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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. |
A long solenoid has `200 turns per cm` and carries a current `i`. The magnetic field at its centre is `6.28xx10^(-2) weber//cm^(2)`. Another long soloenoid has `100 turns per cm` and it carries a current `(i)/(3)`. The value of the magnetic field at its centre isA. `1.05 xx 10^(-2) Weber//m^(2)`B. `1.05 xx 10^(-5) Weber//m^(2)`C. `1.05 xx 10^(-3) Weber//m^(2)`D. `1.05 xx 10^(-4) Weber//m^(2)` |
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Answer» Correct Answer - A `(B_(2))/(B_(1)) = (mu_(0)n_(2)i_(2))/(mu_(0)n_(i)i_(1)) rArr (B_(2))/( 6.28 xx10^(-2)) = ( 100xx(i)/(3))/(200xxi)` rArr = (6.28xx10^(-2))/(6) = 1.05 xx10^(-2) Wb//m^(2)` |
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| 152. |
In a region, steady and uniform electric and magnetic fields are present . These two fields are parallel to each other. A charged particle is released from rest in this region . The path of the particle will be aA. helixB. straight lineC. ellipseD. circle |
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Answer» Correct Answer - B The charged particle will move along the lines of electric field ( and magnetic field).Magnetic field will exert no force. The force by electric field will be along the lines of uniform electric field. Hence the particle will move in a straight line. |
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| 153. |
Which particle will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field ?A. `Li^(+)`B. ElectronC. ProtonD. `He^(+)` |
| Answer» Correct Answer - A | |
| 154. |
A charged particle passes through a region that could have uniform electric field only or uniform magnetic field only or both electric and magnetic fields or none of the fields. Some possible paths of motion are mentioned in Coliumn-I with the related information in Column-II |
| Answer» Correct Answer - A(q), B(r), C(r, s), D(p, q) | |
| 155. |
Mixed `He^(+)` and `O^(2+)` ions (mass of `He^(+)`=4 amu and that of `O^(2+)=16` amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same thenA. `He^(+)` ions will be deflected more than those of `O^(2+)`B. `He^(+)` ions will be deflected less than those of `O^(2+)`C. All the ions will be deflected equallyD. No ions will be deflected |
| Answer» Correct Answer - C | |
| 156. |
A battery is connected between two points `A and B` on the circumference of a uniform conducting ring of radius `r` and resistance `R` . One of the arcs `AB` of the ring subtends an angle `theta` at the centre . The value of the magnetic induction at the centre due to the current in the ring isA. proportional to `2(180^(@)-theta)`B. Inversely proportional to `r`C. zero only if `theta=180^(@)`D. zero for all values of `theta` |
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Answer» Correct Answer - D `b_(1)=(mu_(0)I)/(2R)xx(2pi-theta)/(2pi),B_(2)=(mu_(0)I_(2))/(2R)xx(theta)/(2pi)` But `I_(1)(rho(Rtheta))/(A)=I_(2)(rhoR(2pi-theta))/(A)` `impliesI_(1)theta=I_(2)(2pi-theta)jimpliesB_(1)=B_(2)` But direction of `vec(B)_(1)` and `vec(B)_(2)` are opposite so net magnetic field at the centre is zero. |
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| 157. |
A particle of mass `m` and charge `q` is lying at the origin in a uniform magnetic field `B` directed along x-axis. At time t = 0, it is given a velocity `v_0`, at an angle` theta` with the y-axis in the xy-plane. Find the coordinates of the particle after one revolution.A. `(0,0,(2pimv_(0)sin theta)/(qB))`B. `((2pimv_(0)sin theta)/(qB),0,0)`C. `((2pimv_(0)sin theta)/(qB),0,4)`D. `(0,0,0)` |
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Answer» Correct Answer - B Pitch `= (T) ( ` Horizontal Velocity ) `=(2pim)/(qB)(V_(0)sint theta)` So `( x,y,z)` co- ordinates are `{(2pi mv_(0)sintheta)/(qB),0,0}` |
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| 158. |
Among the following the true statement is,A. ammeter is a high resistance galvanometer and voltmeter is a low resistance galvanometerB. ammeter is a low resistance galvanometer and voltmeter is a high resistance galvanometerC. ammeter and voltmeter cannot be distinguished on the basis of their resistance.D. ammeter and voltmeter have same resistance. |
