45406 + Interview Questions in Physics in Class 12 Page 866 InterviewSolution

43251.	A Fresnel's biprism is usedto form the interference fringes.The distance between the source and the biprism is 20 cms and that between the biprism and the screen is 80 cm. Iflambda = 6563 Å and the separation between the virtual sources is 3.6 mm, then the fringe width is :
Answer» `1.82 CM` `0.182 cm` `0.0182 cm` `0.00182 cm` SOLUTION :d = 3.6 MM = 0.36 cm `LAMBDA = 6563 Å` `BETA = (D)/(d)lambda = 0.0182 cm`.

Discussion

43252.	Define current gain in C-B amplifier.
Answer» Solution :The current GAIN is DEFINED as the RATIO of COLLECTOR current to the emitter current at constant collector voltage.

Discussion

43253.	Vibrating tuning fork of frequency v is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75 cm, the intensity of sound changes from a maximum to minimum, if the speed of sound is 350 m/s. Then v is
Answer» 500 HZ 1000 Hz 2000 Hz 4000 Hz Solution :When the piston is moved through a DISTANCE of 8.75 cm, the PATH difference produced is `2 xx 8.75 cm = 17.5 cm`. This must be equal to `lambda/2` for changes from maximum to MINIMUM. `therefore lambda/2 = 17.5 cm, lambda = 35 cm = 0.35 m` `v = vlambda` `therefore v =v/lambda = 350/(0.35) = 1000 Hz`

Discussion

43254.	Obtain the expression for electric field due to an uniformly charge spherical shell.
Answer» Solution :Electric Field DUE to a uniform charged spherical shell : Consider a UNIFORMLY charged spherical shell. Radius - R Total charge - Q (a) At a point outside the shell (`r gt R`) : P is a point ourside the shell at a distance r from the centre. The charge is uniformly distributed on the surface of the sphere. If `Q gt 0`, field point radially outward. If `Q lt 0,` field point readially inward. Applying Gauss LAW `ointvecE.vec(dA) = Q/epsi_(0)"...(1)"` `vecE` and `vec(dA)` are in the same direction. Hence E`oint "dA"=Q/epsi_(0)` But `oint`dA = total area of Gaussian SUFACE `= 4 pi r^2` Substituting in (1) `E.4 pir^2 =Q/epsi_0 (or)E=1/(4 piepsi_0)Q/r^2` In vector from `vecE= 1/(4 pi epsi_0) Q/r^2 hatr` (b) At a point on the suface of the spherical shell ( r = R). Electric field at point on the spherical shell, is r = R `vec E = Q /(4 pi epsi_(0) R^2) hatr` (c) At a point inside the shell ( ` r gt R`): Consider a point P inside the shell at a distance r from the center. `ointvecE.vec(dA) = Q/epsi_(0)"...(1)"` `E. 4 pi r^2 = Q/epsi_0` SINCE Gaussian surface encloses no charge, so Q = 0 `:. "E =0"`

Discussion

43255.	The ratio of the radii of gyration of a circular disc and a circular ring of the same radii about a tangential axis is :
Answer» `1:sqrt(2)` `sqrt(2):1` `sqrt(2):sqrt(3)` `sqrt(5):sqrt(6)` SOLUTION :For DISC `(5)/(4)mr^(2)=mk_(1)^(2)impliesk_(1)=sqrt((5)/(4))r` For ring `(3)/(2)mr^(2)=mk_(2)^(2)""k_(2)sqrt((3)/(2))r` `THEREFORE (k_(1))/(k_(2))=sqrt((5)/(4))/(sqrt((3)/(2)))=sqrt((10)/(12))=sqrt((5)/(6))`

Discussion

43256.	In an a.c. circuit, the applied voltage and flowing current are E=E_(0)sinomegat and I=I_(0)sin(omegat+pi/2) respectively. What is the average power consumed in one cycle in this circuit ?
Answer» Solution :Zero. [HINT : AVERAGE power consumed = `(E_(0))/(sqrt2).(I_(0))/(sqrt2)cospi/2=0`]

