC. 100 VD. 40 V |
| Answer» Correct Answer - A | |
| 161. |
A loss free transformer has `500` turns on its primary winding and `2500` in secondary. The meters of the secondary indicate `200` volts at `8` amperes under these condition. The voltage and current in the primary isA. `100 V, 16 A`B. `40 V, 40 A`C. `160 V, 10A`D. `80 V, 20 A` |
| Answer» Correct Answer - B | |
| 162. |
A cylindrical bar amgnet is kept along the axis of a circular coil. If the magnet is rotated about its axis, thenA. A current will be induced in a coilB. No current will be induced in a coilC. Only an e.m.f. will be induced in the coilD. An e.m.f. and a current both will be induced in the coil |
| Answer» Correct Answer - B | |
| 163. |
A cylindrical bar magnet is rotated about its axis (Figure). A wire is connect from the axis and is made to touch the cylindrical surface through a contact. Then A. a direct current flows in the ammeter `A`.B. no current flows through the ammeter `A`.C. an alternating sinusoidal current flows through the ammeter `A` with a time period `T = (2pi)/(omega)`.D. a time varying non-sinosoidal flows through the ammeter `A`. |
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Answer» Correct Answer - B When cyllindrical bar magnet is rotated about its axis, no charge in flux linked with the circuit tajes place consequently no emf induced and hence, no current flows through the ammeter `A` |
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| 164. |
The self inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance ofA. `2 mu F`B. `1 mu F`C. `8 mu F`D. `4 mu F` |
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Answer» Correct Answer - B `v = (1)/(2pi sqrt(LC))` `50 = (1)/(2pisqrt(10 C))` `rArr 2500 = (1)/(40 pi^(2)C)` `rArr C = (1)/(2500 xx 40 xx 10)F = 10^(-6)F = 1 mu F` |
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| 165. |
An ideal coil of 10H is connected in series with a resistance of `5(Omega)` and a battery of 5V. 2second after the connections is made, the current flowing in ampere in the circuit isA. `(I-e)`B. `e`C. `e^(-1)`D. `(I-e^(-1))` |
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Answer» Correct Answer - D `I_(0) = (E)/(R ) = 1A , tau = (L)/(R ) = (10)/(5) = 2s, :. (I = 1-e^(-1))A` |
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| 166. |
An inductor `(L = 100 mH)`, a resistor `(R = 100 Omega)`, and a battery `(W = 100 V)` are initially connected in series as shown by the figure. After a long time the battery is disconnected after short circuiting the points `A` and `B`. The currents in the circuit `1 mm` after the short circuit is A. `0.1 A`B. `1 A`C. `(1)/(e) A`D. `e A` |
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Answer» Correct Answer - C `I = I_(0)e^(-Rt//L)` `= (E)/(R )e^(-Rt//L)` `= (100)/(100) e^(-) (100)/(100 xx 10^(-3)) xx 10^(-3)` `= (1)/(e)A` |
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| 167. |
The time constant of an inductor is `tau_(1)`. When a pure resistor of `R Omega` is connected in series with it, the time constant is found to decrease to `tau_(2)`. The internal resistance of the inductor isA. `(R tau_(2))/(tau_(1)-tau_(2))`B. `(R tau_(1))/(tau_(1)-tau_(2))`C. `(R(tau_(1)-tau_(2)))/(tau_(1))`D. `(R(tau_(1)-tau_(2)))/(tau_(2))` |
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Answer» Correct Answer - A `tau_(1) = (L)/(r )," "tau_(2) = (L)/(R+r)` Solve for from the above equations |
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| 168. |
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. `100 mA`B. `67 mA`C. `6.7 mA`D. `0.67 mA` |
