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.
| 151. |
A voltmeter reads V volts in an A.C. circuit then V isA. peak value of the voltageB. peak value of the currentC. r.m.s. value of the currentD. r.m.s. value of the voltage |
| Answer» Correct Answer - D | |
| 152. |
In LCR series circuit , an alternating emf e and current `i` are given by the equations `e = 100sin (100 t) ` volt . `i=100 sin (100 t + (pi)/(3))` mA The average power dissipated in the circuit will beA. 100 WB. 10 WC. 5 WD. 2.5 W |
|
Answer» Correct Answer - D `overline(P)=(e_(0)I_(0))/(2)cosphi` `=(100xx100xx10^(-3))/(2)"cos"(pi)/(3)` `=(10)/(3)xx(1)/(2)=(10)/(4)` =2.5 W. |
|
| 153. |
Lights of a car become dim when the starter is opereterd. Why?A. IncreasesB. DecreasesC. Remains sameD. First increases then decreases |
| Answer» Correct Answer - B | |
| 154. |
What is the use of a motor starter ? |
| Answer» A motor starter is a varialbe resistance. When the motor is switched on, the starter offers maximum resistance so that a small current flows through the motor coil in the absence of back e.m.f. This prevents damage to the motor, When it is switched on. | |
| 155. |
There are two coils `A` and `B` as shown in Figure. A current starts flowing in `B` as shown, when `A` is moved towards `B` and stops when `A` stops moving. The current in `A` is counterclockwise. `B` is kept stationary when `A` moves. We can infer that A. there is a constant current in clockwise direction in AB. there is a varying current in AC. there is no current in AD. there is a cinstant current in the counter-clockwise direction in A |
|
Answer» Correct Answer - D When A stops moving the current in B become zero, possible only if the current in A is constant. If the current in A would be variable, then there must be an induced emf (current) in B even, if A stops moving. |
|
| 156. |
There are two coils `A` and `B` as shown in Figure. A current starts flowing in `B` as shown, when `A` is moved towards `B` and stops when `A` stops moving. The current in `A` is counterclockwise. `B` is kept stationary when `A` moves. We can infer that A. there is a constant current in the clockwise direction in AB. there is a varying current in AC. there is no current in AD. there is a constant current in the counter clockwise direction in A |
|
Answer» Correct Answer - D When A is moved towards B, current starts flowing in B. This is due to the variation of magnetic flux `(phi)`. When A stops moving, the current in B becomes zero. This is possible only if the current in A is constant. If the current in A is variable, there will be an induced current in B, even if A stops moving. (Option d) |
|
| 157. |
A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer isA. `(BR)/(A)`B. `(AB)/(R)`C. `ABR`D. `(B^(2)A)/(R^2)` |
|
Answer» Correct Answer - B Initial flux in the ring `phi_(1)=BA` Since the ring is suddenly squeezed to zero area, final flux `phi_(f)=0`. Change of flux `dphi = phi_(i)-phi_(f)=phi_(i) = BA` Induced e.m.f. = `-(d phi)/(dt)` and induced current `i=(|e|)/(R)` `:. I = (dq)/(dt) = 1/R (d phi)/(dt)` `:. dq = 1/R d phi :. int dq = 1/R int d phi` `:. q = (phi)/(R) = (BA)/(R)`. |
|
| 158. |
A flexible wire bent in the form of a circle is place in a uniform magnetic field perpendicularly to the plane of the coil. The radius of the coil changes as shown in Figure. The graph of magnetude of induced emf in the coil is represented by A. (a) B. (b) C. ( c) D. (d) |
|
Answer» Correct Answer - B (b) In the `r- t` graph, it is clear that from `a to b` there is no change in radius and hence no change in area and magnetic flux. Same is the situaion from `c to d`. Now, `|e| = (d)/(dt)(phi)` `|e| = B(d)/(dt)(pir^(2)) = Bpi2r(dr)/(dt)` Since `r prop t `:.` (dr)/(dT) = constant` ``:.` |e| prop r` |
|
| 159. |
A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer isA. `(BR)/(A)`B. (b) `(AB)/(R )`C. ( c) `ABR`D. (d) `B^(2)A//R^(2)` |
|
Answer» Correct Answer - B (b) `Q = (Deltaphi)/(R ) = (phi_(2) - phi_(1))/(R ) = (BA - 0)/(R ) = (BA)/(R )` |
|
| 160. |