| Answer» Correct Answer - B | |
| 159. |
An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion wil beA. Straight line along the x-directionB. A circule in the xz-planeC. A circule in the yz-planeD. A circle in the xy-plane |
| Answer» Correct Answer - B | |
| 160. |
A light body is hanging at the lower end of a vertical spring . On passing current in the spring, the bodyA. rises upB. goes downC. no changeD. oscillates up & down |
| Answer» Correct Answer - 1 | |
| 161. |
Two streams of electrons are moving parallel to each other in the same direction. TheyA. attract each otherB. repel each otherC. cancel the electric field of each otherD. cancel the magnetic field of each other |
| Answer» Correct Answer - 2 | |
| 162. |
A charged particle enters into a uniform magnetic field with velocity `v_(0)` perpendicular to it , the length of magnetic field is `x=sqrt(3)/(2)R`, where `R` is the radius of the circular path of the particle in the field .The magnitude of charge in velocity of the particle when it comes out of the field is A. `2v_(0)`B. `v_(0)//2`C. `sqrt(3)v_(0)//2`D. `v_(0)` |
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Answer» Correct Answer - D `sin theta=(d)/(R)=(sqrt(3))/(2)impliestheta=60^(@)` `vec(v)_(f)=v cos hat(t)+v sin thetahat(j),vec(v)_(i)=vhat(i)` |
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| 163. |
A uniform wire of resistance `12Omega` is bent in the form of a square. A cell of `emf6V` having negligible innternal resistance connected across the diagonal of the square. The magnetic induction at its centre `(` in tesla `)`. |
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Answer» Correct Answer - 1 `B_(1)=(mu_(0)i)/(4pir)(sintheta_(1)+sintheta_(2))` |
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| 164. |
A wire of length `10cm` is bent into an arc of a circle such that it subtends an angle of 1 radian at the centre. If a current of `1A` is passed through the wire, the magnetic induction at the centre of the circle will beA. `2xx10^(-4)` teslaB. `1xx10^(-6)` teslaC. `1xx10^(-4)` teslaD. `2xx10^(-6)` tesla |
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Answer» Correct Answer - 2 `B=(mu_(0)itheta)/(4pir)` |
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| 165. |
Two wires `A` and `B` are of lengths `40cm` and `30cm`. A is bent into a circle of radius `r` and `B` into an arc of radius `r`. A current `i_(1)` is passed through `A` and `i_(20)` through `B`. To have the same magnetic inductions at the centre, the ratio of `i_(1):i_(2)` isA. `3:4`B. `3:5`C. `2:3`D. `4:3` |
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Answer» Correct Answer - 1 `B_(1)=B_(2)` |
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| 166. |
A circular loop of radius 3 cm is having a current of 12.5 A. The magnitude of magnetic field at a distance of 4 cm on its axis isA. `5.65 xx 10^(-5)T`B. `5.27 xx 10^(-5)T`C. `6.54 xx 10^(-5)T`D. `9.20 xx 10^(-5)T` |
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Answer» Correct Answer - A `B=(mu_(0)IR^(2))/(2(R^(2)+x^(2))^(3//2))` Here, I = 12.5 A, R = 3 cm = `3xx10^(-2)m` x = 4 cm = `4xx10^(-2)m` `therefore" "B=(4pixx10^(-7)xx12.5xx(3xx10^(-2))^(2))/(2[(3xx10^(-2))^(2)+(4xx10^(-2))^(2)]^(3//2))=5.65xx10^(-5)T` |
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| 167. |
A semicircular wire of radius `r`, carrying a current `I`, is placed in a magnetic field of magnitude `B`. The force acting on itA. can never be zeroB. can have the maximum magnitude `2Bir`C. can have the maximum magntidue `Bipir`D. can have the maximum magnitude `Bir` |
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Answer» Correct Answer - B To find the Ampere force on a conductor of any shape, replace the conductor by an imaginary straight conductor joining the two ends of the given conductor. |
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| 168. |
