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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.
| 101. |
A domain in ferromagnetic iron is in the form of a cube of side length `10^-4m`. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is `55g//"mole"`, and its density is `7*9g//cm^3`. Assume that each iron atom has a dipole moment of `9*27xx10^(-24)Am^2`. |
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Answer» The volume of the cube domain is `V=(10^(-6) m)^(3)=10^(-18)m^(3)=10^(-12)cm^(3)` Its mass is volumexx density `=7.9g cm^(-3)xx10^(-12)=7.9xx10^(-12)g` it is given that Avagadro number (6.023xx`10^(23)`) of iron atoms have a mass of 55g. Hence, the number of atoms in the domain is `N=(7.9xx10^(-12)xx6.0233xx10^(23))/(55)` the maximum possible dipole moment `m_("max")` is achieved for the (unrealisec) when all the atomic moments are perfectly alligned. Thus, `m_("max")=(8.65xx10^(10))xx(9.27xx10^(24))` `=8.0xx10^(-13)A m^(2)` the consequent magnetisation is `M_("max")=m_("max")//` Domain volume `=8.0xx10^(-113)Am^(2)//10^(-18)m^(3)` `=8.0xx10^(5)Am^(-1)` |
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| 102. |
Identify the mismatched pair. `{:(,"Hard magnet","Alnico"),(,"Soft magnet","Soft iron"),(,"Bar magnet","Equivalent solenoid"),(,"Electromagnet","Loud speaker"):}`A. `{:(,"Hard magnet","Alnico"):}`B. `{:(,"Soft magnet","Soft iron"):}`C. `{:(,"Bar magnet","Equivalent solenoid"):}`D. `{:(,"Electromagnet","Loud speaker"):}` |
| Answer» Correct Answer - A | |
| 103. |
A domain in ferromagnetic iron is in the form of a cube of side length `1 mu m`. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron 55g/mole and is density is `7.9 g//cm^(3)`. Assume that each iron atom has a dipole moment of `9.27 xx 10^(-24) A m^(2)`. |
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Answer» The volume of the cubic domain is `V = (10^(-6)m)^(3) = 10^(-18) m^(3) = 10^(-12) cm^(3)` Its mass is volume `xx` density `= 7.9 g cm^(-3) xx 10^(-12) cm^(3) = 7.9 xx 10^(-12)g` It is given that Afagadro number `(6.023 xx 10^(23))` of iron atoms have mass of 55g. Hencem the number of atoms in the domain is `N = (7.9 xx 10^(-12) xx 6.023 xx 10^(23))/(55)` `= 8.65 xx 10^(10)` atoms The maximum possible dipole moment `m_("max")` is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus, `m_(max) = (8.65 xx 10^(10)) xx (9.27 xx 10^(-24))` `= 8.0 xx 10^(-13) Am^(2)` The consequent magnetisation is `M_("max") = m_("max")`/Domain volume `= 8.0 xx 10^(-13) A m^(2)//10^(-18) m^(3)` `= 8.0 xx 10^(5) Am^(-1)`. |
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| 104. |
Temperature above which a ferromagnetic substance becomes paremagnetic is calledA. neutral temperatureB. Curie temperatureC. inversion temperatureD. critical temperature |
| Answer» Correct Answer - B | |
| 105. |
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond curie temperature, then it will showA. anti ferromagnetismB. no magnetic propertyC. diamagnetismD. paramagnetism |
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Answer» Correct Answer - D Above Curie temperature, ferromagnetic material become paramangetic. |
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| 106. |
A domain in ferromagnetic iron is in the form of a ube of side length `2mum` then the number of iron atoms in the domain are `("Molecular mass of iron"= 55gmol^(-1) and "density"=7.92 g cm^(-3))`A. `6.92xx10^(12)"atoms"`B. `6.92xx10^(11)"atoms"`C. `6.92xx10^(10)"atoms"`D. `6.92xx10^(13)"atoms"` |
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Answer» Correct Answer - B The volume of the cubic domain `V=(2mu m)^(3)=(2xx10^(-6)m)^(3)` `=8xx10^(-18)m^(3)=8xx10^(-12)cm^(3)` and mass = volume `xx` density `= 8 xx 10^(-12) cm^(3) xx 7.9 g cm^(-3)` `=63.2xx10^(-12)g` Now the Avagadro number `(6.023xx10^(23))` of iron atoms have a mass of 55g. Hence the number of atoms in the domain are `N=(63.2xx10^(-12)xx6.023xx10^(23))/(55)=6.92xx10^(11)"atoms"` |
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| 107. |
In question number 91, maximum value of magnetisation of the given domain is (Dipole moment of an iron atom `9.27 xx 10^(-24)A m^(2))`A. `8.0xx10^(5)Am^(-1)`B. `6.0xx10^(4)Am^(-1)`C. `8.0xx10^(3)Am^(-1)`D. `6.0xx10^(3)Am^(-1)` |
