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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. |
Three small spheres each carrying a positive charge Q are placed on the circumference of a circle of radius to form an equilateral triangle. The electric field intensity at the centre of the circle will be. |
| Answer» Answer :D | |
| 2. |
In the diffraction due to a single-slit experiment, the aperture of the slit is 3mm. If monochromatic light of wavelength 620 mm is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. the distance between the slit and the screen is 1.5 m. |
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Answer» Solution :Here aperture of the slit `a=3mm=3xx10^(-3)m`, wavelength of light `lamda=620nm=620xx10^(-9)m` and DISTANCE of screen from the slit, `D=1.5m` `therefore` Distance of first ORDER minima from central maxima point on the screen `x=(lamdaD)/(a)` and distance of 3rd order maxima from central maxima point `x.=((2xx3+1)lamdaD)/(2a)=(7lamdaD)/(2a)` `therefore` Separation between first order minima and THIRD order maxima `Deltax=x.-x=(7lamdaD)/(2a)-(lamdaD)/(a)=(5lamdaD)/(2a)=(5xx6200xx10^(-9)xx1.5)/(2xx3xx10^(-3))=7.75xx10^(-3)m=7.75mm` |
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| 3. |
Two like poles of strength m_(1) and m_(2)are fardistance apart the energy required to bring them r_(0)distanceapart is |
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Answer» `(mu_(0))/(4PI)(m_(1)m_(2))/(r_(0))` `=(mu_(0)m_(1)m_(2))/(4pir_(0))-0=(mu_(0)m_(1)m_(2))/(4pir_(0))` |
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| 4. |
किसी सीधी सड़क पर 144 km/h की चाल से जाती हुई कार को 200 मीटर की दूरी के अंदर ही रोक दिया जाता है। कार को रुकने में कितना समय लगेगा? |
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Answer» 5s |
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| 5. |
A moving snowball is completely melted by its impact against a wall. The speed (in ms^(-1)) of snow bal is: |
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Answer» `8 CDOT 2xx10^(2) m//s` |
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| 6. |
Among the following quantities , the quantity whose dimension is independent of mass and length is |
| Answer» Answer :D | |
| 7. |
What did the poet see the children doing? |
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Answer» CHILDREN COMING out of their homes |
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| 8. |
In Exercise. 11 obtain the frequency of revolution of the electron in tis circular orbit. Does the answer depend on the speed of the electron ? Explain . |
| Answer» Solution :Frequency in Exercise 11. `UPSILON = (qB)/(2pim) = (1.6 XX 10^(-19)xx6.5 xx 10^(-4))/(2 xx pi xx 9.1 xx 10^(-31))=18 MHz` | |
| 9. |
In a Young's double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case |
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Answer» there shall be alternate interference patterns of red and blue. In Young.s original DOUBLE slit experiment, monochromatic light with definite frequency was used. Hence the two slits behave LIKE coherent SOURCES and they can produce stationary interference pattern on the screen. But here light WAVES emitted by two slits have different frequencies and hence different colours. Hence phase difference between them does not remain constant and so they behave like non-coherent sources. Hence they can not produce any stationary interference on the screen and so we do not obtain interference fringes. |
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| 10. |
Draw a neat labelled diagram of an ac generator or dynamo. |
Answer» SOLUTION : It WORKS on the principle of ELECTROMAGNETIC induction i.e.,when a coil continuously rotates in a magnetic field the magnetic flux associated with it keeps on CHANGING thus and emf is induced in it |
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| 11. |
Assuming the average energy of electrons to be 3/5 of the Fermi energy estimate the pressure of the electron gas in a metal. Do the calculations for aluminium. |
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Answer» `W=3/5Nh^(2)/(2m)(3/(8pi))^(2//3)N^(2//3)V^(-2//3)` Differentiating, we obtain `dW=-2/3xx3/5h^(2)/(2m)(3/(8pi))^(2//3)N^(5//3)V^(-5//3)dV=-(2H^(2))/(5m)(3/(8pi))^(2//3)n^(5//3)dV=-2/5n epsi_(F)dV` The pressure of the electron gas is `P=2/5n epsi_(F)` |
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| 12. |
4muF capacitor and a resistance 2.5M Omega are in series with 12V battery, Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given In ( 2 ) = 0.693] |
