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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.
| 301. |
Two streams of electrons are moving parallel to each other in the same direction. TheyA. Do not exert any force on each otherB. Repel each otherC. Attract each otherD. Get rotated to be perpendicular to each other |
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Answer» Correct Answer - B |
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| 302. |
Current I si following along the path `ABCD`, along the four edges of the cube (figure-a) creates a magnetic field in the centre at the center of the cube current I flowing along the path of the six edges `ABCDHEA` . A. `sqrt((3)/(2)) B_(0)` Towards corner `G` .B. `sqrt3 B_(0)` Towards corner `E`C. `sqrt((3)/(2)) B_(0)` Towards conrner `H` .D. `sqrt3 B_(0)` Towards corner `F` . |
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Answer» Correct Answer - D Each edge share current for two loops so cur rent for one loop is `(1)/(2)` and its field at centre of loop is `(B_(0))/(2)` There are sin loops `oversetrarr(B) =((B_(0))/(2) + (B_(0))/(2))hatk` `oversetrarrB = B_(0) hati + B_(0) hatj +B_(0hat)hatk` `B =sqrt3 B_(0)` towards `F` . |
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| 303. |
The magnetic flux through each of give faces of a neutral playing dice is given `phi_(B)=+-NWb, where N(=1 t o 5)` is the number of spots on the face. The flux is positive (out -ward) for `N` even and negative (inward) for `N` odd What is the flux thogh sixth face of the die ?A. `3Wb`B. `4Wb`C. `-3Wb`D. `-4 Wb` |
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Answer» Correct Answer - A `phi_("net") = 0` `phi_(1) + phi_(2) + phi_(3) + phi_(4) + phi_(5) + phi_(6) =0` `-1 + 2 -3 + 4 -5 + phi_(6) = 0` . |
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| 304. |
In the following hexagons made up of two different `P` and `Q` current enters and leaves from points `X` and `Y` respectively In which case the magnetic field at its centre is not zero .A. B. C. D. |
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Answer» Correct Answer - A Because current will be different in fifferent branch . |
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| 305. |
Consider three quantities (`x= E/B, y= (sqrt(1/((mu_0)(epsilon_0)))) and z=(l/(CR))`. Here, `l` is the length of a wire, C is a capacitance and R is a resistance. All other symbols have standard meanings.A. `x,y` have the same dimensionsB. `y,z` have the same dimensionsC. `z,x` have the same dimensionsD. none of the three pairs have the same dimen sions . |
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Answer» Correct Answer - A,B,C Because all will have diversion of velocity `LT^(-1)` . |
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| 306. |
Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment isA. `2:pi`B. `pi:2`C. `pi:4`D. `4:pi` |
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Answer» Correct Answer - C |
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| 307. |
When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved, the magnetic field at its centre will becomeA. Four timesB. Same as at `Q`C. HalfD. Double |
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Answer» Correct Answer - B |
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| 308. |
A proton of energy `200 MeV` enters the magnetic field of `5 T`. If direction of field is from south to north and motion is upward, the force acting on it will beA. ZeroB. `1.6xx10^(-10)N`C. `3.2xx10^(-8)N`D. `1.6xx10^(-6)N` |
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Answer» Correct Answer - B |
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| 309. |
The sensitiveness of a moving coil galvanometer can be increased by decreasingA. The number of turns in the coilB. The area of the coilC. The magnetic fieldD. The couple per unit twist of the suspension |
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Answer» Correct Answer - D |
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| 310. |
A long wire `AB` is placed on a table. Another wire `PQ` of mass `1.0 g` and length `50cm` is set to slide on two rails `PS` and `QR`. A current of `50 A` is passed through the wires. At what distance above `AB`, will the wire `PQ` be in equilibrium? A. `25 mm`B. `50 mm`C. `75 mm`D. `100mm` |
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Answer» Correct Answer - A |
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| 311. |
Comment on the state of magnetisation of a substance whose atoms contain odd number of electrons. |
| Answer» An atom which contains odd number of electrons will have atleast one electron left unpaired. So it has some permanent magnetic dipole moment. Hence the atom must have a paramagnetic character. | |
| 312. |
