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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 4951. |
Asseration: An artificial satellite with a metal surface is moving above the earth in a circular orbit. A current will be induced in satellite if the plane of the orbit is inclined to the plane of the equator. Reason: The current will be induced only when the speed of satellite is more than 8 km//sec |
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Answer» If both ASSERATION and reason are true and reason ISTHE correct explanation of assertion. |
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| 4952. |
Two point sources distant 0.1m away viewed by a telescope. The objective is covered by a screen having a hole of 1mm width. If the wavelength fo the light used is 6550 A^(0), then the maximum distance at which the two sources are seen just resolved, will be nearly |
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Answer» `125.0m` |
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| 4953. |
In the circuit, all capacitor are identical, each of capacity 2muFand they are infinite in number. If AB is connected to a battery of 10V then the charge drawn from the battery is : |
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Answer» `40 MUC` |
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| 4954. |
Assertion: vecE=E_(x)hati+E_(y)hatj+E_(z)hatk, vecV xx vecE=0 Resion E_(x), E_(y), E_(z) is independent. |
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Answer» If both ASSERTION and REASON are TRUE and reason is the CORRECT EXPLANATION of assertion. |
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| 4955. |
Power deliverd to a load of impedance Z_(L) is optimum when the impedance of transmission line Z is : |
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Answer» greater than `Z_(L)` |
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| 4956. |
The wavelength of H_alphaline in hydrogen spectrum was found 6563Å in the laboratory. If the wavelength of same line in the spectrum of a milky way is observed to be 6586Å, then the recessional velocity of the milky way will be |
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Answer» `0.105 XX 10^6 ms^(-1)` |
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| 4957. |
An electric dipole consisting of two opposite charges of 2xx10^-6 C each separated by a distance of 3cm is placed in an electric field of 2 xx10^5N/C. the maximum torque on the dipole is will be |
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Answer» `12x10^-1` NM |
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| 4958. |
Using Ampere's circutial law, obtain the expression for the magnetic field due to a long solenoid a point inside the solenoid on its axis. |
Answer» SOLUTION : `oint vec(B)*vec(DL)= mu_(0)Sigmai` `int_(a)^(b) vec(B)*vec(dl)+int_(b)^(c)vec(B)*vec(dl)+ int_(c)^(d)vec(B)*vec(dl)+int_(d)^(a)vec(B)*vec(dl)=mu_(0)I(NH)` `Bh + 0 + 0 + 0 = mu_(0) I (nh)` `B= mu_(0)N I` |
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| 4959. |
Define the term .mutual inductance. between the two coils. Obtain the expression for mutual inductance ofa pair of long coaxial solenoids each of length l and radii r_(1) and r_(2) (r_(2) gt gt r_(1)). Total number of turns in the two solenoids are N_(1) and N_(2) respectively. |
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Answer» Solution :Mutual inductance, between a PAIR of coils, equals the magnetic flux, linked with ONE of them DUE to a change of unit current flowing in the other. Alternatively: The mutual inductance, for a pair of coils, equals the emf induced, in one of them, when the current in the other coil is changing at a unit rate.  Let a current `I_(2)` flow through the outer coil. The magnetic field due to thiscurrent ` = mu_(0) (mu_(2))/(l) xx I_(2)` The resulting magnetic flux linked with the INNER coil`= phi_(12) = N_(1).(mu_(0) (mu_(2))/(l) xx l_(2)) xx pi r_(1)^(2)` ` = ( mu_(0) (N_(1) N_(2))/(l), pi r_(1)^(2))l_(2)` `M_(12)I_(2)` `:.M_(12) = mu_(0)(N_(1)N_(2))/(l) . pi r_(1)^(2)` |
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| 4960. |
A block of mass m is placed on a surface with a vertical cross section given by y=x^(3)/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is : |
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Answer» `1/2m` |
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| 4961. |