What will be the reading in the voltmeter and ammeter of the circuit shown in fig. |
|
Answer» Inpedance of the circuit, `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt(45^(2) + (4 - 4)^(2)) = 45 Omega` Reading of ammeter, `I_(v) = (E_(v))/(Z) = (90)/(45) =2 A` Reading of Voltmete, `V = (X_(L) - X_(C )) I_(v)` `= (4 - 4) 2 = 0` |
|
| 161. |
When a coil is connected to a Leclanche cell, its resistance is found to be R. What is the effect on its resistance, if it is conncected to an a.c. source?A. it will decreaseB. it will remain the sameC. it will be zeroD. it will increase |
|
Answer» Correct Answer - D The coil has an inductance L and a resistance R. Hence with a.c. source its effective resistance (reactance) will be `sqrt(R^(2)+omega^(2)L^(2))`. Thus it is more than its d.c. resistance. (for d.c. L=0). With a Leclanche cell, we get its d.c. resistance . |
|
| 162. |
A circular insulated copper wire loop is twisted to form two loops of are A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane ( of the paper). A uniform magnetic `vecB` point into the plane of the paper . At t=0, the loop starts rotation about the common diameter as axis with a constant angular velocity `omega` in the magnetic field. Which of the following options is /are correct ? A. The net emf induced due to both the loops is proportional to `cos omega t`B. The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paperC. The amplitude of the maximum net emf induced due to both the loops is equl to the amplitude of maximum emf induced in the smaller loop aloneD. The emf indued in the loop is proportional to the sum of the area of the two loops. |
| Answer» Correct Answer - B::C | |
| 163. |
Shows a long current carrying wire and two rectangular loops moving with of velocity `v`. Find the direction of current in each loop. |
|
Answer» In loop (i) no emf will be induced because there is no flux change. In loop (ii) emf will be induced because the coil is moving in a region of decreasing magnetic field inward in direction. Therefore to oppose the flux decrease in inward direction, current will be induced seen that its magnetic field will be inwards. For this direction of current should be clockwise. |
|
| 164. |
In a series L-C-R circuit, the values of L and C are so adjusted that the maximum current flows through the circuit. In this case, if the P.D. across L is 200V, then the P.D. across the capacitance will beA. more than 200 VB. less than 200 VC. equal to 200 VD. P.D. across the resistance |
|
Answer» Correct Answer - C In series resonance, P.D. across L = P.D. across C = 200V . |
|
| 165. |
LetC be the capacitance of a capacitor discharging through a resistor R. Suppose `t_1` is the time taken for the energy stored in the capacitor to reduce to half its initial value and `t_2` is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio `t_1//t_2` will beA. 2B. 1C. `(1)/(2)`D. `(1)/(4)` |
|
Answer» Correct Answer - D As `U = (Q^(2))/(2C) = ((Q_(0) e^(-t//tau))^(2))/(2C)` `= (Q_(0)^(2))/(2C) e ^(-2t//tau) = U_(0) e^(-2t//tau)` As when `t = t_(1), U = U_(0)//2`, so `(U_(0))/(2) = U_(0) e^(-2t_(1)//tau)` `(1)/(2) = e^(-(2t_(1))/(tau))` or `t_(1) = (tau)/(2) 1n 2` `(Q_(0))/(4) = Q_(0) e^(-t_(2)//tau)` `t_(2) = tau 1n 4 = 2 tau 1n 2` `(t_(1))/(t_(2)) = (1)/(4)` |
|
| 166. |
Two coils having self-inductances, `L_(1)=5mH` and `L_(2)=1mH`. The current in the coil is increasing of same constant rate at a certain instant and the power supplied to the coils is also same, Find the ratio of (i) induced voltage (ii) current (iii) energy stored in two coils at that instant |
|
Answer» Given, `L_(1)=5mH` and `L_(2)=1mH` (i) As we know, induced voltage is given by , `e=(Ldi)/(dt)` `rArr" "(e_(1))/(e_(2))=(L_(1)(di//dt))/(L_(2)(di//dt))=(L_(1))/(L_(2))=(5)/(1)=5:1` …(i) (ii) Power in the coil is given by P = ei Here, `P_(1)=P_(2)rArr" "e_(1)i_(1)=e_(2)e_(2)rArr(i_(1))/(i_(2))=(e_(2))/(e_(1))` Using Eq. (i), we can write, `(i_(1))/(i_(2))=(1)/(5)=1:5` (iii) Energy stored in a coil is given by `U=(1)/(2)Li^(2)` `rArr" "(U_(1))/(U_(2))=((1//2)L_(1)i_(1)^(2))/((1//2)L_(2)i_(1)^(2))=(L_(1))/(L_(2))((e_(2))/e_(1))^(2)=(5)/(1)((1)/(5))^(2)=1:5` |