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Answer» Correct Answer - A Net dipole moment, `M_(N)=Nxxm` `=6.9xx10^(11)xx9.27xx10^(-24)=6.4xx10^(-12)Am^(2)` `"Net magnetisation"=(6.4xx10^(-12))/(8xx10^(-18))=8xx10^(5)Am^(-1)` |
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| 108. |
A magnetising field of `1600 Am^(-1)` produces a magnetic flux `2.4 xx 10^(-5)` weber in an iron bar of cross-sectional area `0.2 cm^(2)` . The susceptibility of the iron bar will beA. 1788B. 1192C. 596D. 298 |
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Answer» `mu = (B)/(H) = (phi)/(AH)` `= (2.4 xx 10^(-5))/(0.2 xx 10^(-4) xx 16 xx 10^(2))` `= (2.4 xx 10^(-5))/(16 xx 2 xx 10^(-3))` `= (1.2)/(16) xx 10^(-2)` `(3)/(4) xx 10^(-3)` `= 0.75 xx 10^(-3)` Wb/m A `mu_(r) = (0.75 xx 10^(-3))/( 4 pi xx 10^(-7))` `mu_(r) = (75)/(4pi ) xx 10^(2)` `= 5.9713376` `mu_(r) = 597` `chi_(m) = mu_(r) -1` `chi_(m) = 597-1` `chi_(m) = 596` |
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| 109. |
A closely wound solenoid of 2000 turns and area of cross-section `1.6 xx 10^(-4)m^(2)`, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the magnetic moment associated with the solenoid ? |
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Answer» `N = 2000, A = 1.6 xx 10^(-4) m^(2), I = 4 "amp" = M = ?` As m = NIA `:. M = 2000 xx 4 xx 1.6 xx 10^(-4) = 1.28 JT^(-1)`. |
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| 110. |
A closely wound solenoid of 2000 turns and area of cross section `1.6xx10^-4m^2`, carrying a current of `4amp.` is suspended through its centre allowing it to turn in a horizontal plane: (a) What is the magnetic moment associated with the solenoid? (b) What are the force and torque on the solenoid if a uniform horizontal magnetic field of `7*5xx10^-2T` is set up at an angle of `30^@` with the axis of the solenoid? |
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Answer» Number of turns on the solenoid, n=2000 Area of cross-section of the solenoid, A=`1.6xx10^(-4)m^(2)` Current in the solenoid, I=4A (a) The magnetic moment along the axis of the solenoid is calculated as: M=nAI `=2000xx1.6xx10^(-4)xx4` `=1.28Am^(2)` (b) Mangetic field `B=7.5xx10^(-2)T` angle between the magnetic field and the axis of the solenoid, `theta=30^(@)` Torque, r=`MB sin theta` `=1.28xx7.5xx10^(-7)sin30^(@)` `=4.8xx10^(-2)Nm` Since the magnetic field is uniform, the force on the solenoid is zero, the torque on the solenoid is `4.8xx10^(-2)Nm`. |
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| 111. |
At magnetic poles of earth, angle of dip isA. ZeroB. `45^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - C Theory based |
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| 112. |
At place, the magnitudes of the horizontal component and total intensity of the magnetic field of the earth are 0.3 and 0.6 Oersted respectively. The value of the angle of dip at this place will beA. `60^(@)`B. `45^(@)`C. `30^(@)`D. `0^(@)` |
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Answer» Correct Answer - A `B_(H)=0.3` Oersted, `I=0.6` Oversted We have `B_(H)=I cos varphiimpliescosvarphi=B_(H)/I=0.3/0.6 =1/2` `:. varphi=60^(@)` |
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| 113. |
A bar magnet having a magnetic moment of `1.0 xx 10^4 J T^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B = 4 xx 10^(-5) T` exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction `60^@` from the field.A. `0.2J`B. `2.0J`C. `4.18J`D. `2xx10^(2)J` |
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Answer» Correct Answer - A Magnetic moment of bar `M=10^(4)J//T` `B=mu_(0)/(4pi).(2sqrt(2)M)/d^(3)` Hence, work done `W=vec(M).vec(B)` `=10^(4)xx4xx10^(-5)xxcos 60^(@) =0.2J` |
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| 114. |
A magnet having magnetic moment of `2.0 xx 10^(4) (J)/(T)` is free to rotate in a horizontal plane where magnetic field is `6 xx 10^(-5)` T . Find the work done to rotate the magnet slowly from a direction parallel to field to a direction `60^(@)` from the field . |
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Answer» `W = MB [ cos theta_(1) - cos theta_(2)]` `= 2.0 xx 10^(4) xx 6 xx 10^(-5) [ cos 0^(@) - cos 60^(@)]` `= 2 xx 10^(4) xx 6 xx 10^(-5) [ 1 - (1)/(2)]` `= (12)/(2) xx 10^(-1)` `= 0.6 J` |