A chargedmetallic spere A is suspended by a nylon thread another charged metallic spere b heldby an insulating centres is 10 cmc and dare then removed and b is brought closer to a to a what is txpectd repulsionofa on the basis of columb law sphereaandc speres b and dhave indentical sizes ignore the sizes ofa and b in comparision to the separation between theircentres |
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Answer» Solution :Letthe originalcharge on spere a be q and that on b beq at a distance R beteen theircentres the MAGNITUDE of the `F=(1)/(4piepsilon_(0))(QQ)/(r^(2))` neglecting the sizes of sperhers aand b in comparionsto r when an identical but unchanged spere c touches a the chargesredisitribute on a and c and by symmetry each sprere CARRIES a charge `q//2`similary after d touchesb the rredistributedcharge on each isof the electrosatic force on each is `F=(1)/(4pi epsilon_(0))((q//2)(q//2))/(r//2)^(2)=(1)/(4piepsilon_(0))(qq)/(r^(2))=F` Thus the electrostaticforce on a due to b remainsunaltered |
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| 4962. |
An object is projected horizontally from a top of the tower of height h. The line joining the point of projection and point of striking on the ground makes an angle 45^@ with ground, then with what velocity the object strikes the ground |
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Answer» `SQRT((11gh)/(2))` |
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| 4963. |
An LCR series circuit consists of a resistance of 10omega, a capacitor of reactance 60 Omega and an inductor coil. The circuit is found to resonate when put across a 300 V, 100 Hz supply. The inductance of the coil is (take pi= 3) |
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Answer» 0.1 H |
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| 4964. |
A solid conducting sphere of radius .a. is surrounded by a thin uncharged concentric conducting shell of radius 2a. A point chargae q is placed at a distance 4a from common centre of conducting sphere and shell. The inner sphere is then grounded The charge on solid sphere is |
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Answer» `-(Q)/(2)` |
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| 4965. |
(A): Coulomb force between charges is central force (R ): Coulomb force depends an medium between charges |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT explanation of .A. |
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| 4966. |
The angular speed of the electron in the n^(th) Bohr orbit of the hydrogen atom is proportional to...... |
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Answer» N Angular velocity `omega=(nh)/(2pimr^(2))` `:.r^(2)=(n^(4)h^(4)epsi_(0)^(2))/(pi^(2)m^(2)E^(4))` `omega=(pime^(4))/(2h^(3)in_(0)^(2)).(1)/(n^(3))` `:.omegaprop(1)/(n^(3))` |
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| 4967. |
A current I_(1) carrying wire AB is placed near another long wire CD carrying current I_(2). If wire AB is free to move, it will have |
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Answer» ROTATIONAL MOTION only |
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| 4968. |
A convex lens is placed in water. a.Will there be any change in its focal length?B.Give reason. |
| Answer» Solution :There will be an increases in the focal LENGTH ofthe convex lens. This is because the refractive INDEX of GLASS with respect to water is LESS than the refractive index of glass with respect to air. | |
| 4969. |
A magnet NS is suspended from a spring and M while oscillates, the magnet moves in and out of the coil C. The coil is connected to a galvanometer G. Then as the magnet oscillates |
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Answer» G shows deflection to the left and right with CONSTANT AMPLITUDE |
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| 4970. |
Ratio of amplitude of two coherent sources is 5:2. Ratio of intensity of fringes of constructive and destructive interference for stationary interference of these wave is..... |
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Answer» `(49)/(9)` `:. ((E_(1)+E_(2))^(2))/((E_(1)-E_(2))^(2))=((7)^(2))/((3)^(2))=(49)/(9)` `:. (I_(max))/(I_(min))=(49)/(9)` |
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| 4971. |
Which of the following units is different from other ? |
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Answer» ELECTRON volt |
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| 4972. |
In each fission of U^235, 200 MeV of energy is released. If a reactor produces 100MW power the rate of fission in it will be |
