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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.
| 1151. |
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. 5HB. 4HC. 3HD. 2H |
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Answer» Correct Answer - D Given `L_(P)=2.4 H, L_(S)=10H` Let `L_(1) and L_(2)` be the two inductance. We use the laws of series and parallel resistances for `L_(1)` and `L_(2)`. then `L_(1)+L_(2)=10 and 1/(L_1)+1/(L_2)=1/2.4` `:. 2.4 = (L_(1)L_(2))/(L_(1)+L_(2)) :. 2.4 = (L_(1)L_(2))/(10)` `:. L_(1)L_(2)= 24` But `L_(1)-L_(2) = sqrt((L_(1)+L_(2))^(2)-4L_(1)L_(2))` `sqrt(100-4 xx 24)` `sqrt(100-96) = 24` `L_(1)-L_(2) = 2H`. |
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| 1152. |
Figure-5.55 shows a small circular coil of area A suspended from a point O by a string of length l in a uniform magnetic induction B in a uniform magnetic induction B in horizontal direction. If the coil is set into oscillation like a simple pendulum by displacing it a small angle `0_(0)` as shown, find EMF induced in coil as a function of time. Assume the plane of coil is always in plane of string. |
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Answer» Correct Answer - `(1)/(2)Ba theta_(0)^(2)sin(2sqrt((g)/(l)t))` |
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| 1153. |
Two inductances connected in parallel are equivalent to a single inductance of `1.5H` and when connected in series are equivalent to a single inductance of 0.8 H. The difference in their inductance isA. 3 HB. 7.5 HC. 2 HD. 4 H |
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Answer» Correct Answer - D `therefore` In parallel, `L_(P)=(L_(1)L_(2))/(L_(1)+L_(2))` `(L_(1)+L_(2))1.5=L_(1)L_(2)` and in series, `L_(s)=L_(1)+L_(2)=8` `therefore" "L_(1)L_(2)=12` `(L_(1)-L_(2))^(2)=(L_(1)+L_(2))^(2)-4L_(1)L_(2)` `L_(D)=sqrt((8)^(2)-4xx12)` `L_(D)=4 H` |
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| 1154. |
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 secondB. `(R)/(L)`secondC. `(L)/(R)`secondD. `(1)/(LR)`second |
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Answer» Correct Answer - C Time constant, `tau=(L)/(R)` It is the time interval during which the current after opening an inductive circuit falls to `37%` of its maximum value. |
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| 1155. |
The conducting circular loops of radii `R_(1) and R_(2)` are placed in the same plane with their centres coinciding. If `R_(1) gt gt R_(2)`, the mutual inductance M between them will be directly proportional toA. `R_(1)//R_(2)`B. `R_(2)//R_(1)`C. `R_(1)^(2)//R_(2)`D. `R_(2)^(2)//R_(1)` |
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Answer» Correct Answer - D Mutual inductance between two coils in the same plane with their centres coinciding is given by `M=(mu_(0))/(4pi)((2pi^(2)R_(2)^(2)N_(1)N_(2))/(R_(1)))` henry |
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| 1156. |
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. `R//L`B. `R//L`C. `L//R`D. `1//LR` |
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Answer» Correct Answer - C `i_(0)=E//R` `i=i_(1)e^(-(Rt)/(L))` `0.37i_(0)=i_(0)e^(-(Rt)/(L))` `e^(-(Rt)/(L))=(1)/(0.37)=2.7` `(Rt)/(L)=1n(2.7) implies (Rt)/(L)=1 implies t=(L)/(R)` |
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| 1157. |
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 sec`B. `(R )/(L)`secC. `(L)/(R )`secD. `(1)/(LR)`sec |
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Answer» Correct Answer - C When battery disconnected current through the circuit start decreasing exponentially according to`i=i_(0)e^(-Rt//L)`. `implies0.37i_(0)=i_(0)e^(-Rt//L)implies0.37=(1)/(e)=e^(-Rt//L)impliest=(L)/(R )` |
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| 1158. |
Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. Durig this period, the two loopsA. attract each otherB. repel each otherC. do not exert any force on each otherD. attract or repel each other depending on the sense of the current |