Show that the magnetic field along the axis of a current carrying coil of radius r at a distance x from the centre of the coil is smaller by the fraction `3x^2//2r^2` than the field at the centre of the coil carrying current. |
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Answer» The magnetic field induction, at the centre of circular coil of n turns, radius r, carrying current I is `B_0=(mu_0)/(4pi)(2pinI)/(r)=(mu_0nI)/(2r)` Magnetic field induction at a point on the axis of the circular coil carrying current is given by `B_0=(mu_0)/(4pi)(2pinIr^2)/((r^2+x^2)^(3//2))=(mu_0nIr^2)/(2(r^2+x^2)^(3//2))` `:. B/B_0=(r^3)/((r^2+x^2)^(3//2))=r^3/r^3[1+x^2/r^2]^(-3//2)=[1-3/2x^2/r^2+...]=[1-3/2x^2/r^2]` Fractional decrease in magnetic field `=(B_0-B)/(B_0)=1-B/B_0=1-[1-3/2x^2/r^2]=(3x^2)/(2r^2)` |
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| 313. |
A small cylindrical soft iron piece is kept in a galvanometer so thatA. A radial uniform magnetic field is producedB. A uniform magnetic field is producedC. There is a steady deflection of the coilD. All of these |
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Answer» Correct Answer - D |
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| 314. |
S is the surface of a lump of magnetic material.A. Lines of `vecB` are necessarily continuous across SB. Some lines of `vecB` must be discontinuous across sC. Lines of `vecH` are necessarily continuous across SD. Lines of `vecH` can not be continuous across S |
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Answer» Correct Answer - A::D The lines of magnetic field induction `vecB` are necessarily continuous across the surfaces S of a lump of magnetic material. Outside the lump of magnetic material, `H=B//mu_0` and inside the lump of magnetic material, `H=B//mu_0mu_r` where `mu_r` is the relative permeability of material. Thus the lines of H cannot all be continuous across surface S. |
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| 315. |
A permanent magnet in the shape of a thin cylinder of length `10cm` has `M=10^6A//m`. Calculate the magnetisation current `I_M`. (Here M is the intensity of magnetisation). |
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Answer» Here, `l=10cm=0*1m, M=10^6A//m` As `M=(I_M("magnetisation current"))/(l("length")) :. I_M=Mxxl=10^6xx0*1=10^5A` |
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| 316. |
A 250-turns recantagular coil of length 2.1 cm and width 1.25 cm carries a current of `85 muA` and subjected to magnetic field of strength `0.85 T`. Work done for rotating the coil by `180^(@)` against the torque isA. `9.1muJ`B. `4.55muJ`C. `2.3muJ`D. 1.5`mu`J |
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Answer» Correct Answer - a |
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| 317. |
Two protons parallel to each other, keeping distane r between them, both moving with same velocity `vecV` Then the ratio of the electric is magnetic force of interaction between them is .A. `c^(2)//V^(2)`B. `2c^(2)//V^(2)`C. `c^(2)//2V^(2)`D. None |
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Answer» Correct Answer - A `F_(e) = (Kq^(2))/(r^(2)) = (1)/(4pi in_(0)) (q^(2))/(r^(2))` `F_(m) = (mu_(0))/(4pi) (q_(1)q_(2)V^(2))/(r^(2)) = (mu_(0))/(4pi) (q^(2)V^(2))/(r^(2))` `(F_(e))/(F_(m)) = (1)/(mu_(0) in_(0)) . (1)/(V^(2)) = (C^(2))/(V^(2))` . |
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| 318. |
What is the magnitude of the magnetic force per unit length on a wire carrying a current of 8 A making an angle of `30^@` with the direction of a uniform magnetic field of `0*15T`? |
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Answer» Here, `I=8A`, `theta=30^@`, `B=0*15T`, `F=?`, `l=1m`. `F=Bilsintheta=0*15xx8xx1xxsin30^@=0*15xx8xx1xx(1//2)=0*6Nm^-1` |
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| 319. |
What information would you wish to have about the galvanometer before you convert the galvanometer into an ammeter or voltmeter? |
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Answer» Before converting the galvanometer into an ammeter or voltmeter we require the following two informations. (i) The resistance of galvanometer and (ii) the current for full scale deflection of galvanometer. |
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| 320. |
Which one has lowest resistance, ammeter, voltmeter and galvanometer? Explain. |
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Answer» Ammeter has the lowest resistance. A galvanometer is converted into an ammeter by using a suitable low resistance shunt in parallel with the galvanometer. The effective resistance of the network when resistors are connected in parallel becomes less than the individual resistance. Hence resistance of ammeter is less than that of galvanometer. Voltmeter is formbed by connecting a suitable high resistance in series with the galvanometer. Since the effective resistance of the net-work increases when the resistors are connected in series, hence the resistance of voltmeter is more than that of galvanometer. |