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Answer» `3.125xx10^18` PER MIN |
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| 4973. |
Explain the formation of stationary waves in an air column enclosed in open pipe . Derive the equations for the frequencies of the harmonics produced. |
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Answer» Solution :A pipe, which is opened at both ends is calledopen pipe.When a sound waves is sent through a open pipe , which gets reflected by the earth . Then incident and reflected waves are in same frequency,travelling in the opposite DIRECTIONS aresuper imposedstationary waves are formed. Harmonics in open pipe `:` To form the STATIONARY wave in open pipe, which has to anti nodes at two ends of the pipe with a node between them . `:.` The vibrating length ( l) `=` half of the wavelength `((lambda_(1))/(2))` `l = ( lambda_(1))/(2) implies lambda_(1) = 2l` fundamentalfrequency `v_(1) = (v)/( lambda_(1))` where v is velocity of sound in air `v_(1) =(upsilon)/( 2l) = v `...(1) For second harmonic ( first OVERTONE)will have one more node and antinode than thefundamental. If `lambda_(2)`is wavelength of second harmonic`l = ( 2lambda_(2))/(2) implies lambda_(2) =(2l)/(2)` If `'v_(2)'`is frequency of second harmonic then `v_(2) = ( upsilon)/( lambda_(2))= ( upsilon xx 2)/( 2l) =2v ` `v_(2)= 2v `...(2) Similarly for third harmonic( second overtone ) will have three nodesand four ANTINODES as shown in above FIGURE. If ` lambda _(3)` is wave length of third harmonic ` l = ( 3lambda_(3))/(2)` If `' v_(2)'` is frequency of third harmonic then `v_(3) = ( upsilon)/( lambda_(3)) = ( upsilon xx 3)/( 2l) = 3 v ` `v_(3) =3v `...(3)  Similarly we can find the remaining or higher harmonicfrequencies i.e.`v_(3), v_(4)` etc, can be determined in the same way. Therefore the ratio of the harmonicfrequencies in the open pipe can be written as given below. `v: v_(1): v_(2)= 1:2:3.....` |
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| 4974. |
An electron traveling in a uniform electric field passes from a region of potential V_1 to a region of higher potential V_2. Then |
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Answer» no change TAKES place in velocity component parallel to interface of TWO regions. `vsina=usintheta`, but `v cosagtucostheta` so, `alttheta, v=u` 
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| 4975. |
A current of 10 A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due north on the table is |
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Answer» `2xx10^(-5)` T , acting downwards. |
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| 4976. |
Explain -"for photoelectric emission frequency of incident radiation should be greater than threshold frequency". |
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Answer» Solution :For different metals,value of THRESHOLD frequency is different. Different photosensitive substance have different response to incident radiation. Selenium is more SENSITIVE than zinc and copper ifferent response to lightt of different wavelength `to` ultraviolet light gives rise to photoelectric effect in copper whereas green or red light does not. If frequency of incident radiation exceeds the threshold frequency,the photoelectric EMISSION start INSTANTANEOUSLY even if intensity of incident radiation is less. Emission start in time of `10^(-9)`s or less. |
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| 4977. |
If there is no difference of pressure on the two sides of the surface, then the liquid surface is |
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Answer» Concave |
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| 4978. |
The focal lengths of the objective and the eye - piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye - piece is at infinity. The two lenses are thin. The distances in centimeters of the object and image produced by the objective measured from the objective respectively. |
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Answer» `2.4 " and " 12` |
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| 4979. |
A uniform magnetic field gets modified as shown, when two spectimens X and Y are placed in it. i. Identify the two specimens X and Y . ii. State the reason for the behaviour f the field lines in X and Y. |
