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InterviewSolution
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. |
To convert galvanometer into ammeter which one of the following is connected with the coil:A. small resistance in seriesB. large resisance in seriesC. small resistance in parallelD. large resistance in parallel |
| Answer» Correct Answer - C | |
| 152. |
The magnietic force on a charge particle moving in the field does no work becauseA. kinetic energy of the charged particle does not changeB. the charge of the particle remains sameC. the magnetic force is parallel to velocity of the particleD. the magentic force is parallel to magnetic field |
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Answer» Correct Answer - C The magnetic force on a charged particle moving in the field does no work because the magneitc force is parallel to velocity of the particle |
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| 153. |
The figure shows three situations when an electron moves with velocity `vec(v)` travels through a uniform magnetic field `vec(B)`. In each case, what is the direction of magnetic force on the electron A. positive z axis negative x axis positive y axisB. negative z axis negative x axis and zeroC. positive z axis positive y axis and zeroD. negative z axis positive x axis and zero |
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Answer» Correct Answer - B According to fleming left hand rule in and magnetic force on the electron will be directed in negative Z axis and negative X axis respectively velocity of elctron and direction of magnetic field are anti parallel so no force will act on electron |
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| 154. |
To send 10% of the main current through a moving coil galvanometer of resistance `99omega`, the shunt required is –A. `9.9 Omega`B. `10 Omega`C. `11 Omega`D. `9 Omega` |
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Answer» Correct Answer - C Required resistance which should be added in series so that , the current in the circuit is given by Initial resistance = Final resistane i.e., `G=((Gs)/(G+s))+R` `therefore R=G-((GS)/(G+S))=(G^2)/(G+S)`. |
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| 155. |
A cyclotron can accelerateA. `beta`- particlesB. high velocity `gamma`- raysC. `alpha`- particlesD. high velocity x - rays |
| Answer» Correct Answer - C | |
| 156. |
In a cyclotron, a magnetic field of `2*4T` is used to accelerate protons. How rapidly should the electric field between the dees be reversed? The mass and the charge of protons are `1*67xx10^(-27)kg` and `1*6xx10^(-19)C` respectively.A. `1.342xx10^(-69)`B. `2.342 xx10^(-8)`C. `2.3645xx10^(-10)`D. none of these |
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Answer» Given magnetic field B= 1.4 `wbm^(-2)` charge of protons q=`1.6xx10^(-19) C` mass of proton m `=1.67xx10^(-27)`kg `T=(pim)/(Bq)=(3.142 xx1.67xx10^(-27))/(1.4xx1.6xxx10^(-19))` `T= 2.342xx10^(-8)` S |
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| 157. |
A cyclotron is operating at a frquency of `12xx10^6Hz`. Mass of deutron is `3.3 xx10^-19` coulomb. To accelerate deutrons the magnetic field isA. 0.016 teslaB. 0.16 teslaC. 16 teslaD. 1.6 tesla |
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Answer» Correct Answer - D `v =(q B)/(2 pi m)` `therefore B =(2 pi mv )/(q)` `=(2xx3.14xx3.3 xx10^-27xx12xx10^6)/(1.6xx10^-19)` `= 1.6 T`. |
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| 158. |
A beam of ions with velocity ` 2xx 10^(5) ` m/s enters normally into a uniform magnetic field of ` 4 xx 10^(-2)` T. if the specific charge to the ions is ` 5xx 10^(7) ` C/kg, the radius of the circular path described will beA. 0.10 mB. 0.16 mC. 0.20 mD. 0.25 m |
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Answer» Correct Answer - A `r=(mv)/(q B)=(2xx10^5)/(5xx10^7xx4xx10^-2)=0.1m` |
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| 159. |
A narrow beam of protons and deutrons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them isA. `1:2`B. `1:1`C. `2:1`D. `1:3` |
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Answer» Correct Answer - B Since the radius of circualr path of charge particle of an magnetic field `r=(mv)/(qB)=(p)/(qB)` now the radius of circular path of charge particle of given mometum p and magnetic field B is given by `r prop (1)/(q)` But charge on both charge particles protons and deutones is same therefore `(r_(p))/(r_(d))=(q(d))/(q_(p))=1/1` |
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| 160. |
