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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.
| 2751. |
Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. what is the justification ? |
| Answer» Solution :REASON is exactly same as in part (d). In most of the optical instruments size of INSTRUMENT is much greater than the wavelength of light. As a result, DIFFRACTION of light has negligible effect and we may apply principles of ray optics WITHOUT any hesitation. | |
| 2752. |
How is it that the reverse current in Zener diode starts increasing suddenly at a relatively low breakdown voltage of 5 volt or so ? |
| Answer» Solution :Zener diode is fabricated by HEAVILY doping both p- and n-sides of the p-n junction. As a result, the depletion REGION formed is very THIN, even less than `10^(6)` m, and consequently the electric field at the junction is extremely high (of the order of `5 XX 10^(6)` V/m or even more) even for a small REVERSE bias voltage of about 5 V and breakdown voltage has a low value of about 5 V. When the reverse bias voltage is equal to Zener (or breakdown) voltage `["i.e., "V_(R) ge V_(Z) ]` ,the electric field is high enoughto pull valence electrons from the host atoms on p-side of the junction, which are then accelerated to n-side . These electrons account for high current observed at the breakdown stage. | |
| 2753. |
If an aC ammeter reads I ampere in an aC circuit, the peak value of the current is |
| Answer» ANSWER :D | |
| 2754. |
Steradian is the solid angle subtended at the centre of the sphere by its surface whose area is equal to ____ of the sphere. |
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Answer» the RADIUS |
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| 2755. |
Six lead-acid secondary cells each of emf 2.0 V and internal resistance 0.015 Omegaare joined in series to provide a supply to a resistance 8.5Omega. What are the current drawn from the supply and its terminal voltage? |
| Answer» SOLUTION :`1.4 A , 11.9 V ` | |
| 2756. |
The density of a nucleus varies with mass number A as |
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Answer» `A^2` |
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| 2757. |
Employing Eq. (6.4g), find at T=0 (a) the maximum kinetic energy of free electrons in a metal if their concentration is qual to n, (b) the mean kinetic energy of free electrons if their maximum kinetic energy T_(max) is known. |
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Answer» SOLUTION :(a) From the fromula `DN=(sqrt(2)m^(3//2))/(pi^(2) ħ^(3))E^(1//2)dE` the maximum value `E_(max)` of `E` is determined in terms of `n` by `n=(sqrt(2)m^(3//2))/(pi^(2) ħ^(3)) int_(0)^(E_(max))E^(1//2)dE` `=(sqrt(2)m^(3//2))/(pi^(2) ħ^(3))(2)/(3)E_(max)^(3//2)` or `E_(max)^(3//2)=( ħ^(2)/(2M))^(3//2)(3 pi^(2)n)` `E_(max)=( ħ^(2))/(2m)(3PI^(2)n)^(2//3)` (b) Mean `K.E lt E gt` is `lt E geint_(0)^(E_(max))Edn//int_(0)^(E_(max))dn` `=int_(0)^(E_(max))E^(3//2)//int_(0)^(E_(max))E^1//2dE(2)/(5)E_(max)^(5//2)//(2)/(3)E_(max)^(3//2)=(3)/(5)E_(max)` |
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| 2758. |
What is a rectifier? Explain the working of semi-conductor diode as a full waverectifier with a necessary circuit diagram. Also give the input and output waveforms for the same |
Answer» Solution : Rectifier : Rectifier is an electronic device which converts AC into DC Fig shows the circuit diagram of a FULL wave rectifier. It makes use of two diodes `D_1` and `D_2`with a CENTRE TAPPED transformer. `R_L`is the load resistor across which the output is taken. When VOLTAGE at A with respect to the centre tap at any instant is positive, at that instant voltage at B being out of phase will be negative. So `D_1`GET formed forward biased and conducts, `D_2`get reverse biased and does not conduct. Hence during the half cycle we get an output voltage when voltage at A becomes negative with respect to centre tap the voltage at B would be positive. So diode `D_1`. does not conduct and `D_2`conducts. Hence during negative half cycle also, we get an output voltage. Thus output voltage is obtained during both positive and negative half cycles of ac. 