Discussion

43257.	Rotation of plane of polarization under influence of magnetic field called Faraday effect was conformed by
Answer» Newton.s CORPUSCULAR theory Huygen.s WAVE theory Electromagnetic theory Quantum theory Answer :C

Discussion

43258.	The figure shows a schematic diagram showing the arrangement of Youngs'sDouble Slit Experiment Identify the correct statement(s) if the source slit S moved closer to S_(1)S_(2), i.e., the distance l decreases
Answer» NOTHING happens to fringe pattern fringe pattern MAY GETS LESS sharp fringe width REMAINS unchanged fringe pattern may dissapear Answer :B::C::D

Discussion

43259.	An electron has an intrinsic angular momentum (spin) whose component in an arbitrary direction is one half of the Planck's constant, i.e. L_2=h//2=5.25xx10^(-35) J .s. Making use of the fact that the speed of light in vacuum is the maximum attainable, prove that a model in which the spin of an electron is due to the rotation about its axis is not feasible.
Answer» Solution :The moment of the ball.s momentum is `L = Iomega = 2/(5)mr^2v/r=2/5` mvr ,where v is the orbital VELOCITY on the equator. SINCE `v ltc, L lt2/5` , whence `rgt(5L)/(2mc) , "or " r gt (5h)/(4 mc)=4.8 xx10^(-13)m` This dimension does not agree with experimental data, according to which the effective electron radius is two orders of magnitude less.

Discussion

43260.	Suppose a pure Si crystal has 5xx10^(28) atoms m^(-3).It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n_(i)=1.5xx10^(16)m^(-3).
Answer» Solution :Here, thermally generated electrons, `n_(i)=1.5xx10^(16)m^(-3)` 1 ppm = 1 part per million `=10^(6)` `therefore` Electron number density due to As atom, `n_(D)=(5xx10^(28))/(10^(6))=5xx10^(22)m^(-3)` Number of electron due to doping. `n_(i)` is neglected compare to `n_(D)` `therefore n_(e )=n_(D )=5xx10^(22)m^(-3)` Now `n_(i)^(2)=n_(e )n_(H)` `therefore n_(h)=(n_(i)^(2))/(n_(e ))` `=((1.5xx10^(16))^(2))/(5xx10^(22))` `therefore n_(h)=4.5xx10^(9) m^(-3)`

Discussion

43261.	The number of neutrons released during the fission reaction is : _0^1n+_92^235U to _51^133Sb + _41^99Nb + neutrons
Answer» 1 92 3 4 Answer :D

Discussion

43262.	A single conservative force F(x) acts on particle that moves along the x-axis. The graph of the potential energy with x isgiven. At x = 5m, the particle has a kinetic energyof 50 J and its potential energy is related to position 'x' as U = 15 + (x - 3)^(2) Joule, where x is in mete then :
Answer» The MECHANICAL energy of system is 69 J The mechanical energy of system is 19 J At `x = 3`, the KINETIC energy of PARTICLE in minimum the maximum value of kinetic energy is 54 J Solution :`U = 15 + (x - 3)^(2)` `U = (5) = 19` & `KE (5) = 50` Total mechanical energy `= 50 + 19 = 69J` `U_("min") -= U (x - 3) = 15 J` `KE_("max") = 69 - U_("min") = 69 - 15 = 54 J`

Discussion

43263.	What is the condition of constructive interference ?
Answer» SOLUTION :`phi = 2NPI` [where N = 0,1,2.3........] path DIFFERENCE, x = n`lambda`.

Discussion

43264.	When a forward bias is applied to a p-n junction, it……
Answer» raises the POTENTIAL barrier. REDUCES the majority carrier current to zero. lowers the potential barrier. NONE of the above. SOLUTION :lowers the potential barrier. When we will apply forward bias to p-n junction the applied voltage is opposed by the potential barrier. Hence, the potential barrier across the junction gets REDUCED.