A 1.25 kW heater works on a 220 V mains supply. What current rating would a suitable fuse have ?A. `2` AB. `5` AC. `10` AD. `13` A |
| Answer» Correct Answer - C | |
| 161. |
An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, thenA. path of both will be straight lineB. both are equally curvedC. the path of proton shall be less curved than that of electronD. the path of proton shall be less curved than that of electron |
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Answer» Correct Answer - B The radius of circular path traced by electron and proton is given by `r=(mv)/(qB)=(p)/(qB)=sqrt(2mKE)/(qB)` `therefore r prop v prop p prop sqrt(k)` As both proton and electron have same momentum hence both will have the same circular path i.e both are equally curved |
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| 162. |
A proton and a deuteron both having the same kinetic energy enter perpendicularly in to uniform magnetic field B for motion of proton and deuteron on ciruclar path of radius `R_(p)` and `R_(d)` respectively the correct statement isA. `R_(d)=sqrt(R_(p))`B. `R_(d)=R_(p)//sqrt(2)`C. `R_(d)=R_(p)`D. `R_(d)=2R_(p)` |
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Answer» Correct Answer - A `(mv^(2))=qvB` for proton `R_(p)=(mv)/(qB)=sqrt(2m_(p)E)/(qB)` for deuteron `R_(d)=sqrt(2m_(d)E)/(qB)` `rarr (R_(d))/(R_(p))=sqrt(m_(d))/(m_(p))=sqrt(2)` `rarr R_(d)=sqrt(2R_(p)` |
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| 163. |
When a galvanometer of resistance G iws converted into an ammeter of range IA then the current passing through shunt (S) isA. `I_s=((G)/(G+S))I`B. `I_s=((S)/(S+G))I`C. `I_s=((SI=G)/(S))`D. `I_s=SxxG` |
| Answer» Correct Answer - A | |
| 164. |
An electron having charge `1.6xx10^(-19)C` and mass `9xx10^(-31)` kg is moving with `4xx10^(6)ms^(-1)` speed in a magnetic field `2xx10^(-1)` tesla in circular orbit. The force acting on electron and the radius of the circular orbit will beA. `12.8xx10^(-13)B,11xx10^(-3)m`B. `1.28 xx10^(-14)N, 11xx10^(-3)m`C. `12.8xx10^(-13)N,11xx10^(-3)m`D. `1.28xx10^(-13)N,11xx10^(-4)m` |
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Answer» Correct Answer - D Force F=evB `=1.6xx10^(-19)xx4xx10^(6)xx2xx10^(-1)` `=12.8 xx10^(-13)N` Radius of circular orbit `r=(mv)/(Bq)` `=(9xx10^(-31)xx4xx10^(6))/(2xx10^(-1)xx1.6xx10^(-19))` lt `=1.1xx10^(-4)` |
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| 165. |
A magnetic field exerts no force onA. an electric charge moving perpendicular to its directionB. an unmagnetised iron barC. a stationary electric chargeD. a magnet |
| Answer» Correct Answer - C | |
| 166. |
A magnetic field `4xx10^(-3)` kT exerts a force `(4 hat I + 3 hat j) xx10^(10)` N on a particle having a charge `10^(-9)` C and going on the XY plane The velocity of the particle isA. `-75 hat I + 100 hat j`B. `-100 hat I + 75 hat j`C. `25hat I + 2 hat j`D. `2 hat I + 25 hat i` |
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Answer» Correct Answer - A From lorentz force `F=q(vxx B)` and `B=4xx10^(-3) hatk T` solving we get a=-75 , b =100 `rarr` v=-75 `hat I +100 hatj` |
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| 167. |
Currents of `10A, 2A` are passed through two parallel wires `A` and `B` respectively in opposite directions. If the wire `A` is infinitely long and the length of the wire `B` is 2 metre, the force on the conductor `B`, which is situated at `10cm` distance from `A` will beA. `8xx10^(-7)`NB. `8xx10^(-5)N`C. `4xx10^(-7)N`D. `4xx10^(-5)N` |
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Answer» Correct Answer - B As force on the conductor B ltbvrgt `F=(mu_(0))/(2pi)(2i_(1)i_(2))/(r )` `=(10^(-)xx2xx10xx2)/(0.1)xx2=8xx10^(-5) N` |
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| 168. |
An electron is accelerated by a potential difference of 12000 volts. It then enters a uniform magnetic field of `10^(-3)T` applied perpendicular to the path of electron. Find the radius of path. Given mass of electron `=9xx10^(-31)kg` and charge on electron `=1.6xx10^(-19)C`A. 36.7 mB. 36.7 cmC. 3.67 mD. 3.67 cm |
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Answer» Correct Answer - A Radius `r=sqrT(2mK)/(qB)=(1)/(B)sqrt(2mV)/(q)` `=1/10^(-3)(2xx9xx10^(-31)xx1200)/(1.6xx10^(-19))` =0.367 m =36.7 cm |
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| 169. |
A current loop in a magnetic fieldA. experience a troque whether the field is uniform or non unifrom in all orientationsB. can be in equilibrium in one orientatinC. can be equilbrium in two orientations both the equilibrium states are unsatableD. can be in equilibrium in two orientation one stable while the other is unstable |