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| 2759. |
The net field vecE due to theuniformly charged rod at P makes angles theta_(1) and theta_(2) with AP and BP respectively. Then find the value of theta_(1)//theta_(2) |
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| 2760. |
Find the binding energy and binding energy per nucleon of nucleus""_(83)Bi^(209). Given : mass of proton = 1.0078254 u. mass of neutron = 1.008665 u. Mass of ""_(83)Bi^(209)= 208.980388u |
| Answer» SOLUTION :1639.38 MEV and 7.84 MeV/Nucleon | |
| 2761. |
A short bar magnet of magnetic moment 25JT^(-1) is placed with its axis perpendicular to earth's field direction. At what distance from the centre of the magnet,the resultant field is inclined at 45^(@) with earth's field, if H=0.4xx10^(-4)T. |
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Answer» 5 m |
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| 2762. |
A parallel plate capacitor (figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^(-1).a. What is the rms value of the conduction current ?b.Is the conduction current equal to the displacement current?c.Determine the amplitude of B at a point 3.0 cm from the axis between the plates. |
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Answer» Solution :`R=6.0 cm = 6xx10^(-3)m, C-100 pF = 100xx10^(-12)T` `V_(rms)=230V, omega = 300 rad s^(-1)` a.`X_(C )=(1)/(C omega)` `I_(rms)=(V_(rms))/(X_(C ))=V_(rms)xx C omega = 230xx100xx10^(-12)xx300=6.9xx10^(-6)A=6.9 MU A` B. Yes c.`B=(mu_(0)rI)/(2PI R^(2))`, But `I_(0)=sqrt(2)I_(rms)=sqrt(2)xx6.9xx10^(-6)=9.76xx10^(-6)A` `therefore B = (2xx10^(-7)xx3xx10^(-2)xx9.76xx10^(-6))/((6xx10^(-2))^(2))=1.63xx10^(-11)T` |
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| 2763. |
5 कुलम आवेश के दो बराबर तथा विपरीत आवेशों के बीच की 5.0 सेमी दूरी है। इसका वैद्युत द्विध्रुव-आघूर्ण है: |
| Answer» Answer :A | |
| 2765. |
Which one of the following criteria is the basis to measure the development of a country according to UNDP? |
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Answer» PER CAPITA income |
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| 2766. |
y(x,t) = (0.8)/([(4x + 5t)^2 + 5])represents a moving pulse, where x and y are in metres and t' is inseconds, then |
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Answer» pulse is moving in +ve x DIRECTION |
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| 2767. |
Two pipes closed at one end, 51 cm and 52 cm long, produce 3 beats per second when they are sounded together in their fundamental modes. Ignoring the end correction, calculate the speed of sound in air. |
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Answer» SOLUTION :Data ` L_(1) = 0.51 m, L_(2) = 0.52 m` beat frequency = 3Hz Since ` L_(1) lt L_(2) , n_(1) gt n_(2)` Beat frequency= `n_(1) - n_(2) = 3Hz ` ` n_(1) = v/(4L_(1)) and n_(2) = v/(4L_(2))` ` v/4 (1/L_(1) - 1/L_(2)) =3` The SPEED of sound in AIR. ` v = (12L_(1)L_(2))/(L_(2)-L_(1)) = (12(0.51)(0.52))/(0.52 - 0.51) = 12 xx 51 xx 0.52 = 318 .3 ` m/s |
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| 2768. |
A bulb of 100W - 250V is connected to 250V mains. If another similar bulb is connected to it in series , the total power consumption is : |
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Answer» 100 W |
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| 2769. |
Foe tuning radio and T.V. circuits, series resonance circuits are used as : |
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Answer» reactifier circuits |
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| 2770. |
A particle performing linear S.H.M. has a maximum displacement of 0.1 m. Its acceleration at a distance of 0.03 m from its position is 0.12 m//s^2. What is its velocity at a distance of 0.06 m from its mean position ? |
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Answer» SOLUTION :`v=omegasqrt(a^2-x^2) = SQRT(AC c^n/x)xsqrt(a^2-x^2)` `sqrt0.12/0.13xxsqrt((0.1)^2-(0.06)^2)` `2sqrt(0.01-0.0036)` `2sqrt0.0064=2xx0.08=0.16m/s.` |
