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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. |
An electron moves around the nucleus in a hydrogen atom of radius 0.51 Å , with a velocity of 2 xx 10^(5) m s^(-1) . Calculate the following : (i) the equivalent current due to orbital motion of electron, (ii) the megnetic field produced at the centre of the nucleus, (iii) the magnetic associated with the electron. |
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Answer» Solution :It is given that `r = 0.51 Å = 5.1 xx 10^(-11) m, v = 2 xx 10^(5) ms^(-1) , and q = e = 1.6 xx 10^(-19)C` `l = (ev)/(2 pi r) = (1.6 xx 10^(-19) xx 2 xx 10^(5))/(2 xx 3.14 xx 5.1 xx 10^(-11)) = 10^(-4)A` (ii) Magnetic field at the centre of the nucleus `B = (mu_0 I)/(2r) = (4 pi xx 10^(-7) xx 10^(-4))/(2 xx 5.1 xx 10^(-11)) = 1.25 T`. (iii) The magnetic moment ASSOCIATED with the electron `m = IA = I pir^2 = 10^(-4) xx 3.14 xx (5.1 xx 10^(-11))^2 = 8.2 xx 10^(-25) A m^(2)`. |
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| 2. |
A combination is formed by connecting alternately n number of equidistant parallel pltes. If C be the capacitance for any two consecutive plates, then the capacitance of the whose system will be |
| Answer» ANSWER :C | |
| 3. |
Explain the following features of compton scattering of light by matter: (a) the increase in wavelength Delta lambda is independent of the nature of the scattering substance, (b) the intensity of the displaced component of scattered light grows with the increasing angle of scattering and with the diminishing atomic number of the substance, (c) the presence of a non-displaced in the scattered radiation. |
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Answer» Solution :(a) Compton scattering is the scattering of light by free electrons. (The free electrons are the electrons whose binding is much smaller than the typical energy transfer to the electrons). For this reason the increase in wavelength `Delta LAMBDA` is independent of the nature of the scattered substance. (B) This is because the effective number of free electrons increase in both cases. with increasing ANGLE of scattering, the energy transfered to electrons increases. With diminishing atomic number of the substance the bindingenergy of the elecrtons decreases. (c) The presence of a non-dispalced component in the scattered radiation is due to scattering from strong bound (inner) electrons as well as nuclei. For scattering by these the atom ESSENTIALLY recoils as a WHOLE and there is very little enrgy transfer. |
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| 4. |
A block is floating in a liquid as shown in figure. Suppose. W = weight of block P = Pressure at bottom of block F = Upthrust on the block Now suppose container starts moving upward with some positive acceleration. The new values are suppose W',P' and F'. Then |
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Answer» `W'gtW` 
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| 5. |
If the length of second's pendulum is increased by 2%, how many seconds it loses per day ? |
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Answer» 3927 s `:.""(T.)/(T)=sqrt((102)/(100))impliesT.=2[1+(2)/(100)]^(1//2)` or `""T.=2[1+(1)/(100)]=2+(2)/(100)` Loss in TIME `=2+(2)/(100)-2=(2)/(100)` s. `:.`Loss in time for 2 s `=(2)/(100)` Loss in timefor DAY `=(2)/(100)xx(86400)/(2)=864` s. Hencecorrectchoice is (d). |
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| 6. |
A man of height 'h' is walking away from a street lamp with a constant speed 'v'. The height of the street lamp is 3h. The rate at which the length of the man's shadow is increasing when he is at a distance 10h from the base of the street lamp is : |
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Answer» v/2 |
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| 7. |
Anearly model for an atom consider it to have a positively charged point inucleus ofcharges ze surronded by awhole is neutral for this model what is the electric fieldat a distance r from the nucleus |
