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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A force of 6 kg and another 8 kg can be applied to produce the effect of a single force equal to |
| Answer» Answer :D | |
| 2. |
In Bernoulli's theorem which of the following is conserved |
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Answer» Mass |
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| 3. |
A boat taken two hours to travel 8 km and back in still water lake. If the velocity of water is 4km/h,what is the time taken for going up stream of 8 km and coming back ? |
| Answer» Answer :B | |
| 4. |
Using Bohr model, calculate the electricitycurrent created by the electron when the H atom is in the ground state |
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Answer» SOLUTION : In ground state electron has orbital radius `a_(1)`and SPEED `v_(0)` From `v=r omega` `v_(0)=a_(0) omega` `omega=(2pi)/(T)` `v_(0)=a_(0)((2pi)/(T))` `T=(a_(0)2pi)/(v_(0))` Current formed, `1=(E)/(T)` `=(e)/([(2pia_(0))/(v_(0))])` `=(ev_(0))/(2pi a_(0))` `=(1.6xx10^(-19)xx2.187xx10^(6))/(2pixx0.53xx10^(-10))` `=1.05xx10^(-3)` `=10.5mA` |
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| 5. |
The band gap at 300K of crystals A, B, C, D is 5.5, 1.2, 0.67 and 0.1 units respectively. Answer the fol lowing questions based on this information.In the above case, which crystal will have the highest electrical conductivity at 300K? Why? |
| Answer» Solution :CRYSTAL D. Because it has least BAND GAP ENERGY | |
| 6. |
Interference was observed in interference chamber when air was present. Now, the chamber is evacuated and the same light is used. A careful observer will see: |
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Answer» no interference |
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| 7. |
A coil of 800 turns and 50 cm^(2) area makes 10 rps about an axis in its own plane in a magnetic field of 100 gauss perpendicular to this axis. What is the instantaneous induced emf in the coil? |
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Answer» Solution :`A=50cm^(2)=50 XX 10^(-4)m^(2)` n=10 rps, N=800 B= 100 gauss= `100xx10^(-4)T=10^(-2)T` Now, `epsilon=epsilon_(0) sin omegat=NBA omega sin omegat` `=800 xx 10^(-2) xx 50 xx 10^(-4) xx 2pi xx 10sin(20PI t)` or `epsilon-2.5 sin(2pi t)` volt |
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| 8. |
A bodyis floating such that it is completely immersed on the surface of a liquid . The densityof the bodyis same as thatof the liquid. The bodyis quasistaticallypushed downupto small distance . What will happen to the body? |
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Answer» It willslowlycomebackto itsearlierposition |
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| 9. |
A soccer ball, at rest on the ground, is kicked with an initial velocity of 10 m//s at a launch angle of 30^(@). Calculate its total flight time, assuming that air resistance is negligible. |
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Answer» 0.5 s `v_(y) overset("set")=0 implies v_(0_(y))=GT = 0 implies t=(v_(0_(y)))/(g)` The total flight time is EQUAL to twice the following value : `T= 2t= 2(v_(0_(y)))/(g)=2(v_(0)sin theta_(0))/(g)=(2(10 m//s)sin 30^(@))/(10 m//s^(2))=1 s` |
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| 10. |
A beam of non-relatitivistic chagred particles moves withoutdeviation through the region of space A (fig) where there are transervemutuallyperpendicularelectric and magnetic fieldswith streghtE and induction B. Whenthe magnetic field is swichted off, the trace of the beam on the screenSshifts by Delta pi. Knowing the distances a and b, findthe speficchargeq//m of the particles. |
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Answer» Solution :When there is no deviation, `-q vec(E) = q(vec(v) xx vec(B))` or, in scalar from, `E = vB ("as "vec(v) _|_ vec(B))` or, `v = (E)/(B)` .....(1) Now, when the magnetic fieldis swiched on, let the deciationin the field be `x`. Then. `x = (l)/(2) ((q vB)/(m)) t^(2)`, where `t` is the time required to pass through this region. also, `t = (a)/(v)` Thus `x = (1)/(2) ((qvB)/(m)) ((a)/(v))^(2) = (1)/(2) (q)/(m) (a^(2) B^(2))/(E)` ....(2) For the regionwhere thefield is absent, velocity in upward direction `= ((qvB)/(m))t = (q)/(m) aB` ....(3) Now, `Delta x -x = (q aB)/(m) t'` `= (q)/(m) (aB^(2)b)/(E)` when `t' = (b)/(v) = (BB)/(E)` ....(4) From (2) and (4) `Delta x - (1)/(2) (q)/(m) (a^(2) B^(2))/(E) = (q)/(m) (a B^(2) b)/(E)` or, `(q)/(m) = (2 E Delta x)/(a B^(2) (a + 2b))` |
