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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. |
Assertion" When diamagnetic material is placed in a non-uniform magnetic Held, it tends to move from stronger to the weaker part of the magnetic field. Reason: Diamagnetic materials possess strong magnetism.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is ture but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C Electrons in an atom orbiting around nucleus possess orbital angular momentum. these electrons are equivalent to current-carrying loop an thus possess orbital magnetic moment. Diamagnetlc substances are the ones in which resultant magnetic moment in an atom is zero. When magnetic field is applied, those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up. Thus, the substance develops a net magnetic moment in direction opposite to that of the applied field. |
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| 152. |
The magnetic susceptibility of a metrial of a rod is 299. Permeatbility of vaccum `mu_(0)`A. `3771xx10^(-7)Hm^(-1)`B. `3771xx10^(-5)Hm^(-1)`C. `3770xx10^(-6)Hm^(-1)`D. `3771xx10^(-8)Hm^(-1)` |
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Answer» Correct Answer - A `"The given" chi_(m)=299` `and " " mu_(0)=4pixx10^(-7)Hm,mu=?` `"We know that", mu=mu_(0)(1+chi_(m))` `therefore " " 4pixx10^(-7)(1+299) Rightarrow =4xx(22)/(7)xx10^(-7)xx300` `" " =(2.6400)/(7)xx10^(-7)=3771.4xx10^(-7)Hm^(-1)or ` `" "cong3771xx10^(-7)Hm^(-1)` |
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| 153. |
A compass needle placed at a distance r from a short magnet in tan A position showns a deflection of `60^(@)`. If the distance is increased to `r(3)^(1//3)`, then the deflection of the compass needle is:A. `30^(@)`B. `60^(@)xx(3)^(1//3)`C. `60^(@)xx(3)^(2//3)`D. `60^(@)xx(3)^(3//3)` |
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Answer» Correct Answer - A We know that for `tan A` position `mu_(0)/(4pi)(2M)/r^(3)=H tan theta` `:. Tan theta prop 1/r^(3)implies(tantheta_(2))/(tantheta_(1))=(r/(r(3)^(1/3)))^(3)=1/3` `:. tantheta_(2)=(tantheta_(1))/3=(tan 60^(@))/3=sqrt(3)/3=1/sqrt(3)` |
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| 154. |
Two tangent galvanometer have redii `7.5 cm` and `10 cm`, number of turns are 15 and 10 and resistances are `8 Omega` and `12 Omega`. They are joined in parallel in circuit. If deflection in one is `60^(@)` the deflection in second galvanometer is :A. `45^(@)`B. `30^(@)`C. `40^(@)`D. `35^(@)` |
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Answer» Correct Answer - B `r_(1)=7.5 cm, r_(2)=10cm` `n_(1)=15, n_(2)=10` `R_(1)=8, R_(2)=12` `theta_(1)=60^(@), theta_(2)=?` `m_(0)/2n_(1)/r_(1)xxI_(1)=tantheta_(1)` `mu_(0)/2n_(2)/r_(2)xxI_(2)=tantheta_(2)` `:. (tantheta_(2))/(tantheta_(1))=n_(2)/n_(1)xxr_(1)/r_(2)xxI_(2)/I_(1)` But `I_(1)=R_(2)/(R_(1)+R_(2))xxV/R_(p)` `I_(2)=R_(2)/(R_(1)+R_(2))xxV/R_(p)` `I_(2)/I_(1)=R_(1)/R_(2)` From equation (iii), `(tantheta_(2))/(tantheta_(1))=10/15xx75/10xx8/12=1/3` `:. tantheta_(2)=sqrt(3)/3=1/sqrt(3)` `:. theta_(2)=30^(@)` |
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| 155. |
In hydrogen atom , the electron moves in an orbit of radius 0.5 Å , making `10^(16)` cps . Calculate magnetic moment associated with the orbital motion of electron ? |
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Answer» M = `ixxA` =` q xx f xx pir^(2)` = `1.6 xx 10^(-19) xx 10^(16) xx 3.14 xx 25 xx 10^(-22)` `= 1.6 x 3.14 xx 25 xx 10^(-25)` `= 125. 6 xx 10^(-25) A m^(2)` `= 1.26 xx 10^(-23) A m^(2)` |
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| 156. |
Two tangent galvanometer A and B are identical except in their number of turns. They are connected in series. On passing a current through them, deflections of `60^(@) and 30^(@)` are produced. The ration of the number of turns in A and B isA. `1:3`B. `3:1`C. `1:2`D. `2:1` |