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| 115. |
The intensity of magnetic field is H and moment of magnet is M. The maximum potential energy isA. `MH`B. `2MH`C. `3MH`D. `4 a MH` |
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Answer» Correct Answer - A Potential energy `U=-MB costheta` `implies U_(max)=MH(at theta=180^(@))` |
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| 116. |
A bar magnet of magnetic moment `200 A-m^(2)` is suspended in a magnetic field of intensity `0.25 N//A-m`. The couple required to deflect it through `30^(@)` isA. `50 N-m`B. `25 N-m`C. `20 N-m`D. `15 N-m` |
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Answer» Correct Answer - B `tau=MB sintheta` `tau=200xx0.25xxsin 30^(@)=25 Nxxm`. |
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| 117. |
Magnetic intensity for an axial point due to a short bar magnet of magnetic moment `M` is given byA. `mu_(0)/(4pi)xxM/d^(3)`B. `mu_(0)/(4pi)xxM/d^(2)`C. `mu_(0)/(2pi)xxM/d^(3)`D. `mu_(0)/(2pi)xxM/d^(2)` |
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Answer» Correct Answer - C `B_(a)=mu_(0)/(4pi)(2M)/d^(3)=mu_(0)/(2pi)M/d^(3)` |
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| 118. |
A rectangular loop of sides 10cm and 5cm carrying a current I of 12A is placed in different orienctations as shown in the figure If there is a uniform magnetic field of 0.3 T in the positive z-direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium .A. (2)and(4), respectivelyB. (2)and(3), respectivelyC. (1)and(2), respectivelyD. (1)and(3), respectively |
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Answer» Correct Answer - B `I=12A, vec(B)=0.3hatk T`, `A=10xx5cm^(2)=50xx10^(-4)m^(2)` `vec(M)=I Ahat(n)=12xx50xx10^(-4)hatn Am^(2)` `6xx10^(-2)hat(n)Am^(2)` `"Here", vec(M)_(1)=6xx10^(-2)hati Am^(2), vec(M_(2))=6xx10^(-2)Am^(2)` `vec(M_(3))=-6xx10^(-2)hatj Am^(2), vec( M_(4))=-6xx10^(-2)hatk Am^(2)` `vec(M_(2))` is parallel to `vec(B)`, it means potential energy is minimum, therefore in orientiation (2) the loop is in stbale equilibrium. `vec(M _(4))` is antiparallel to `vec(B_(4))`, it means potential energy is maximum therefore in oriention (4) the loop is in unstbale equilibirum. |
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| 119. |
A magnetic needle lying parallel to a magnetic field requires `W units` of work to turn it through `60^(@)`. The torque needed to maintain the needle in this position will beA. `sqrt(3)W`B. `W`C. `sqrt(3)/2W`D. `2W` |
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Answer» Correct Answer - A `W=MB(costheta_(1)-costheta_(2))=MB(cos 0^(@)-cos 60^(@))` `=MB(1-1/2)=(MB)/2` and `tau=MB sin theta=MB sin 60^(@)=MBsqrt(3)/2` `:. Tau=((MB)/2)sqrt(3)impliestau=sqrt(3)W` |
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| 120. |
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 the assertion and reason are true and reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true is true but reason is false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - D The permeability of a ferromagetic material is not independent of magnetic field, `vec(B)=K_(m)vec(B_(0))`. `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 magnetization `vec(M)` increases linearly with `B_(0)`. Even at small value of the hysteresis curve, magnetic permeability is greater for lower field. |
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| 121. |
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 the assertion and reason are true and reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true is true but reason is false.D. If both the assertion and reason 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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| 122. |
Assertion: Magnetism is relativistic. Reason: When we move along with the charge so that there is no motion relative to us, we find no magnetic field associated with the charge.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 A magnetic field is produced by the motion of electric charge. Since motion is relative, the magnetic field is also relative. |
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| 123. |