Discussion

43265.	What percentage (approximate) of original radioactive substance left after 4 half-Life?
Answer» 0.08 0.06 0.04 0.03 Answer :B

Discussion

43266.	Substances which do not contain free electrons and which do not conduct electricity but transmit electric effects are called ____.
Answer» ANSWER :DIELECTRICS

Discussion

43267.	The magnitude of the electric field on the surface of a sphere of a sphere of radius r having a unifrom surface charge density sigma is
Answer» `sigma//epsilon_(0)` `sigma//2epsilon_(0)` `sigma//2epsilon_(0)R` `sigma//2epsilon_(0)r` ANSWER :A

Discussion

43268.	A car is moving with speed u. Driver of the car sees red traffic light. His reaction time is t, then find out the distance travelled by the car after the instant when the driver decided to apply brakes. Assume uniform retardation 'a' after applying brakes.
Answer» SOLUTION :`UT+(U^(2))/(2A)`

Discussion

43269.	In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is 4mm. The screeon on which the pattern is displaced is 2m from the slit and wavelength of light used is 6000Å. Calculate width of the slit and width of the central maximum.
Answer» SOLUTION :GIVEN: `lambda=6000Å`, `D=2m` CONSIDER that `X_(1)` and `X_(2)` be the distance between FIRST minima on the right and left respectively of central maximum. `X_(1)+X_(2)=4mm` `X_(1)+X_(2)=(2lambdaD)/(d)` `4xx10^(-3)=(2xx6xx10^(-7)xx2)/(d)` `d=6xx10^(-4)m` `d=0.6mm` `:. ` Width of central maximum, `W=X_(1)+X_(2)` `=4xx10^(-3)m` `=4mm`.

Discussion

43270.	A P^(32) radionuclide with half-life T= 14.3 days is prodiced in a reactor at a constant rate q= 2.7.10^(9) nuclei per second. How soon after the beiginning of production of that radionuclide will its activity be equal to A= 1.0.10^(9) dis//s?
Answer» Solution :`(DN)/(dt)= g(dN)/(dt)=underset("supply")underset (uarr) g-lambda underset(decay)underset(darr)N` We see that `N` will approach a constant value `(g)/(lambda)`. This can also be PROVED directly. MULTIPLY by `e^(lambda t)` and write `(dN)/(dt)e^(lambda t)+lambdae^(lambda t)N="ge"^(lambda t)` Then `(d)/(dt) (Ne^(lambda t))= "ge"^(lambda t)` or `Ne^(lambda t)=(g)/(lambda)e^(lambda t)+ CONST` At `t=0` when the production is started, `N=0` `0= (g)/(lambda)+constant` Hence `N=(g)/(lambda)(1-e^(-lambda t))` Now the activity is `A= lambdaN= g(1-e^(-lambda t))` From the problem `(1)/(2.7)= 1-e^(-lambda t)` This gives `lambda t= 0.463` so `t=(4.463)/(lambda)=(0.463xxT)/(0.693)=9.5 days` Algebraically `t=-(T)/(In 2) In(1-(A)/(g))`

Discussion

43271.	What is skip zone or skip area.
Answer» Solution :There is a ZONE in between where there is no reception of electromagnetic waves NEITHER ground nor sky, CALLED as SKIP zone or skip area.

Discussion

43272.	A body having a moment of inertia about its axis of rotation equal to 3 kg m^(-2) is rotating with angular velocity of 3 rad s^(-1). Kinetic energy of this rotating body is same as that of a body of mass 27 kg moving with velocity v. The value of v is
Answer» `1ms^(-1)` `0.5 MS^(-1)` `2MS^(-1)` `1.5 ms^(-1)` SOLUTION :`(1)/(2) mv^(2) = (1)/(2) I OMEGA^(2)` `(1)/(2) xx 27 xx v^(2) = (1)/(2) xx 3 xx 3^(2)` v = 1/s

