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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. |
Fraunhofer's lines are due to which phenomenon? |
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Answer» |
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| 2. |
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment. Write down the expression for its magnifying power. |
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Answer» SOLUTION :A ray DIAGRAM showing image formation by an astronomical telescope in near POINT POSITION is shown in Fig. 9.51. The magnifying power of telescope in near point position `m=-f(0)/f_(e) (1+f_(e)/D)` 
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| 3. |
A body cools at the ratio of 1.2^(@)C//min when its temperature is more than that of the surrounding by 40^(@)C. The rate of cooling of the body when its temperature is more than that of surrounding by 25^(@)C will be |
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Answer» `0.75^(@)C`/MIN `(R_(2))/(R_(1))=((theta_(2)-theta_(0)))/((theta_(1)-theta_(0)))=(25)/(40)=(5)/(8)` `R_(2)=0.75`. |
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| 4. |
To turn galvanometer into ammeter, _______ |
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Answer» GREATER RESISTANCE should be JOINED in series. |
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| 5. |
The condition for the first resonance in a resonating air column closed at one side is, |
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Answer» `l+e=lamda/2` |
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| 6. |
Two cells of e.m.f. 's E_1 and E_2are connected in series in a circuit . Let r_1 and r_2 be the internal resistance of the cells. Find the current through the circuit. |
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Answer» SOLUTION :The CELLS are in SERIES. E= `E_1 + E_2` `r_1 and r_2 ` are the INTERNAL resistancescurrent`i=(E_1 + E_2)/((r_1 + r_2))` |
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| 7. |
The work done in rotating a bar magnet of magnetic moment M from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field B is |
| Answer» ANSWER :D | |
| 8. |
A capacitor C_1 of capacitance 5muF is charged to a potetntial of 100V and another capacitor C_2 of capacitance 8muF is charged to 50V. The positive and negative plates are mutually connected. Energy loss will be |
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Answer» `3.11xx10^(-13)J` |
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| 9. |
What is the minimum height to which a gamma quanta source containing excited Zn^(67) nuclei has to be raised for the gravitational displacement of the Mossbauer line to exceed the line width itself, when registered on the Earth's surface? The registered gamma quanta are known to have an energy epsilon= 93keV and appear on transition of Zn^(67) nuclei to the ground state, and the mean lifetime of the exited state is tau=14 mu s. |
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Answer» Solution :The natural LIFE time is `GAMMA=( ħ)/(tau)= 4.7xx10^(-10)eV` Thus the condition `deltaE_(gamma)GE Gamma" implies" (gh)/(c^(2))ge(Gamma)/(epsilon)=( ħ)/(tau epsilon)` or `h ge(c^(2) ħ)/(tau epsilong)= 4.64` meter. ( `h` here is height of the place, not plank's constant.) |
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| 10. |
The source of electromagnetic waves can be a charge |
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Answer» moving with a CONSTANT velocity |
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| 11. |
A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will : |
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Answer» SEPARATE the red colour PART from the green and blue colours `therefore mu=sqrt2` `therefore mu=1.414` `1.39 LT 1.414,1.44 gt 1.414,1.47 gt 1.414` `mu_(red) lt mu,mu_(green) gt mu, mu_(blue) gt mu` Therefore, only red colour will not do tota internal reflection. |
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| 12. |
What modification has to be made in the galva nometer if he still wants to use the galvanometer in place of the ammeter? |
| Answer» SOLUTION :USE a SHUNT RESISTANCE. | |
| 13. |
Which of the following is not a unit of electric field? |