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Answer» Solution :The charge distributionfor this model of the atom is asthe total chargein the uniform sperical of charge z e + NEGATIVE harge is neutral this immedialtely gives us the negative charge density p since we MUST have `(4piR^(3))/(3) p= 0-Ze` to fin d the electricfieldE(r ) at apointP whichis a distance r awayfrom the nucleuswe usegausslawits direction is alongtheradius vector rfrom the origin to the pointthe OBVIOUS gaussian surface is a spherical surface centredto the point P the obviousgaussian surfaceis a spherical surfacecentred at the nucleus we CONSIDER twosituation namely `rgtR` and `rlt R` (i) `r gt R`the electricflux `Phi`enclosedbythe spherical surface is `Phi =E(r )xx4 pi r^(2)` where E(r )is the magnitude of the electricfieldat r this is because the FIELD at any pointon the sphericaland has the same magnitude at all points on the surface i.e `q=Ze+(4pir^(3))/(3) P` substituting for the charge density p obtained earlier we haveM (ii)`r gt R`in this case the totalenclosed by the gaussiansperical surface is zerosince the atom is neutral thus from gauss law at r =R both cases given the same result E =0 |
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| 8. |
Microwaves are clectromagnetic waves with frequency in the range of |
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Answer» MICRO hertz |
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| 9. |
At t = 0, a source of sonic oscillations S and an observer O start moving along x and y axes witn velocity 5 m/s and 10 m/s respectively. The figure gives their position at t = 0. The frequency of sonic oscillations of source is 1000 Hz. Obtain the frequency of signals recieved by the observer after 5 seconds. Speed of sound is 330 m/s |
| Answer» ANSWER :C | |
| 10. |
Radioactive decay takes from nucleus and obeys____law. |
| Answer» SOLUTION :CURIE, RUTHERFORD | |
| 11. |
Ultraviolet radiations are more effective for causing photoelectric emission than visible light. |
| Answer» Solution :TRUE: ENERGY of ultraviolet light PHOTON is more than that of visible light. So photoelectric emission can TAKE place easily. | |
| 12. |
A thin copper ring of radius a is charged with q units of electricity. Calculate the electric field at a point at a distance x from the centre and on the axis of the ring. Use the result to find the electirc field due to a charged disc of surface density of charge epsilon. |
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Answer» |
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| 13. |
In YDSE , ten maxima of two pattern(Interference ) is formed in central maxima of single slit pattern (diffraction ) If separtionbetween two slits is 1mm, then what is the width of each slit ? |
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Answer» `0.2` MM |
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| 14. |
A certain charge 2Q is divided at first into two parts q_(1) and q_(2). Later the charges are placed at a certain distance. If the force of interaction between two charges is maximum then Q/(q_(1)) = ________. |
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Answer» 1 `F alphaq_1 q_2` ` alpha q_1(2 Q -q_1)` ` F alpha2q_1 Q-q_(1)^2` `(dF)/(dq_1) =0implies 2Q-2q_1=0 thereforeq_1=Q` `(q_1)/(Q)=1` |
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| 15. |
In case of induced electric field due to a time-dependent magnetic field in a cylindrical region, which of the statement is correct? |
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Answer» It OBEYS the Gauss' law for electric flux. |
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| 16. |
Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate? |
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Answer» The number of free conductioin ELECTRONS is significant in C but small in Si and Ge. |
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| 17. |
If the terminal speed of a sphere of gold (density 19·5 kg/m^(3)) is 0.2 m/s in a viscous liquid (density 1.5 kg/m^(3)), find the terminal speed of sphere of silver (density 10.5 kg//m^(3)) of same size in same liquid. |
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Answer» 0·2 m/s `therefore(v_(2))/(v_(1))=((o_(2)-sigma)/(p_(1)-sigma))` as r and N are same`. `thereforev_(2)=((10.5-1.5)/(19.5-1.5))xx0.2=0.1ms^(-1)` THUS correct choice is ©. |
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| 18. |
If the equivalent capacity between A and B in the circuit is 12 muF, the capacity C is |
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Answer» `5muF` |
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| 19. |
Define a wavefront. Use Huygen's principle to show diagrammatically the propagation of a wave-front from the instant t_(1)=0 to a later time t_(2)=t. |
