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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.
| 201. |
Which of the two curves shown below has lesser time constant. |
| Answer» Correct Answer - curve 1 | |
| 202. |
In a transformer, number of turns in the primary coil are 140 and that in the secondry coil are 280. If current i primary ciol is 4A, then that in the secondary coil isA. `4A`B. `2A`C. `6A`D. `10A` |
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Answer» Correct Answer - B `(V_(p))/(V_(s))=(i_(s))/(i_(p))impliesi_(s)=4xx(140)/(280)=2A` |
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| 203. |
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 balance the centrifugal forces and a steady state will be reached.In the steady state `m_(e)omega^(2)x= E` `V_(P)-V_(Q)=underset(x=0)overset(x=l)int barE.bar(dx)=underset(0)overset(l)int(m_(e)omega^(2)x)/edx=(m_(e)omega^(2)l^(2))/(2e)` |
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| 204. |
A rod of length `l` is rotating with an angular speed `omega` about one of its ends which is at a distance a from an infinitely long wire carrying current `i`. Find the emf induced in the rod at the instant shown in . |
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Answer» `E=int(mu_(0)i)/(2pi(a+rcos theta))xx(romega)*(dr)` `E=(mu_(0)omegai)/(2pi) int_(0)^(l)( r)/(a+r cos theta) drimplies E=(mu_(0)I omega)/(2pi cos theta)[l-(a)/(cos theta)ln""((a+l cos theta)/(a))]` |
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| 205. |
A rod of length `l` is rotating with an angular speed `omega` about one of its ends which is at a distance a from an infinitely long wire carrying current `i`. Find the emf induced in the rod at the instant shown in . |
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Answer» `Eint(mu_(0)i)/(2pi(a+r cos theta))xx(romega).(dr)` `E=(mu_(0)omegai)/(2pi)underset(0)overset(l)intr/(a+r cos theta)dr` `E=(mu_(0)iomega)/(2pi cos theta)[l-a/(cos theta)"ln" ((a+l cos theta)/a)]` |
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| 206. |
Find the velocity of the moving rod at time t if the initial velocity of the rod is v and a constant force F is applied on the rod. Neglect the resistance of the rod. |
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Answer» At any time , let the velocity of the rod be v . Applying Newtons law: F-iIB=ma…..(1) Also `Biv=i_(1)R=(q)/( c)` Applying Kcl, `i=i_(1)+(dq)/(dt)=(BlV)/( R)+(D)/(Dt)(BIvC) or i=(B l V)/( R)+B l C a ` Putting the value of `i` in eq. (1), `F-(B^(2)l^(2)V)/( R)=(m+B^(2)l^(2)C)a=(m+B^(2)l^(2)C)(dv)/(dt)` `(m+B^(2)l^(2)C)(v)/(F-(B^(2)I^(2)v)/( R))=dt` Integrating both sides, and solving we get `v=(FR)/(B^(2)l^(2))(1-e^((tB^(2)l^(2))/(R(m+CB^(2)l^(2)))))` |
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| 207. |
Two rods, each of length `L = 0.6 m`, are rotating in same plane about their ends `C_(1)` and `C_(2)`. The distance between `C_(1)` and `C_(2)` is 1.201 m. The rods rotate in opposite directions with same angular speed `omega` and they are found to be in position shown in the Figure at a given instant of time. The ends `C_(1)` and `C_(2)` are connected using a conducting wire. There exists a uniform magnetic field perpendicular to the figure having magnitude `B = 5.0 T`. At what angular speed `omega` do we expect to see sparks in air? The dielectric breakdown of air happens if electric field in it exceeds `3 × 10^(6) Vm^(–1)`. |
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Answer» Correct Answer - `omega = 1667 rads^(-1)` |
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| 208. |
A vertical cylindrical region has a horizontal radial magnetic field inside it. A wire ring is made of a wire of cross section S, density d and resistivity `rho`. Radius of the ring is r. The ring is kept horizontal with its centre on the axis of the cylindrical region and released. The field strength at all points on the circumference of the ring is B. At a certain instant velocity of the ring is v downward. Find (a) Current in the ring and (b) Acceleration of the ring at the instant. |
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Answer» Correct Answer - (a) `(BvS)/(rho)` (b) `g - (B^(2)v)/(rhod)` |
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| 209. |