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Answer» Solution :i. X is a diamagnetic substance. Y is a PARAMAGNETIC substance ii. When a diamagnetic bar is placed in an EXTERNAL field, the fieldlines are repelled or `""` When a paramagnetic bar is placed in an external field , the field LINES get concentrated INSIDE the material and the field is increased. |
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| 4980. |
The space betweenthe platesof a parallel -platecapacitoris filled upwith inhomogneous poorlyconductingmedium whoseresistivity varieslinearlyin the directionperpendicularto theplates. The ratio of the maximumvalueof resistivity to the minimumone is equalto eta The gapwidthequals d. Find thevolume densityof the chargein the gap if a voltage V is appliedto the capacitor. epsilon is assumedto be 1 everywhere. |
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Answer» Solution :As in the PREVIOUS problem `E_(X) = C rho (x) = C (rho_(0) - rho_(1) x)` where `rho_(0) + rho_(1) d eta rho_(0)` or,`rho_(1) = ((eta - 1) rho_(0))/(d)` By intergation `V = int_(0)^(d) C rho (x)dx = Crho_(0) d (1 + (eta - 1)/(2)) = (1)/(2) C rho_(0) d (eta - 1)` Thus `C = (2V)/(rho_(0) d (eta + 1))` Thus volume density of charge present in the medium `= (DQ)/(sdx) = epsilon_(0) dE (x)//dx` `= (2 epsilon_(0) V)/(rho_(0) d (eta + 1)) xx ((eta - 1) rho_(0))/(d) = (2epsilon_(0) V(eta - 1))/((eta + 1) d^(2))` |
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| 4981. |
A block of mass m_(2) is placed on a horizontal table and another block of mass m_(1) is placed on top of it. An increasing horizontal force F = at is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is μ_(1) and that between the lower block and the table is μ_(2), then what is the maximum possible value of μ_(1)//μ_(2)? |
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Answer» `(m_(2))/(m_(1))` As `m_(2)` never moves, so friction acting in between the blocks is always less than or equal to friction acting between lower block and table. `thereforef_(1)lef_(2)` or `mu_(1)m_(1)glemu_(2)(m_(1)+m_(2))g` or `(mu_(1))/(mu_(2))le(m_(1)+m_(2))/(m_(1))` `rArr(mu_(1))/(mu_(2))le1+(m_(2))/(m_(1))` So, MAXIMUM possible value of `(mu_(1))/(mu_(2))=1+(m_(2))/(m_(1))`. 
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| 4982. |
From the surface of the moon, one can see the stars even when the sun is shining brightly while we cannot do so from the surface of the earth. Why? |
| Answer» Solution :PRESENCE of dark lines due to existence of ABSORBING MATERIAL on CHROMOSPHERES is sun. | |
| 4983. |
In which oeientation a dipole placed in a uniform electric field is in (i) stable , (ii)unstable equilibrium? |
| Answer» Solution :When dipole is (i) parallel to FIELD , (II)antiparallel to the field. | |
| 4984. |
A coil of inductance 0.5 henry is connected to a 18 volt battery. Calculate the initial rate of growth of current. |
| Answer» SOLUTION :`36 As^(-1)` | |
| 4985. |
The length of a seconds' pendulum on the earth is 60m. What will be the length of a seconds' pendulum on the surface of moon? Given : the value of g on the surface of moon ? Given : the value of g on the surface of moon? Given : the value of g on the surface of moon is 1//6th of value of g on the surface of earth. |
| Answer» ANSWER :C | |
| 4986. |
An electric dipole of length' 1 cm is at 30^@mth the electric field of strength 10^4NC^-1 If it experiences a. torque of 10sqrt2Nm.the potential energy of the dipole is |
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Answer» `0.245 J` |
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| 4987. |
Light waves can be polarized while sound waves cannot. Why? |
| Answer» SOLUTION :Only TRANSVERSE WAVES can be polarized. SOUND waves are logitudinal in nature. | |
| 4988. |
In photoelectric emission from a metal, work function phi_(0) and threshold frequency v_(0) are related as _____. |
| Answer» SOLUTION :`phi_(0)=hv_(0)` | |
| 4989. |
Two nicol prisms are inclined to each other at an angle 30^(@). If I is the intensity of ordinary light incident on the first prism, then the intensity of light emerges from the second prism will be |
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Answer» `3I"/"4` |
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| 4990. |
The relation betweenalpha and betawith usual notation is |
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Answer» `BETA= ( ALPHA )/( (1- alpha))` |
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| 4991. |
Two points P and Q lie on either side of an axis XY as shown. It is desired to produce an image of P at Q using a spherical mirror, with XY as the optic axis. The mirror must be |