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Answer» `NC^(-1) ` |
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| 14. |
Calculate the resistance between points A and B for the following networks : |
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Answer» |
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| 15. |
The human eye has an approximate angular resolution of phi = 5.8 xx 10^(-4) rad and a typical photprinter prints a minimum of 300 dpi (dots per inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots ? |
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Answer» 10 CM `L = (2.54)/(300) cm APPROX 0.84 xx 10^(-2) cm` At a distance of Z cm this subtends an angle. `phi -l/z therefore z = (l)/(phi) = (0.84 xx 10^(-2)cm)/(5.8 xx 10^(-4)) approx 14.5 cm`. |
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| 16. |
Two independent monochromatic sources of light cannot produce a sustained interference pattern. Give reason. |
| Answer» Solution :The phase DIFFERENCE between the light waves originating from two indepent MONOCHROMATIC sources will change RAPIDLY with time. | |
| 17. |
We see the sun a few minutes _________ sunset or _________ sunrise [Fill in the blank]. |
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| 18. |
An accelerated electron is absorbed by a proton at rest and a neutron is formed. Write the reaction equation. Assuming that the resulting neutron remains at rest, calculate the minimum kinetic energy of the electron at which the reaction is possible. |
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Answer» `p_(0)=p_(v),epsi_(e)+epsi_(0p)=epsi_(on)+epsi_(v)` Since the rest mass of a neutrino is zero, its energy is `epsi_(v)=p_(v)c=p_(e)c=sqrt(epsi_(e)^(2)-epsi_(0e)^(2))` Hence `epsi_(e)=epsi_(0p)=epsi_(0n)+sqrt(epsi_(e)^(2)-epsi_(0e)^(2))` Solving this equation, we obtain for the total energy of an electron `epsi_(e)=((epsi_(0n)-epsi_(0p))^(2)+epsi_(0e)^(2))/(2(epsi_(0n)-epsi_(0p)))` The KINETIC energy of the electron is `K_(e)=epsi_(e)-epsi_(0e)=((epsi_(0n)-epsi_(0p)-epsi_(0e))^(2))/(2(epsi_(0n)-epsi_(0p)))` Substituting the respective VALUES, we obtain the desired result. |
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| 19. |
A string of length L consists of two distinct sections. The left half has linear mass density mu_(1) = mu_(0)//2 , while the right half has linear mass densitymu_(2) = 3 mu_(0) Tension in the string is F_(0). The time required for a transverse wave pulse to travel from one end of the string to the other is: |
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Answer» `L/4 SQRT(mu_(0)/F_(0)) (sqrt(2) + sqrt(6))` |
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| 20. |
In an aluminium there is a hole of diameter 2 m and is horozontally mounted on a stand. Onto this hole an iron sphere of diameter 2.004 m is resting. Initial temperature of this system is 25^(@)C. Find at what temperature the iron sphere will fall down through the hole in sheet. The coefficient of linear expansion for aluminium and iron are 2.4xx10^(-5) and 1.2xx10^(-5) respectively. |
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Answer» Solution :As value of coefficient of linear expansion for ALUMINIUM is more than that for iron, it expends FASTER then iron. So at some higher temperature when diameter of hole will exactly BECOME equal to that of iron sphere, the sphere will pass thorugh the hole. Let if happen at some higher temperature T. Thus we have at this temperature T, `("diameter of hole")_(Al)=("diameter of sphere")_("iron")` `2[1+alpha_(Al)(T-25)]=2.004[1+alpha_("iron")(T-25)]` `2alpha_(Al)=(T-25)=0.004+2.004alpha_(iron)(T-25)` or `t=(0.004/(2alpha_(Al)-2.004alpha_("iron"))+25).^(@)C` or `T=0.004/(2.xx2.4xx10^(-5)-2.004xx1.2xx10^(-5))+25` or `T=191.7^(@)C` |
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| 21. |
Which of the following radioactive substance is used in archeological survey? |
| Answer» Answer :A | |
| 22. |
A microscope is focussed on a coin lying at the bottom of a beaker. The micorscop is now raised up by 1 cm. To what depth should the water be poured in to the beaker so that coin is again in focus (mu_(w)=4//3) |
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Answer» 1 CM REAL, inverted, magnified, formed in the GLASS |