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| 11. |
If Coulomb's law is represented by F=kq_(1)q_(2) r^(n) , then n = ...... |
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Answer» `1/2` |
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| 12. |
The band gap at 300K of crystals A, B, C, D is 5.5, 1.2, 0.67 and 0.1 units respectively. Answer the fol lowing questions based on this information.What is the most common unit to measure band gap? |
| Answer» SOLUTION :ELECTRON VOLT (EV) | |
| 13. |
A voltmeter having a resistance of 1800 Omega is employed to measure the potential difference across 200 Omega resistance, which is connected, to dc power supply of 50 V and internal resistance 20 Omega. What is the approximate percentage change in the potential difference across 200 Omega resistance as aresult of connecting the voltmeter across it? |
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Answer» `2.2%` `= 50 - 4.6 = 45.4 V` Now, `V_(2) = 50 - (50)/(180) xx 20 = 44.4 V` percentage change `= (V_(1) - V_(2))/(V_(1)) xx 100` `= 2.27` (also see the question) |
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| 14. |
An electric bulb of 100 W converts 3% of electrical energy into light energy.If the wavelength of light emitted is 6625 Å,the number of photons emitted in 1 s is……….. (h=6.625xx10^(-34))JS |
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Answer» Solution :`LAMBDA=6625Å` `=6.625xx10^(-7)m` `c=3xx10^(8)m//s` ENERGY obtained from 100 W bulb in 1s=100 J Energy converted into light energy from 3% of electric energy =3J Now E=nhf `E=(NHC)/(lambda) ` `thereforen=(Elambda)/(hc)` `=(3xx6.625xx10^(-7))/(6.625x10^(-34)xx3xx10^(8))` `therefore n=10^(19)` |
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| 15. |
Nuclear fission is the phenomenon of splitting of a heavy nucleaus into two or more ligther nuclei. Nuclear fusionis the phenomenon of fusing two or more ligther nuclei to form a single heavy nucleus. In both these processes, certain mass (Deltam) disappears, which appears in the form of nuclear energy, E=(Deltam)c^(2). The release of energy is so sudden that it cannot be controlled. This causes havoc. Nuclear fission is the basis of the an atom bomb and nuclear fusion is the basis of a hydrogen bomb. A powerful device, called Nuclear Reactor has been developed, where in nuclear energy produced is utilised for constructive purposes. (i) How much energy is released in the fission of one nucleus of U(235) and in how much time? (ii) Give an estimate of devastation potential of an atomic explosion. (iii) Should we ban nuclear research in this field ? Give your views briefly. |
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Answer» SOLUTION :(i) form the fission of a single nucleus of `._(92)U^(235), 200 MeV` energy is released in `10^(-9)` second i.e., almost instantly. (ii) First atomic explosion occurred in 1945 when an atom bomb was dropped on the city of Hiroshima, japan. This resulted in the killing of about 66000 people and grievous injury to an equal number. `2//3` rd of city STRUCTURES were smashed and so on. (iii) No, the reserch must go on. But we must investigate wasys and means for harnessing the enormous amount of nuclear energy for peacful PURPOSE. Nuclear reactor is an outcome of that reseach. |
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| 16. |
If E, M, J and G respectively denote energy, mass angular momentum and universal gravitational constant, the quantity, which has the same dimensions as the dimensions of (EJ^(2))/(M^(5)G^(2)) |
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Answer» Time ` E = [ML^(2)T^(-2)], J = [ML^(2)T^(-1)], M = [M], G = [M^(-1)L^(3)T^(-2)]` On subtituting above formulae in `EJ^(2)/M^(5)G^(2)`, `EJ^(2)/M^(5)G^(2) = ([ML^(2)T^(-2)] [M^(2)L^(4)T^(-2)])/([M]^(5)[M^(-2)L^(6)T^(-4)]) = [M^(3)L^(6)T^(-4)]/[M^(3)L^(6)T^(-4)] = M^(0)L^(0)T^(0)` = Dimensionless |
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| 17. |