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Answer» converging |
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| 4992. |
Which one of the following pair of particles move with same velocity along the same circular path in a uniform magnetic field ? |
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Answer» electron, PROTON |
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| 4993. |
A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 rad. s^(-1) . If the horizontal component of earth's magnetic field is 0.2 xx 10^(4) T, then the emf developed between the ends of the conductor is |
| Answer» Solution :`50 mu V` | |
| 4994. |
A high tension (HT) supply of, say, 6kV must have a very large internal resistance. Why? |
| Answer» Solution :Because, if the circuit is SHORTED (ACCIDENTALLY), the current drawn will EXCEED safety limits, if INTERNAL resistance is not large. | |
| 4995. |
A force F of constant magnitude acts on a block of mass 'm' and t=0,initially along positive x-direction. The direction of force F starts rotating with a constant angular velocity omega in anticlockwise sense. Then the magnitude of velocity of the block as a function of time 't' is ( initial block was at rest ) |
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Answer» `(F)/( m OMEGA ) SIN ((omegat)/( 2))` |
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| 4996. |
The first member of the Balmer series of hydrogen atom has wavelength of 656.3 nm. Calculate the wavelength and frequency of the second member of the same series. Given, c=3xx10^(8)m//s. |
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Answer» Solution :`1/lambda_1=R(1/n_1^2-1/n_2^2)` For I member of Balmer series, `n_1=2` & `n_2=3, lambda_1` = 6563 Å `1/lambda_1=R(1/4-1/9)=(5R)/36` For II member of Balmer series, `n_1`=2 and `n_2`=4 `1/lambda_1=R(1/4-1/16)=(3R)/16` Arriving at `lambda_2`= 4861Å `v=c//lambda_2=6.1xx10^14` Hz Given `lambda_0`=6563 Å=`6.536xx10^(-7)` m `lambda_beta` =? `f_beta` = ? `c=3xx10^(8) ms^(-1)` `rArr 1/lambda_1 =R(1/n_1^2-1/n_2^2)` For I member of Balmer series, `n_1=2,n_2=3` `therefore 1/(6xx563xx10^(-7))=R(1/2^2-1/3^2)` i.e., `10^7/6.563=R(5/36)` HENCE `R=36/5xx10^7/6.563` i.e., `R=1.097xx10^7 m^(-1)` For II member of Balmer series `x_1=2, x_2=4` `therefore 1/lambda_2=1.097xx10^7(1/2^2-1/4^2)` i.e., `1/lambda_2=(1.097xx10^7xx12)/64` Hence `lambda_2=64/(1.097xx12xx10^7)` m i.e., `lambda_2=4.8617xx10^(-7)` m `lambda_2`=4861.7Å and frequency , `Y_2=c/lambda_2=(3xx10^8)/(4.8617xx10^(-7))` i.e., `Y_2=0.6171xx10^15` Hz = `6.171xx10^14` Hz |
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| 4997. |
Explain the phenomenon of total internal reflection. |
Answer» Solution :Consider a point object in the denser medium. There will be cone of rays striking the interface. For a ray of light OA, it SUFFERS NORMAL refraction. For any ANGLE of incidence less than critical angle such as ray OB, the refracted ray bends away from teh normal. However for a particular angle of incidence called critical angle, the refracted ray of light grazes the interface. Therefore critical angle for the given pair of media and given wavelength of light may be defined as the angle of incidencein the denser mediumfor which the refractedray is at right angles to the normal drawn at the point of incidence. Fora ray of light such as OD, whose angle of incidence is greater than the criticalangle, there will be no refractionof light, instead the lightbounces BACK intothe same densermedium. This phenomenonof light is knownas the totalinternal reflection. Conditions for the TOTAL internalreflection. (a) Object shouldbe in the densermedium and viewed through rarer medium. (b) Angle of incidence in the denser medium should be more than the critical angle for the given pair of mediaand wavelengthof light. |
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| 4998. |
Which of the following law is used in the Millikan's method for the determination of charge |
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Answer» AMPERE's law |
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| 4999. |
Which transistor configuration is widely used in the circuit and why? |
| Answer» Solution :C.E CONFIGURATION is WIDELY USED because of the HIGH voltage and power GAINS. | |
| 5000. |
To a germanium crystal equal number of aluminium and indium atoms are added. Then |
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Answer» it remains an intrinsic SEMICONDUCTOR |
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