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Answer» SOLUTION :a) Charging current `i=(V_(0))/(R)e^(-(t)/(RC))` `:.` POTENTIAL difference across R is `V_(R) =iR=V_(0)e^(-(t)/(RC))` `:.` Potential difference across .C. is `V_(C) =V_(0) - V_(R) = V_(0)-V_(0)e^(-(t)/(RC))= V_(0)(1-e^(-(t)/(RC)))` but given `V_(C) = 3V_(R)`, we get `1-e^(-t//RC)` or `1=4e^(-t//RC)` `e^((t)/(RC))=4 implies (t)/(RC)=ln4 implies t= 2RC ln2` `t = 2.5 xx 10^(6)xx 4 xx 10^(-6)xx2 xx 0.693` or t = 13.86 sec |
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| 13. |
The distance between two point sources of light is 24 cm. Find out where you would place a converging lens of focal length 9 cm, so that the images of both the sources are formed at the same point. |
| Answer» SOLUTION :6 CM from ONE of the SOURCE | |
| 14. |
फलक केंद्रित इकाई कोशिका में a तथा r में सम्बन्ध |
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Answer» a=2r |
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| 15. |
In Wheatsone bridge experiment as shown in figure- |
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Answer» KEY `K_(1)` should be PRESSED first and then `K_(2)` |
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| 16. |
A generator produces a voltage that is given by V = 240 sin 120t, where t is in second the frequency of voltage is ........ Hz. |
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Answer» 20 `omega=120` `therefore f=120/6.28` `therefore 2pif =120` `therefore f APPROX` =19 Hz `therefore f=120/(2PI)` |
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| 17. |
Two coherent point sources S_(1) and S_(2) vibrating in phase emit light of wavelength lambda. The separation between the sources is 2lambda. Consider a line passing through S_(2) and perpendicularto line S_(1) S_(2). Find the position of farthest and nearest minima. . |
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Answer» `(7lamda)/12` |
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| 18. |
A series combination of n identical cells has two cells P and Q with reverse polarities.IF emf of each cell is e and internal resistance is r, what is the potential difference across each of the cells P and Q? |
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Answer» |
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| 19. |
A step-up transformer operates on a 230 V line and supplies a load of ampere. The ratio of the primary and secondary windings is 1:25. The current in the primary is |
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Answer» Solution :`(E_(P))/(E_(S))=(N_(P))/(N_(S))=E_(S)=(E_(P)xx NS)/(N_(P))rArr E_(S)=5750 V` `I_(S)=2 amp, P_(S)=2xx5750 W` `therefore I_(P)=(P_(P))/(V_(P))=(P_(S))/(V_(P))=(2xx5750)/(230)=50 A`. |
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| 20. |
If epsilon_(0) and mu_(0) are the permittivity and permeability of free space and are the corresponding quantities for a medium, then refractive index of the medium is |
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Answer» INSUFFICIENT information |
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| 21. |
A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes |
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Answer» INCLINED at 45° to the MAGNETIC field |
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| 22. |
A balll is projected with 20sqrt(2) m/s at angle 45^@ with horizontal, the angular velocity of the particle at highest point of its journey about point of projection is |
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Answer» `0.1 rad//s` |
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| 23. |
If a diamond plate of refractive index 2.5 is introduced in path of second beam to bring the central band to original position then its thickness is |
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Answer» `1 xx 10^(-5) cm` `THEREFORE t = (lambda)/(mu_(d) -1) = (6000 xx 10^(-8))/(2.5 - 1) = 4 xx 10^(-5)` cm. |
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| 24. |
Discuss the process of nuclear fission and its properties. |
Answer» Solution :When uranium nucleus is bombarded with a neutron. It breaks up into two smaller nuclei of comparable masses with the release of energy.  The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is CALLED nuclear fission. The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of MAGNITUDE greater than the energy released in chemical reactions. Uranium undergoes fission reaction in 90 different WAYS. The most common fission reactions of `""_(92)^(235)"U"` nuclei are shown here. `""_(92)^(235)"U + ""_(0)^(1)"n"` to `""_(92)^(236)"U^(ast)"` to `.(56)^(141)"Ba" + ""_(36)^(92)"Kr" + 3_(0)^(1)n + Q` `""_(92)^(235)"U"` + `""_(0)^(1)"n" to ""_(92)^(236)"U^(ast)" to .(54)^(140)"Xa" + ""_(38)^(94)"Sr" + 2_(0)^(1)n + Q` Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is aborbed by the uranium nuclei, the mass number increases by one and goes to an excited state `""_(92)^(236)"U^(ast)"`. But these excited state does not last longer than `10^(-12)s` and average. 2.5 neutrons are emitted. |