| Answer» Correct Answer - D | |
| 1159. |
The adjoining figure shows two bulbs `B_(1)` and `B_(2)` resistor `R` and an inductor and `L`. When the switch `S` is turned off A. Both `B_(1)` and `B_(2)` die out promptlyB. Both `B_(1)` and `B_(2)` die out with some delayC. `B_(2)` dies out promptly but `B_(1)` with some delayD. `B_(1)` dies out promptly but `B_(2)` with some delay |
| Answer» Correct Answer - C | |
| 1160. |
The adjoining figure shows two bulbs `B_(1)` and `B_(2)` resistor `R` and an inductor and `L`. When the switch `S` is turned off A. The bulb `B_(2)` lights up earlier than `B_(1)` and finally both the bulbs shine equally brightB. `B_(1)` lights up earlier and finally both the bulbs acquire equal brightnessC. `B_(2)` lights up earlier and finally `B_(1)` shines brighter than `B_(2)`D. `B_(1)` and `B_(2)` light up together with equal brightness all the time |
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Answer» Correct Answer - C In L-R circuit current grows exponentially. So, `B_(2)` lights up earlier and finally `B_(1)` shines brighter than `B_(2)`. |
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| 1161. |
Figure represents two bullbs `B_(1)" and " B_(2)`resister R and inductor L. When the switch S in turned off,then A. Both `B_(1)` and `B_(2)` die out promptlyB. Both `B_(1)` and `B_(2)` die out with some delayC. `B_(1)` dies out promptly but `B_(2)` with some delayD. `B_(2)` dies out promptly but `B_(1)` with some delay |
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Answer» Correct Answer - C Current in `B_(1)` will promptly become zero while current in `B_(2)` will slowly tend to zero. |
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| 1162. |
A capacitor of capacitance `10 muF` is connected to an oscillator giving an output voltage `epsilon = (10 V) sin omegat`. Find the peak currents in the circuit for `omega = 10 s^(-1), 100 s^(-1), 500 s^(-1), 1000 s^(-1)`. |
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Answer» Here, `C = 10 mu F = 10 xx 10^(-6)` `= 10^(5) F` As `E = 10 sin omega t :. E_(0) = 10 V, omega = 10 rad//s` `I_(0) = (E_(0))/(X_(C )) = (E_(0))/(1//omega C) = E_(0) omega C` `= 10 xx 10 xx 10^(-5) = 10^(-3) A = 1 mA` |
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| 1163. |
What is the natural frequency of LC circuit ? What is the reactance of this circuit at this frequency ? |
| Answer» `v = (1)/(2 pi sqrt(LV))`. Reactance of the circuit at this frequency is zero. | |
| 1164. |
Does the steady state current in RL circuit depend upon L ?` |
| Answer» No, it depends only on R, and not on L. | |
| 1165. |
A circular coil `P` of `100` turns and radius `2cm` is placed coaxially at the center of another circular coil `Q` of `100` turns and radius `20cm` . Calculate (a) the mutual inductance of the coils (b) the induced emf in coil `P` when the current in the coil `Q` decreases from `5A` to `3A` in `0.04` sec (Take `pi^(2)=10)` |
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Answer» Correct Answer - `(a) 3.94xx10^(-4)H (b) 19.72xx10^(-3)V (c ) 19.72xx10^(-3)Wb//s (d) 9.86xx10^(-5)C` |
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| 1166. |
The figure-5.298 shows a specific RL circuit, the time constant for this circuit is: A. `(L)/(2R)`B. `(2L)/(R)`C. `(2R)/(L)`D. `(R)/(2L)` |
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Answer» Correct Answer - B |
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| 1167. |
A circular ring of diameter 20cm has a resistance `0.01Omega` How much charge will flow through the ring if it is rotated from positon perpendicular to the uniform magnetic field of B=2T to a position parallel to field?A. 4CB. 628CC. 3.14CD. 25.12C |
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Answer» Correct Answer - C |