What is the force experienced by a semicircular wire of radius R when it is carrying a current l and is placed in a uniform magnetic field of induction B as shown in figure. ? A. ZeroB. `l(2R) B`C. `l(R)B`D. `B(l) pi R` |
| Answer» Correct Answer - A(a) zero , is the correct answer for the above question | |
| 169. |
Which one of the following is not correct about Lorentz Force?A. In presence of electric field `vecE(r)` ang magnetic field `vecB(r)` the force on a moving electric charge is `vecF=q[vecE(r)+vecvxx vecB(r)]`B. The force, due to magnetic field on a negative charge is opposite to that on a positive charge.C. The force due to magnetic field becomes zero if velocity and magnetic field are parallel or anti- parallel.D. For a static charge the magnetic force is maximum. |
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Answer» Correct Answer - D If charge is not moving then the magnetic force is zero. Since `vecF_(m)=q(vecv xxvecB)` As, `vecv=0`, for stationary charge`" "therefore vecF_(m)=0` |
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| 170. |
When a magnetic compass needle is carried nearby to a straight wire carrying current, then I. the straight wire cause a noticeable deflection in the compass needle. II. the alignment of the needle is tangential to an imaginary circle with straight wire as its centre and has a plane perpendicular to the wireA. (I) is correctB. (II) is correctC. both (I) and (II) are correctD. neither (I) nor (II) is correct |
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Answer» Correct Answer - C Both points are correct and these are the result of experments done by Danish physicist Hans Christian Oersted in 1820. |
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| 171. |
A particle having a charge `q=1C` and mass `m=0.5kg` is projected on a rough horizontal plane `(x,y)` from a point `(15m,0,0)` with a initial velocity `vec(v)=20j`. In space a uniform electric field `vec(E)=25(-hat(k))` and magnetic filed `vec(B)=10(-hat(k))`. The acceleration due to gravity `g=10sec^(-2)` and co`-` efficient of friction `mu=0.5` . The particle moves on a spiral path and reaches to the origin . The time taken by particle to reach origin is found `yxx10sec`. Find `y`. |
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Answer» Correct Answer - 2 `m(dv)/(dt)=-mu(mg+qE)` `(mv^(2))/(r)=qvBimpliesr=(mv)/(qB)(dv)/(dt)=(qB)/(m)xx(dr)/(dt)` `int_(0)^(t)dt=-(1)/(mu)" "int_(a)^(0)(qB)/(mg+qE)dr,impliest=20sec` `B=(mu_(0)I)/(4piR)ln(sqrt(2)+1)impliesn=1` |
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| 172. |
An infinitely long wire of uniform cross section carries a current whose current density varies as `j=br^(alpha)`. Where `b=(10^(7))/(pi)`unit,` alpha=.1`unit and `r` the distance from axis. The magnetic induction at a distance `0.25` from axis is given by `(nxx5)/(3)x10^(-2)` tesla. Find `n` |
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Answer» Correct Answer - 5 `I=int(2pix)dx=int_(0)^(r)bx^(alpha)2pixdx=(2pibr^(a+2))/(a+2)` `ointB2pir=mu_(0)(2pibr^(a+2))/(alpha+2)=(mu_(0)br^(a+1))/(alpha+2)impliesn=5` |
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| 173. |
What is the form of magnetic field lines due to a straight current-carrying conductor?A. Straight linesB. EllipticalC. CircularD. Parabolic |
| Answer» Correct Answer - 3 | |
| 174. |
A long straight wire of circular cross- section carries a current along its length. On the axis inside the wire, it follows thatA. strength of electric and magnetic fields are zeroB. strength of electric field is zero but magnetic field is non`-` zeroC. strength of electric and magnetic field non`-`zeroD. strength of electric field is non`-`zero but magnetic field is zero. |
| Answer» Correct Answer - 1 | |
| 175. |
In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of radius `a`. Net magnetic field at centre will be A. `B=(mu_(0)i)/(2pia)tan .(pi)/(n)`B. `B=(mu_(0)ni)/(2pia)tan .(pi)/(n)`C. `B=(2mu_(0)ni)/(pia)tan .(pi)/(n)`D. `B=(mu_(0)ni)/(2a)tan .(pi)/(n)` |