Answer» Solution : A wavefront is defined as a surface of CONSTANT phase. Thus, a wavefront is a locus of all those points in space which are vibrating in EXACTLY same phase at any one particular instant of time. Let AB be the position of wavefront at time `t_(1)=0`. Points `A_(1),B_(1),C_(1)`, . . on the wavefront act as SOURCE of light and emit wavelets which travel with speed of light v and will COVER a distance s=vt in time t. draw spheres with points `A_(1),B_(1),C_(1),` . . . as centre and s=vt as the RADIUS. then, common envelope A.B. in the forward direction gives the new wavefront at time `t_(2)=t` as shown in figure. |
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| 20. |
An аudio signal given by e_(s) = 15 sin 2pi (100, 000t) modulates a carrier wave given by e_(s) = 60 sin 2pi(100,000t). If calculatePercent modulation |
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Answer» Solution :SIGNAL Amplitude , B= 15 CARRIER amplitude A= 60 `m= (B)/(A)= (15)/(60)= 0.25` `:. "Percentage MODULATION"= 0.25 xx 100=25%` |
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| 21. |
An аudio signal given by e_(s) = 15 sin 2pi (100, 000t) modulates a carrier wave given by e_(s) = 60 sin 2pi(100,000t). If calculateFrequency spectrum of the modulated wave. |
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Answer» Solution : By comparing the given equations of signal and carrier with their standard FORM `e_(s)= E_(s)sin OMEGA _(s)t= E_(s)sin2pif_(s)t` `e_(C)= E_(c)sinomega_(c)t= E_(c) sin2pif_(c)t` we have signal frequency `F_(s)= 2000Hz` and carrier frequency `f_(c)= 100, 000Hz` The frequencies present in modulated wave (i) `f_(c)= 100, 000Hz= 100kHz` (ii) `f_(c)-f_(s)= 100,000-2000= 98kHz` (iii) `f_(c)+ f_(s)= 100kHz+ 2kHz=102 kHz` Therefore, frequency spectrum of modulated wave extends from 98kHz to102 kHz is called BAND width. |
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| 22. |
A magnetic field can be produced by ______ . |
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Answer» A moving CHARGE |
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| 23. |
A ray of light enters in denser medium from rarer medium. Speed of light in rarer medium is double than that in denser medium, what is the critical angle for total internal reflection ? |
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Answer» `30^@` `thereforesinC=1/n=1/2``thereforeC=30^@` |
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| 24. |
Describe the motion of a charged particle in a cyclotron if the frequency of the oscillating electric field were doubled. |
| Answer» Solution :The charged PARTICE will accelerate and decelerate alternately. As a result the charged PARTICLE will continue to revolve inside the dees of cyclotron in a CIRCULAR path of constant radius. | |
| 25. |
Show that for the particle whose coordinate incertainty is Deltax=lambda//2pi, where lambda is its de Broglie wavelength, the velocity uncertainty is of the same order of magnitude as the particle's velocity itself. |
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Answer» <P> SOLUTION :If `Deltax=lambda//2pi=(2piħ)/(P).(1)/(2pi)=(ħ)/(P)=(ħ)/(mv)`Thus `Deltavunderset(~)gt(ħ)/(mDeltax)=v` Thus `Deltav` is of the same ORDER as `v`. |
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| 26. |
If N is the number of turns in a coil, the value of self inductance varies as prop _____ |
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Answer» `N^0` In `L=(mu_0N^2A)/l , (mu_0A)/l` CONSTANT `therefore L prop N^2` |
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| 27. |
A prism of angle 60^(0) produces angle of minimum deviation of 40^(0). What is its refractive index ? Calcualte the angle of incidence. |
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Answer» SOLUTION :Here : `A=60^(0),D=40^(0),n=?,i=?` we have `n=(Sin((A+D)/(2)))/(Sin((A)/(2)))` `=(Sin((60+40)/(2)))/(Sin((60)/(2)))` `=(Sin(50))/(Sin(30))=(0.7660)/(0.5000)` n = 1.532 At MINIMUM DEVIATION position `i=(A+D)/(2)=(60+40)/(2)=50^(0)` |
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| 28. |
A step index fibre has a relative refractive index difference of 0.9%.Estimate the critical angle at the core -cladding interface,when the core index is 1.46.find the numerical aperture,if the source to fiber medium is air. |