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| 1168. |
A loop shown in the figure is immersed in the varying magnetic field `B=B_0t`, directed into the page. If the total resistance of the loop is `R`, then the direction and magnitude of induced current in the inner circle is A. Cockwise `(B_(0)(pia^(2)-b^(2)))/(R)`B. Anticlockwise `(B_(0)pi(a^(2)+b^(2)))/(R)`C. Clockwise `(B_(0)(piomega^(2)+4b^(2)))/(R)`D. Clockwise `(B_(0)(4b^(2)-pia^(2)))/(R)` |
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Answer» Correct Answer - C |
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| 1169. |
A long solenoid has `500` turns. When a current of `2A` is passed through it, the resulting magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid isA. 1.0 HB. 4.0 HC. 2.5 HD. 2.0 H |
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Answer» Correct Answer - A We have, `Nphi_(B)=Li` Self-inductance, `L=(Nphi_(B))/(i)=(500xx4xx10^(-3))/(2)=1H` |
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| 1170. |
A long solenoid has `500` turns. When a current of `2A` is passed through it, the resulting magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid isA. 2 HB. 1.5 HC. 1 HD. 0.5 H |
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Answer» Correct Answer - C The total magnetic flux `(phi)` linked with the solenoid is given by `phi = 500 xx 4 xx 10^(-3) = 2 Wb` But for a coil `phi = Li :. L = (phi)/(i) = (2Wb)/(2A) = 1H`. |
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| 1171. |
Figure. shows three rigions of magnetic field each of area `A`, and in each rigion, magnitude of mqagnetic field decreases rate `alpha`. If `vec(E)` is the induced electric field, then the vlue of the line `ointvec(E).dvec(r )` along the given loop is equal to A. (a) `alphaA`B. (b) `-alphaA`C. ( c) `3alphaA`D. (d) `-3alphaA` |
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Answer» Correct Answer - B (b) As anticlockwise direction is positive, so area vector out wards is positive. So net flux through the given loop is `phi = -BA - BA + BA = -BA` `ointvec(E).vec(dr) = -(dphi)/(dt) = -(d(-BA))/(dt) = A(dB)/(dt) = A(-alpha)` `rarr` `ointvec(E).vec(dr) = - alphaA` |
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| 1172. |
A series circuit consists of a resistance, inductance and capacitance. The apllied voltage and the current at any instant are given as e=141.4 cos (300t-10^(@))` and `i=5 cos (3000t-55^(@))` The inductance is 0.01H. Calculate the values of the resistance and capacitance. |
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Answer» Correct Answer - `33.33 mu F` |
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| 1173. |
Find the natural frequency of oscillation of a circuit containing an inductance of 400 mH and a capacity of `40 mu F`. To which wavelength will its response be maximum ? How long will the oscillations last ? |
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Answer» `v = ? L = 400 mH = 400 xx 10^(-3) H`, `C = 40 mu F = 40 xx 10^(6) F, lambda = ?` `v = (1)/(2 pi sqrt(LC)) = (1)/(2 pi sqrt(400 xx 10^(-3) xx 40 xx 10^(-6)))` `= (1000)/(8 pi) = 39.8 Hz` `lambda = (c )/(v) = (3 xx 10^(8))/(39.8) m = 7.5 xx 10^(6) m` When there is no loss of energy, oscillations will last for infinite time. |
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| 1174. |
Find the natural frequency of a circuit containing inductance of `100 mu` H and a capacity of 0.01 mu F`. To which wavlength, its response will be maxmum ? For how long will the oscillations continue ? |
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Answer» Here, E = 100 sin 314 t. In a capacitor. Current leads the applied e.m.f. by a phase angle of `90^(@)` `:. I = I_(0) sin (314 t + 90^(@)) = (E_(0))/(1//omega C) cos 314 t` ` = 100 xx 314 t xx 637 xx 10^(-6) cos 314 t` `= 20 cos 314 t` Instantaneous power = EI `= 10 sin 314 t xx 20 cos 314 t` `= 1000 (2 sin 314 t cos 314 t)` `= 1000 (sin 628 t)` watt Max. energy stored ` = (1)/(2) CE_(0)^(2)` ` = (1)/(2) xx 637 xx 10^(-6) (100)^(2) = 3.185 J` |