|
| 167. |
In a series circuit `C=2muF,L=1mH` and `R=10 Omega`, when the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will beA. `1:2`B. `5:1`C. `1:5`D. `1:1` |
|
Answer» Correct Answer - B Maximum current flows in a series L-C-R circuit, when the condition of resonance is satisfied. In that case, `X_(L) = X_(C) and Z = R and I = I_(max) = V/R = V/10 A` Energy stored in the inductor `E_(L) = 1/2 LI_(max)^(2) = 1/2 L ((V)/(10))^(2)` ........(1) and the energy stored in the capacitor `E_(C) = 1/2 CV^(2)` .......(2) `:. (E_L)/(E_C) = (1/2L((V)/(10))^(2))/(1/2CV^(2)) = L/100C` `=(10^(-3))/(100 xx 2 xx 10^(-6)) = 1000/200 = 5/1`. |
|
| 168. |
A non-conducting ring of radius r has charge per unit length `lambda`. A magnetic field perpendicular to plane of the ring changes at rate Db/dt. Torque experienced by the ring is A. `lambdapir^(3)(dB)/(dt)`B. `lambda2pir^(3)(dB)/(dt)`C. `lambda^(2)(2pir)^(2)(dB)/(dt)`D. zero |
|
Answer» Correct Answer - A `e= (dphi)/(dt)= pir^(2)(dB)/(dt)rArr E(2 pi r) = pir^(2)(dB)/(dt)` `E= (1)/(2) r (dB)/(dt)rArr dF = lambdadSE` `rArr " "d tau= rlambdaDSE rArr tau = lambdarEintdS = lambdarE2 pi r` |
|
| 169. |
An inductive circuit a resistance of `10ohm` and an inductance of `2.0` henry. If an AC voltage of `120`volt and frequency of `60Hz` is applied to this circuit, the current in the circuit would be nearlyA. 0.32AB. 0.016AC. 0.48AD. 0.80A |
|
Answer» Correct Answer - B |
|
| 170. |
An inductive circuit a resistance of `10ohm` and an inductance of `2.0` henry. If an AC voltage of `120`volt and frequency of `60Hz` is applied to this circuit, the current in the circuit would be nearlyA. 1.6 AB. 0.16 AC. 2.6 AD. 6.2 A |
|
Answer» Correct Answer - B `I=(e)/(Z)` `=(e)/(sqrt(R^(2)+X_(L)^(2)))=0.16A`. |
|
| 171. |
A `120` volt `AC` source is connected across a pure inductor of inductance `0.70` henry. If the frequency of the source is `60Hz`, the current passing through the inductor isA. 4.55AB. 0.355AC. 0.455AD. 3.55A |
|
Answer» Correct Answer - A |
|
| 172. |
A triangle loop as shown in the figure is started to being pulled out at `t=0` from a uniform magnetic field with a constant velocity `v`.Total resistance of the loop is constant and equals to `R`.Then the variation of power produced in the loop with time will be: A. linearly increasing with time till whole loop comes outB. increases parabolically till whole loop comes outC. P `alphat^(3)` till whole loop come outD. will be constant with time. |
| Answer» Correct Answer - B | |
| 173. |
A rectangular loop is being pulled at a constant speed `v`, through a region of certain thickness `d`, in which a uniform magnetic field `B` is set up. The graph between position `x` of the right hand edge of the loop and the induced e.m.f. `E` will be B. C. D. |
|
Answer» Correct Answer - B As `x` increase so `(dB)/(dt)` increase i.e., induced emf (e) is negative. When loop completely entered in the magnetic field, emf=0 when it exit out `x` increase but `(dB)/(dt)` decrease i.e., `e` is positive. |
|
| 174. |
shows a conducting loop being pulled out of a magnetic field with a speed v. Which of lthe four plots shown in may represent the power delivered by the pulling agent as a function of the speed v ? (##HCV_VOL2_C38_E01_022_Q01##) (##HCV_VOL2_C38_E01_022_Q01##)A. AB. BC. CD. D |
| Answer» Correct Answer - B | |
| 175. |
shows a conducting loop being pulled out of a magnetic field with a speed v. Which of lthe four plots shown in may represent the power delivered by the pulling agent as a function of the speed v ? A. aB. bC. cD. d |
|
Answer» Correct Answer - b |
|
| 176. |
Fig. shows a conducting loop being pulled out of a magnetic field with a constant speed `v`.Which of the four plots shown in fig. may represent the power delivered by the putting agent as a function of the constant speed `v`. A. `A`B. `B`C. `C`D. `D` |
| Answer» Correct Answer - B | |
| 177. |