The magnetic potential at a point on the axial line of a bar magnet of dipole moment `M` is `V`. What is the magnetic potential due to a bar magnet of dipole moment `M/4` at the same point ?A. `4V`B. `2V`C. `V/2`D. `V/4` |
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Answer» Correct Answer - D Magnetic potential at a distance d from the bar magnet on its axial line is given by `V=mu_(0)/(4pi).M/d^(2) implies V prop M implies V_(1)/V_(2)=M_(1)/M_(2)` `implies V/V_(2)=M/(M//4) implies V_(2)=V/4` |
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| 124. |
Due to a small magnet intensity at a distance `x` in the end on position is `9` Gauss. What will be the intensity at a distance `(x)/(2)` on broad side on position?A. `9` GaussB. `4` GaussC. `36` GaussD. `4.5` Gauss |
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Answer» Correct Answer - C In C.G.S. `B_("axial")=9=(2M)/x^(3)` …(i) `B_("equaterial")=M/(x/2)^(3)=(8M)/x^(3)` …(ii) From equations (i) and (ii) `B_("equaterial")=36` Gauss |
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| 125. |
Which of the following material is used in making the core of a moving coil galvanometer?A. CopperB. NickelC. IronD. Both (a) and (b) |
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Answer» Correct Answer - C Soft iron is used in making the core of moving coil galvonometer, because it has high initial permeability and low hysteriss loss. |
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| 126. |
Materials suitable for permanent magnet, must have which of the following properties?A. High retantivity low coercivity and high permeabilityB. Low retantivity low coercivity and low permeabilityC. Low retantivity high coercivity and low permeabilityD. High retantivity high coercivity and high permeability |
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Answer» Correct Answer - D Materials suitbale for permanent magnetis should have retentivity, high coercivity and high permeability. |
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| 127. |
Which of the following is the most suitable mateial for making permanent magnet ?A. SteelB. Soft ironC. CopperD. Nickel |
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Answer» Correct Answer - A Steel has more retentivity and coercivity, so it is used for making premanent magnet. |
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| 128. |
A permanent magnet attractsA. Diamagnetic substanceB. Paramagnetic substanceC. Ferromagnetic substanceD. Both (2) & (3) |
| Answer» Correct Answer - D | |
| 129. |
A superconductor exhibits perfectA. ParamagneticB. FerromagneticC. DiamagneticD. Both (1) & (2) |
| Answer» Correct Answer - C | |
| 130. |
Domains for motion is not necessary feature ofA. ParamagnetismB. DiamagnetismC. FerromagnetismD. Both (1) & (2) |
| Answer» Correct Answer - D | |
| 131. |
Consider a sort magnetic dipole o f magnetic length 10cm. Find its geometric length. |
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Answer» Geometric length of a magnet is `(6)/(5)` times its magnetic length . `therefore` Geometric length = `(6)/(5) xx 10 = 12` cm . |
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| 132. |
Assertion: Magnetic resonance imaging (MRI) is a useful diagnostic tool for producing images of various parts of human body. Reason: Protons of various tissues of the human body play a role in MRI.A. If both the assertion and reason are true and reason is a true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true is true but reason is false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - B Spin technology, called magnetic resonance imaging (MRI), has been applied to medical diagnostics with great success. The protons of the various tissues of the human body are situated in many different local magnetic environments. When the body, or part of it, is immersed in a strong external magnetic field, these environmental differences can detected by spin-flip techniques and translated by computer processing into an image resembling those produced by X-rays. Hence, option `(b)` is ture. |
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| 133. |