The distance travelled by an object along a straight line in time t is given by s = 3 -41 + 5t^(2) , the initial velocity of the objectis |
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Answer» 3 unit `(DS)/(dt) =-4+10t` `U= (ds)/(dt)|_(t=0)=-4` unit |
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| 18. |
How does the maximum kinetic energy of electrons emitted vary with the work function of the metal . |
| Answer» SOLUTION :The maximum kinetic ENERGY of EMITTED electrons `k_max = 1/2 m(v_max)^2` = HV - `W_0`. | |
| 19. |
(a)a conductor a witha cavitygiven a charge Q show that the entire charge must appear on the outer surface of theconductor (b) another condluctor B with charge q is sensitive instrument is to be shielded from the strong electrosatitic fields in its environmentsuggest a possible way |
| Answer» Solution :Theforce is `10^(-2)`in the negativez direction that is in the direction of decreasing ELECTRIC fieldyoucan check that this is ALSO THEENERGY of the DIPOLE torque is zero | |
| 20. |
Define Electric dipole. |
| Answer» Solution :Two equal and opposite CHARGES SEPARATED by a SMALL distance constitute an ELECTRIC DIPOLE | |
| 21. |
A : Radio waves diffract pronouncedly around the sharp edges of the buildings than visible light waves. R : Wave length of radio waves is comparable to the dimension of the edges of the building. |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 22. |
Statement I: The north pole of a bar magnet is moving towards a closed circular coil along its axis. As a result the direction of induced current in the front face of the coil will be clockwise. Statement II: Any incident connected with electromagnetic induction obeys Lenz's law. |
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Answer» |
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| 23. |
How can eddy currents be reduced in a metallic conductor ? |
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Answer» Solution :(i) To reduce eddy currents the conducting (metallic) parts are built of a LARGE number of thin layers, known as LAMINATIONS, SEPARATED by an insulating material like lacquer. The plane of laminations is arranged parallel to the magnetic field so that they CUT ACROSS the eddy current paths. Thus, effectively increases the resistance of the possible paths to the flow of eddy currents and eddy currents are minimised. (ii) Eddy currents may also be reduced by punching holes or rectangular slots in metallic conductors, as these reduce the area available to the flow of eddy currents. |
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| 24. |
A uniform disc of mass m and radius R is pivoted at point P and is free rotate in vertical plane. The centre C of disc is initially in horizontal position with P as shown in figure. If it is released from this position, then its angular acceleration when the line PC is inclined to the horizontal at an angle theta is |
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Answer» `(2G COS THETA)/(3R)` |
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| 25. |
In a portion of some large electrical network, currents icertain branches are known. The values of (V_(A)-V_(B)) and (V_(c) - V_(o))are X and Y, respectively, where X and Y are |
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Answer» `X=29 V, Y = 26 V` |
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| 26. |
In a rectangular circuit ABCD a capacitor having capacitance C = 20 muF is charged to a potential difference of 100 V. Resistance of circuit is R = 10 Omega. Another rectangular conducting loop PQRS is kept side by side to the first circuit with sides AB and PQ parallel and close to each other. The length and width of rectangle PQRS is a = 10 cm and b = 5 cm respectively and AB as well as BC is large compared to a. PQ is located near the centre of side AB with d = 5 cm. The loop PQRS has 25 turns and the wire used has resistance of 1 Omega m^(–1). The switch S is closed at time t = 0. Neglect self inductance of loops. (a) Find the current in ABCD at t = 200 mus. (b) Find the current in PQRS at t = 200 mus. |
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Answer» |
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| 27. |