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| 1175. |
Find the natural frequency of a circuit jcontaininig inductance of `100 mu H` and a capacity of `0.1 mu F`. To which wavelength will its respone be maximum? How long will the oscillations last ?` |
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Answer» Here,`v = ? L = 100 mu H = 10^(-4) H` `C = 0.01 mu F = 10^(-8) F` `v = (1)/(2 pi sqrt(LC)) = (1)/(2 pi sqrt(10^(-4) xx 10^(-8))) = (10^(6))/(2 pi) Hz` `v = 159.2 xx 10^(3) Hz = 159.2 kHz` `lambda = (C )/(v) = (3 xx 10^(8))/(159.2 xx 10^(3)) = 1.884 xx 10^(3) m` As no loss of energy is assumed, therefore, the oscillations will continue for infinite time. |
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| 1176. |
A 20 watt, 50 V lamp is connected in series to a.c. mains of 250 V, 50 Hz. Calculate to value of capacitor to run the lamp. |
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Answer» Here, ` p = 20` watt, `V = 50` volt `E_(v) = 250 V, C = ?, v = 50 Hz` `I = (P)/(V) = (20)/(50) = 0.4 A` `R = (V)/(I) = (50)/(0.4) = 125 ohm` `Z = (E_(v))/(I_(v)) = (E_(v))/(I) = (250)/(0.4) = 625 ohm` `X_(C ) = sqrt(Z^(2) - R^(2)) = sqrt(625^(2) - 125^(2)) = 250 sqrt 6` As `X_(C ) = (1)/(omega C) = (1)/(2 pi v C) = 250 sqrt 6` `:. C = (1)/(2 pi v xx 250 sqrt 6)` `= (1)/(2 xx 3.14 xx 50 xx 250 sqrt 6)` farad `C = 0.052 xx 10^(-4) F 5.2 mu F` |
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| 1177. |
A `20 V` 5 watt lamp is used in ac main `220 V` and frequency 50 c.p.s. What pure resistance should be included in place of the above passive elements so that the lamp can run on its rated voltage? |
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Answer» Correct Answer - `(a) 4.0 mu F (b) 2.53 H ( c) 720 Omega; (d) Using L or C will be more economical` |
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| 1178. |
The magnetic flux passing perpendicular to the plane of a coil given by `phi=6t^(2)+4t+3` Where `phi` is in milliweber and t is in seconds. What is the e.m.f. induced in the coil at t=1 sec?A. 4 mVB. 8 mVC. 12 mVD. 3 mV |
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Answer» Correct Answer - D `phi = 6t^(2) + 4t + 3` `:. E = (d phi)/(dt) = 12t+4` at `t =1, e=12+4 = 16 mV`. |
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| 1179. |
A circuit containing an 80 mH inductor and a `250 mu F` capacitor in series is connected to 240 V, `100 rad//s` supply. The resistance of the circuit is negligible. (i) Obtain rms value of current. (ii) What is the total average power consumed in the circuit ? |
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Answer» Here, `L = 80 mH = 80 xx 10^(-3) H`, `C = 250 mu F = 250 xx 10^(-6) F, E_(v) = 240 V`, `omega = 100 rad//s` `X_(L) = omega L = 100 xx 80 xx 10^(-3) H = 8 Omega` `X_(C ) = (1)/(omega C) = (1)/(100 xx 250 xx 16^(-6)) = 40 Omega` `Z = X_(C ) -X_(L) = 40 - 8 = 32 Omega` `I_(v) = (E_(v))/(Z) = (240)/(32) = 7.5 A` As `R = 0`, therefore power cosume is zero. |
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| 1180. |
A `100 mu F` capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductor, as a result of which 5 A current flows through the inductance. Calculate the value of inductance. |
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Answer» Here, `C = 100 mu F = 10^(-4) F`, `V = 50` volt, `I = 5 A, L = ?` As energy stored in inductor = energy stored in capacitor, `:. (1)/(2) LI^(2) = (1)/(2) CV^(2)` `L = (CV^(2))/(I^(2)) = (10^(-4) (50)^(2))/((5)^(2)) = 0.01 H` |
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| 1181. |
A circuit consists of a resistance 10 ohm and a capacitance of `0.1 mu F` If an alternating e.m.f. of 100 V. 50 Hz is applied, calculate the current in the circuit. |