Assertion: Time period of vibration of a pair of magnets in sum position is always smaller than in difference position. Reason:`T=2pi sqrt(I//MH), where symbols have their standard meaning.A. If both Assertain and Reason are true and Reason is the correct explanation of AssertainB. If both Assertain and Reason are true but Reason is not correct explantion of AssertainC. If Assertion is true but Reason is falseD. If Assertion is false but Reason in true. |
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Answer» Correct Answer - B The time period in sum position, `T_(s)=2pi sqrt((I_(s))/(M_(s)H))` `where, " "I_(s)+I_(1)+I_(2) and M_(s)=M_(1)+M_(2)` ` " " T_(s)=2pi sqrt((I_(1)+I_(2))/((M_(1)+M_(2))H))` Time period in difference position, ` " " T_(d)=2pi sqrt((I_(1)+I_(2))/((M_(1)-M_(2))H)) Rightarrow T_(s)lt T_(D)` |
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| 134. |
If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field?A. 0.075NmB. 0.080NmC. 0.081NmD. 0.091Nm |
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Answer» Correct Answer - C `Here, M=0.65JT^(-1), B=0.25T, theta=30^(@)` `therefore tau=mB sin theta=0.65xx0.25xxsin 30^(@)` `=0.65xx0.25xx(1)/(2)=0.08125Nm` =0.081 Nm |
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| 135. |
If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field? |
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Answer» Magnetic field strength, `B=0.25T` M agnetic moment, `M=0.6T^(-1)` The angle `theta`, between the axis of the solenoid and the direction of the applied field is `30^(@)`. Therefore, the torque acting on the solenoid is given as: `tau=MB sin theta` `=0.6xx0.25sin30^(@)` `=7.5xx10^(-2)J` |
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| 136. |
If the solenoid is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of `30^(@)` with the direction of applied field ? Here `M = 0.6 J T^(-1)` |
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Answer» Here `M = 0.6 J T^(-1)` `B = 0.25 T r = ? q = 30^(@)` As `r = m B sin theta :. r = 0.6 xx 0.25 sin 30^(@) = 0.075 N.m`. |
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| 137. |
Interestellar space has an extremely weak magnetic field of the order of `10^(-12)T`. Can such a weak field be of any significant consequence ? Explain. |
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Answer» When a charged particle moves in a magnetic field, it is deflected along a circular path such that `BeV = (mV^(2))/(r) :. r = (mV)/(Be)` When B is low, r high i.e., radius of curvature of path is very large. Therefore, over the gigantic inter stellar distance, the deflection of charged particles becomes less noticeable. |
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| 138. |
The unit of magnetic susceptiblity isA. HB. Wb/mC. A/mD. None of these |
| Answer» Correct Answer - D | |
| 139. |
The relation connecting B, H nad I in SI system isA. B=H+1B. B=H-1C. `B=mu_(0)(H+1)`D. `b=mu_(0)(H-I)` |
| Answer» Correct Answer - C | |
| 140. |
The time period of viberation of two magnets in sum position (magnets placed with similar poles on one sides one above the other) is 3s. When polarity of weaker magnet is reversed the combination makes 12 oscillations per minutes. What is the ratio of magnetic moments of two magnets? |
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Answer» `Here, T_(1)=3s,T_(2)=(1)/(12)min=(60)/(12)=5s` `" " (M_(1))/(M_(2))=(T_(2)^(2)+T_(1)^(2))/(T_(2)^(2)-T_(1)^(2)) therefore (M_(1))/(M_(2))=(5^(2)+3^(2))/(5^(2)+3^(2))=(34)/(16)=(17)/(8)` |
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| 141. |
A bar has a magnetic moment equal to `5xx10^(-5) weberxxm`. It is suspended in a magnetic field which has a magnetic induction (B) equal to `8pixx10^(-4)` tesla. The magnet vibrates with a period of vibration equal to 15 sec. The moment of intertia of the magnet isA. `22.5 kgxxm^(2)`B. `11.25xxkgxxm^(2)`C. `5.62xxkgxxm^(2)`D. `7.16xx10^(-7)kgm^(2)` |
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Answer» Correct Answer - D Time period of a magnet `T=2pisqrt(I/(MB))` or `I=(T^(2)MB)/(4pi^(2))=(225x5xx10^(-5)xx8pixx10^(-4))/(4pi^(2))` `:. I=7.16xx10^(-7) kg m^(2)` |
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| 142. |
At a certain place a magnet makes 30 oscillations per minute. At another place where the magnetic field is double, its time period will beA. `4 sec`B. `2 sec`C. `1/2 sec`D. `sqrt(2) sec` |