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Answer» Correct Answer - 2 Magnetic field at the centre due to one side `B_(1)=(mu_(0)i)/(2pir)sin theta`, where `r=a cos theta` `:. B_(1)=(mu_(0)i)/(2piacos theta)sin theta=(mu_(0)i)/(2pia)tan theta , :. B_(n e t)=nB_(1)` |
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| 176. |
A long straight wire is kept along x-axis. It carries a current i in the positive x-direction. A proton and an electron are placed at (0, a,0) and (0, -a, 0) respecitively. The proton is imparted an initial velocity v along +z axis and the electron is imparted an initial velocity v along +x-axis. The magnetic forces experienced by the two particles at the instant areA. `(mu_(0)iev)/(2pia) hat(i) (mu_(0)iev)/(2pia) hat(i)`B. `0, (-mu_(0)iev)/(2pia) hat(j)`C. `0, (mu_(0)iev)/(2pia) hat(j)`D. `0, 0` |
| Answer» Correct Answer - B | |
| 177. |
A wire of length l is formed into a circular loop of one turn only and is suspended in a magnetic field B. When a current i is passed through the loop, the maximum torque experienced by it isA. `((1)/(4pi)) Bil`B. `((1)/(4pi)) l^(2)iB`C. `((1)/(4pi)) B^(2)il`D. `((1)/(4pi)) Bi^(2)l` |
| Answer» Correct Answer - B | |
| 178. |
The resistance of an ideal voltmeter isA. zeroB. infinityC. finite, very smallD. finite and large |
| Answer» Correct Answer - 2 | |
| 179. |
A current carrying coil tends to set itselfA. parallel to an external magnetic field.B. parallel to its own magnetic fieldC. perpendicular to the external magnetic field.D. perpendicular to the geographic meridian |
| Answer» Correct Answer - 3 | |
| 180. |
A rectangular coil `20cmxx20cm` has `100` turns and carries a current of `1 A`. It is placed in a uniform magnetic field `B=0.5 T` with the direction of magnetic field parallel to the plane of the coil. The magnitude of the torque required to hold this coil in this position isA. ZeroB. 200 NmC. 2 NmD. 10 Nm |
| Answer» Correct Answer - C | |
| 181. |
The sensitivity of a galvanometer of resistance `990Omega` is increased by `10` times. The shunt used isA. `100 Omega`B. `120 Omega`C. `110Omega`D. `50 Omega` |
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Answer» Correct Answer - 3 `S=(G)/((n-1))` |
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| 182. |
The restoring couple in the moving coil galvanometer is due toA. current in the coilB. magnetic field of the magnet.C. material of the coil.D. twist produced in the suspension wire. |
| Answer» Correct Answer - 4 | |
| 183. |
The restoring couple in the moving coil galvanometer is due toA. Current in the coilB. Mangetic field of the magnetC. Material of the coilD. Twist produced in the suspension wire |
| Answer» Correct Answer - D | |
| 184. |
The area of the coil in a moving coil galvanometer is `15cm^(2)` and has 20 turns. The magnetic induction is `0.2T` and the couple per unit twist of the suspended wire is `10^(-6) Nm` per degree. If the deflection is `45^(@)`, the current passing through it isA. `75xx10^(-4)A`B. `7.5xx10^(-4)A`C. `0.75xx10^(-4)A`D. `750xx10^(-4)A` |
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Answer» Correct Answer - 1 `tau theta=BAIn` |
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| 185. |
A vertical rectangular coil of sides `5 cm xx 2 cm` has 10 turns and carries a current of `2A`. The torque ( couple) on the coil when it is placed in a uniform horizontal magnetic field of `0.1T` with its plane perpendicular to the field isA. `4xx10^(-3)N-m`B. ZeroC. `2xx10^(-3)N-m`D. `10^(-3)N-m` |
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Answer» Correct Answer - 2 `tau=BAIN cos theta` |
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| 186. |
The coil in a MCG has an area of `4cm^(2)` and `500` turns. The intensity of magnetic induction is `2T`. When a current of `10^(-4)A` is passes through it, the deflection is `20^(@)`. The couple per unit twist is `(N-m)`A. `3xx10^(-6)`B. `2xx10^(-6)`C. `4xx10^(-6)`D. `5xx10^(-6)` |
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Answer» Correct Answer - 2 `tau theta=BAIn` |
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| 187. |