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Answer» As the current and the other factors are sence same for both the galvonometers, so `N alpha tan theta Rightarrow (N_(1))/(N_(2))=(tan 60^(@))/(tan 30^(@)) or (N_(1))/(N_(2))=(sqrt3)/(1//sqrt3)=3:1` |
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| 157. |
Assertion: Magnetic moment of helium atom is zero. Reason: All the electron are electron are paired in helium atom orbitals.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A Helium atom has paired electrons so their electron spin are opposite to each other and hence its net magnetic moment is zero. |
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| 158. |
Assertion: Reduction factor (K) of a tangent galvanometer helps in reduction to current. Reason: Reduction factor increases with increase of current.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - C The reduction factor of tangent galvanometer is `K=B_(H)/G=B_(H)xx(2r)/(nmu_(0))` Thus reduction factor of a tangent galvanometer depends upon the geometry of its coil. It increases with increase of radius and decreases with increase in number of turn of the coil of the galvanometer. |
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| 159. |
Assertion: The permeability of a ferromagnetic material is independent of the magnetic field. Reason: Permeability of a material is a constant quantity.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D The permeability of a ferromagnetic materialis not independent of magnetic field, `vec(B)=K_(m)vec(B_(0))`. Is applied field. The total magnetic field vec(B)inside a ferromagnet may be `10^(3)` or `10^(4)` times the applied field `B_(0)`. The permeability `K_(m)` of a ferromagnetic material is not constant, neither the field `vec(B)` nor the magnetizatio `vec(M)` increases linearly with `vec(B)`, even at small value of `B_(0)`. From the hysteresis curve, magnetic permeability is greater for lower field. |
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| 160. |
Is the permeability of a ferromagnetic material independent of the magnetic field ? If not, is it more for lower or higher fields ? |
| Answer» No, permeability of a ferromagnetic material is not independent of magnetic field. As a clear from the hysteresis curve, `mu` is greater for lower fields. | |
| 161. |
Assertion: When a magnet is brought near iron nails, only translatory force act on it. Reason: The field due to magnet is generally uniformA. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D In general, the field due to a magnet is non-uniform. Therefore, it exerts both, a net force and a torque on the nails which will translate and also rotate the nails before striking to north pole of magnet with their induced south poles and vice-versa. |
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| 162. |
Assertion: A paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled. Reason: The magnetisation does not depend on temperature.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - C A paramagnetic sample display greater magnetisation when cooled, this is because at lower temperature, the tendency to disrupt the alignment of dipoles (due to magnetising field) decreases on account of reduced random thermal motion. |
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| 163. |
A small bar magnet has a magnetic moment `1.2 A-m^(2)`. The magnetic field at a distance `0.1 m` on its axis will be: `(mu_(0)=4pixx10^(-7)T-m//A)`A. `1.2xx10^(-4)T`B. `2.4xx10^(-4)T`C. `2.4xx10^(4)T`D. `1.2xx10^(4)T` |
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Answer» Correct Answer - B `B=mu_(0)/(4pi).(2M)/d^(3) implies B=10^(-7)xx(2xx1.2)/(0.1)^(3)=2.4xx10^(-4)T` |
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| 164. |
The torque required to hold a small circular coil of 10 turns, area `1mm^(2)` and carrying a current of `((21)/(44))A` in the middle of a long solenoid of `10 ^(3)"turns"//"m"` carrying a current of 2.5A, with its axis perpendicular to the axis of the solenoid is A. `1.5xx10^(-6)Nm`B. `1.5xx10^(-8)Nm`C. `1.5xx10^(6)Nm`D. `1.5xx10^(8)Nm` |