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Answer» Correct Answer - D `T=2pisqrt(I/(MB_(H))), :. T_(1)/T_(2)=sqrt((B_(H))_(2)/((B_(H))_(1))) impliesT_(2)=T_(1)sqrt((B_(H))_(1)/((B_(H))_(2)))` Here `n_(1)=30 "oscillation"//"min"=1/2 "oscillation"//"sec"` `:. mu/(4pi).m^(2)/r^(2)=50 gm-wt` `:. T_(2)=2sqrt(B_(H)/(2B_(H)))=2xx1/sqrt(2)=sqrt(2) sec` |
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| 143. |
The strength of the magnetic field in which the magnet of a vibration magnetometer is oscillating is increased 4 times its original value. The frequency of oscillation would then becomeA. Twice its original valueB. Four times its original valueC. Half its original valueD. One-fourth its original value |
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Answer» Correct Answer - A Frequency `v prop sqrt(B_(H))` |
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| 144. |
The time period of oscillation of a magnet in a vibration magnetometer is `1.5` seconds. The time period of oscillation of another of another magnet similar in size, shap and mass but having one-fourth magnetic moment than that of first magnet, oscillating at same place will beA. `0.75 sec`B. `1.5 sec`C. `3 sec`D. `6 sec` |
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Answer» Correct Answer - C `T prop 1/sqrt(M) impliesT_(1)/T_(2)=sqrt(M_(2)/M_(1))implies1.5/T_(2)=sqrt((M_(1)//4)/M_(1))=1/2` `implies T_(2)=3sec ` |
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| 145. |
The period of oscillations of a magnetic needle in a magnetic field is 1.0 sec. If the length of the needle is halved by cutting it, the time perood will beA. `1.0 sec`B. `0.5 sec`C. `0.25 sec`D. `2.0 sec` |
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Answer» Correct Answer - B `T=2pisqrt(I/(MB))=2pisqrt((wl^(2)//12)/("Pole strength" xx2lxxB))` `:.T propsqrt(Wl)` `:.T_(2)/T_(2)=sqrt(w_(2)/w_(1)xxl_(2)/l_(1))=sqrt((w_(1)//2)/w_(1)xx(l_(1)//2)/l_(1))=1/2` `impliesT_(2)=T_(1)/2=0.5s` |
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| 146. |
Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ? |
| Answer» The magnetic permeability of a ferromagnetic material `mu gt gt 1`. That is why the field lines meet this medium normally. | |
| 147. |
The distance of two points on the axis of a magnet from its centre is `10 cm` and `20 cm` repectively. The ratio of magnatic intensity at these points is `12.5 : 1`. The length of the megnet will beA. 5cmB. 25cmC. 10cmD. 20cm |
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Answer» Correct Answer - C `B_(1)/B_(2)=d_(1)/d_(2)((d_(2)^(2)-l^(2))/(d_(2)^(1)-l^(2)))^(2) rArr 12.5/1=10/20((400-l^(2))/(100-l^(2)))^(2)` `rArr l=5 cm` Hence, length of magnet `=2l=10 cm` |
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| 148. |
Ratio of magnetic intensities for an axial point and a point on broad side-on position at equal distance d from the centre of magnet will be or The magnetic field at a distance d from a short bar magnet in longitudinal and transverse positions are in the ratioA. `1:01`B. `2:03`C. `2:01`D. `3:02` |
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Answer» Correct Answer - C `B_(1)=(2M)/d^(3).B_(2)=M/d^(3), :. B_(1)/B_(2)=2:1` |
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| 149. |
Assertion: Diamagnetism is universal, it is present in all materials Reason: Field due to induced magnetic moment is opposite to the magnetising field.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 - A Diamagnetism is when a material acquires amagnetisation in the opposite direction to the applied magnetic field. As a magnetic field will always induce magnetism, diamagnetism will occur in all substances. The magnitude may be small and be overcome by other para or ferromagnetic properties. |
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| 150. |
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 nine of the above directions |
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Answer» Correct Answer - C A diamagnetic material in a magnetic field moves from stronger to the weaker parts of the field. |
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