Two long conducting rods suspended by means of two insulating threads as shown in Fig. are connected at one end to a charged capacitor through a switch S, which initially open. At the other end , they are connected by a loose wire. The capacitor has charge Q and mass per unit length of the rod is `lambda`. The effective resistance of the circuit after closing the switch is R. Find the velocity of each rod when the capacitor is discharged after closing the switch. (Assume that the displacement of rods during the discharging time is negligible.) A. `(mu_(0)Q^(2))/(4pilambdadRC)`B. `(mu_(0)Q)/(4pilambdadRC)`C. `(mu_(0)Q^(3))/(4pilambdadRC)`D. `(mu_(0)Q^(4))/(4pilambdadRC)` |
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Answer» Correct Answer - A The impulsive force due to ( antiparallel ) discharging current is equal to the change in momentum or `lambdaupsilon=int_(0)^(oo)Fdt=int_(0)^(oo)(mu_(0)I^(2))/(2pid)dt` where `I=` discharging current Now `q=QE^(t//RC)" "implies|I|=(Q)/(RC)e^(-t//RC)` Magnetic force per unit length between two current carrying wires is `(F)/(l)=(mu_(0)I_(2)I_(2))/(2pid)" "=(mu_(0))/(2pid)((Q)/(RC)e^(-t//RC))^(2)` Impulse due to this force `int_(0)^(oo)(mu_(0))/(2pid)(Q^(2))/(R^(2)C^(2))=e^(-2t//RC)dt` Hence `lambdaupsilon=|int_(0)^(oo)(mu_(0))/(2pid)(Q^(2))/(R^(2)C^(2))e^(-2t//RC)dt|=(mu_(0)Q^(2))/(4pidRC)` or `" "upsilon=(mu_(0)Q^(2))/(4pilambda dRC)`. |
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| 188. |
A conducting loop carrying a current `I` is placed in a uniform magnetic field ponting into the plane of the paper as shown. The loop will have a tendency to A. contractB. expandC. move towards positive `x-` axisD. move towards negative `x-` axis |
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Answer» Correct Answer - B Since force on any section of wire will be outward so the loop will have a tendency to expand. |
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| 189. |
A conductor (shown in the figure) carrying constant current `I` is kept in the `x-y` plane in a uniform magnetic field ` vec(B)`. If ` vec(F)` is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are) A. If `vec(B)` is along `hat (z)` , `F prop (L+R)`B. If `vec(B)` is along `hat (x)` ,F = 0`C. If `vec(B)` is along `hat (y)` , `F prop (L+R)`D. If `vec(B)` is along `hat (z)` , `F = 0` |
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Answer» Correct Answer - A::B::C ` vec(F) = I [(int vec( dl ))xx vec(B)]` If ` vec(B)` is along `z` then ` vec(F)` = I[(2L+ 2R)hat(i) xxB hat (x)]` option ` [A]` is correct If ` vec(B)` is along ` vec(x)` then `vec(F) = I[(2L+ 2R)hat(i) xxB hat (i)] = 0` If `vec(B)` is along ` vec(y)` then ` vec(F) = I[(2L+ 2R)hat(i) xx hat (j)]` Option `(b) and (c)` are also correct |
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| 190. |
Consider two identical galvanometers and two identical resistors with resistance `R`. If the internal resistance of the galvanometers `R_(c) lt R//2`, which of the following statement(s) about any one of the galvanometers is (are) true?A. The maximum voltage range is obtained when the two resistors and one galvanometer are connected in seriesB. The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series , and the second galvanometer is connected in parallel to the first galvanometerC. The maximum current range is obtained when all the components are connected in parallelD. The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors |
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Answer» Correct Answer - A::C The range of voltmeter `V` is given by the experession ` V = I_(g) [R_(c) + (R_(c)+R+R)]` `V` is max inthis case as `RHS` is maximum. Thus (a) is correct. The range of ammeter `I` is given by the expression ` I = (I_(g) R_(c))/(R_(eq))+I_(g)` Where ` (1)/(R_(eq)) = (1)/(R_(c))+(1)/(R)` Here `R_(eq)` is minimum and therefore `I~ is maximum. Thus(c) is the correct option. |
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| 191. |
Which of the following properties of magnetic line of force differs them from electrostatic lines of force ?A. These are closed curvesB. These can emit or teminate at any angle to the surfaceC. Total flux linked with a close surface is always zeroD. All of these |