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| 2771. |
What are limitation of dimensional analysis. |
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Answer» Solution :Following are the limitations of DIMENSIONAL ANALYSIS. (i) It gives no information about pure numerics. (ii) The dimensional analysis can not be used to derive a relation if the physical quantity depends upon more that three factors because we can get only three EQUATIONS by equations the POWERS of M.L and T. |
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| 2772. |
The figure shows a schematic diagram showing the arrangement of Young.s Double Slit Experiment If the distance d is varied,then indentify the correct statement |
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Answer» The angularwidth does not change |
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| 2773. |
A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E_(0) and a resistance r_(1). An unknown e.m.f E is balaned at a length l of the potentiometer wire. The e.m.f E will be given by : |
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Answer» `(LE_(0)r)/((r+r_(1))l)` POTE ntial gradient `sigma = (Ir)/(L)` But I `= (E_(0))/(r + r_(1))` `therefore sigma = (E_(0)r)/((r + r_(1))L)` `therefore` emf E = `sigma l = (E_(0)rl)/((r+ r_(1))L)` |
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| 2774. |
The figure shows a schematic diagram showing the arrangement of Young.s Double Slit Experiment If the distance D is varied , then choose the correct statement (s) |
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Answer» The ANGULAR FRINGE width does not change |
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| 2775. |
A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity v. At this instant. |
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Answer» the ELECTRIC forces on both the particles cause IDENTICAL accelerations |
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| 2776. |
The output (X) of the logic circuit shown in figure will be |
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Answer» `X= barbar(A).barbar(B)` |
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| 2777. |
(A) : The tangent galvanometer can be made more sensitive by increasing the number of turns of its coil. (R) : For a given P.D current through galvanometer is inversely proportional to the number of turns of coil. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 2778. |
A man of 80 kg mass is standing on the rim of a circular platform rotating about its axis. The platform with the man on it rotates at 12.0 r.p.m. How will the system rotate, if the man moves to the platform's centre? What work will the man perform in changing his position? The mass of the platform is 200 kg and its radius is 1.2 m. |
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| 2779. |
A stone is projected fromthepointof a groundin such adirection soas to hita birdon thetop of atelegraphpostof heighth and then attain the maximum height 2h above the ground . Ifat theinstant of projection,the bird were to fly away horizontally with a unifrom speed. Find the ratio between the horizontal velocitiesof thebird and stone, if thestone still hitsthe brid while ascending . |
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Answer» Solution :Letthe stoneis projected with a velcoityu at an ANGLE `theta`WITHTHE horizontal , we have ` (2h) = ((u sin theta)^(2))/(2g) oru sin theta = 2 sqrt(gh)` Supposet is thetimetaken by thestoneto reachthe HEIGHTH abovethe ground. Then `h = u sin theta t - (1)/(2) "gt"^(2) ( or) ("gt"^(2))/(2) - u sin theta+ h = 0` As we have`u sin theta = 2sqrt(gh) "" THEREFORE ("gt"^(2))/(2) - 2sqrt(gh) t+ h = 0` Solvingaboveequationfor t, we get `t = (2sqrt(gh) pm sqrt((2sqrt(gh))^(2) - 4 xx (g)/(2)h))/(2 xx (g)/(2)) = (2sqrt(gh) pm sqrt(2gh))/(g)` Let ` t_(1) = sqrt((2h)/(g))(sqrt(2)-1) and t_(2) = sqrt((2h)/(g))(sqrt(2)+1)` Where `t_(1)` and `t_(2)` correspond to P and Q in the figure. Supoosev is the horizontalvelcoityof the bird. Then`PQ = vt_(2)` For the motion of stone PQ =` u cos theta (t_(2) - t_(1)) "" therefore (t_(2) - t_(1))` whichgives `(v)/(u cos theta) = (t_(2) -t_(1))/(t_(2)) =(sqrt((2h)/(g)) (sqrt(2)+1) - (sqrt(2)-1))/(sqrt((2h)/(g))(sqrt(2)+1)) = (2)/(sqrt(2) +1)` 