The SI unit of inductance, the henry, can be written asA. weber/ampereB. volt-second/ampereC. `joule/(ampere)^(2)`D. ohm-second |
|
Answer» Correct Answer - A::B::C::D (a) `L=(phi)/i or henry =(weber)/(ampere)` (b) `e=-L((di)/(dt))` `:. L=-(e)/((di)//(dt)) or henry =(volt-second)/(ampere)` (c) `U=1/2 (li_(2))` `:. L=(2U)/(i^2) or henry = (joule)/(ampere^2)` (d) `U=1/2(Li_(2))=(i^2)Rt` `:. L=R.t or `henry =ohm-second``. |
|
| 178. |
A non-conducting ring having q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field `B = 4t^(2)` is switched on at time t = 0. Mass of the ring is m and radius is R. The ring starts rotating after 2 s, the coefficient of friction between the ring and the table isA. `(4qmR)/g`B. `(2qmR)/g`C. `(8qR)/(mg)`D. `(qR)/(2mg)` |
|
Answer» Correct Answer - C `ointE.dl=-(d phi)/(dt) rArr Exx2piR=piR^(2)(dB)/(dt) rArr E=R/2xx8t=R8 rArr (qE)R=(mumg)R rArr mu=(8qR)/(mg)` |
|
| 179. |
A triangle loop as shown in the figure is started to being pulled out at `t=0` from a uniform magnetic field with a constant velocity `v`.Total resistance of the loop is constant and equals to `R`.Then the variation of power produced in the loop with time will be: A. linearly increasing with time till whole loop come outB. increases parabolically till whole loop comes outC. `P alpha t^(3)` till whole loop come outD. will be constant with time |
| Answer» Correct Answer - B | |
| 180. |
A small square loop of wire of side l is placed inside a large square loop of wire of side `L(Lgtgtl)`. The loops are co-planer and their centres coincide. The mutual inductance of the system is proportional toA. `l/L`B. `(l^(2))/L`C. `L/l`D. `(L^(2))/l` |
|
Answer» Correct Answer - B `phi=Mi` `(4mu_(0)i)/(4pi(L//2))[1/(sqrt(2))+1/(sqrt(2))]xxl^(2)=Mi` `M=(2sqrt(2) mu_(0)l^(2))/(piL)` `M prop (l^(2))/L` |
|
| 181. |
A metal rod of resistance `20 Omega` is fixed along a diameter of a conducting ring of radius `0.1 m` and lies on `x-y` plane. There is a magnetic field `vec(B) = (50 T)vec(k)`. The ring rotates with an angular velocity `omega = 20rad s^(-1)` about its axis. An external resistance of `10 Omega` is connected across the center of the ring and rim. The current external resistance isA. `1/4A`B. `1/2A`C. `1/3A`D. zero |
|
Answer» Correct Answer - C `(BomegaR^(2))/2=10 i/2+10 i rArr ((50)(20)(0.1)^(2))/2=(30i)/2 i=1/3A` |
|
| 182. |
A small square loop of wire of side l is placed inside a large square loop of wire of side `L(Lgtgtl)`. The loops are co-planer and their centres coincide. The mutual inductance of the system is proportional toA. `l//L`B. `l^(2)//L`C. `L//l`D. `L^(2)//l` |
|
Answer» Correct Answer - B Resonant frequency in an LCR series circuit is `v_(r) = (1)/(2 pi sqrt(LC))` To reduce `v_(r)` , c can be increased, by adding another capacitor in parallel to the first. Choice (b) is correct. |
|
| 183. |
A rectangular loop with a sliding connector of length `l = 1.0 m` is situated in a uniform magnetic field `B = 2T` perpendicular to the plane of loop. Resistance of connector is `r = 2 Omega`. Two resistances of `6Omega` and `3Omega` are connected as shown in . The external force required to keep the connetor moving with a constant velocity `v = 2 m s^(-1)` is A. 6 NB. 4 NC. 2 ND. 1 N |
|
Answer» Correct Answer - c |
|
| 184. |
A non-resistive inductor is connected across a fully charged capacitor and the L-C circuit is set oscillating at its natural frequency. What is the value of the current when the charge on the capacitor has the maximum value of `100 mu C`?A. ZeroB. InfinityC. `10 mu A`D. `100 mu A` |
|
Answer» Correct Answer - A When the capacitor is fully charged, the current in the L-C circuit is zero. |
|
| 185. |
In the following electrical network at `t lt 0`, key is placed on (1) till the capacitor got fully charged. Key is placed on (2) at `t = 0`. Time when the energy in both the capacitor and inductor will be same for the first time is A. `(pisqrt(LC))/(r)`B. `(pisqrt(LC))/(2)`C. `(5pisqrt(LC))/(4)`D. `(5pisqrt(LC))/(2)` |
|
Answer» Correct Answer - B::C |
|
| 186. |
An inductor may store energy inA. its electric fieldB. its coilsC. its magnetic fieldD. both in electric and magentic fields |