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| 321. |
The figure shows a circular loop of radius `a` with two long parallel wires `(numbered 1 and 2)` all in the plane of the paper . The distance of each wire from the centre of the loop is `d`. The loop and the wire are carrying the same current `I`. The current in the loop is in the counterclockwise direction if seen from above. (q) The magnetic fields(B) at `P` due to the currents in the wires are in opposite directions. (r) There is no magnetic field at `P`. (s) The wires repel each other. (5) Consider `dgtgta`, and the loop is rotated about its diameter parallel to the wires by `30^(@)` from the position shown in the figure. If the currents in the wire are in the opposite directions, the torque on the loop at its new position will be ( assume that the net field due to the wires is constant over the loop).A. `(mu_0I^2a^2)/(d)`B. `(mu_0I^2a^2)/(2d)`C. `(sqrt3mu_0I^2a^2)/(d)`D. `(sqrt3mu_0I^2a^2)/(2d)` |
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Answer» Correct Answer - B When `d gt gt a`, then magnetic field at O due to currents in wires 1 and 2 will be `B=2xx(mu_0)/(4pi)(2I)/(d)=(mu_0I)/(pid)` Torque on the loop `=BIAsin30^@` `=(mu_0I)/(pid)xxIxx(pia^2)xx1/2=(mu_0I^2a^2)/(2d)` |
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| 322. |
A current of `5A` is flowing through a 10 turns circular coil of radius `7cm`. The coil lies in x-y plane. What is the magnitude and direction of dipole moment associated with it? If this coil were placed in a uniform external magnetic field directed along the x-axis, in which plane would be coil lie in equilibrium? `pi=22//7`. |
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Answer» Here, `I=5A, N=10, r=7xx10^-2m` `M=NIA=NI(pir^2)` `=10xx5xx22/7(7xx10^-2)^2=0*77Am^2` The direction of `vecM` is perpendicular to the plane of the coil, along Z-axis. Torque on the current loop, `tau=MB sin theta` where `theta` is angle between `vecM` and `vecB`. For stable equilibrium, `tau=0 :. theta=0^@`. Hence `vecB` should be perpendicular to plane of the coil. Hence in the condition of stable equilibrium, coil will be in Y-Z plane. |
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| 323. |
Magnetic field due to `0.1 A` current flowing through a circular coil of radius `0.1`m and `1000` turns at the centre of the coil isA. `2xx10^(-1)T`B. `4.31xx10^(-2)T`C. `6.28xx10^(-4)T`D. `9.81xx10^(-4)T` |
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Answer» Correct Answer - C |
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| 324. |
If a current is passed through a spring then the spring willA. Gets compressedB. Gets expandedC. OscillatesD. Remains unchanged |
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Answer» Correct Answer - A |
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| 325. |
A circular coil having `N` turns is made from a wire of length `L` meter. If a current `I` ampere is passed through it and is placed in amagnetic field of `B` Tesla, the maximum torque on it isA. Directly proportional to `N`B. Inversely proportional to `N`C. Inversely proportional to `N^(2)`D. Independent of `N` |
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Answer» Correct Answer - A |
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| 326. |
A metallic loop is placed in a magnetic field. If a current is passed through it, thenA. The ring will feel a force of attractionB. The ring will feel a force of repulsionC. It will move to and fro about its centre of gravityD. None of these |
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Answer» Correct Answer - D |
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| 327. |
A circular coil of radius 4 cm has 50 turns. In this coil a current of `2 A` is flowing. It is placed in a magnetic field of `0.1 "weber"//m^(2)`. The amount of work done is rotation it through `180 ^(@)` from its equilibrium position will beA. `0.1J`B. `0.2J`C. `0.4J`D. `0.8J` |
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Answer» Correct Answer - A |
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| 328. |
Magnetic dipole moment of rectangular loop isA. Inversely proportional to current in loopB. Inversely proportional to area of loopC. Parallel to plane of loop and proportional to area of loopD. Perpendicular to plane of loop and proportional to area of loop |
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Answer» Correct Answer - D |
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| 329. |
If `m` is magnetic moment and `B` is the magnetic field, then the torque is given byA. `vecm.vecB`B. `(|vecm|)/(|vecB|)`C. `vecmxxvecB`D. `|vecm|.|vecB|` |
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Answer» Correct Answer - C |
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| 330. |