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| 25. |
The shortest wavelength of X-rays emitted from an X- ray tube depends upon : |
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Answer» NATURE of the gas in the tube or `eV prop (1)/(lambda)` |
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| 26. |
Define salf-inductance of a coil. Obain the expression for the energy stored in an inductor L connected across a source of emf. |
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Answer» Solution :Selfinductanceis defined as the magenetic flux linked with the coil when UNIT current flows through it  Let current I flow through a,coil at any instant, then the magnetic flux `phi` , linked with the coil is found to be proportional to the strength of the current(I). `phi alpha Ior , phi = LI` Where L = self inductence , a constant of proportionality. Let, an indicator of inductance L. be connected across a battery when current I flowes throuth through the inductor , an emf `in` is induced in it. This induced emf is given by ` in = - L (dI)/(dt)`...(i) The-ve sign have implies that `'in`' opposes the passage of current I in the conductor. To drive the current through the inductor against the induced emf`'in`', the external valtage is applied . External valtage = - `in` From (i) ` in = L (dl)/(dt)` Let an infinitesimal change dq be driven through the inductor . So , the work DONE by the external suppley is given by, `domega = omega dq or domega= L (dl)/(dt) dq ` `domega = LdI(dq)/(dt) , rArr d omega = LIdItherefore(dq)/(dt) = 1` Total wark done to maintain the MAXIMUM value of current `(I_(0))`through the inductor is `intdomega = underset(0)overset (I)overset(0)intLIdI rArromega = L ((I^(2))/(2))_(0)^(I_(0))` `omega = L (I_(0)^(2))/(2) rArromega = (1)/(2)LI _(0)^(2)` This work done is STORED as the magnetic field energy u in the inductor and n`= 1//2L I^(2)` . |
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| 27. |
In an a.c. circuit the voltage across an inductance and resistance, joined in series, are 120 V and 160 V respectively. The total voltage applied across the combined circuit is ………….. |
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Answer» |
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| 28. |
A cube is arranged such that its length, breadth and height are along X, Y and Z directions, One of its corners is situated a the origin. Length of each side of the cube is 25cm. The components of electric field are E_(x) = 400 sqrt2 N//C, E_(y) = 0 and E_(z)= 0 respectively. Find the flux coming out of the cube at one end. |
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Answer» 25 |
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| 29. |
In each situation of Column I, a physical quantity related to orbiting electron in hydrogen -like atom is given. The terms 'Z' and 'n' given in Column II have usual meaning in Bohr's theory. Match the quantities in Column I with the terms they depend on it Column II. . |
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Answer» a-p,s , b-q,s , c-q,s , d-q,s |
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| 30. |
Select and write the corrcet answer : Which of the following would permit finer detail to be examined using a microscope of given numerical aperture ? |
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Answer» YELLOW light |
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| 31. |
Explain what do you mean by magnetic Lorentz force. |
| Answer» Solution :It is the force ACTING on a CHARGE MOVING in a MAGNETIC field. Magnetic Lorentz force is given by `vecF=q(vecvxxvecB)`. It acts perpendicular to the velocity vector and the magnetic field. | |
| 32. |
The mercury in a household glass-tube thermometer has a volume of 500mm^(3)(=5.0xx10^(-7)m^(3)) at T=19^(@)C. The hollow column within which the merucry can rise or fall has a cross-sectional area of 0.1mm^(2)(=1.0xx10^(-7)m^(2)). Ignoring the volume expansion of the glass, how much will the mercury rise in the thermometer when its temperature is 39^(@)C ? (The coefficient of volume expansion of mercury is 1.8xx10^(-4)//""^(@)C.) |
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Answer» SOLUTION :First let's figure out by how much the volume of the MERCURY increases. `DeltaV=betaV_(0)DeltaT=(1.8xx10^(-4))/(""^(@)C)(5.0xx10^(-7)m^(3))(39^(@)C-19^(@)C)=1.8xx10^(-9)m^(3)` Now, since volume=cross-sectional area `xx`height, the CHANGE in height of the mercury column will be `Deltah=(DeltaV)/(A)=(1.8xx10^(-9)m^(3))/(1.0xx10^(-7)m^(2))=1.8xx10^(-2)m=1.8m` |