| Answer» Correct Answer - D | |
| 192. |
Statement-1 Electrostatic field lines are discontinuous at the surface of a conductor Statement-2 Magnetic field lines produced by permanent magnets are never discontinuousA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-9B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-9C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 193. |
Statement-1 Two parallel wire carrying current in same direction attract each other, whereas two proton beams moving parallel to each other repel each other Statement-2 Wire carrying current is electrically neutral, therefore it is experiencing only magnetic attraction while the electric force of repulsionn between protons is more than the magnetic attractionA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-8B. Statement-1 is True, Statement-2, is True, Statement-2 is NOT a correct explanation for Statement-8C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 194. |
Three very long straight thin wires are connected parallel to each other through a battary of negligible internal resistance. The resistance of the wired are `2Omega,3 Omega` and `4 Omega`. The ratio of distance of middle wire from the first and thrid wires if resusltant magnetic force on the middle wire is zero is `:`A. `2:1`B. `3:4`C. `2:3`D. `3:5` |
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Answer» Correct Answer - 1 As the wires are parallel, `I_(1):I_(2):I_(3)=(1)/(2):(1)/(3):(1)/(4)=6:4:3` Force betweent first second & second, their wires are equal `(F)/(l)=(mu_(0)I_(1)I_(2))/(2pir)` |
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| 195. |
A current of `1A` is flowing through a straight conductor of length `16cm`. The magnetic induction `(` in tesla `)` at a point `10cm` from the either end the wire is `:`A. `(8)/(3)xx10^(-6)`B. `(1)/(6sqrt(2))xx10^(-5)`C. `(1)/(6sqrt(3))xx10^(-6)`D. `(sqrt(3))/(6)xx10^(-6)` |
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Answer» Correct Answer - 1 `B=(mu_(0)I)/(4pir)(sinphi_(1)+sinphi_(2))` |
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| 196. |
Two long parallel wires are separated by a distance of 2m. They carry a current of `1A` each in opposite direction. The magnetic induction at the midpoint of a straight line connecting these two wires isA. zeroB. `2xx10^(-7)T`C. `4xx10^(-5)T`D. `4xx10^(-7)T` |
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Answer» Correct Answer - 4 `B=B_(1)+B_(2)` |
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| 197. |
A moving coil galvanometer `A` has 200 turns and resistance `100 Omega`. Another meter `B` has 100 turns and resistance `40 Omega`. All the other quantities are same in both the cases. The current sensitivity ofA. `B` is double as that of `A`B. `A` is 2 times of `B`C. `A` is 5 times of `B`D. `B` is 5 times of `A`. |
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Answer» Correct Answer - B `(theta)/(i)=(NAB)/(C)` |
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| 198. |
If the direction of the initial velocity of a charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will beA. a straight lineB. an ellipseC. a circleD. a helix |
| Answer» Correct Answer - 4 | |
| 199. |
A TG has 50 turns each of diameter `15 cms`. When a certain current is passed through its coil, the deflection of the needle is `45^(@)`. If `B_H)=4 pi muT` the value of current isA. `77.44mA`B. `44.77mA`C. `74.74mA`D. `60 mA` |
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Answer» Correct Answer - 4 `I=((2r)B_(H))/(mu_(0)n).tan theta` |
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| 200. |
A vertical straight conductor carries a current vertically upwards. A point `P` lies to the east of it at a small distance and another point `Q` lies to the west at the same distance. The magnetic field at `P` isA. Greater than at `Q`B. Same as at `Q`C. Less than at `Q`D. Greater or less than at `Q` depending upon the magnetic field of the current. |
| Answer» Correct Answer - 1 | |