A faulty thermometer has its fixed points marked 5^(@)C and 95^(@)C. The thermometer reads the temperature of body as 59^(@)C. The correct temperature on Celsius scale is: |
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Answer» `59^(@)C` `=("TRUE reading - lower point.")/("range")` `(59-5)/(95 cdot 5)=(T-0)/(100)` `(54)/(90)=(T)/(100) rArr T=(54)/(90) xx100 =60^(@)C` Thus, correct choice is (b). |
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| 28. |
Force acting per unit area of each plate |
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Answer» `(1)/(2)mu_(0)J^(2)` |
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| 29. |
Give object and image distances, type, size and magnification for object placed in front of spherical mirror. |
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Answer» Solution :Concave mirror : (i) Object position : (u = `infty`) Image position : at PRINCIPAL focus F Image type : Real and inverted Image size : Highly diminished Magnification :m `lt lt ` -1  (i) Object position : Slight far from .C. `(infty gt u gt R)` Image position : Between F and C Image type : Real and inverted Image size : Diminished Magnification :m `lt`-1  (iii) Object position : At .C. (u = 2F) Image position : At .C. Image type : Real and inverted Image size : Same as of object Magnification : m = -1  (iv) Object position : Between .C. and .F. (f `lt` u `lt` 2f) Image position : Far from .C. Image type : Real and inverted Image size : Magnified Magnification : m `gt` -1  (v) Object position : At .F. (u = f ) Image position : At `infty` Image type : Real and inverted Image size : Highly magnified Magnification : m `gt gt` -1  (vi) Object position : Between F and P (0 `gt` u `gt` f) Image position : BEHIND mirror Image type : Virtual and erect Image size : Magnified Magnification : m `gt` +1  Convex mirror :(VII) Object position : At `infty` (u =`infty`) Image position : At F Image type : Virtual and erect Image size : Highly diminished Magnification : m `lt lt `+1  (viii) Object position : At any distance on axis Image position : Between P and F Image type : Virtual and erect Image size : Diminished Magnification : m `lt`+1 
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| 30. |
Find the lifting force of a balloon of 2 xx 10^4 m^3 capacity filled with helium at the surface of the Earth and at an altitude of 10 km above sea level. The balloon's envelope is open underneath. For data on the properties of the Earth's atmosphere see xi 26.10, Table. |
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Answer» `F = (rho t - rho_(He))Vg` |
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| 31. |
What will be the molar specific heat at constant volume of an ideal gas consisting of rigid diatomic molecules? |
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Answer» `(3)/(2)R` `therefore` Total energy of MOLECULE of one mole of a diatomic gas will be `C_(V)=(dU)/(dT)=(d)/(dT)((5)/(2)RT)=(5)/(2)R` |
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| 32. |
A wave disturbance in a medium is described by y(x,t) = 0.02 cos(50pit + pi/2), where x and y are in meter and t is in second. |
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Answer» A NODE occurs at X = 0.15m |
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| 33. |
In a parallel plate capacitor with air between the plates, each plate has an area of 6 xx 10^(-3) m^(2) and the distance between the plates is 3mm. Calculate the capcitanceof the capacitor. If this capacitor is connected to a 100V supply, what is the charge on each plate of hte capacitor? |
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Answer» Solution :`A=6 xx 10^(-3) m^2` `d=3mm =3 xx 10^(-3)m` `C=(epsi_(0)A)/(d)=(8.85 xx 10^(-12) xx 6 xx 10^(-3))/(3 xx 10^(-3))=17.70 xx 106(-12)=17.7pF` `Q=CV=17.7 xx 10^(-12) xx 100=1.77 xx 10^(-9)C=1.77nC` |
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| 34. |
Consider the situration shown in figure. The width of each plate is b and distance between the plates is d. The capacitor plates are rigidly clamped in the laboratory and connected to a battary of emf E. All surfaces are frictionless. Calculate the value of M for which the electric slab will stay in equilibrium. |