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Answer» Correct Answer - D `I= I_(0e)^((1)/(tau)), tau = (L)/(R )` `= (15)/(150) e^((-1 xx 10^(-3))/(1 //5 xx10^(3))) = 0.67 mA` |
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| 169. |
A coil having resistance `15 Omega` and inductance `10H` is connected across a `90` Volt `dc` supply. Determine the value of current after `2sec`, What is the energy stored in the magnetic field at that instant. |
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Answer» Give that , `R = 15 Omega, L = 10H, E = 90 "Volt"` Peak value of current `I_(0) = (E)/(R ) = (90)/(15)A = 6A` also, `tau_(L) = (L)/(R ) = (10)/(15) = 0.67 sec` Now, `I = I_(0)(I-e^((-Rt)/(L)))`, After `2 sec`, `I = 6[1-e^(-2//0.67)] = 6[1-0.05] = 5.7A` Energy stroed in the magnetic field `U = (1)/(2)LI^(2) = (1)/(2) xx 10 xx (5.7)^(2)J = 162.45 J`. |
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| 170. |
A cell of `1.5V` is connected across an inductor of `2mH` in series with a `2 Omega` resistor. What is the rate of growth of current immediately after the cell is switched on. |
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Answer» `E = L(dI)/(dt) + IR`, therefore, `(dI)/(dt) = (E- =IR)/(L)` `E = 1.5 "Volt", R = 2 Omega, L = 2mH = 2 xx 10^(-3)H` When the cell is switched on, `I = 0` Hence `(dI)/(dt) = (E)/(L) = (1.5)/(2 xx 10^(-3))As^(-1) = 750 As^(-1)` |
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| 171. |
In the diagram shown if a bar magnet is moved along the common axis of two single turn coils A and B in the direction of arrow A. Current is induced only in A & not in BB. Induced currents in A & B are in the same directionC. Current is induced only in B and not in AD. Induced currents in A & B are in opposite directions |
| Answer» Correct Answer - D | |
| 172. |
A square coil of side `0.5 m` has movable sides. It si placed such that its plance is perpendiuclar to uniform magnetic field of induction `0.2T`. If all the sides are allowed to move with a speed of `0.1 m//s` for `4` sec outwards, average induced `emf` isA. ZeroB. `0.01V`C. `0.028 V`D. `0.072 V` |
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Answer» Correct Answer - D `e = B((dA)/(dt))` |
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| 173. |
In an `LR`-circuit, time constant is that time in which current grows from zero to the value (where `I_(0)` is the steady state current)A. `0.63 I_(0)`B. `0.50 I_(0)`C. `0.37 I_(0)`D. `I_(0)` |
| Answer» Correct Answer - A | |
| 174. |
A conductor of `3 m` in length is moving perpendicularly to magnetic field of `10^(-4)` tesla with the speed of `10^(2)m//s`, then the e.m.f. produced across the ends of conductor will beA. `0.03 "volt"`B. `0.3 "volt"`C. `3 xx 10^(-3)` voltD. 3 volt |
| Answer» Correct Answer - B | |
| 175. |
An e.m.f. of `12"volts"` is produced in a coil when the current in it changes at the rate of `45 amp//minute`. The inductance of the coil isA. 0.25 henryB. 1.5 henryC. 9.6 henryD. 16.0 henry |
| Answer» Correct Answer - D | |
| 176. |
Two different coils have self-inductances `L_(1) = 8 mH and L_(2) = 2 mH`. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are `i_(1), V_(1) and W_(1)` respectively. Corresponding values for the second coil at the same instant are `i_(2), V_(2) and W_(2)` respectively. Then:A. `(i_(1))/(i_(2)) = (1)/(4)`B. `(i_(1))/(i_(2)) = 48`C. `(W_(2))/(W_(1)) = 4`D. `(V_(2))/(V_(1)) = (1)/(4)` |
| Answer» Correct Answer - A::C::D | |
| 177. |
The equivalent inductance of two inductances is `2.4` henry when connected in parallel and `10` henry when connected in series. The difference between the two inductance isA. 2 henryB. 3 henryC. 4 henryD. 5 henry |
| Answer» Correct Answer - A | |
| 178. |