Discussion

43273.	Let f: Rrarr R be defined by f (x) = X-2 AA XepsilonR, which of the follwing is f^(-1)(x)-
Answer» x-2 x+1 x-1 x+2 Answer :D

Discussion

43274.	At a particular height, the velocity of an ascending body is vecu. The velocity at the same height while the body falls freely is
Answer» `2vecu` `-vecu` `vecu` `-2vecu` Solution :VELOCITY of ascending BODY at height (H) `= vecu` Velocity of body under free FALL at height H=- Velocity of ascending body at height H=` -vecu`

Discussion

43275.	Zener diode use for ….
Answer» amplification rectification stablisation OSCILLATOR Solution :stablisation Some reverse bias voltage and current FLOWS from the ZENER diode and so the voltage of the load RESISTANCE is MAINTAINED and that is why it is used in te voltage regulator.

Discussion

43276.	In the above question disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is :
Answer» `(2Iomega)/(3t)` `(9Iomega)/(2t)` `(9Iomega)/(4T)` `(3Iomega)/(2t)` Solution :LET `omega`. be the angular VELOCITY when the two discs GET combined. Then by conservation of angular momentum, `(I+2I)omega.=Ixx2omega+2Iomega` or `omega.=(4omega)/(3)""...(i)` Now `omega.=omega+at=omega+(tau)/(2I)xxt` (where `tau` is the average FRICTIONAL torque) or `(4omega)/(3)-omega=(tau)/(2I).t` `tau=(2Iomega)/(3t)`

Discussion

43277.	A magnetic needle suspended freely
Answer» a) orients itself in a definite direction b) REMAINS in a direction c) BECOME itself in a definte direction d) become extical with N-pole down Answer :A::C::D

Discussion

43278.	What area of the ground below can be photographed at one time by a camera of focal length 45cm and plate size 3cmxx3cm kept in a satellite at height of 1500km ?
Answer» ANSWER :`100xx100km`

Discussion

43279.	(A): If a conducting medium is placed between two charges, then electric force between them becomes zero. (R ): Reduction in a force due to introduced material is inversely proportional to its dielectric constant.
Answer» Both .A. and .R. are true and .R. is the CORRECT explanation of .A. Both .A. and .R. are TRU and .R. is not the correct explanation of .A. A. is true and .R. is false A. is false and .R. is true Answer :A

Discussion

43280.	A comet of mass 10^(8)kg travels around the sun in an elliptical orbit . When it is closed to the sun it is 2.5 xx 10 ^(11)m away and its speedis 2 xx 10 ^(4) m s^(-1)Find the change in kinetic energy when it is farthest from the sun and is 5 xx 10 ^(10)m away from the sun
Answer» `38 xx10^(8) J` `48 xx10^8J` `58 xx10^8 J` `56xx10^8J` ANSWER :B

Discussion

43281.	The layer responsible for reflection of ratio wave is
Answer» MESOSPHERE photosphere troposphere ionosphere Answer :D

Discussion

43282.	A Fresnel's biprism arrangement is set with sodium light (lambda = 5893 Å) and in the field of the eyepiece we get 62 fringes. How many fringes shall we get if we replace the source by mercury lamp using , (a) green filter (lambda=5461 " "A^(@)) (b) violet filter (lambda = 4358 A^(@))
Answer» Solution :The fringe width p given by : p = DX/d ... (1) In the given PROBLEM, D and d are constant and hence there will be a change in fringe widthnwhen there is a change in the wavelength of light USED. When the source-is replaced by mercury light and” a green filter is used (the wavelengthsay `lambda_(1)= 5461 Å`) Let fringe width in this case be `beta_(1)` The width of field of view = 62 `beta` From equations (1) and (2), `beta/beta_(1) = lambda/lambda_(1) ...(3)` The width of field of view = 62 `beta` Suppose using a green filter, the field of view contains `n_(1)` FRINGES. Then width of field of view =`n_(1)beta_(1)` `62 beta = n_(1)beta_(1)` or `n_(1)=62(beta//beta_(1))= 62 (lambda//lambda_(1))=62 xx (5893xx10^(-8))/(546 xx 10^(-8))=67` (b)Letthe field of view contains `n_(2)` fringes using a violet filter wavelength. `(4358xx10^(-8))/(4358xx10^(-8)) =84`