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Answer» Correct Answer - B Here, For small circular coil, Number of turns, `N` = 10, Area , `A`=1 `"mm"^(2)` = `1xx10^(-6)" m"^(2)` Current, ` I_(1)=(21)/(44)A ` For a long solenoid, Number of turns per metre, `n=10^(3)` per m Current, `I_(2)=2.5` A Magnetic field due to a long solenoid on its axis is `B=mu_(0)nI_(2)" "...(i)` Magnetic moment of a circular coil is ` M= NAI_(1)" "...(ii)` ` vec(tau)=vec(M) xx vec(B) ` ` tau=MBsintheta=MB" "(because theta=90^(@)("Given"))` `tau=(NA I_(1))(mu_(0)nI_(2))" "("Using"(i)and (ii)) ` ` tau=10xx1xx10^(-6) xx(21)/(44 )xx4xx(22)/(7)xx10^(-7)xx10^(3)xx2.5` `=1.5xx10^(-8)" N m"` |
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| 165. |
A circular coil of 25 turns and radius of 12 cm is placed in a uniform magnetic field of 0.5 T normal to the plane of coil. If the current in the coil is 5 A, then total torque experienced by the coil isA. 1.5NmB. 2.5NmC. 3.5NmD. zero |
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Answer» Correct Answer - D Here, n=25turns, r=12cm, B=0.5T Since the coil is placed in uniform magnetic field normal to the plane of the coil. Hence the anlge between magnetic moment and magnetic field direction is zero `(i.e. theta=0)` `therefore pi=mB sin theta=mB sin theta` `Rightarrow tau=0` |
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| 166. |
The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = `0.5 Am^(2))`A. 0.07JB. 0.08JC. 0.09JD. 0.1J |
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Answer» Correct Answer - C Since the most stable position is at `theta=0` and the most unstable position is at `theta=180^(@)`, then the work done is given by `w=underset(theta=0^(@))overset(theta=180^(@))inttau(theta)d theta=underset(theta=0^(@))overset(theta=180^(@))int mB sin theta d theta=-mB[cos theta]_(0)^(180^(@))` `=-mB[cos 180^(@)-cos 0^(@)]=-mB[-1-1]` `=-mB[-2]=2mB` `therefore W=2xx0.50xx0.09=0.09J` |
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| 167. |
What is work done on dipole to rotate it form stable equilibrium position `(theta_(1) = 0^(@))` to unstable equilibrium `(theta_(2) = 180^(@))` ? |
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Answer» w = MB `(cos theta_(1) - cos theta_(2))` `= MB [cos 0^(@) - cos 180^(@)]` =` MB[1-(-1)]` w = 2 MB |
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| 168. |
A straight solenoid of length 50 cm has 1000 turns per metre and mean cross-sectional area of `2 xx 10^(-4) m^(2)` . It is placed with its axis of `30^(@)` with uniform magnetic field of 0.32 T . Find the torque acting on the solenoid when a current of 2 ampere is passed through it . |
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Answer» Torque of solenoid is given by `tau = MB sin theta` = (NiA) `B sin theta` `= 500 xx 2 xx 2 xx 10^(-4) xx 0.32 xx (1)/(2)` = 0.032 Nm |
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| 169. |
A closely wound solenoid of 750 turns and area of cross section of `5xx 10^(4)m^(2)` carries a current of 3.0 A. Its associated magnetic moment isA. `4.12JT^(-1)`B. `3.12JT^(-1)`C. `2.12JT^(-1)`D. `1.13JT^(-1)` |
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Answer» Correct Answer - D Here, n=750 turns, `A=5xx10^(-4)m^(2), I=3A` Then, magnetic moment `m=nIA=750xx3xx5xx10^(-4)=11250xx10^(-4)` `=1.125JT^(1)=1.13JT^(1)"along the axis of solenoid"` |
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| 170. |
A solenoid of cross-sectional area `2xx 10^(-4)m^(2)` and 900 turns has `0.6A m^(2)` magnetic moment. Then the current flowing through it isA. 2.4AB. 2.34mAC. 3.33AD. 3.33mA |
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Answer» Correct Answer - C Here, N=900 turns `A=2xx10^(-4)m^(2), m_(s)=0.6Am^(2)` The magnetic moment of solenoid `m_(s)=NIA` The current flowing through the solenoid is `I=(m_(s))/(NA)=(0.6)/(900xx2xx10^(-4)=3.33A` |