An inductance `L` and a resistance `R` are first connected to a battery. After some time the battery is disconnected but `L` and `R` remain connected in a closed circuit. Then the current reduces to `37%` of its initial value inA. RL secB. `(R)/(L)` secC. `(L)/(2)` secD. `(1)/(LR)` sec |
| Answer» Correct Answer - C | |
| 179. |
A `mu F` capacitor is charged by a `400 V` supply through `0.1 M Omega` resistance. The time taken by the capacitor to develop a potential difference of `300 V` is : (Given `log_(10) 4 = 0.602`)A. `2.2 sec`B. `1.1 sec`C. `0.55 sec`D. `0.48 "secs"` |
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Answer» Correct Answer - D `tau = (L)/(R ) = RC` `R = sqrt(L//C) = sqrt(8//2 xx 10^(-6)), R = 2 K Omega` |
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| 180. |
In electromagnetic induction, the induced charge in a coil is independent ofA. Change in the fluxB. TimeC. Resistance of the circuitD. None of the above |
| Answer» Correct Answer - C | |
| 181. |
The formula for induced e.m.f. in a coil due to change in magnetic flux through the coil is (here A = area of the coil, B = magnetic field)A. `e =-A.(dB)/(dt)`B. `e =-B.(dA)/(dt)`C. `e =-(d)/(dt)(A.B)`D. `e =-(d)/(dt)(A xx B)` |
| Answer» Correct Answer - C | |
| 182. |
The magnetic flux through a coil is varying according to the relation `phi = (5 t^(3) + 4 t^(2) + 2t - 5)` Wb. Calculate the induced current through the coil at `t = 2` s if resistiance of coil is 5 ohm. |
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Answer» `phi = 5t^(3)+4t^(2)+2t-5` `|e| = (d phi)/(dt) = 15 t^(2)+8t+2` at `t = 2 sec, e = 78V` `i xx 5 = 15 xx 4+8xx2+2 rArr i = 15.6A` |
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| 183. |
Lenz’s law is expressed by the following formula (here e = induced e.m.f., `phi` = magnetic flux in one turn and N = number of turns)A. `e = -phi(dN)/(dt)`B. `e = -N(dphi)/(dt)`C. `e = -(d)/(dt)((phi)/(N))`D. `e = N(dphi)/(dt)` |
| Answer» Correct Answer - B | |
| 184. |
An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along positive z-axis. The direction of induced current as seen from the z-axis will beA. Clockwise of the + ve z axisB. Anticlockwise of the + ve z axisC. ZeroD. Along the magnetic field |
| Answer» Correct Answer - C | |
| 185. |
A magnet is dropped down an infinitely long vertical copper tubeA. The magnet moves with continuously increasing velocity and ultimately acquires a constant terminal velocityB. The magnet moves with continuously decreasing velocity and ultimately comes to restC. The magnet moves with continuously increasing velocity but constant accelerationD. The magnet moves with continuously increasing velocity and acceleration |
| Answer» Correct Answer - A | |
| 186. |
A metal ring is held horizontally and bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnetA. Equal to that due to gravityB. Less than that due to gravityC. More than that due to gravityD. Depends on the diameter of the ring and the length of the magnet |
| Answer» Correct Answer - B | |
| 187. |
Assertion : A bar magnet is dropped into a long vertical copper tube. Even taking air resistance as negligible, the magnet attains a constant terminal velocity. If the tube is heated, the terminal velocity gets increased. Reason : The terminal velocity depends on eddy current produced in bar magnet.A. If both assertion and reason are t rue and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false |
| Answer» Correct Answer - B | |
| 188. |
In what form is the energy stored in an inductor or A coil of inductance L is carrying a steady current i . What is the nature of its stored energyA. MagneticB. ElectricalC. Both magnetic and electricalD. Heat |