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Answer» `(epsilon_0bE^2(K-1))/(2dg)` |
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| 35. |
What are type of transistors ? |
| Answer» Solution :JUNCTION TRANSISTOR are of two TYPES : (a) n-p-n transistor (B) p-n-p transistor | |
| 36. |
If energy can neither be created nor destroyed, what happens to the so much energy spent against friction ? |
| Answer» Solution :The ENERGY spent against friction is DISSIPATED in the FORM of heat which is not available for WORK. | |
| 37. |
Using Huygen's construction, explain reflection of a plane wave by a plane surface. |
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Answer» Solution :Laws of reflection (i) Ist Law. It states that angle of incidence is equal to angle of reflection. (ii) 2ND Law. It states that the incident ray, the reflected ray and the normal, at the point of incidence all lie in the same plane. First law of reflection. Let XY be a plane reflecting surface and AB be a plane wavefront incident on the surface as shown in the figure. According to Huygens. principle, every point on wavefront AB is a source of SECONDARY wavelets and the time during which wavelet from B reaches at C, the reflected wavelet from A WOULD arrive at D.  i.e. `t=(BC)/(v)=(AD)/(v)` or BC = AD, ...(i) where v is the velocity of light in the medium. In rt `/_ DeltaABC` , `sin i=(BC)/(AC)` or `BC = AC sin i` ...(ii) In rt `/_ Delta ADC, sin r=(AD)/(AC)` or AD = AC sin r ...(iii) Putting Eqs, (ii) and (iii) in (i), we GET AC sin i =AC sin r or sin i = sin r or i = r i.e: Angle of incidence = angle of reflection This is first law of reflection. Second law of reflection. From the figure, we find that, the incident ray, the reflected ray and the normal all lie on the same plane XY. This proves second law of reflection. |
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| 38. |
Explain why a mariner 's compass does not work inside a submarine ? |
| Answer» SOLUTION :Due to the presence of geomagnetic field a mariner's COMPASS indicates . But the iron covering of a submarine ACTS as a magnetic screen . As a result , geomagnetic field cannot penetrate to the INTERIOR of a submarine . Thus , a mariner's compass does not WORK inside a submarine . | |
| 39. |
In the above problem. At what distance from the centre will the first maximum be located ? |
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Answer» 0.021 CM on the ONE SIDE and 0.028 cm on the other side |
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| 40. |
A clock face has negative charges -q, -2q, -3q,.....-12q fixed at the position of the corresponding numerals on the dial. The clocj hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial? |
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Answer» Solution :Let distance of each charge from unit charge at centre 'O' = r. Resultant ELECTRIC field of each charge, `E=1/(4pi epsi_(0)) (6q)/r^(2) [ :' -6q-(-12 q)]` Let OX be the reference AXIS, The angles of resultant fields with OX-axis shown.  Resultant field along OX-axis `=(0+1/2+sqrt(3)/2+1+sqrt(3)/2+1/2)i=(2+sqrt(3))i` Resultant field along OY-axis `=(1+sqrt(3)/2+1/2+0-1/2-sqrt(3)/2) hat(j)=1 hat(i)` `:.` Resultant electric field, `E_(R_((OH)))=(2+sqrt(3)) hat(i)+1 hat(j)` The DIRECTION of resultant field (OH) is given by, `tan theta=(|OY|)/(|OX|)` `=tan theta=1/(2+sqrt(3))=tan 15^(@)` `implies theta=15^(@)`, with OX-axis. `:.` The hour hand SHOWS at the centre of the dial is at 9.30. |
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| 41. |
The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What is the height of jump by the same person on the planet B ? |
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Answer» `(2//9)`m i.e., `SQRT(2g.h.)=sqrt(2gh)` `sqrt(2xxgxxh.)=sqrt(2xx9gxx2)` `:. 2h.=36 rArr h.=18`m. |
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| 42. |
Give expression for the force of a current carrying conductor in a magnetic field. |