Discussion

43283.	The conductor ABCD has the shape shown. It lies in the yz plane, with A and e on the y-axis. When it moves with a velocity v in a magnetic field B, and emf e is induced between A and E .
Answer» E = 0, if v is in the y- direction and B is in the x-direction e = 2Bav, if v is in the Z-direction and B is in the x-direction e = B`LAMBDA`v, if v is in the z- direction and B is in the x- direction e = B`lambda`v, if v is in the x-direction and B is in the z-direction Answer :A::C::D

Discussion

43284.	Two point charges q_(A) =3 muC and q_(B) = -3 muCare located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges ? (b) If a negative test charge of magnitude 1.5 xx 10^(-9)C is placed at this point, what is the force experienced by the test charge ?
Answer» SOLUTION :(a) Electrical field at C DUE to `q_(A)` is, `vecE_(CA) = (k\|q_(A)\|)/r_(CA)^(2) hati` (From A to C)………..(1) Electric field at C due to `q_(B)` is, `vecE_(CB) = (k\|q_(B)\|)/r_(CB)^(2).hati` (From C to B) `therefore vecE_(CB) = vecE_(CA)` `(therefore \|q_(A)\| = \|q_(B)\|)` and `r_(CA) = r_(CB)`) Now resultant electric field at C would be, `vecE_(C) = vecE_(CA) + vecE_(CB)` `=2vecE_(CA)` (`therefore` From equation (2)) `=2(kq_(A))/(r_(CA)^(2)) hati` `therefore vecE_(C) = 5.4 xx 10^(6) N//C hati` (From `q_(A)` to `q_(B)`) (b) Electric force EXERTED on `q_( C)`, placed at C (midpoint of `bar(AB)`) `vecF_(e) = q_( C) vec(E)c` `=(1.5 xx 10^(-9))(5.4 xx 10^(6))hati` `therefore vecF_(e) = 8.1 xx 10^(-3) Nhati` (From `q_(A)` to `q_(B)`)

Discussion

43285.	In Fig , if h_1-h_0=1.0 cm, h_0 = 100 cm , and t =20^@C ,what is the value of gammafor the liquid ?
Answer» ANSWER :`gamma =1.0 [(100)(20)=500xx10^(-6)//""^@C`

Discussion

43286.	A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is _______ .
Answer» B 2B 4B `B/2` Solution :A LONG solenoid carrying a current produces magnetic field B along its AXIS. Now if the current I is doubled and the NUMBER of turns N per cm is halved the magnetic field `B.=mu_(0)n/2xx2I=Bmu_(0)nI=B`

Discussion

43287.	Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer» <P> Solution :Consider a long straight wire NM with current I flowing from N to M as shown in Fig. Let P be the point at a distance a from point . Consider an element of LENGTH dl of the wire at a distance l from point O and `hat r` be the vector joining the element dl with the pointP. Let `theta` be the angle between `vec(dl)" and " hatr`. Then , the magnetic field at P due to the element is `vec(dB) = (mu_(0)I)/(4pi) (vec(dl))/r^(2) sin theta(" unit vector perpendicular to " vec(dl) and VECR)` The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors `vec(dl)" and " hat r ` ( let it be `hat n`) . The net magnetic field can be determined by integrating equation with PROPER limits . From the Figure, in a right angle triangle PAO, `TAN ( pi - theta) =a/l` ` l = - a/ (tan theta) ( "since "tan(pi - theta) = - tan theta)` ` l = - a cot theta and r = a cosec theta` Differentiating , `dl = a cosec .^(2) theta d theta` `vec(dB) =( mu_(0)I)/(4 pi) ((a cosec. ^(2) theta d theta))/ ((a cosec theta )^(2)) sin theta hat n ` ` vec(dB) = (mu_(0)I)/(4 pi) ((a cosec . ^(2) theta d theta ))/((a^(2) cosec ^(2) theta)) sin theta hatn ` ` = (mu_(0)I)/(4 pi a) sin theta d theta hat n` This is the magnetic field at a point P due to the current in small elemental length . Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. `theta`. Therefore, the net magnetic field at the point P can be obtained by integrating`vec(dB)` by varying the angle from `theta = varphi _(1) " to " theta = varphi_(2) ` is `vecB = (mu_(0)I)/(4 pi a) underset(varphi_(1))overset(varphi_(2))int sin theta d theta hatn = (mu_(0)I)/(4 pi a) ( cos varphi_(1) - cos varphi_(2)) hat n ` For a an infinitely long straight wire , `varphi_(1) = 0 " and " varphi_(2) = pi ` , the magnetic field is ` vecB = (mu_(0)I)/ (2 pi a) hat n` Note that here `hat n` represents the unit vector from the point O to P .