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| 2780. |
A ball isthrown vertically upwards with velocity of 25m/s from the top of a tower of height 30m . How long will if travel before it hits the ground ? |
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Answer» 6s |
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| 2781. |
24 cells, each of emf 1.5V and internal resistance is 2Omega connected to 12Omega series external resistance. Then, |
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Answer» in order to SEND the maximum current through a `12OMEGA` resistor, CONNECT the two rows of 12 cells in SERIES |
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| 2782. |
The focal length of convex lens is 10 cm. Find its magnifying power when it is used as a magnifying glass to form the image at (i) near point and (ii) far point. |
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| 2783. |
Dimension of MI is _____. |
| Answer» SOLUTION :[`M^1L^2T^0`] | |
| 2784. |
What is the difference between noise and musical sound? |
| Answer» SOLUTION :The SOUND which produces a frying effect is called noise while the sound which produces a PLEASING effect is called a MUSICAL sound. | |
| 2785. |
A conductor with respectivityrho bounds on a dielectric with permittivity epsilon. At a certain point Aat theconductor's surface the electric displacementequals D the vectorD beingdirectedaway from theconductorand formingan anglealpha with the normalof thesurface. Find the surfacedensityof chargeson theconductor at thepoint Aand the current densityin the conductor in the vicinitity of the same point. |
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Answer» Solution :The dielectric ends in a CONDUCTOR. It is given that on one side (the dielectric side) the electric displacement`D`is as SHOWN. Within the conductor, at any point `A`, there can be no normalcomponenetof electric FIELD. For if there weresuch a field, a currentwill flowtowards despositing chargethere which in turn will set upcounteringelectricfieldcausingthe normalcomponenetto vanish. They by Gauss' theorem, we easily derive `sigma = D_(n) = D cosalpha` where `sigma` in the surface charge density at `A`. The tangential COMPONENET is determind from the circularion theorem `ointvec(E) . d VEC(R ) = 0` It must be continous across the surface of the conductor there is a tangentail electric field of magnitude, `(D sinalpha)/(epsilon epsilon_(0)) at A` This implies a current by Othm's law of `j = (D sinalpha)/(epsilon epsilon_(0) rho)` 
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| 2786. |
Does the emf represnet a force or potential energy or work done per unit charge or potential differences? |
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Answer» SOLUTION :The term emf is a misnomer. In a closed circuit, emf is a source of external force (infact a source of energy) which compels the current CARRIERS (electrons or ions) to move in a definite direction (i.e., from lower potential energy to higher potential energy). The electromotive force is associated with an arrangement or mechanism which can supply energy or does work to move the electric CHARGE from lower potential energy to higher potential energy. Such an arrangment is called a source of emf, which may be a cell, a battery, a generator or dynamo. EMF of a cell is defined as the energy supplied by the cell to drive a unit POSITIVE charge once around the complete circuit. It is also defined as the maximum potential difference between two electrodes of a cell when no current is DRAWN from the cell. |