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| 33. |
When a man stands inside a shallow pond the depth of water at that place appears _________ and the depth of other places appear comparatively ________ [Fill in blank]. |
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Answer» |
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| 34. |
There is a disc of mass M and radius R. A pont charge +Q and of mass m is fixed at the centre of the disc. The disc is held on a fixed horizontal rough suface . Another point charged +q is fixed on the surface such that disance between +q and +Q is (4R)/3. Now the disc is set free. If the disc does not loose contact with the ground at the same intent the relation must hold is Mgt eta m. Find eta. (Use (27Qq)/(256pi epsilon_(0)R^(2))=3mg) |
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Answer» `gt( 27Qq)/(256piepsilon_(0)R^(2))` `(M+m)g gt 3MG` `Mgt2mrarreta=2` 
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| 35. |
Two nonconducting plates A and B of radii 2R and 4R, respectively, are kept at distance x and 2x from the point charge q. A surface cutout of a nonconducting shell A is kept such that is center coincides with the point charge. Each plate and spherical surface carries surface charge density sigma. If F_A,F_B,and F_C are the forces on plate A, plate B, and spherical surface C due to charge q, respectively, then |
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Answer» `F_A = F_B = F_C` Flux `=phi=intE.ds` Force `=intsigmaE.ds=sigmaintE.ds=sigmaphi` As flux due to charge q in each plate is same force will be same. For C we have TAKEN the HORIZONTAL component of electric FIELD , but for As force due to ONE plate on another plate is same, hence `phi` due to 1 in another and due to 2 in 1 will be same. When x and R increased `phi` in plate due to charge remains same, hence force due to charge on plates does not change. |
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| 36. |
An ionised H-molecule consists of an electron and two protons. The protons are separatedby a small distance of the order of angstrom. In the ground state, |
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Answer» the electron would not move in circular orbits |
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| 37. |
Two nonconducting plates A and B of radii 2R and 4R, respectively, are kept at distance x and 2x from the point charge q. A surface cutout of a nonconducting shell A is kept such that is center coincides with the point charge. Each plate and spherical surface carries surface charge density sigma. If phi_1 is flux through surface of (B) due to electric field to (A) and phi_2 is the flux through (A) due to electric field of (B), then |
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Answer» `phi_1 = phi_2` Flux `=PHI=intE.ds` Force `=intsigmaE.ds=sigmaintE.ds=sigmaphi` As flux due to charge q in each plate is same force will be same. For C we have taken the horizontal component of electric field , but for As force due to one plate on another plate is same, hence `phi` due to 1 in another and due to 2 in 1 will be same. When x and R increased `phi` in plate due to charge remains same, hence force due to charge on plates does not change. |
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| 38. |
Two nonconducting plates A and B of radii 2R and 4R, respectively, are kept at distance x and 2x from the point charge q. A surface cutout of a nonconducting shell A is kept such that is center coincides with the point charge. Each plate and spherical surface carries surface charge density sigma. If distance x and radius R are doubled so that F_A becomes F'A,F_B becomes F'B, then the correct option is |
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Answer» `F_BgtF'_B` Flux `=phi=intE.ds` Force `=intsigmaE.ds=sigmaintE.ds=sigmaphi` As flux DUE to charge Q in each plate is same force will be same. For C we have taken the HORIZONTAL component of electric field , but for As force due to one plate on another plate is same, hence `phi` due to 1 in another and due to 2 in 1 will be same. When x and R increased `phi` in plate due to charge remains same, hence force due to charge on plates does not CHANGE. |
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| 39. |
Let E be the electric field of magnitude 6.0 xx 10^(6) N C^(-1) and B be the magnetic field magnitude 0.83 T. Suppose an electron is accelerated with a potential of 200 V, will it show zero deflection ? If not, at what potential will it show zero deflection . |