Answer» SOLUTION :1. As shown in figure, consider a conductor PQ of length l are of cross-section A, carrying current I along + ve y-direction. The field `vecB` ACTS along + ve z-direction.  2. The electrons drift towards left with velocity `vecv_(d)`. 3. Each electron experience force along + ve X-axis which is given by, `vecf=-e(vecv_(d)xxvecB)` 4. If n is the number of free electrons per unit volume, then total number of electrons in the conductor is, `N=nxx"Volume"=nAl` 5. Total force on the conductor is, `vecF=Nvecf=nAl[-e(vecv_(d)xxvecB)]=nAe(-(lvecv_(d)xxvecB))` But `Ivecl` represents a current element vector in the direction of current so we can take `-lvecv_(d)=v_(d)vecl` `thereforevecF=nAe(v_(d)veclxxvecB)=nAev_(d)(veclxxvecB)` but `nAev_(d)` = current I `thereforevecF=I(veclxxvecB)` and magnitude F = `IlBsintheta` where `theta` is angle between `vecBandI`. 6. This equation can be applicable for straight conducting rod. 7. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dl and summing, `thereforevecF=sum_(i=1)^(n)vec(dl)_(i)xxvecB` where, i = 1, 2, 3, ...., n 8. Direction of this force can be given by Flemming.s left hand rule. 9. Special Cases : (i) If `theta=0^(@)or180^(@)` then, `F=IlBsin0^(@)=0""[becausesin0^(@)=0andsin180^(@)=0]` Thus, a current carrying conductor placed PARALLEL to the direction of the MAGNETIC field does not experience any force. (ii) If `theta=90^(@)`, then `F=IlBsin90^(@)` `thereforeF=IlB""[becausesin90^(@)=1]` Thus, a current carrying conductor placed perpendicular to the direction of a magnetic field experience a maximum force. |
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| 43. |
A linear objectof height 10 cm is kept in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm. The image formed is |
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Answer» magnified and erect Radius of curvature (R ) = 15 cm Distance of object (U) `= -10 cm` We KNOW that, `f=(R )/(2)=(15)/(2)cm` According to mirror formula, `(1)/(f)=(1)/(u)+(1)/(v)` `(1)/(-(15)/(2))=(1)/(-10)+(1)/(v)` `rArr (1)/(v)=(-2)/(15)+(1)/(10)` `rArr (1)/(v)=(-4+3)/(30)` `rArr (1)/(v)=(-1)/(30)` `rArr v=-30 cm` MAGNIFICATION (m) `= (h_(i))/(h_(o))=(-v)/(u)` `= (h_(i))/(10)=(-(-30))/(-10)` `h_(i)=-3` |
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| 44. |
The equivalent resistance between A and B in the arrangement of resistance as shown is |
Answer» SOLUTION :d. ,<br> `-= r` (By Wheatstone balanced bridge concept).)
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| 45. |
The de Broglie wave corresponding to a particle of mass m and velocity upsilon has a wavelength associated with it |
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Answer» <P>`(H)/(m UPSILON)` |
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| 46. |
Knowing the decay constant lambda of a nucleus, find: (a) the probability of decay of the nucleus during the time from 0 to t, (b) the mean lifetime tau of the nucleus. |
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Answer» Solution :(a) The probability of survival(i.e., not decaying) in TIME `t` is `e^(-lambda t)`. Hence the probability of decay is `1-e^(- lambdat)` (b) The probaility that the particle decays in time `DT` around time `t` is the difference `e^(-lambda t)-e^(-lambda(t+dt))= e^(-lambdat)[1-e^(e lambda dt)]= lambda e^(-lambda t)dt` Therefore the mean LIFE time is `T= int_(0)^(oo)t lambdae^(-LAMBDAR)dt// int_(0)^(oo) lambdae^(-lambdat)dt=(1)/(lambda) int_(0)^(oo)xe^(-x)dx//int_(0)^(oo)e^(-x)dx=(1)/(lambda)` |
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| 47. |
The sensitivity of a M.C.G. will increase if |
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Answer» the restoring torque per unit angular displacemetn is increased. |
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| 48. |
Two charges 2 µC and –2 µC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface? |
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Answer» SOLUTION :(a) The plane normal to AB and passing through its mid-point has zero POTENTIAL EVERYWHERE. (b) Normal to the plane in the DIRECTION AB. |
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| 49. |
What does a DC voltmeter measure when connected across a load resistor in an ac circuit? |
| Answer» SOLUTION :ZERO (VOLTAGE) READING. | |
| 50. |
Suggest a method to shield a certain region of space from magnetic fields. |
| Answer» SOLUTION :By USING a FERROMAGNETIC case. PUT an IRON ring in the magnetic field inside the ring field will be zero. | |