(a) When a small magnet is moved towards a sole- noid, an emf is induced in it. However, if a magnet is moved inside the hole of a toroid, no emf is induced. Explain. (b) A wire is kept horizontal along North-South direction. It is allowed to fall freely. How much emf will be induced across the ends of the wire due to presence of Earth’s magnetic field. |
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Answer» Correct Answer - Zero |
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| 210. |
A conducting square loop of side length a is held vertical with the help of a non conducting rod of length L. The rod is made to rotate in horizontal circle about a vertical axis through its end. The loop rotates with the rod while its plane always remains perpendicular to the rod. The resistance of the loop is R and angu- lar speed of the rod is `omega`. There is a uniform horizontal magnetic field B in the entire space. Find the average rate of heat dissipation in the loop during one rotation. |
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Answer» Correct Answer - `(a^(4)B^(2)omega^(2))/(2R)` |
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| 211. |
A horizontal conducting loop of radius R is fixed in air. A uniformly charged rod having charge Q on it is held vertically above the conducting loop at a height h above it. The rod is released and it begins to fall along the axis of the loop. Calculate the emf induced in the conducting loop. |
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Answer» Correct Answer - Zero |
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| 212. |
A short circuited coil is moved towards a fixed magnet at a constant velocity. When the coil is at a distance x from the magnet it experiences a magnetic force F. Now the number of turns in the coil is doubled and the experiment is repeated with the coil moving with same constant velocity towards the magnet. What will be the magnetic force on the coil when it is at a distance x from the magnet? |
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Answer» Correct Answer - `2F` |
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| 213. |
A short circuited coil is moved towards a fixed magnet at a constant velocity. When the coil is at a distance x from the magnet it experiences a magnetic force F. Now the number of turns in the coil is doubled and the experiment is repeated with the coil moving with same constant velocity towards the magnet. What will be the magnetic force on the coil when it is at a distance x from the magnet? |
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Answer» Correct Answer - `2F` |
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| 214. |
A square coil of 600 turns, each side 20cm, is placed with its plane inclined at `30^(@)` to a uniform magnetic field of `4.5xx10^(-4)Wbm^(-2)`, Find the flux through the coil. |
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Answer» Correct Answer - `5.4xx10^(-3) Wb` |
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| 215. |
A coil of area `500cm^(2)` and having `1000` turns is held perpendicular to a uniform field of `0.4` gauss. The coil is turned through `180^(@)` in `1//10sec` . Calculate the average induced e.m.f.A. 0.02 VB. 0.04 VC. 1.4 VD. 0.08 V |
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Answer» Correct Answer - B Here, `A=500cm^(2)=500xx10^(-4)m^(2)` `N=1000,B=0.4G=0.4xx10^(-4)T` `t=(1)/(10)s` When the coil is held perpendicular to the field, the normal to the plane of the coil makes an angle of `0^(@)` with the field B. `:.` initial flux, `phi_(1)=BAcos0^(@)=BA` final flux, `phi_(2)=BAcos180^(@)=-BA` Average induced emf, `epsi=-N((phi_(2)-phi_(1))/(t))=-N((-BA-BA)/(t))` `=(2NBA)/(t)=(2xx1000xx0.4xx10^(-4)xx500xx10^(-4))/(1//10)=0.04V` |
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| 216. |
A coil of area `500cm^(2)` and having `1000` turns is held perpendicular to a uniform field of `0.4` gauss. The coil is turned through `180^(@)` in `1//10sec` . Calculate the average indued e.m.f. |
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Answer» `A=500cm^(2)=500xxI0^(-4)m^(2)=5xxI0^(-2)m^(2)` `N=I000` , `B=0.4` gauss `=0.4xxI0^(-4)T=4xxI0^(-5)T` `Deltat=(I)/(I0)s` `phi_(I)=NBAcos0=NBA` `phi_(2)NBAcosI80^(@)=-NBA` `|Deltaphi|=2NBA` `|bar(e)|=(|Deltaphi|)/(Deltat)=(2NBA)/(Deltat)=(2xxI000xx4xxI0^(-2))/(I//I0)` `=0.4V` |
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| 217. |
A coil of area `0.04 m^(2)` having 1000 turns is suspended perpendicular to a magnetic field of `5.0xx10^(-5) Wbm^(-2)`. It is rotated through `90^(@)` in 0.2 s. Calculate the average emf induced in it. |