| Answer» Correct Answer - C | |
| 187. |
Assertion If current shown in the figure is increasing, then Reason IF current passing through an inductor is constant, then both ends of the inductor are at same potential.A. If both Assertion and Reason are corrent and Reason is the corrent explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is ture. |
|
Answer» Correct Answer - B IF moving from left to right, current is increasing, then battery (induced emf `=L.(di)/(dt)`) will produce right to left current i.e., its positive terminal is on left hand side or `V_(A)gtV_(B)`. Further `i` = costant `rArr(di)/(dt)=0or V_(L)=0`. |
|
| 188. |
Match the following : `{:(" Column I"," Column II"),("(A) Steady electric field","(p) Can accelerate a stationary charge"),("(B) Steady magnetic field","(q) Can accelerate a moving charge"),("(C) Time varying magnetic field","(r) Can change the speed of a charge"),("(D) Induced electric field","(s) Forms closed loops"):}`, |
| Answer» Correct Answer - A::B::C::D | |
| 189. |
STATEMENT - 1 : By making suitable arrangement, a magnetic field can cause a stationary charge to accelerate. and STATEMENT - 2 : A constant magnetic field cannot interact with stationary chrages.A. Statement-1 is True, Statement-2, is True, Statement-2 is a correct explanation for Statement-5B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 190. |
STATEMENT - 1 : Energy associated with current configuration is a non-additive quantity. and STATEMENT - 2 : In case of two or more than two current elements, mutual interaction energy also comes into play.A. Statement-1 is True, Statement-2, is True, Statement-2 is a correct explanation for Statement-7B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-7C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 191. |
Is flux linked with all the inductors in parallel same? |
| Answer» Correct Answer - No | |
| 192. |
STATEMENT - 1 : Whenever there is a change in flux linked with a coil in a circuit, an induced current is set up in it. and STATEMENT - 2 : Whenever there is a change in flux linked with a coil in a circuit, an induced emf is set up in it.A. Statement-1 is True, Statement-2, is True, Statement-2 is a correct explanation for Statement-6B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-6C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - D | |
| 193. |
The magnetic flux linked with coil is proportional toA. voltageB. currentC. length of coilD. resistance of coil |
| Answer» Correct Answer - B | |
| 194. |
If any surface is parallel to magnetic lines of forces, then magnetic flux linked with it isA. very small but not zeroB. infiniteC. large but not finiteD. zero |
| Answer» Correct Answer - D | |
| 195. |
Magnetic flux through a coil depends uponA. number of turnsB. areaC. magnetic fieldD. all of these |
| Answer» Correct Answer - D | |
| 196. |
The phenomenon of electromagnetic induction was discovered byA. FlemingB. OerstedC. FaradayD. Henry |
| Answer» Correct Answer - C | |
| 197. |
A circular coil of radius R is moving in a magnetic filed B with a velocity v as shown in the figure. Find the emf across the diametrically opposite points A and B. |
|
Answer» emf=`BVI_("effective")` =2RvB |
|
| 198. |
A circular coil of radius R is moving in a magnetic filed B with a velocity v as shown in the figure. Find the emf across the diametrically opposite points A and B. |
|
Answer» `emf=BVI_(effective)` `=2 R vB` |
|
| 199. |
An electromagnetic eddy current brak a consists of a disc of conductivity `sigma` and thickness d rotating about axis through its centre between rectangular poles of face area A at a distance from the centre from the centre . Calculate the torque tending to show down the disc . |
|
Answer» Correct Answer - `B^(2)Awr^(2) sigmad` |
|
| 200. |
In circular coil, when no. of turns is doubled and resistance becomes `(1)/(4)th` of initial, then inductance becomesA. `4`timeB. `2` timesC. `8` timesD. no change |
|
Answer» Correct Answer - A `LpropN^(2)` |
|