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Answer» SOLUTION :Electric field , E = `6.0 xx 10^(6) N C^(1)` and magnetic field , B = 0.83 T Then, v = `(E)/(B) = (6.0 xx 10^(6))/(0.83) = 7.23 xx 10^(6) ms^(-1)` when an electron goes with this velocity, it shows null deflection.SINCE the accelerating potential is 200 V , the electron acquires kinetic energy because of this accelerating potential Hence, `(1)/(2) mv^(2) = eV rArr v = sqrt((eV)/(2m))` Since the mass of the electron, m = 9.1`xx 10^(-31)` kg and charge of an electron, |q| = e = 1.6 `xx 10^(-19) C` The velocity due to accelerating potential 200 V `v_(200) = sqrt( (2(1.6 xx 10^(-19))(200))/((9.1 xx 10^(-31) )))= 8.39 xx 10^(6) m s^(-1)` Since the speed `v_(200) gt `v, the electron is DEFLECTED towards direction of Lorentz force. So, in order to have null deflection , the potential , we have to supply is V = `(1)/(2) (mv^(2))/(e ) = ((9.1 xx 10^(-31)) xx (7.23 xx 10^(6))^(2) )/(2 xx (1.6 xx 10^(-19)) ` V = 148.65 V |
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| 40. |
The graph of the total number of alpha -particles scattered at different angles in a given interval of time for alpha-particles scattering in the Geiger-marsden experiment is given by (Number of scattered particles taken on y-axis and scattering angle theta taken on x-axis) |
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Answer»
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| 42. |
If the susceptibility of dia, para , and ferro magnetic materials are X_d,X_p,X_f respectively , then |
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Answer» `X_d LT X_p lt X_f` |
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| 43. |
Describe briefly, with the help ofa diagram, the role of the two important processes involved in the formation of a p-n junction. |
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Answer» Solution :Two important PROCESSES involved in the FORMATION of a p-n junction are setting up of DIFFUSION a Current and drift current. These leads to formation of: (i) depletion region, and (II) barrier potential. For these processes, SEE Short Answer Question Number 5. |
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| 44. |
Thefour arma os a Wheatstone bridge have the following resistances: AB =100Omega, BC =10 Omega, CD=5Omega and DA=60Omega A galvanometer of 15Omega resistacne is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 Vis maintained across AC. |
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Answer» Solution :Considering the MESH BADB, we have `100I _(1) + 15I _(g) - 60 I _(2) =0` or `20 I _(1) + 3I _(g) -12 I _(2) =0` Considering the mesh BCBD, we have `10 (I _(1) - I _(g)) - 15 I _(g) - 5 (I _(2) + I _(g)) =0` `10I _(1) - 30 I _(g) - 5I _(2) =0` `2I _(1) - 6I _(g) - I _(2) =0` Considering the mesh ABCDA. `60I _(2) + 5 (I _(3) + I _(g)) =10` ` 65 I _(2) +5 I _(g) =10` ` 13 I _(2) + I _(g) =2` Multiplying `(3.84b)` by 10 `20I _(1) - 60I _(g) -10I _(2)=0` From eqs. `(3.84d) and (3.84a)` we have `63I_(g) - 2I _(2) =0` `I _(2) = 31. 5I_(g)` Substtuting the vlaue of `I _(2)` into Eq. `[3.84(C)],` we get `13 (31. 5I_(g)) + I _(g) =2` `410.5 I_(g) =2` `I _(g) = 4. 87 mA.` |
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| 45. |
The 6563A^(@) line emitted by hydrogen atom in a star is found to be red shifted by 5A^(@). The speed with which the star is recedingfrom the earth is |
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Answer» `17.3 XX 10^(3)m//s` |
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| 46. |
Kepler's second law regarding constancy of areal velocity of a planetisa consequence of the law of conservation of: |
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Answer» energy Thus correct CHOICE is (c ). |
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| 47. |
In a Young.s doubles slit experiment the slit separation is 0.5m from the slits. For a monochromatic light of wavelength 500nm, the distance of 3^(rd) maxima from 2^(nd) minima on the other side is |
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Answer» 2.25mm D REQUIRED to workout DATA insufficient. Moreover .d. is huge value 0.5m !! (should be in mm) No answer.s match |
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| 48. |
Three resistances of 3 Omegaeach are connected in the form of an equilateral triangle. The effective resistance between any two corners is ________. |
| Answer» Solution :`2OMEGA`. The ARRANGEMENT is a parallel combination of resistance of 3`OMEGA` and 6`Omega` | |
| 49. |
The time period of pendulum at temperature t_(1)^(@)C is T_(1) s. Its time periodat t_(2)^(@)C is T_(2) s. If coefficient of linear expansion of material of pendulum is alpha, then increase in time period is : |
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Answer» `alpha(t_(2)-t_(1))` |
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