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| 171. |
A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid ? |
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Answer» `l = 22.5 cm = 22.5 xx 10^(-2) m = (45)/(2) xx 10^(-2)m` `N = 900, I = 0.8A, H = ?` `H = (NI)/(l) = (900 xx 0.8)/((45)/(2)xx10^(-2))` `H = (900)/(45) xx 0.8 xx 10^(2) xx 2` `:. H = 3200 Am^(-1)`. |
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| 172. |
A bar magnet of length 0.1 m and with a magnetic moment of `5Am^(2)` is placed in a uniform a magnetic field of intensity 0.4T, with its axis making an angle of `60^(@)` with the field. What is the torque on the magnet ? |
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Answer» Given, `2l = 0.1 m, m = 5A - m^(2), B = 0.4T, theta = 60^(@)`. Torque, `T = mB sin theta = 5 xx 0.4 xx sin 60^(@) = 2 xx (sqrt(3))/(2)` `:. T = 1.732 N - m` |
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| 173. |
A wire of length l is bent in the form a circular coil of some turns. A current I flows through the coil. The coil is placed in a uniform magnetic field B. The maximum torqur on the coil can beA. `(iBl^(2))/(4pi)`B. `(iBl^(2))/(pi)`C. `(iBl^(2))/(2pi)`D. `(2iBl^(2))/(pi)` |
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Answer» Correct Answer - A Let `N` be the number turns and `R` the redius of the coil. Then, `l=2piRN` or `R=l/(2piN) …(i)` Now magnetic moment of the coil is `M=NiA=Ni(piR^(2))` `=(Nipi)(l/(4pi^(2)N^(2)))=(il^(2))/(4piN)` Maximum value of `M` can be `M_(max)=(il^(2))/(4pi) at N=1` `:. tau_(max)=M_(max)B sin 90^(@)=(iBl^(2))/(4pi)`. |
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| 174. |
The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:A. `tan^(-1)(sqrt(3)/2)`B. `tan^(-1)(sqrt(3))`C. `tan^(-1)(sqrt(3)/sqrt(2))`D. `tan^(-1)(2/sqrt(3))` |
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Answer» Correct Answer - A `:. tam delta=V/H` `tan 45^(@)=V/(H cos 30^(@))` (Divide `(1)` and `(2)`) `delta=tan^(-1)(sqrt(3)/2)` |
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| 175. |
A dip cicrle is at right angles to the magnetic meridian. What will be the apparent dip ?A. `0^(@)`B. `30^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - D | |
| 176. |
A magnet oscillating in a horizontal plane has a time period of 2 seconds at a place where the angle of dip is `30^(@)` and 3 seconds at another place where the angle of dip is `60^(@)`. The retio of resultant magnetic field at the two places isA. `(4sqrt(3))/7`B. `4/(9sqrt(3))`C. `4/(4sqrt(3))`D. `9/sqrt(3)` |
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Answer» Correct Answer - C `T prop 1/sqrt(B_(H))=1/sqrt(B cos varphi)implies T_(1)/T_(2)=sqrt((B_(2) cos varphi_(2))/(B_(2) cos varphi_(1)))` `impliesB_(1)/B_(2)=T_(2)^(2)/T_(1)^(2)xx(cos varphi_(2))/(cos varphi_(1))=(3/2)^(2)xx(cos 60)/(cos 30)implies B_(1)/B_(2)=9/(4sqrt(3))` |
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| 177. |
A magnet suspended at `30^@` with magnetic meridian makes an angle of `45^@` with the horizontal. What shall be the actual value of the angle of dip? |
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Answer» `tan 45^(@) = (tan delta)/(cos 30^(@))` `1 = (2 tan delta)/(sqrt3)` `delta = tan^(-1) ((sqrt3)/(2))` |
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| 178. |
When the `N`-pole of a bar magnet points towards the south and S-pole towards the north, the null points are at theA. Magnetic axisB. Magnetic centreC. Perpendicular divider of magnetic axisD. N and S poles |
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Answer» Correct Answer - A Theory based |
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| 179. |