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Answer» Correct Answer - 0.01 V |
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| 218. |
A coil of area `500cm^(2)` and having `1000` turns is held perpendicular to a uniform field of `0.4` gauss. The coil is turned through `180^(@)` in `1//10sec` . Calculate the average induced e.m.f. |
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Answer» Correct Answer - [0.04V] |
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| 219. |
A coil of area 500 `cm^(2)` having 1000 turns is put perpendicular to a magnetic field of intensity `4xx10^(-5)`T. if it is rotated by `180^(@)` in 0.1 s, the induced emf produced isA. 20 mVB. 40 mVC. 60 mVD. 80 mV |
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Answer» Correct Answer - B `e=(dphi)/(dt)=(nAB)/(dt)(costheta_(1)-costheta_(2))` `e=(2xx1000xx500xx10^(-4)xx10^(-5))/(0.1)=40mV` |
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| 220. |
A coil of 500 turns and mean radius 10 cm makes 3000 rotations per minute about one of its diameter in a uniform magnetic field of induction `4xx10^(-2)Wb//m^(2)` perpendicular to its axis of rotation. The rms current through the coil if its resistance is 100 `Omega`, isA. 4.1AB. 1.4AC. 4.4AD. 1.1A |
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Answer» Correct Answer - B `e_(0)=nAB2pif` `I_(rms)=(e_(rms))/(R)` `=(e_(0))/(Rsqrt(2))=(nAB2pif)/(Rsqrt(2))` `=(500xx3.14xx10^(-2)xx4xx10^(-2)xx2xx3.14xx50)/(100xx1.414)` `=1.4A` |
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| 221. |
A copper disc of radius 10 cm placed with its plane normal to a uniform magnetic field completes 1200 rotations per minute. If induced emf between the centre and the edge of the disc is 6.284 mV, Find the intensity of the magnitude field. Take `pi=3.142` |
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Answer» Correct Answer - `10^(-2)T` |
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| 222. |
A horizontal solenoid is connected to a cell and a switch, Fig. A Copper ring with its axis along the axis of the solenoid is placed on a frictionless track. What happen to the ring, as the switch is closed ? A. move towards the solenoidB. remain stationaryC. move away from the solenoidD. move towards the solenoid or away from it depending on which terminal (positive or negative) of the battery is connected to the left end of the solenoid |
| Answer» Correct Answer - C | |
| 223. |
A highly conucting ring of radius `R` is perpendicular to and concentric with the axis of a long solenoid as shown in fig. the ring has a narrow gap of width `d` in its circumference. The solenoid has cross sectional area `A` and a uniform internal field of magnitude `B_(0)`. Now beginning at `t=0`, the solenoid current is steadily increased to so that the field magnitude at any time `t` is given by `B(t)=B_(0)+alphat` where `alphagt0`. Assuming that no charge can flow across the gap, the end of ring which has excess of positive charge and the magnitude of induced e.m.f. in the ring are respectively A. `x,Aalpha`B. `XpiR^(2)a`C. `Y,pA^(2)alpha`D. `Y,piR^(2)a` |
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Answer» Correct Answer - A Since the current is increasing, so inward magnetic flux linked with the ring also increase (as viewed from left side). Hence induced current in the ring is anticlockwise, so end `x` will be positive. Induced emf `|e|=A(dB)/(dt)=A(d)/(dt)=A(d)/(dt)(B_(0)+alphat)implies|e|=Aalpha` |
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| 224. |
The acceleration of the charge on the gap faces will cease when the total electric field within the ring becomes zero. For this to happen, the electric field `E_(0)` in the gap isA. (a) `E_(0) = (abeta)/(delta)`B. (b) `E_(0) = (2a beta)/(delta)`C. ( c) `E_(0)` is dependent of `R` for `R gt sqrt((a)/( pi))`D. ( d) `E_(0)` is independent of `R` for `R gt sqrt((a)/( pi))` |
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Answer» Correct Answer - A::D (a,d): `E = (|xi|)/(delta) = (abeta)/(delta)` This expression is independent of `R` as long as the radius of the ring exceeds the radius `sqrt((a)/(pi))` of the solenoid. |
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| 225. |
An all-metal aeroplane flies horizational at 600 km per hours at a place where the vertical induction is `4xx 10^(-5)` tesla . If the wing -span is 10 m , will be the resulting p.d between the tips of the wings ? |