Discussion

43288.	A positively charge moving along the positive x-direction with a speed traces a non-linear trajectory OA in the x-y plane after passing O as shown in the figure. The combination of electric (vecE) and magnetic field (vecB) fields that would lead to the trajectory OA is (a, b, c are non-zero positive constants),
Answer» `vecE=0 , vecB=ahatj+chatk` `vecE=ahati, vecB=bhatj+chatk` `vecE=0 , vecB=bhatj+chatk` `vecE=ahati, vecB=bhati-chatk` ANSWER :D

Discussion

43289.	Two long wires are hanging freely. They are joined first in parallel and then in series and then are connected with a battery. In both cases, which type of force acts between the two wires
Answer» ATTRACTION FORCE when in PARALLEL and REPULSION force when in series Repulsion force when in parallel and attraction force when in series Repulsion force in both cases Attraction force in both cases Answer :A

Discussion

43290.	A convace mirror forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shiff of object is 6 cm, the shift of the screen and focal length of mirror are
Answer» 36 CM, 36 cm 36 cm, 16 cm 72 cm, 36 cm None of these Answer :A

Discussion

43291.	The energy flux of sunlight reaching the surface of the earth is 1.388xx10^(3)W//m^(2). How many photons(nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer» SOLUTION :`E=1.388xx10^(3)WM^(-2), lambda=550nm=550xx10^(-9)m` `E=NH upsilon="nh"(c )/(lambda), n=(Elambda)/(hc)= (1.388xx10^(3)xx550xx10^(-9))/(6.63xx10^(-34)xx3xx10^(8))=3.8xx10^(21)" photon"//m^(2)//s`

Discussion

43292.	Assertion :- If magnetic flux through any circuit is changes then an emf in induced in it . This emf is equal to negative of rate of changes of flux . Reason :- Direction of induced emf is always such that it always opposes the cause by which it is produced.
Answer» If the Assertion & REASON are TRUE& the Reason is a CORRECT explanation of the Assertion . If both Assertion & Reason are True & the Reason is not a correct explanation of the Assertion. If Assertion is True but the Reason is Fasle. If both Assertion and Reason are False . ANSWER :C

Discussion

43293.	In an interial reference frameK there are twouniformmutulaly perpendicularfields, an electricfieldof strengthE = 40 kV//m anda magnetic fleid induction B = 0.20 mT. Find theelectricstrengthE' (or the magnetic induction B') in the referenceframeK' where onlyone field, electricor magnetic, is observed.
Answer» SOLUTION :In this case, `vec(E).vec(B) = 0`, as the FIELDS are mutually perpendicular.Also, `E^(2) - c^(2) B^(2) = -20 xx 10^(8) ((V)/(m))^(2)` is `-ve`. Thus, we can finda frame, in which `E' = 0`, and `B' = (1)/(c) sqrt(c^(2) B^(2) - E^(2)) = B sqrt(1 - (E^(2))/(c^(2) B^(2))) = 0.20 sqrt(1 - ((4 xx 10^(4))/(3xx10^(8)xx2xx10^(-4)))^(2)) = 0.15 mT`