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Answer» Correct Answer - D For parallel M is stable and for antiparallel M is unstable |
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| 170. |
A rectangular loop carrying a current ` i` is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop . If steady current ` I` is established in the wire as shown in the figure , A. rotate about an axis parallel to the wireB. move away from the wireC. move towards the wireD. remain statioinary |
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Answer» Correct Answer - C Wire placed close to each other carry current in the same direction and hence attract wires placed far apart carry current in opposite direction and hence repel each other but here attraction is stroing and replusion is week so loop moves towards the wire |
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| 171. |
A current loop in a magnetic fieldA. can be in equilibrium in one orientationB. can be in equilibrium in two orientation both the equilibrium states are unstableC. can be in equilibrium in two orientation one stable while the other is unstableD. experiences a torque whether the field is uniform or non uniform in all orientations. |
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Answer» Correct Answer - C `vec(tau)= vec(M)xxvec(B)rArr tau=MB sinalpha`. `alpha = 0^@` corresponds to stable equilibrium , `alpha = 180^@` corresponds to unstable equilibrium. |
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| 172. |
when a proton is released from rest in a room, it starts with an initital acceleration `a_0` towards west . When it is projected towards north with a speed `v_0,` it moves with an initial acceleration `3a_0` towards west. Find the elecrtic field and the minimum possible magnetic field in the room.A. `(ma_0)/(2)"west",(2ma_0)/(ev_0)"down"`B. `(ma_0)/(e)"east",(3ma_0)/(ev_0)"up"`C. `(ma_0)/(e)"east",(3ma_0)/(ev_0)"down"`D. `(ma_0)/(e)"west",(3ma_0)/(ev_0)"up"` |
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Answer» Correct Answer - A `ma_0= eErArr E=(ma_0)/( e)` and `(e _0 B+ e E)/(m)=3a_0` `rArr ev_0B=3ma_0-e E =3ma_0 -ma_0=2ma_0` `rArr B=(2ma_0)/(ev_0)`. |
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| 173. |
A galvanometer having a coil resistance of `60 Omega` shows full scale defection when a current of `1.0 A` passes thoguth it. It can vbe convered into an ammeter to read currents up to `5.0 A` byA. `240Omega`in parallelB. `15Omega`in seriesC. `240Omega`in seriesD. `240Omega`in parallel |
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Answer» Correct Answer - D `S=(I_g.G)/(I-I_g)` `=(1xx60)/(5-1)=(60)/(4)` ` =15 Omega` in parallel. |
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| 174. |
A particle of charge `q` and mass `m` moves in a circular orbit of radius `r` with angular speed `omega`. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends onA. `q//m`B. `2q//m`C. `q//2m`D. `q^2//2m` |
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Answer» Correct Answer - C Magnetic moment is given by , `M=IA=(q)/(T) pi r^2=(2pi)/(T)(qr^2)/(2)=(omega qr^2)/(2)` Angular momentum `=L=I omega=mr^2omega` `therefore (M)/(L)=(omegaq r^2//2)/(mr^2omega)=(q)/(2m)`. |
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| 175. |
Magnetic field due to a ring having n turns at a distance `x` on its axis is proportional to (if `r =` radius of ring)A. `(r)/((x^(2)+r^(2))`B. `(r)/((x^(2)+r^(2))^(3//2))`C. `(nr^(2))/((x^(2)+r^(2))^(3//2))`D. `(n^(2)r^(2))/((x^(2)+r^(2))^(3//2))` |
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Answer» Correct Answer - C Magnetic field due to a circular coil having n turns at a distance x on its axis is give by `B=(mu_(0))/(4pi)*(2pi n ir^(2))/((x^(2)+r^(2))^(3//2))` `:. B prop=(nr^(2))/((x^(2)+r^(2))^(3//2))` |
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| 176. |
A current of `0.1 A` circulates around a coil of 100 turns and having a radius equal to `5 cm`. The magnetic field set up at the centre of the coil is (`mu_(0)=4pixx10^(-7)` weber/amper-metre)A. `5pixx10^(-5)T`B. `8pixx10^(-5)T`C. `4pixx10^(-5)T`D. `4pixx10^(-5)T` |
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Answer» Correct Answer - C Magnetic field at the centre of circular coil of N turns is given by `B=(mu_(0)Ni)/(2r)=(4xx10^(-7)xx100xx0.1)/(2xx5xx10^(-2))=4pxx10^(-6)T` |
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| 177. |