| Answer» Correct Answer - A | |
| 189. |
In what form is the energy stored in an inductor or A coil of inductance L is carrying a steady current i . What is the nature of its stored energyA. MagneticB. ElectricalC. Both magnetic and electrical D. Heat |
| Answer» Correct Answer - A | |
| 190. |
Assertion: The quantity `L//R` possesses dimensions of time. Reason: To reduce the rate of increases of current through a solenoide should increase the time constant `(L//R)`.A. If both assertion and reason are t rue and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false |
| Answer» Correct Answer - B | |
| 191. |
The magnetic flux linked with a vector area `vec(A)` in a uniform magnetic field `vec(B)` isA. `vec(B) xx vec(A)`B. ABC. `vec(B).vec(A)`D. `(B)/(A)` |
| Answer» Correct Answer - C | |
| 192. |
Assertion: Magnetic flux is a vector quantity Reason: Vlaue of magnetic flux can be positive, negative or zeroA. Both `A` and `R` are true and `R` is the correct explanation of `A`B. Both `A` and `R` are true and `R` is not the correct explanation of `A`C. `A` is true but `R` is falseD. `A` is false but `R` is true. |
| Answer» Correct Answer - D | |
| 193. |
A triangular wire frame (each side =2m) is placed in a region of time variant magnetic field `dB//dt = (sqrt3) T//s`. The magnetic field is perpendicular to the plane of the triangle and its centre coincides with the centre of triangle. The base of the triangle AB has a resistance `1(Omega)` while the other two sides have resistance `2(Omega)` each. The magnitude of potential difference between the points A and B will be A. `0.4 V`B. `0.6 V`C. `1.2 V`D. None |
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Answer» Correct Answer - B `P.D` in the loop `= int vec(E).d vec(l) = (d phi)/(dt) = A(dB)/(dt)` `i = (Delta V)/(R_(T))" " R_(T) = 5 Omega , V_(AB) = iR_(AB)` |
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| 194. |
Which of the following is not an application of eddy currentsA. Induction furnaceB. Galvanometer dampingC. Speedometer of automobilesD. X -ray crystallography |
| Answer» Correct Answer - D | |
| 195. |
plane of eddy currents make an angle with the plane of magnetic lines of force equal toA. `40^(@)`B. `0^(@)`C. `90^(@)`D. `180^(@)` |
| Answer» Correct Answer - C | |
| 196. |
Eddy currents are produced in a matterial when it isA. A metal is kept in varying magnetic fieldB. A metal is kept in the steady magnetic fieldC. A circular coil is placed in a magnetic fieldD. Through a circular coil, current is passed |
| Answer» Correct Answer - A | |
| 197. |
A coil of `40 H` inductance is connected in series with a resistance of `8` ohm and the combination is joined to the terminals of a `2 V` battery. The time constant of the circuitA. 40 secondsB. 20 secondsC. 8 secondsD. 5 seconds |
| Answer» Correct Answer - D | |
| 198. |
A coil of `40 H` inductance is connected in series with a resistance of `8` ohm and the combination is joined to the terminals of a `2 V` battery. The time constant of the circuitA. `40`B. `20`C. `5`D. `0.2` |
| Answer» Correct Answer - `t = (L)/(R )` | |
| 199. |
A varying current at the rate of `3A//s` in a coil generates an e.m.f. of `8mV` in a nearby coil. The mutual inductance of the two coils isA. `2.66 mH`B. `2.66 xx 10^(-3) mH`C. `2.66 H`D. `0.266 H` |
| Answer» Correct Answer - A | |
| 200. |
Two coil are placed close to each other. The mutual inductance of the pair of coils depends upon.A. The currents in the two coilsB. The rates at which currents are changing in the two coilsC. Relative position and orientation of the two coilsD. The materials of the wires of the coils |
| Answer» Correct Answer - C | |