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| 23. |
Who were the other visitors? |
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Answer» Squirrels |
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| 24. |
A wheel with 10 metallic spokes each 0.5 m long rotated with a speed of 120 "rev"/"min" in a plane normal to the horizontal component of earth's magnetic field B_h at a place. If B_h = 0.4G at the place, what is the induced emf between the axle and the rim of the wheel ? (1 G = 10^(-4) T) |
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Answer» 0V `epsilon=(B omegaR^2)/2`…(i) (Here B=0.4 , `G=0.4 xx10^(-4)` T) `omega=120 "rev"/"MIN"=(240pi)/60"rad"/s` `omega=4pi` rad/s,R=0.5 m, N=10 `THEREFORE epsilon=(0.4xx10^(-4)xx4xx3.14xx(0.5)^2)/2` [ `because` From eqn. (i)] `=(0.4xx4xx3.14xx0.25xx10^(-4))/2` `=0.628xx10^(-4)` `=62.8xx10^(-6)` `=62.8 muV` |
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| 25. |
200 Mev energy is released when one nucleus of ""^(235)Uundergoes fission. Find the number of fissions per second required for producing a power of 1 mega watt. |
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Answer» `3.125 xx10^(14)` |
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| 26. |
In a double slit experiments, the two slits are 1 mm apart and the screen is placed in away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double sli within the central maxima of single sli pattern? |
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Answer» `0.2` mm `d sin theta_(10)= 10 lambda` `:. Sin theta_(10)=(10 lambda)/(d)""...(1)` Now, angular WIDTH of central maximum by SINGLE slit `:. Sin2theta_(1)-(2lambda)/(d.)""...(2)` but `sin theta_(10)=sin 2 thea_(1)` `(10 lambda)/(d)=(2lambda)/(d.)` `:.d=0.2 mm` |
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| 27. |
Suppose that 64 raindrops combine into a single drops. Calculate the ratio of the total surface of 64 drops to that of single drop. (S.T. of water = 0.072 N/m) |
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Answer» 4 : 1 |
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| 28. |
If the atom ""_(100)Fm^(257) follow the Bohr's model and the radius of ""_(100)Fm^(257) is n tince the Bohr radius, then find n : |
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Answer» 100 Here, n=1, Z=1 `therefore r_(H)=(h^(2) epsi_(0))/(pi me^(2))` THUS, `r_(n)=(n^(2))/(Z) r_(H)` Here, n=5 for OUTER most orbit, Z=100 `r_(n)=(5^(2))/(100) r_(H) rArr r_(n)=1/4 r_(H)` But `r_(n) =n r_(H) ("given")` `n=1/4` |
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| 29. |
A prism of a refracting angle 60^(@) is made with a material of refractive index mu. For a certain wavelength of light, the angle of minimum deviation is 30^(@). For this wavelength, the value of mu of material is |
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Answer» 1.82 |
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| 30. |
In frequency modulation |
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Answer» the AMPLITUDE of the modulated wave varies as FREQUENCY of the CARRIER wave |
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| 31. |
As shown in figure, four electric charges are placed on vertices of a square and there is a free electron on it. Where this free electron will move ? |
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Answer» it will move toward A. |
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| 32. |
The minimum speed with which a body must be thrown to reach a height of (R )/(4) above the surface of earth is : |
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Answer» `sqrt(g (R )/(2))` `rArr v=sqrt((2GM)/(5R))=sqrt((2)/(5R)*gR^(2)) = sqrt((2)/(5)gR)` Thus CORRECT choice is (C ). |
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| 33. |
There is one isolated atom of iron which emits K_alpha X-rays of energy 6.4 keV . With what kinetic energy, the iron atom will recoil If mass of iron atom is 9.3xx10^(-26) kg. |