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Answer» Correct Answer - 0.067 V |
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| 226. |
Two ends of an inductor of inductance L is connected to two parallel conducting rails. A conducting wire of length (that is equal to separation between the rails) can slide on the rails without friction. The wire has mass m. It is projected with a velocity `v_(0)` parallel to the rails (see Figure). Neglect self inductance and resistance of the loop. (a) Find velocity of the wire as a function of time. (b) Write current through the wire at time t. (c) Find speed of the wire as a function of its displacement. (d) Is the current in the conductor zero when it stops? If no, find this current . (e) Will the conductor move after it stops? |
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Answer» Correct Answer - (a) `v = v_(0) cos omegat` (b) `i=v_(0) sqrt((m)/(L))sin omega t` where `omega = (Bl)/(sqrt(mL))` (c) `v = sqrt(v_(0)^(2) -(B^(2)l^(2))/(mL)x^(2))` (d) No, `v_(0) sqrt((m)/(L))` (e) Yes |
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| 227. |
A conducting rod is moving with a constant velocity v over the parallel conducting rails which are connected at the ends through a resistor R and capacitor C as shown in the figure. Magnetic field B is into the plane. Consider the following statements. i. Current in loop `AEFBA` is anti clockwise, ii. Current in loop `AEFBA` is clockwise iii Current through the capacitor is zero iv Energy stored in the caspacitor is `1/2 CB^2L^2v^2` Which is the following options in correct?A. Statement i and iii are correctB. Statemetn ii and iv are correctC. Statement i,iii and iv are correctD. None of these |
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Answer» Correct Answer - A |
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| 228. |
A pair of long conducting rails are held vertical at a separation `l = 50 cm`. The top ends are connected by a resistance of `R = 0.5 Omega` and a fuse (F) of negligible resistance. A conducting rod is free to slide along the rails under gravity. The whole system is in a uniform horizontal magnetic field B = 1.5 T as shown. Resistance of the rod and rails are negligible and the rod remains horizontal as it moves. The rod is released from rest. Find the minimum mass of the rod that will ensure that the fuse blows out. It is known that rating of the fuse is 4 A. |
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Answer» Correct Answer - `0.3 kg` |
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| 229. |
A flat coil, in the shape of a spiral, has a large number of turns N. The turns are wound tightly and the inner and outer radii of the coil are a and b respectively. A uniform external magnetic field (B) is applied perpendicular to the plane of the coil. Find the emf induced in the coil when the field is made to change at a rate `(dB)/(dt)`. |
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Answer» Correct Answer - `(piN)/(3) (a^(2) + b^(2) +ab) (dB)/(dt)` |
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| 230. |
Loop `A` of radius `rgtgtR` moves toward loop `B` with a constant velocity `V` such a way that their planes are always parallel. What is the distance between the two loops `(x)` when the induced emf in loop `A` is maximum? A. `R`B. `R/sqrt2`C. `R/2`D. `R(1-1sqrt2)` |
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Answer» Correct Answer - C `phi_(A)=(mu_(0)ipiR^(2))/(2pi(R^(2)+x^(2))^(3//2))pir^(2)` `E_(A)=-(dphi)/(dt)=(mu_(0)ipi)/2R^(2)r^(2)(-3//2)(R^(2)+x^(2))^(5//2)2x(v)` `E_(A)` is maximum when `(dE_(A))/(dx)=0` `rArr d/(dx)x/(R^(2)+x^(2))^(5//2)=0` or `(R^(2)+x^(2))^(-5//2)-(5x)/2(R^(2)+x^(2))^(3//2)` 2x=0 or `R^(2)+x^(2)-5 x^(2)=0` or `x=R/2` |
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| 231. |
Loop `A` of radius `rgtgtR` moves toward loop `B` with a constant velocity `V` such a way that their planes are always parallel. What is the distance between the two loops `(x)` when the induced emf in loop `A` is maximum? A. RB. `( R)/(sqrt(2))`C. `(R )/(2)`D. `R(1-(1)/(sqrt(2)))` |
| Answer» Correct Answer - C | |
| 232. |
A close loop is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved keeping the radius of the loop unchanged, the electrical power dissipated would be :A. halvedB. the sameC. doubledD. quadrupled |
| Answer» Correct Answer - B | |
| 233. |