Discussion

43294.	Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant 'K' ?
Answer» SOLUTION :Potential difference between the plates of capacitor V = q/C Work done to add additional charge dg on the capacitor ` dw = V xx dq` ` therefore ` Total energy stored in the capacitor ` U = INT d w = int_0^Q dq = 1/2 (Q^2)/(C )` When battery is disconnected, (i)Energy stored will be DECREASED or energy stored = 1/K TIMES the initial energy (ii) Electric field WOULD decrease or E. = E/K

Discussion

43295.	The current gain beta is
Answer» `BETA =I_e/I_c` `beta =I_c/I_e` `beta =I_b/I_c` `beta =I_c/I_b` ANSWER :D

Discussion

43296.	A metallic ring is connected to a rod oscillates freely like a pendulum. If now a magnetic field is applied in horizontal direction so that the pendulum now swings through the field, the pendulum will
Answer» keep OSCILLATING with the old TIME period keep oscillating with a smaller time period keep oscillating with a LARGER time period come to REST very soon Answer :D

Discussion

43297.	A charged particle oscillates about its mean equilibrium position with a frequency of 10^(9) Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer» will have FREQUENCY of `10^(9)` Hz will have frequency of `2xx10^(9)` Hz will have a wavelength of 0.3 m. fall in the region of radiowaves. SOLUTION :Hence frequency of charged PARTICLE is given as `10^(9)` Hz. Hence, wavelength `lambda=(UPSILON)/(v)=(3xx10^(8))/(10^(9))=0.3 m` Hence (A) and (C ) are correct. `10^(9)` Hz is in range of radiowaves. Hence, option (D) is correct.

Discussion

43298.	A tank containing water has an orifice in one vertical side . If the center of orifice is 4.9 m below the surface level inside a tank , the velocity of discharge is
Answer» a)4.9m/s B)9.8m/s c)2.45m/s d)0m/s Answer :B

Discussion

43299.	A body is projected horizontally from the top of a hill with a velocity of 9.8 m/s. What time elapses before the vertical velocity is twice the horizontal velocity ?
Answer» `0.5` SEC 1 sec 2 sec `1.5` sec Answer :C

Discussion

43300.	An ammeter consists of a480Omega resistance connected in parallel to a20Omegashunt. The maximum current which can be measured by this ammeter is :
Answer» 1.25A 0.192A 0.14A 0.125A Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

A Fresnel's biprism is usedto form the interference fringes.The distance between the source and the biprism is 20 cms and that between the biprism and the screen is 80 cm. Iflambda = 6563 Å and the separation between the virtual sources is 3.6 mm, then the fringe width is :

Define current gain in C-B amplifier.

Obtain the expression for electric field due to an uniformly charge spherical shell.

The ratio of the radii of gyration of a circular disc and a circular ring of the same radii about a tangential axis is :

In an a.c. circuit, the applied voltage and flowing current are E=E_(0)sinomegat and I=I_(0)sin(omegat+pi/2) respectively. What is the average power consumed in one cycle in this circuit ?

Rotation of plane of polarization under influence of magnetic field called Faraday effect was conformed by

The figure shows a schematic diagram showing the arrangement of Youngs'sDouble Slit Experiment Identify the correct statement(s) if the source slit S moved closer to S_(1)S_(2), i.e., the distance l decreases

Suppose a pure Si crystal has 5xx10^(28) atoms m^(-3).It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n_(i)=1.5xx10^(16)m^(-3).

The number of neutrons released during the fission reaction is : _0^1n+_92^235U to _51^133Sb + _41^99Nb + neutrons

A single conservative force F(x) acts on particle that moves along the x-axis. The graph of the potential energy with x isgiven. At x = 5m, the particle has a kinetic energyof 50 J and its potential energy is related to position 'x' as U = 15 + (x - 3)^(2) Joule, where x is in mete then :

What is the condition of constructive interference ?