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Answer» Solution :Using conservation of linear momentum we can understand that the linear momentum of iron atom MUST be equal in magnitude but opposite in direction to that of linear momentum of photon so that total linear momentum remains zero after the emission of photon because there is no external force ACTING on the system . Linear momentum of photon can be written as follows : `p=E//c=(6.4xx10^3xx1.6xx10^(-19))/(3xx10^8)=3.41xx10^(-24) kg m s^(-1)` Linear momentum of iron is also same in magnitude so its kinetic energy can be written as follows : `K=p^2/(2m)=(3.41xx10^(-24))^2/(2xx9.3xx10^(-26))=(11.63xx10^(-48))/(18.6xx10^(-26))=6.25xx10^(-23)` J `=(6.25xx10^(-23))/(1.6xx10^(-19))`EV =`3.9xx10^(-4)` eV |
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| 34. |
A solenoid is of length 20 cm and consists of 1000 turns. When a current of 2A is passed through the coil of the solenoid, what is the value of magnetic induction at the centre of the solenoid. DATA: i=5A, N=1000 l=20 cm =0.2 m mu_@=4pixx10^(-7) ((Wb)//A To find:B=? |
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Answer» Solution :Magnetic FIELD B at the CENTRE of the SOLENOID: `B=mu_@xx(N/l)xxi` `(4xx3.14xx10^(-7)xx1000xx2)/(0.2)`=0.1256T. |
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| 35. |
Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends. |
| Answer» SOLUTION :ElectricaJ CONDUCTIVITY of a PURE semiconductor depends on (i) the energy gap (band gap) e.g., between the VALENCE band and the conduction band (ii) the TEMPERATURE. | |
| 36. |
What is the working principle of a cream separator ? |
| Answer» Solution :When the milk is churned in a vessel, CREAM being LIGHTER moves towards the axis of the vessel and COLLECTS in the MIDDLE. | |
| 37. |
A point particle carries a +ve charge +q and is maintained at a distance h above a large conductive uncharged plate with earth connected. The electric force of interaction between the charge and the plate is best described as: |
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Answer» ZERO FORCE |
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| 38. |
A ball reaches a rocket at 60 m/s along + X direction, and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50gm and the contact time is 0.02 second, the force exerted by the racket on the ball is |
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Answer» 300 N along + X direction |
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| 39. |
Given quantity of water is boiled by an electric heater in 5 min. If supply voltage of heater reduces to half then time taken to boil the same quantity or water will be ...... min. (Assume the resistance of the beater remaining constant) |
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Answer» 40 H = `I^(2) RT= (V^(2))/(R^(2)) xx Rt= (V^(2))/(R) t ` In t =`(HR)/(V^(2))`, HR REMAINS same. `therefore t PROP (1)/(V^(2))` `therefore (t_(2))/(t_(1)) = ((V_(1))/(V_(2)))^(2) ` = `((V)/((V)/(2)))^(2)` = 4 `therefore t_(2)= t_(1) xx 4 = 5 xx 4 `= 20 MIN |
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| 40. |
The diagram shows a uniformly chargedbemisphere of radius R. It has volume charge density rho. If the electric field at a point 2R distance below its centre ? |
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Answer» `(RHOR)/(6epsilon_(0))+E` |
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| 41. |
which of the following can be made into crystal- |
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Answer» A bacterium |
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| 42. |
For a current loop we can define magnetic moment mu . |
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Answer» `mu` is a vector quantity |
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| 43. |
A potentiometer wire of length 1.0 m has a resistance of 15 ohm. It is connected to a 5 V source in series with a resistance of 5 Omega. Determine the emf of the primary cell which has a balance point at 60 cm. |
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Answer» Solution :Current `I=V/(R_1+R_2)` `=5/(15+5)`=0.05 A Potential drop across the potentiometer wire V = IR= 0.25 × 15 = 3.75 V Potential gradient, `k=V/t=3.75/1.0 =3.75` V/m `therefore`Unknown emf of the CELL = KI = 3.75 × 0.6 = 2.25 V. |
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| 44. |
An observer is at 2 m from an isotropic point source of light emitting 40 W power. What are the rms values of the electric and magnetic fields due to the source at the position of the observer ? [c=3xx10^(8)ms^(-1), epsilon_(0)=8.854xx10^(-12)C^(2)N^(-1)m^(-2)] |