An electric motor runs a `D.C.` source of e.m.f. `200 V` and draws a current of `10A`. If the efficiency is `40%`, then ressistance of the armature is:A. `5Omega`B. `12Omega`C. `120Omega`D. `160Omega` |
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Answer» Correct Answer - B Here `eta=(e)/(E)xx100` `(40)/(100)=(e)/(E)=(4)/(10)=(2)/(5)` `e=(2E)/(5)=(2xx200)8/(5)=80V` But `I+(E-e)/(R )` :. `10=(200-80)/(R )` :. `R=12Omega` |
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| 234. |
An electric motor runs a `D.C.` source of e.m.f. `200 V` and draws a current of `10A`. If the efficiency is `40%`, then ressistance of the armature is:A. `5Omega`B. `12 Omega`C. `120Omega`D. `160 Omega` |
| Answer» Correct Answer - B | |
| 235. |
A motor runs at 20 V. The resistance of the armature is 11 ohm. If the back emf produced is 198 V, when at full speed, then calculate the current through armature, when (i) motor is just switched on and (ii) motor is at full speed. |
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Answer» Here, `V = 220 V, R = 11 Omega, E = 198 V` Current at full speed `I = (V - E)/(R )` `= (220 - 198)/(11) = (22)/(11) = 2 A` When motor is just switched on, `E = 0` `:. I = (V)/(R ) = (220)/(11) = 20 A` |
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| 236. |
The number of turns in the coil of an ac genrator is `5000` and the area of the coil is `0.25m^(2)`. The coil is rotate at the rate of `100 "cycles"//"sec"` in a magnetic field of `0.2W//m^(2)`. The peak value of the emf generated is nearlyA. 786 kVB. 440 kVC. 220 kVD. 157 kV |
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Answer» Correct Answer - D Peak value of emf generated in generator is given by `e_(0)=omegaNBA` `=(2piv)NBA` `=2xx3.14xx100xx5000xx0.2xx0.25` = 157 kV |
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| 237. |
The number of turns in the coil of an ac genrator is `5000` and the area of the coil is `0.25m^(2)`. The coil is rotate at the rate of `100 "cycles"//"sec"` in a magnetic field of `0.2W//m^(2)`. The peak value of the emf generated is nearlyA. `786 kV`B. `440 kV`C. `220 kV`D. `187.1 kV` |
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Answer» Correct Answer - D `e_(0)=omegaNBA=(2piv)NBA` `=2xx3.14xx1000xx5000xx0.2xx0.25=175kV` |
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| 238. |
Armature current in dc motor will be maximum whenA. Motor has acquired maximum speedB. Motor has acquired intermediate speedC. Motor has just strarted movingD. Moving is switched off |
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Answer» Correct Answer - C Motor e.m.f. equation `E_(b)=V-I_(a)R_(a)` At starting `E_(b)=0`, so `I_(a)` will be maximum. |
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| 239. |
The armature of a `DC` motor has `20Omega` resistance. It draws a current of `1.5 A` when run by `200 V DC` supply The value of back emf induced in it will beA. `150V`B. `170V`C. `180V`D. `190V` |
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Answer» Correct Answer - B Back emf = Applied voltage potential drop across armature coil `=200-iR` `=200-1.5xx20` `=170V` |
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| 240. |
The armature of dc motor has `20Omega` resistance. It draws current of `1.5` ampere when run by `220` volts dc supply. the value of back e.m.f. induced in it will beA. `150 V`B. `170 V`C. `180 V`D. `190 V` |
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Answer» Correct Answer - D `i=(E-e)/(R )implies1.5=(220-e)/(20)impliese=190V` |
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| 241. |
Assertion : An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason : Efficiency of electric motor depends only on magnitude of back emf.A. AB. BC. CD. D |
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Answer» Correct Answer - C From the knowledge of theory, assertion is true but the reason is false. |
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| 242. |
In a circular conducting coil, when current increases from `2A` to `18A` in `0.05` sec., the induced e.m.f. is `20V`. The self-inductance of the coil isA. `62.5mH`B. `6.25mH`C. `50mH`D. none of these |
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Answer» Correct Answer - A `|e|=L(di)/(dt)implies20=Lxx((18-2))/(0.05)impliesL=62.5mH` |
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| 243. |