A current of `0.1 A` circulates around a coil of 100 turns and having a radius equal to `5 cm`. The magnetic field set up at the centre of the coil is (`mu_(0)=4pixx10^(-7)` weber/amper-metre)A. `4pi xx1-^-5`teslaB. `8pixx10^-7` teslaC. `4xx10^-5` teslaD. `2xx10^-5` tesla |
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Answer» Correct Answer - A `B=(mu_0nI)/(2a)=(4pixx10^-7xx100xx0.1)/(1xx5xx10^-2)=4pixx10^(-5)T` |
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| 178. |
A long solenoid is formed by winding 20 turns `cm^-1`. What current is necessary to produce a magnetic field of 20 mT inside the solenoid?A. 8.0 AB. 4.0 AC. 2.0 AD. 1.0 A |
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Answer» Correct Answer - A `B=mu_(0)nI` `because I=(B)/(mu_0n)=(20xx10^-3xx10^7)/(4pixx20xx10^2)=8A` |
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| 179. |
The magnetic field is made radial in a M.C.G.A. to make field strongerB. to make field weakerC. to make scale linearD. to reduce its resistance |
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Answer» Correct Answer - C If resistance of galvanometer is very large, momentary current produce oscillations with decreasing amplitude. |
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| 180. |
In radial magnetic field, the plane of the coil of M.C.G. for any position is alwaysA. perpendicular to the lines of forceB. parallel to the lines of forceC. inclined with `45^@` with lines of forceD. none of these |
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Answer» Correct Answer - B In radial magnetic field , plane of the coil is always parallel to magnetic induction. `therefore theta=0` Thus , `tau_n= n IAB cos theta= n IAB cos 0` = n IAB (maximum). |
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| 181. |
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field `vec(B)=B_(0)hat(K)`A. they have equal z components of momentaB. they must have equal chargesC. they necessarily represent a particle antiaritcl pairD. the charge to masss ratio satisfy `(e )/(m)_(1)+(e )/(m)_(2)=0` |
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Answer» Correct Answer - D In a uniform magnetic field the two charged particles will traverse identicla helical path in a completely oppsoset sense if the charge / mass ratio of these two articles is ame and charges on them are of opposite character in this situation `(e//m)_(1) +(e//m)_(2)Q` holds good |
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| 182. |
The strength of the magnetic field in a long solenoid having 5000 turns per meter is `3.14X10^(-2)T`. The current flowing through the solenoid isA. 2AB. 3AC. 4AD. 5A |
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Answer» Correct Answer - D `I=(B)/(mu_0n)=(3.14xx10^-2)/(4pixx10^-7xx5000)=5A` |
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| 183. |
Which one of the following statements is not true ?A. In a house circuit, lamps are used in parallel.B. Switches, fuses and circuit breakes should be placed in the neutral wireC. An electric iron has its earth wire connected to the metal case to prevent the user receiving a shockD. When connecting a three-core cable to a 13 A three-pin plug, the red wire goes to the live pin |
| Answer» Correct Answer - B | |
| 184. |
A car headlamp of 48 W works on the car battery of 12 V. The correct fuse for the circuit of this car headlamp will be :A. `5` AB. `10` AC. `3` AD. `13` A |
| Answer» Correct Answer - A | |
| 185. |
The phenomenon of electromagnetic induction is :A. the process of charging a body.B. the process of generating magnetic field due to current passing through a coil.C. producing induced current in a coil due to relative motion between a magnet and the coil.D. the process of rotating a coil of an elecric motor. |
| Answer» Correct Answer - C | |
| 186. |
Which of the following is not the unit of magnetic induction ?A. maxwellB. gaussC. oerstedD. weber/`"meter"^2` |
| Answer» Correct Answer - A | |
| 187. |
Magnetic effect of current was discovered byA. FlemingB. FaradayC. AmpereD. Oersted |
| Answer» Correct Answer - D | |
| 188. |
1 maxwell = ............weberA. `10^-8`B. `10^-4`C. `10^-6`D. `10^-3` |
| Answer» Correct Answer - A | |
| 189. |
There are 3 voltmeters A,B,C having the same and `5,000 Omega` respectively. The best voltmeter amongest them is the one whose resistance isA. `5,000Omega`B. `10,000Omega`C. `15,000Omega`D. all are equally good |
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Answer» Correct Answer - C The voltmeter of alrge resistance meausres accurate value of potential difference because for ideal voltmeter , its resistance is infinity. |