When a forward bias is applied to a p-n junction, it……

What percentage (approximate) of original radioactive substance left after 4 half-Life?

Substances which do not contain free electrons and which do not conduct electricity but transmit electric effects are called ____.

The magnitude of the electric field on the surface of a sphere of a sphere of radius r having a unifrom surface charge density sigma is

A car is moving with speed u. Driver of the car sees red traffic light. His reaction time is t, then find out the distance travelled by the car after the instant when the driver decided to apply brakes. Assume uniform retardation 'a' after applying brakes.

A P^(32) radionuclide with half-life T= 14.3 days is prodiced in a reactor at a constant rate q= 2.7.10^(9) nuclei per second. How soon after the beiginning of production of that radionuclide will its activity be equal to A= 1.0.10^(9) dis//s?

What is skip zone or skip area.

A body having a moment of inertia about its axis of rotation equal to 3 kg m^(-2) is rotating with angular velocity of 3 rad s^(-1). Kinetic energy of this rotating body is same as that of a body of mass 27 kg moving with velocity v. The value of v is

Let f: Rrarr R be defined by f (x) = X-2 AA XepsilonR, which of the follwing is f^(-1)(x)-

At a particular height, the velocity of an ascending body is vecu. The velocity at the same height while the body falls freely is

Zener diode use for ….

In the above question disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is :

A magnetic needle suspended freely

What area of the ground below can be photographed at one time by a camera of focal length 45cm and plate size 3cmxx3cm kept in a satellite at height of 1500km ?

(A): If a conducting medium is placed between two charges, then electric force between them becomes zero. (R ): Reduction in a force due to introduced material is inversely proportional to its dielectric constant.

A comet of mass 10^(8)kg travels around the sun in an elliptical orbit . When it is closed to the sun it is 2.5 xx 10 ^(11)m away and its speedis 2 xx 10 ^(4) m s^(-1)Find the change in kinetic energy when it is farthest from the sun and is 5 xx 10 ^(10)m away from the sun

The layer responsible for reflection of ratio wave is

A Fresnel's biprism arrangement is set with sodium light (lambda = 5893 Å) and in the field of the eyepiece we get 62 fringes. How many fringes shall we get if we replace the source by mercury lamp using , (a) green filter (lambda=5461 " "A^(@)) (b) violet filter (lambda = 4358 A^(@))

The conductor ABCD has the shape shown. It lies in the yz plane, with A and e on the y-axis. When it moves with a velocity v in a magnetic field B, and emf e is induced between A and E .

In Fig , if h_1-h_0=1.0 cm, h_0 = 100 cm , and t =20^@C ,what is the value of gammafor the liquid ?

A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is _______ .

Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.

Two long wires are hanging freely. They are joined first in parallel and then in series and then are connected with a battery. In both cases, which type of force acts between the two wires

A convace mirror forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shiff of object is 6 cm, the shift of the screen and focal length of mirror are

The energy flux of sunlight reaching the surface of the earth is 1.388xx10^(3)W//m^(2). How many photons(nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Assertion :- If magnetic flux through any circuit is changes then an emf in induced in it . This emf is equal to negative of rate of changes of flux . Reason :- Direction of induced emf is always such that it always opposes the cause by which it is produced.

Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant 'K' ?

The current gain beta is

A metallic ring is connected to a rod oscillates freely like a pendulum. If now a magnetic field is applied in horizontal direction so that the pendulum now swings through the field, the pendulum will

A charged particle oscillates about its mean equilibrium position with a frequency of 10^(9) Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

A tank containing water has an orifice in one vertical side . If the center of orifice is 4.9 m below the surface level inside a tank , the velocity of discharge is

A body is projected horizontally from the top of a hill with a velocity of 9.8 m/s. What time elapses before the vertical velocity is twice the horizontal velocity ?

An ammeter consists of a480Omega resistance connected in parallel to a20Omegashunt. The maximum current which can be measured by this ammeter is :