Two conducting sphere of radius R are placed at a large distance from each other. They are connected by a coil of inductance L, as shown in the figure. Neglect the resis- tance of the coil. The sphere A is given a charge Q and the switch ‘S’ is closed at time `t = 0`. Find charge on sphere B as a function of time. At what time charge on B is `(Q)/(2)` ? |
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Answer» Correct Answer - `q=(Q)/(2)[1-cos(sqrt((2K)/(LR))t)]; t=(pi)/(2)sqrt((LR)/(2K))` where `K=(1)/(4pi in_(0))` |
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| 244. |
A uniform magnetic field B exists perpendicular to the plane of the Figure in a circular region of radius b with its centre at O. A cir- cular conductor of radius `a ( lt b)` and centre at O is made by join- ing two semicircular wires ABC and ADC. The two segments have same cross section but different resistances `R_(1)` and `R_(2)` respectively. The magnetic field in increased with time and there is an induced current in the conductor. (a) Find the ratio of electric fields inside the conductors ABC and ADC. (b) Explain why the electric field in two conductors is different despite the fact that the magnetic field is symmetrical. |
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Answer» Correct Answer - (a) `(E_(1))/(E_(2))=(R_(1))/(R_(2))` |
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| 245. |
An inductor `20 mH`, a capacitor `100muF` and a resistor `50Omega` are connected in series across a source of emf `V=10sin314t`. The power loss in the circuit isA. `1.13W`B. `0.79W`C. `2.74W`D. `0.43W` |
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Answer» Correct Answer - B `V_(0)=10V,omega=314 rad//s` `X_(L)=omegaL=(314)(20xx10^(-3))=6.280` `X_(C)=(1)/(omegaC)=(1)/(314xx100xx10^(-6))=31.84Omega` and `R=50Omega` `Z=sqrt((X_(c)-X_(L))^(2)+R^(2))` `=sqrt((31.84-6.28)^(2)+(50)^(2)=56Omega` Power loss `P=(i_(rms))^(2).R=(i_(0)^(2)R)/(2)` Hence `P=((0.18)^(2)xx50)/(2)=0.81W` |
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| 246. |
A resistor an inductor and a capacitor are connected in series to an ac source An ac voltmeter measures the votage across them as `800 V, 30 V` and `90V` respectively The rms value of the supply voltage is .A. 10 VB. 1 VC. 100 VD. 90 V |
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Answer» Correct Answer - C `e=sqrt(e_(R)^(2)+(e_(C)-e_(L))^(2))=100V` |
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| 247. |
In a transformer, the number of turns in primary and secondary are `500` and `2000` respectively. If current in primary is `48A`, the current in the secondary isA. `12A`B. `48A`C. `192A`D. `144A` |
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Answer» Correct Answer - A `I_(S)=(n_(P))/(n_(S))xxI_(P)=(500)/(2000)xx48=12A` |
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| 248. |
An inductance of 100 mH and a resistance of 100 `Omega` are connected in series and an alternating emf of peak value 100 V, 50 Hz is applied across the combination. The power factor of the circuit isA. 0.954B. 9.54C. 95.4D. 0.845 |
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Answer» Correct Answer - A `X_(L)=2pifL=2xx3.14xx50xx0.1` =31.4`Omega` Power factor`=(R)/(Z)=(R)/(sqrt(R^(2)+X_(L)^(2)))` `=(100)/(sqrt(100^(2)+(31.4)^(2)))=0.954` |
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| 249. |
How does ohmic resistance in a circuit affeated when frequecny of a.c. source in the circuit is doubled ? |
| Answer» Ohmic resistance R remains unaffected. | |
| 250. |
A uniform wound solenoid coil of self inductance `1.8 xx 10^(-4) H` and resistance ohm is broken into two identical coils. These indentical coils are connected in parallel across a 12 V battery of negligible resistance. Calculate the steady state current through the battery and time constant of the circuit. |
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Answer» Here, when the soleniod is broken ino two identical coils, inductance of each of coil becomes half and so is the resistance `:. L_(1) = L_(2) = (1.8 xx 10^(-4))/(2) = 0.9 xx 10^(-4)` henry `R_(1) = R_(2) = 6//2 = ohm`. When these two coils are connected in parallel to a battery, then `(1)/(L) = (1)/(L_(1)) + (1)/(L_(2)) = (L_(2) + L_(1))/(L_(1) L_(2))` `= 0.45 xx 10^(-4) H` and `R = (R_(1) R_(2))/(R_(1) + R_(2)) = (3 xx 3)/(3 + 3) = 1.5 phm` Time constant, `tau = (L)/(R ) = (0.45 xx 10^(-4))/(1.5) = 3 xx 10^(-5) s` The steady state current does not depend upon L. `I_(0) = (E)/(R ) = (12)/(1.5) = 8 A` |
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