A magnet oscillating in a horizontal plane has a time period of 2 seconds at a place where the angle of dip is `30^(@)` and 3 seconds at another place where the angle of dip is `60^(@)`. The retio of resultant magnetic field at the two places isA. `(4sqrt(3))/(7)`B. `(4)/(9sqrt(3))`C. `(9)/(4sqrt(3))`D. `(9)/(sqrt(3))` |
| Answer» Correct Answer - C | |
| 180. |
Lines which represent places of constant angle of dip are calledA. isoclinic lineB. isogonic lineC. isoclinic linesD. isodynamic lines |
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Answer» Correct Answer - C Lines which represent places of constant angle fo dip ar ecalled isoclinic lines. The line joining places of ero dip is called aclinic (or magnetic equator). At all places wupon this line, a freely suspended magnet willl remain horizontal. |
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| 181. |
A line passing through places having zero value of magnetic dip is calledA. Isoclinic lineB. Agonic lineC. Isogonic lineD. Aclinic line |
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Answer» Correct Answer - D Theory based |
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| 182. |
Magnetic dip was measured at various places on earth in one of the following countries . It was found to be zero inA. IndiaB. CanadaC. BrazilD. Scotland |
| Answer» Correct Answer - C | |
| 183. |
Two identical bar magnets of magnetic moment M each are inclined to `60^(@)` as shown in figure . The net magnetic moment of the combination in this position , is A. `sqrt3` MB. `2 M`C. MD. `(M)/(sqrt3)` |
| Answer» Correct Answer - C | |
| 184. |
Two identical short bar magnets at a distance `d_(1) = 2 m ` apart are placed with their magnetic moment parallel to each other . Now keeping same direction of magnetic moment they are placed at `d_(2) = 3` m apart . By which factor force between them changes ?A. `(F_(1))/(F_(2)) = (81)/(16)`B. `(F_(1))/(F_(2))= (16)/(81)`C. `(9)/(8)`D. `(8)/(9)` |
| Answer» Correct Answer - A | |
| 185. |
Force between two identical bar magnets whose centres are r metre apart is `4.8 N`, when their axes are in the same line. If separation is increased to `2r`, the force between them is reduced toA. `2.4 N`B. `1.2 N`C. `0.6 N`D. `0.3 N` |
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Answer» Correct Answer - D In magnetic dipole, force `prop1/r^(4)` Hence, new force `=4.8/2^(4)=4.8/16=0.3 N` |
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| 186. |
A bar magnet `A` of magnetic moment `M_(A)` is found to oscillate at a frequency twice that of magnet `B` of magnetic moment `M_(B)` when placed in a vibrating magneto-meter. We may say thatA. `M_=2M_(B)`B. `M_(A)=8M_B`C. `M_(A)=4M_(B)`D. `M_(B)=8M_(A)` |
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Answer» Correct Answer - C `v=1/(2pi)sqrt((MB_(H))/I)implies v prop sqrt(M)` `implies v_(A)/V_(B)=sqrt(M_(A)/M_(B)) implies2/1=sqrt(M_(A)/M_(B))implies M_(A)=4M_(B)` |
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| 187. |
The magnet of vibration magnetometer is heated so as to reduce its magnetic moment by 36%. By doing this the periodic time of the magnetometer willA. Increases by 67%B. Increases by 25%C. Increases by 12%D. Is not changed |
| Answer» Correct Answer - B | |
| 188. |
Two identical short bar magnets, each having magnetic moment `M`, are placed a distance of `2d` apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them isA. `(mu_(0))/(4pi) (3M)/(d^(3))`B. `(mu_(0))/(4pi) (Msqrt5)/(d^(3))`C. `(mu_(0))/(4pi) (2M)/(d^(3))`D. `(mu_(0))/(4pi) (M)/(d^(3))` |
| Answer» Correct Answer - B | |
| 189. |
Two bar-magnets having their moment of inertia in the ratio `2: 3` oscillate in a horizontal plane with time periods `(5)/(2)` s and `(9)/(2)` s respectively . The ratio of their magnetic moments isA. `27 : 8`B. `54: 25`C. `25: 7`D. `32 : 9` |
| Answer» Correct Answer - B | |
| 190. |