In the circuit shown in figure, the current is to measured. What is the value of the current if the ammeter shown (i) is a galvanometer with a resistance `R_G=60*00Omega`. (ii) is a galvanometer described in (i) but converted to an ammeter by a shunt resistance `r_s=0*02Omega`, (iii) is an ideal ammeter with zero resistance? |
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Answer» Let G be the resistance of ammeter (i.e. galvanometer). Then current in the circuit, `I=(epsilon)/(R+G)=(3*00)/(3*00+60*00)=0*048A`. (ii) When the ammeter (i.e., galvanometer) is shunted with resistance S, its effective resistance, `R_p=(GS)/(G+S)=(60xx0*02)/(60+0*02)~~0*02Omega` Current in the circuit, `I=(epsilon)/(R+R_p)=(3*00)/(3*00+*02)~~0*99A`. (iii) For the ideal ammeter with zero resistance current, `I=(3*00)/(3*00)=1*00A`. |
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| 331. |
The magnetic induction at the centre of a current carrying circular of coil radius `r`, isA. Directly proportional to `r`B. Inversely proportional `r`C. Directly proportional to `r`D. Inversely proportional to `r` |
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Answer» Correct Answer - B |
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| 332. |
A proton and an `alpha-`particle enter a uniform magnetic field moving with the same speed. If the proton takes `25 mu s` to make 5 revolutions, then the periodic time for the `alpha-` particle would beA. `50 mu sec`B. `25 mu sec`C. `10 mu sec`D. `5 mu sec` |
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Answer» Correct Answer - C |
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| 333. |
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 N-m`C. `2 N-m`D. `10 N-m` |
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Answer» Correct Answer - C |
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| 334. |
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. Moves towards `+vex`-axisD. Move towards `-vex`- axis |
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Answer» Correct Answer - B |
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| 335. |
The magnetic field B and the magnetic intensity H in a material are found to be `1*6T` and `1000Am^-1`, respectively. Calculate the relative permeability `mu_r` and susceptibility of the material. What is the nature of the material? |
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Answer» Here, `B=1*6T`, `H=1000Am^-1` `mu_r=B/H=(1*6)/(1000)=0*0016` From `1+chi_m=mu_r` `chi_m=mu_r-1=0*0016-1=-0*9984` As susceptibility is negative and less than one, the material must be diamagnetic in nature. |
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| 336. |
A small bar magnet having magnetic moment of `9xx10^9Wb-m` is suspended at its centre of gravity by a light torsionless string at a distance of `1m` vertically above a long straight horizontally wire carrying a current of `1A`. Find the frequency of oscillation of the magnet about its equilibrium position, assuming that the motion is undamped. The moment of inertia of the magnet is `6xx10^-9kgm^2`. |
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Answer» Here, `M=9xx10^-9Wb-m` `r=1cm=10^-2m`, `i=1A v=?` `I=6xx10^-9kgm^2` Magnetic field strength at `1cm` from the wire carrying current `B=(mu_0)/(4pi)(2i)/(r)` The torque acting on the small magnet in this external magnetic field. `tau=-MB sin theta=-MB theta`, when `theta` is small. If `alpha` is angular acceleration produced by the torque, then `tau=lalpha=-MB theta=-M((mu_0)/(4pi)(2i)/(r))theta` `alpha=-(mu_0)/(4pi)(2iM)/(Ir)theta=-omega^2theta` where `omega^2=(mu_0)/(4pi)(2iM)/(Ir)` `:.` Motion of magnet is simple harmonic and its frequency is `n=(omega)/(2pi)=(1)/(2pi)sqrt((mu_0)/(4pi)(2iM)/(Ir))` `=(1)/(2xx3.14)sqrt((10^-7xx2xx1.0xx9xx10^-9)/(6xx10^-9xx10^-2))` `n=8*7xx10^-4Hz` |
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| 337. |
When is potential energy of a magnetic dipole minimum? |
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Answer» P.E. is minimum, when `theta=0^@`, i.e., dipole is aligned along the field. Minimum `P.E.=-MB cos 0^@` `=-MBrarr minim um`. |
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| 338. |
What is magnetic inclination at a place? |
| Answer» Magnetic inclination at a place is the angle of dip at that place. | |
| 339. |
A magnetic needle placed on a piece of cork is floating on the calm surface of a lake in the northern hemisphere. Would the magnetic needle alongwith the cork move towards north? |
| Answer» No, the magnetic needle will come to rest along N-S direction. | |
| 340. |
When an ammeter is put in series of a circuit, does it record slightly less or more than the actual current in the circuit ? Explain. |
| Answer» When ammeter is put in series of a circuit, it slightly increases the resistance of circuit. Due to it, the current in circuit decreases. As a result of it ammeter records a slightly less current than actual. | |