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Answer» Here, `R = 10 ohm`, `C = 0.1 mu F = 10^(-7) F`, `E_(v) = 100 V, v = 50 Hz, I_(v) = ?` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C) = (1)/(2 pi xx 50 xx 10^(-7))` `= (10^(5))/(pi) = 3.18 xx 10^(4) Omega` `Z = sqrt(R^(2) + X_(C )^(2)) = sqrt(10^(2) + (3.18 xx 10^(4))^(2))` `= 3.18 xx 10^(4)` `I_(v) = (E_(v))/(2) = (100)/(3.18 xx 10^(4)) = 3.142 xx 10^(-3) A` |
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| 1182. |
The dimensional formula for magnetic flux isA. `[phi]=[M^(2)L^(1)T^(-2)A^(-1)]`B. `[phi]=[M^(1)L^(2)T^(-2)A^(-1)]`C. `[phi]=[M^(1)L^(-1)T^(2)A^(1)]`D. `[phi]=[M^(1)L^(2)T^(2)A^(-1)]` |
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Answer» Correct Answer - B `phi = BA and R = BIl` `:. [B] = [(F)/(Il)]=[(M^(1)L^(1)T^(-2))/(A^(1)L^(1))] = [M^(1)L^(0)T^(-2)A^(-1)]` `:. [phi] = [BA] = [M^(1)L^(0)T^(-2)A^(-1)xxL^(2)] = [M^(1)L^(2)T^(-2)A^(-1)]`. |
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| 1183. |
What is the dimensional formula of `sqrt(LC)` ? |
| Answer» `[M^(0) L^(0) T^(1)]` | |
| 1184. |
The equivalent quantity of mass in electricity isA. currentB. self inductanceC. potentialD. charge |
| Answer» Correct Answer - B | |
| 1185. |
Name the physical quantity which is measured in weber `amp^(-1)` .A. self inductanceB. mutual inductanceC. magnetic fluxD. both (a) and (b) |
| Answer» Correct Answer - D | |
| 1186. |
Name the physical quantity which is measured in weber `amp^(-1)` . |
| Answer» Self inductance and Mutual inductance. | |
| 1187. |
What is the relation between weber and Maxwell ? |
| Answer» 1 weber `= 10^(8)` Maxell | |
| 1188. |
Magnetic field of `2 xx 10^(-2)` Wb `m^(2)` is acting at right angle to a coil of 1000 `cm^(2)` having 50 turns. The coil is removed from the field in `(1)/(10)` second. Find the magnitude of induced e.m.f. |
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Answer» Here, `B_(1) = 2 xx 10^(-2) Wb m^(2), theta = 0^(@)` `A = 1000 cm^(2) = 1000 xx 10^(-4) m^(2) = 10^(-1) m^(2)`, `N = 50`, `B_(2) = 0, dt = (1)/(10) s, e = ?` `e = (-d phi)/(dt) = - (N (B_(2) - B_(1)) A)/(dt)` `= - (50 (0 - 2 xx 10^(-2)) 10^(-1))/(1//10) = 1.0 V` |
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| 1189. |
A coil intercepts a magnetic flux of `0.2 xx 10^(-2)` Wb in 0.1 s. What is the emf induced in the coil ? |
| Answer» `|e| = (d phi)/(dt) = (0.2 xx 10^(-2))/(0.1) = 0.02 V` | |
| 1190. |
A magnetic field of `2 xx 10^(-2) Wb//m^2` acts at right angles to a coil of area `100 cm^(2)` with 50 turns. The average e.m.f. Induced in the coil is 0.1V, what it is removed from the field in t sed. What is the value of t?A. 1 secB. 0.5 secC. 0.1 secD. 0.01 sec |
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Answer» Correct Answer - C `e = (NBA)/t or t = (NBA)/e` `=(50 xx 2 xx 10^(-2) xx 10^(2) xx 10^(-4))/(1 xx 10^(-1)) = 10^(-1) = 0.1 sec`. |
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| 1191. |
The magnetic flux threading a coil changes from `12 xx 10^(-3)` Wb to `6 xx 10^(-3)` Wb in 0.01. Calculate the induced e.m.f. |
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Answer» Here, `phi_(1) = 12 xx 10^(-3) Wb, phi_(2) = 6 xx 10^(-3) Wb`, `dt = 0.1 s = 10^(2)s, e = ?` `e = (- d phi)/(dt) = (-(phi_(2) - phi_(1)))/(dt)` `= (-(6 xx 10^(-3) - 12 xx 10^(-3)))/(10^(2))` `e = 0.6` volt |
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| 1192. |
A coil having an area `2 m^(2)` is placed in a magnetic field which changes from `1 wb//m_(2)` n interval of 2s. The e.m.f. induced in the coil is: |
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Answer» Here, `A = 2 m^(2)`, `(dB)/(dt) = ((4 - 1) Wb//m^(2))/(2 sec) = (3)/(2) T//s` `e = (d phi)/(dt) = A (dB)/(dt) = 2 x (3)/(2) = 3` volt |
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| 1193. |
What is magnetic flux linked with a coil of N turns and cross section area A held with its plane parallel to the field?A. `(NAB)/(2)`B. NABC. `(NAB)/(4)`D. zero |