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| 190. |
The magnetic field on the axis of a long solenoid having n turns per unit length and carrying a current isA. `mu_(0)ni`B. `mu_(0)n^(2)i`C. `mu_(0)ni^(2)`D. none of these |
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Answer» Correct Answer - A If the solenoid is of infinite length and the point is well inside the solenoid `B_("in")=mu_(0)ni` |
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| 191. |
The resistance of an ideal voltmeter isA. ZeroB. lowC. highD. infinity |
| Answer» Correct Answer - D | |
| 192. |
Which of the following is likely to have the largest resistance ?A. Ammeter of range 1AB. Voltmeter of range 10VC. Moving coil galvanometerD. All will have same resistance |
| Answer» Correct Answer - B | |
| 193. |
The net resistance of a volmeter should be large to ensure thatA. not get overheatedB. not draw excessive currentC. measure large potential differenceD. not change the value of potential difference |
| Answer» Correct Answer - D | |
| 194. |
A rectangular coil of 100 turns each of area `10 cm^2` hangs freely in radial magnetic field. The coil deflects through an angle of `30^@` when current of 0.5 m A is paased through it . If torsional constant of suspension fibre is `2.5 xx10^-9` Nm/rad, then magnetic field will be ,A. `3.6xx10^-5Wb//m^2`B. `2.2xx10^-5Wb//m^2`C. `2.6xx10^-5Wb//m^2`D. `4.2xx10^-5Wb//m^2` |
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Answer» Correct Answer - C `B=(K theta )/(n AI)=(2.5xx10^-9xx5)/(10^2xx10^-3xx5xx6xx10^-4)` |
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| 195. |
A rectangular coil of 20 turns , each of area `10 cm^2`, is suspended freely in a uniform magnetic field of induction `2.5 xx10^-2` T with its plane inclined at `45^@` with the field. The torque acting on the coil when a current of 10 mA is passed through it isA. `5.5xx10^-6Nm`B. `3.5xx10^-6Nm`C. `3.3xx10^-6Nm`D. `5.3xx10^-6Nm` |
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Answer» Correct Answer - B `tau_(n)= l IABcos theta ` `20 xx10^-2xx10^-3xx2.5xx10^-2xx0.7` prop 3.5xx10^6 Nm`. |
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| 196. |
A rectangular coil of a moving coil galvanometer has 50 turns and an area of `12cm^2`. It is suspended in a radial magnetic field of induction `0.025 Wb//m^2`. The torsional constant of the suspension fibre is `1.5 xx10^-9` Nm/degree. The current required to produce the deflection of `5^@` isA. `5muA`B. `2muA`C. `4.5muA`D. `5.5muA` |
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Answer» Correct Answer - A `I=(K)/(n AB)theta= (1.5xx10^-9xx5)/(50xx12xx10^-4xx25xx10^-3)=5mu A ` . |
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| 197. |
A rectangular coil of M.C.G. of 100 turns , each of length 4 cm and breadth 3 cm , is suspended freely in a radial magnetic field of induction `2xx10^-2` N/Am. The torsional constant of the suspension fibre is `2xx10^-8` Nm per radian. The deflection of the coil when a current of `2 mu A` is passed through it is nearly .A. `14^@`B. `12^@`C. `13^@`D. `15^@` |
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Answer» Correct Answer - A `theta = (B n AI)/(K)=(2xx10^-2xx10^2xx12xx10^-4xx2xx10^-6)/(2xx10^-8)` `=0.24 xx(180)/(pi) prop 14^@`. |
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| 198. |
A rectangular coil having 100 turns each of length 1.0 cm and breadth 0.5 cm is suspended in radial magnetic field of induction 0.02 T the torisional constant of suspension fibre is `2xx10^(-8)` Nm / degreee calculate current sensitivity of MCGA. 500 div /AB. 600div /AC. 400 div / AD. none of these |
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Answer» Given n =100 `l=1.0 cm =10xx10^(-2) m` `b=0.5 cm =0.5 xx10^(-2) m` `B=0.002 T` |
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| 199. |
A rectangular coil of a M,.C.G. of effective area `0.05 m^2` is suspended freely in a radical magnet field of induction `0.01 Wb//m^2`. If the torsional constant of the suspension fibre is `5xx10^-9` Nm per degree. When a current of 300 `mu` A is passed through it, deflection of coil isA. `15^@`B. `30^@`C. `45^@`D. `60^@` |
| Answer» Correct Answer - B | |
| 200. |
The radius of circular path of an electron when subjected to a perpendicular magnetic field isA. `(mV)/(be)`B. `(me)/(Be)`C. `(mE)/(Be)`D. `(Be)/(Mv)` |
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Answer» Correct Answer - A Radius of the circular path is given by `r=(mV)/(qB)=(mV)/(eB)` |
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