| Answer» Solution :Here,`mu_(1)=1.46`,`(mu_(1)-mu_(2))/(mu_(1))=(0.9)/(100)=0.009`….or `1-(mu_(2))/(mu_(1))=0.009` or `(mu_(2))/(mu_(1))=1-0.009=0.991` CRITICAL angle, `theta _(C)=sin^(-1)((mu_(2))/(mu_(1)))=sin^(-1)0.991=82.3^(0)`,`mu_(2)=0.991xx mu_(1)=0.991xx1.46=1.45`NA=`sqrt(mu_(1)^(2) -mu_(2)^(2))=sqrt ((1.46)^(2)-(1.45)^(2))=0.17` | |
| 29. |
Callular Mobile radio works on the frequency range is |
| Answer» Answer :A | |
| 30. |
Two identical particles having the same mass m and charges +q and -q separated by a distance d entering a uniform magnetic field vecB as shown in the figure. For what value of d the particles will not collide. |
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Answer» Solution :`F=Bqv=(MV^(2))/r rArrr=(mv)/(BQ)` `dltr_(1)+r_(2),dgt(mV)/(Bq)+(mV_(2))/(Bq)` `dgtm/(Bq)(V_(1)+V_(2))` 
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| 31. |
The period of a satellite in a circular orbit of radius R is T.The period of another satellite in circular orbit of radius 4R is: |
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Answer» 4T So the CORRECT CHOICE is (c ). |
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| 32. |
Two masses of 1 gm and 9 g are moving with equal kinetic energies, the ratio of the magnitudes of their respective linear momenta is |
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Answer» `1:9` |
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| 33. |
A charged particle is moving in a uniform magnetic field. Then ______ . |
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Answer» its MOMENTUM CHANGES but KINETIC ENERGY does not change. |
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| 34. |
A uniformly charged conducting sphere of 2.4 m dimeter has a surface charge density of 80.0 mu C//m^(2). Find the charge on the sphere. |
| Answer» SOLUTION :`-6.67 NC` | |
| 35. |
Distinguish between 'intrinsic' and 'extrinsic semiconductors. |
Answer» SOLUTION :
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| 36. |
emf of battery is 2.2 V. When 5Omega resisitor i connected across the battery, its termina voltage is 1.8 V. In tern al resistance of battery .... Omega. |
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Answer» `(10)/(9)` `EPSILON = V + I r ` `therefore r = (epsilon - V)/(I)` = `(2.2 - 1.8)/((V)/(R)) = (0.4)/((1.8)/(5))= (0.4 xx 5)/(1.8) ` `therefore r = (20)/(18)` = `(10)/(9) Omega` |
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| 37. |
An aeroplane is flying with the velocity of V_(1)=800 kmph relative to the air towards south. A wind with velocity of V_(2)=15 ms^(-1) is blowing from West to East. What is the velocity of the aeroplane with respect to the earth. |
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Answer» `222.7ms^(-1)` |
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| 38. |
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined systemto that stored initially in the single capacitor. |
| Answer» SOLUTION : `2 :1 ` | |
| 39. |
Which constituent radiation of the electromagnetic spectrum is used (i) in radar (ii) to photograph internal part of a human body, and (iii) for taking photographs of the sky during night and foggy conditions? Give one reason for your answer in each case. |
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Answer» Solution :(i) Microwaves are used in the operation of RADAR because microwaves have small wavelengths and are not DIFFERENT or bent by objects of normal diemension comming in its PATH. Due to it, these waves can be SENT in a particular direction as a beam singnal. (II) x-rays are used to photograph internal parts of a human body because x-rays are quite short wavelength. They can easily pass through flash (iii) Infrared rays are used for taking photographs of the sky during night and foggy conditions because these radiations are radialy absorbed by water molecules present in foggy conditions. After absorption of heat, these foggy particles are heater up and there by surrounding is heated. Due to, it the infrared rays can locate the object present there. |
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| 40. |
Find antilog of bar2.8036 |
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Answer» Solution :From the antilog table, FIND the value of 0.80 under 3 and ADD it to the mean difference given under 6. The number obtained is 6362. Now place the decimal point so that the characteristic of its LOG is `bar2`. THEREFORE, ANTI `logbar(2).8036=0.06362` |
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| 41. |