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Answer» Correct Answer - D Magnetic flux linked with a coil `phi=NBcostheta` Since the magnetic field B is parallel to the area A, `phi=NBcostheta``phi=NBcostheta``i.e.,theta=90^(@)`. `phi=0` |
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| 1194. |
A coil having an area of `2m^(2)` is placed in a magnetic field which changes from `2 Wb//m^(2)` to `5Wb//m^(2)` in 3 seconds. The e.m.f. Induced in the coil isA. 4VB. 3VC. 2VD. 1V |
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Answer» Correct Answer - C `e = -(d phi)/(dt) = =(A[B_(2)-B_(1)])/(t) = 2 xx 3/3 = 2V`. |
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| 1195. |
A wire of length 0.1 m moves with a speed of `10 m//s` perpenducular to a magnetic field of induction ` Wb//m^(2)`. Calculate induced e.m.f. |
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Answer» Here, `l - 0.1 m, upsilon = 10 m//s, B = 1 Wb//m^(2), e = ?` As `e = B l upsilon :. e = 1 xx 0.1 xx 10 = 1` volt |
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| 1196. |
An LR circuit with a battery is connected at t=0. Which of the following quantities is not zero just after the connection?A. Current in the circuitB. magnetic field energyC. Power delivered by the batteryD. Emf induced in the inductor |
| Answer» Correct Answer - A | |
| 1197. |
A square coil of `10^(-2) m^(2)` area is placed perpenducular to a uniform magnetic field of `10^(3) Wb//m^(2)`. What is magnetic flux through the coil? |
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Answer» Here, `A = 10^(-2) m^(2), B = 10^(3) Wb//m^(2)` `theta = 0^(@), phi = ?` `phi = BA cos theta = 10^(3) xx 10^(-2) cos 0^(@) = 10` Weber |
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| 1198. |
The coefficient of mutual induction between two circuits is equal to the, emf produced in one circuit when the current in the second circuit isA. kept steady at 1 AB. cut-off ai 1 A levelC. changed at the rate of `1As^(-1)`D. changed from `1 As^(-1)` to `2 As^(-1)` |
| Answer» Correct Answer - C | |
| 1199. |
Faraday established that an e.m.f. can be induced in a coil by changing the amount of magnetic flux `(phi)` linked with the coi. As `phi = Ba cos theta`, therefore, three methods of inducting e.m.f. are by changing magnetic field B or by changing area A or by changing relative orientation `theta` of the coil w.r.t. the magnetic field. Read the above passage and answer the following question : (i) Have you heard of a battery less flashlight ? How does it work? (ii) Can you operate a typical cell phone from the hip movements of the person? Whant do you know about gas-elcetric hybrid autos ? |
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Answer» (i) Yes, a flashlight without a battery can give s light for a few minutes. It is based on EMI. On shaking the flashligth, its permanent internal magnet passes through a coil, inducing some current. Thus current is converted into d.c. and is used for charging a capacitor. The charged capacitor then sends current through light emitting diode (LED), which gives us light. (ii) Yes, a backpack generator is used to convert the hip movements of the person into electricity and produce 5-7 watt of electric power to operate a typical call phone. (iii) Gas electric hybrid autos are the latest innovations in the auto industry. The engine is made smaller than the convertional one. Through the sue of EMI, some energy is supplied to batteries and electric motor. The purpose behind is to get more mileage for every litre of petrol consumed. This would make the auto cost effective. |
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| 1200. |
The back emf in a DC motor is maximum when,A. Motor has picked up maximum speedB. Motor has just started movingC. Speed of motor is still increasingD. Motor has been switched off |
| Answer» Correct Answer - A | |