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Answer» Solution :INTENSITY `I=(P)/(A)` `therefore epsilon_(0)c E_(rms)^(2)=(P)/(4PI R^(2))` `therefore E_(rms)=SQRT((P)/(4pi R^(2)epsilon_(0)c)) "" (P=40 W, R=2m)` `therefore E_(rms)=sqrt((40)/(12.56xx4xx8.85xx3xx10^(-12)xx10^(8)))` `=sqrt((10xx10^(4))/(331.875))` `=17.3 Vm^(-1)` Now, `B_(rms)=(E_(rms))/(c )=(17.3)/(3)xx10^(-8)=5.77xx10^(-8)T`. |
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| 45. |
Calculate the binding energy and binding energy per nucleon (in MeV) of a nitrogen nucleus (""_(7)^(14)N)from the following data: Mass of proton=1.0078 u Mass of neutron=1.00867 u Mass of nitrogen nucleus=14.00307 u |
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Answer» <P> Solution :Calculation of mass defect.`Delta m=Zm_(p)+(A-Z)m_(n)=-M` `Delta =7xx1.00783+7xx1.00876-14.00307` Calculation of binding energy for 1 U(1amu) binding energy `Delta=931.5` MEV For `0.11243u`, Ginding energy `DeltaE=0.11243xx931x931.5 MeV` `=104.72854 MeV` Binding energy per nucloen `=(BE)/(A)=(104.7258)/(14)` `=7.4861.` MeV/ nucloeon |
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| 46. |
Briefly explain the principle and working of electron microscope. |
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Answer» Solution :Principle `:` (i) The wave NATURE of the electron is used in the construction of microscope CALLED electron microscope. (ii) The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under stydy . (iii)Higher magnification as well as higher resolving power can be obtained by employing the shorter wavelengths of `e^(-)`. (iv) Electron microscopes giving magnification more than 2,00,000 TIMES than optical microscope.  Working `:` (i) The construction and working of an electron microscope is SIMILAR to that of an optical microscope except that in electron microscope focussing of electron beam is doen by the electrostatic or magnetic lenses. (ii) The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done ( Figure ) . (iii) The electrons emitted from the source are accelerated by high potential. (iv) The beam is made parallel by magnetic condenser lens. When the beam passs through the SAMPLE whose magnified image is needed, the beam carries the image of the sample. (v) With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. |
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| 47. |
An echo repeats two syllables. If the velocity of sound is 330 m/s. then the distance of reflecting surface is : |
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Answer» 16.5 m `therefore ` TIME TAKEN to speak disyllable = `(2)/(5) `sec If d is the distance of reflector then. ` t = (2 d)/(V) ` `(2)/(5) = (2d)/(330)` `rArr "" d = 66 m ` Hence the corret choice is (C). |
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| 48. |
Column-I contains the path traced by light rays and the positions of object and image, with a reflecting spherical mirror. Choose the possible options for type of mirror and the nature of image rom Column-II to each case in Column- I |
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Answer» |
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| 49. |
A water fountain on the ground sprinkles water all around it. If the speed of water coming out of fountain is v, the total area around the fountain that gets wet is : |
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Answer» `pi v^(4)/G^(2)` `:.` TOTAL area around the fountain that gets wet is `piR_(max)^(2)` `pi(v^(2)/g)^(2)=piv^(4)/g^(2)` |
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| 50. |
Two concentric circular coils, one of small radius r_(1) and the other of large radius r_(2), such that r_(1)ltltr_(2), are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. |
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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 `r_(1)ltltr_(2),B_(2)` may be considered constant over cross-sectional area of the other coil. Hence, flux linking with it is `phi_(1)=pir_(1)^(2)B_(2)=(mu_(0)pir_(1)^(2))/(2r_(2))I_(2)=M_(12)I_(2)` Thus, `M_(12)=(mu_(0)pir_(1)^(2))/(2r_(2))` from equation (3) and (6) `M_(12)=M_(21)=(mu_(0)pir_(1)^(2))/(2r_(2))` |
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