| 341. |
Two substances A and B have relative permeabilities slightly greater and less than unity respectively. What is their magnetic nature? |
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Answer» `mu_r=1+chi` or `chi=mu_r-1` For A, `mu_rgt1` `:. chi` is small and positive. It must be paramagnetic. For B, `mu_rlt1` `:. chi` is small and negative. Hence, B must be diamagnetic. |
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| 342. |
Susceptibility of iron is more than that of copper. What does this statement imply? |
| Answer» This statement implies that iron can be magnetised more easily than copper. | |
| 343. |
How does a magnetic compass behave at a neutral point? |
| Answer» At a neutral point, net magnetic field due to magnet and earth is zero. Therefore, a compass needle may point out in any random direction. | |
| 344. |
How does a current loop behave like a bar magnet? |
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Answer» A current loop behaves as a bar magnet because (i) one face of current loop behaves as a south pole and the other face as north pole. (ii) it possesses a magnetic dipole moment `(M=IA)` and (iii) it experiences a torque in an external magnetic field, which tends to align the axis of the loop along the direction of magnetic field as bar magnet does. |
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| 345. |
Why is phosphor bronze alloy preferred for the suspension wire of a moving coil galvanometer? |
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Answer» The suspension wire of phosphor bronze alloy is preferred in moving coil galvanometer because it has several advantages: (i) Its restoring torque per unit twist is small. Due to it, the galvanometer is very sensitive. (ii) It has great tensile strength so that even if it is thin, it will not break under the weight of the coil suspended from its end. (iii) It is rust resisting. Hence it remains unaffected by the weather conditions of air in which it is suspended. |
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| 346. |
State properties of the material of the wire used for suspension of the coil in a moving coil galvanometer. |
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Answer» The properties of the material of the wire used for suspension of the coil in a moving coil galvanometer are as follows: 1. It should have low torsional constant, i.e., restoring torque per unit twist should be small. 2. It should have high tensile strength. 3. It should be a non-magnetic substance. 4. It should have a low temperature coefficient of resistance. 5. It should be a good conductor of electricity. |
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| 347. |
Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Justify this statement. |
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Answer» Current sensitivity of a galvanometer is given by `I_s=(nBA)/(k)` where n is the no. of turns in the coil, A is the area of each turn of the coil, B is the strength of uniform magnetic field and k is the restoring torque per unit twist of suspension fibre in galvanometer. Voltage sensitivity `V_s=(I_s)/(R)=(nBA)/(kR)` `I_s` can be increased by increasing n and A. But increase of n increases the resistance R of galvanometer. Due to it, `V_s` will decrease. A can not be increased beyond a limit because in that case the entire coil will not be in a uniform magnetic field. Moreover it will make the galvanometer bulky and unmanageable. |
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| 348. |
Explain the action of shunt? |
| Answer» The shunt is a low resistance connected in parallel to a galvanometer to protect it from the strong currents in the circuit. The galvanometer shows full scale deflection with a very small current. If the whole of the strong current in an electric circuit passes through galvanometer, it may get damaged. The shunt allows only a small part of the current to flow through the galvanometer by acting as a by pass to the major part of the current. | |
| 349. |
Three long straight wires `A`, `B` and `C` are carrying current as shown in figure. Then the resultant force on `B` is directed A. Towards AB. Towards CC. Perpendicular to the plane of paper and outwardD. Perpendicular to the plane of paper and inward |
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Answer» Correct Answer - B |
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| 350. |
Assertion: Magnetic moment is a vector quantity, whose direction inside the magnet is, from South to North. Reason: Magnetic lines of force emanate from N-pole and enter into the S-pole.A. both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - A Both, the assertion and reason are true and the latter is correct explanation of the former. |
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