A solenoidal coil has 50 turns per centimetre along its length and a cross-sectional area of 4 xx 10^(-4)m^2 . 200 turns of another wire is wound round the first solenoid co-axially. The two coils are electrically insulated from each other. Calculate the mutual inductance between the two coils. |
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Answer» Solution :` n_1` = 50 turns per cm = 5000 turns per METRE `n_2l = 200,A = 4 xx 10^(-4) m^2` `M = mu_0 n_1 (n_2l) A` ` = 4pi xx 10^(-7) xx 5000 xx 200 xx 4 xx 10^(-4) H` ` = 5.03 xx 10^(-4) H` |
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| 42. |
A potentiometer has a wire of 100cm length and its resistance is 10 Ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40cm length of potentiometer wire, the value of 'E' is |
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Answer» 0.8 V |
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| 43. |
On what factors does the period of oscillation of a bar magnet in a uniform magnetic field depend. |
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Answer» Solution :1) MOMENT of inertia of the rectangular bar magnet (I) 2) Magnetic moment of the bar magnet (M). 3) Uniform magnetic FIELD (B). |
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| 44. |
Poise is the unit of |
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Answer» Pressure |
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| 45. |
कौन- सा राज्य चाय का प्रमुख उत्पादक राज्य है? |
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Answer» केरल |
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| 46. |
The expression relating polarising angle and refractive index is |
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Answer» `nsintheta=1` |
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| 47. |
Two concentric circular coils, one of small radius r_(1) and the other of large radius r_(2), such that r_(1) lt lt r_(2), are placed co-urially with centres coinciding. Obtain the mutual inductance of the arangement. |
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Answer» Solution :Let a current `I_(2)` flow through the outer circular coil. The field at the centre of the coil is `B_(2)=mu_(0) I_(2) // 2r_(2)`. Since the other co-axially placed coil has a very small RADIUS, `B_(2)` may be considered CONSTANT over its cross- sectional area, Hence. `Phi_(1)=pi r_(1)^(2)B_(2)=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)=M_(12)I_(2)` thus, `:.` mutual inductance of solenoid `S_(1)` with respect to `S_(2)` `M_(12)=(mu_(0)pir_(1)^(2))/(2r_(2))` -------(i) but `M_(12) = M_(21) = (mu_(0)pi r_(1)^(2))/(2r_(2))`, NOTE that we calculated `M_(12)` from an approximate value of `Phi_(1)` assuming the magnetic field `B_(2)` to be uniform over the area. `pi r^(2)`. However we can accept this value because `r_(1) lt lt r_(2)` `***` It would have been difficult to calculate the flux through the bigger coil of the nonuniform field due to the current in the smaller coil and hence the mutual inductance `M_(12)`. The equality `M_(12) = M_(21)` is helpful. Note also that mutual inductance DEPENDS solely on the geometry. |
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| 48. |
When p.d between two ends of secondary of full wave rectificationis 100 V then in p-n junctionp.d. is …… between p and n in reverse bias. |
| Answer» ANSWER :B | |
| 49. |
A wire of 5.8 m long, 2 mm diameter carries 750 mA current when 22 mV potential difference is applied at its ends. If drift speed of electrons is found 1.7xx10^(-5) m/s then |
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Answer» currenty DENSITY is `2.4xx10^(+5)A//m^(2)` |
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| 50. |
A storage battery of emf 8.0 V and internal resistance 0.5 Omegais being charged by a 120 V de supply using a series resistor of 15.5 Omega . What is the terminal voltage of the battery during charging ? |
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Answer» Solution :Charging CURRENT is given by formula, `I = (V - epsilon)/(R + r)` = `(120 - 8)/(15.5 + 0.5) ` = ` (112)/(16) ` `THEREFORE ` I = 7A Terminal voltage of battery during its charging `V. = epsilon` + Ir = 8 + (7) (0.5) 8 + 3.5 `therefore V. = 11.5` V By connecting series resistance, we can reduc charging current (with the increase in chargin, time). By REDUCING charging current, we reduce the heat loss in the internal resistance. |
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