Magnets `A` and `B` are geometrically similar but the magnetic moment of `A` is twice that of `B`. If `T_(1)` and `T_(2)` be the time periods of the oscillation when their like poles and unlike poles are kept togather respectively, then `T_(1)/T_(2)` will beA. `1/3`B. `1/2`C. `1/sqrt(3)`D. `sqrt(3)` |
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Answer» Correct Answer - C `T_(Sum)=2pisqrt(((I_(1)+I_(2)))/((M_(1)+M_(2))B_(H)))` `T_(d i ff)=2pisqrt(((I_(1)+I_(2)))/((M_(1)+M_(2))B_(H)))` `implies T_(s)/T_(d)=T_(1)/T_(2)=sqrt((M_(1)-M_(2))/(M_(1)+M_(2)))=sqrt((2M-M)/(2M+M))=1/sqrt(3)` |
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| 191. |
Which of the following is true?A. Diamagnetism is temperature dependentB. Paramagnetic is temperature dependentC. Paramagnetic is temperature independentD. None of these |
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Answer» Correct Answer - B With rise in temperature their magnetic susceptibility decreases, i.e., `chi_(m) prop 1/T`. |
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| 192. |
A diamagnetic material in a magnetic field movesA. From weaker to the stronger parts of the fieldB. Perpendicular to the fieldC. From stronger to the weaker parts of the fieldD. In none of the above directions |
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Answer» Correct Answer - C Theory based |
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| 193. |
If the angular momentum of an electron is `vec(J)` then the magnitude of the magnetic moment will beA. `(eJ)/m`B. `(eJ)/(2m)`C. `eJ 2m`D. `(2m)/(eJ)` |
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Answer» Correct Answer - B As we know for circulating electron magnetic moment `M=1/2evr` and angular momentum `J=mvr` From equation (i) and (ii) `M=(eJ)/(2m)` |
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| 194. |
When a piece of a ferromagnetic sobstance is put in a uniform magnetic field, the flux density inside it is four times the flux density away from the piece. The magnetic permeability of the material isA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - D `mu_(r)=B/B_(0)=4` |
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| 195. |
The magnitude of magentic field , due to a dipole of magnetic moment `2.4 Am^(2)` , at a point 200 cm away from it in the direction making an equal of `90^(@)` with the dipole axis isA. `3 xx 10^(-6) T`B. `3 xx 10 ^(-7) T`C. `3 xx 10^(-8) T`D. `0.3 xx 10^(-7) T` |
| Answer» Correct Answer - C | |
| 196. |
Which of the following equations indicates that magnetic monopole does not exist ?A. `oint vecB * dvec s = 0`B. `oint vecB * d vecs = (Q)/(epsi_(0))`C. `oint vecB * vec(dl) = mu_(0) l + epsi_(0) mu_(0) (dphi)/(dt)`D. `oint vecE * vec(ds) = (q)/(epsi_(0))` |
| Answer» Correct Answer - A | |
| 197. |
STATEMENT-1 : Magnetic fluex linked with a closed surface is always zero . and STATEMENT-2 : Magnetic field can exist without North and South poles .A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement- 1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - A | |
| 198. |
Which one of the following is a non-magneticsubstance?A. IronB. NickelC. CobaltD. Brass |
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Answer» Correct Answer - D Theory based |
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| 199. |
Curie-Weiss law is obeyed by iron at a temperature….A. Below Curie temperatureB. Above Curie temperatureC. At Curie temperature onlyD. At all temperature |
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Answer» Correct Answer - B Theory based |
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| 200. |
For an isotropic medium `B, mu, H` and `M` are related as (where `B`, `mu_(0)`, H and M have their usual meaning in the context of magnetic materialA. `(B-M)=mu_(0)H`B. `M=mu_(0)(H+M)`C. `H=mu_(0)(H+M)`D. `B=mu_(0)(H+M)` |
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Answer» Correct Answer - D Net magnetic induction `B=B_(0)+B_(m)=mu_(0)H+mu_(0)M` |
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