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| 2787. |
The coil in figure carries current i=2.00 A in the direction indicated is parallel to an xz plane, has 3.00 turns and an area of 4.00xx10^(-3) m^(2), and lies in a uniform magnetic field overset-B=(2.00hati-3.00hatj-4.00hatk)mT. What are (a) the orientation energy of the coil in the magnetic field and (b) the torque (in unit-vector notation) on the coil due to the magnetic field? |
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| 2788. |
Maximum density of H_2O is at temperature |
| Answer» Answer :B | |
| 2789. |
What do you mean by the term half life period ? |
| Answer» SOLUTION :It is the TIME TAKEN for HALF of the ORIGINAL substance to disintegrate. | |
| 2790. |
A tuning fork vibrating with frequency of 512 Hz is kept close to the open end of a tube filled with water (figure). The water level in the tube is gradually lowered. When the water level is 17cm below the open end maximum internsity of sound is heard If the room temperature is 20^(@)C calculate |
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Answer» Speed of sound in air at room TEMPERATURE (`a`) For list maximum of intersity the length of the air coloumn `l=(lambda)/(4) implies 4l=4xx17xx10^(-2)m` Hence speed of sound `v=f lambda=512(417xx10^(-2))` `=348.16m//s` (`b`) We know that `v prop sqrt(T)` where temperature (`T`) is in kelvin `(v_(20))/(V_(0))=sqrt((273+20)/(273+0))=sqrt((293)/(273))` `(v_(20))/(V_(0))=sqrt(1.073)=1.03` `v_(0)=(v_(20))/(1.03)=(348.16)/(1.03)=338m//s` (`c`) The resonance will still be observed for `17cm` length of air coloumn above mercuy. However due to more compelete reflection of sound WAVES or mercury. However due to conplete reflection of sound waves at mercury, surface the INTENSITY relected sound increases. |
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| 2791. |
A beam of light is incident on a glass slab of refractive index 1.5 at an angle of incidence of 45o. Find the ratio of the width of refracted beam to the incident beam. |
| Answer» SOLUTION :`1.24 :1` | |
| 2792. |
A monoenergetic (18 keV) electron beam Initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e= 9.11 xx 10^(-31) kg). (Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth's magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.] |
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Answer» Solution :Given : `(K.E)_(e) = 18 xx10^(3)xx1.6xx10^(-19) J = 28.8 xx10^(-16) `J. `B_(H) = 0.04 G = 4XX 10^(-6) T.` SINCE electronsdescribe circular path Bev sin `theta = (mv^(2))/( R) and theta = 90^(@)` ` R = (mv)/(EB)` and `"" KE = 1//2 mv^(2)` ` v = sqrt((2)/(m)(KE))` and `mv = sqrt(2m (KE))` i.e., `"" R= (1)/(eB) sqrt(2m(KE))` i.e., `"" R = sqrt(2xx9.11xx10^(31)xx28.8xx 10^(-16))/(1.6xx10^(-19)xx4xx10^(-6))` R= 11.3 m sin `theta = 0.3//11.3 = 0.2655 ` `cos theta = sqrt(1-sin^(2)theta) = sqrt(1-0.007)= 0.9996` Up or down deflection - R (1-cos0) = 11.3 (1-0.9996) = 0.00398 m `= 3.98 xx 10^(-3)`m `~~ 4.0 mm` |
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| 2793. |
The equation of standing wave of the second resonance can be , |
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Answer» `P_(EX) = 2A sin 2pi (y = 1cm) cos 2pi 9340)t` FIRST NODE will be formed at `y = -1` instead of `y =0` so eqn of standing wave is `P_(ex) = 2A sin 2pi (y + 1cm) cos 2pi (340)t` . |
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| 2794. |
Which of the following is not coherent source |
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Answer» a source along with its virtual IMAGE in the CASE of Lloyd.s single mirror |
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| 2795. |
Who had the fantastic notion of spending £10 on a holiday? |
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Answer» SON of Corn-merchant |
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| 2796. |
In fig. a central particle of charge -q is surrounded by two circular rings of charged particle of radii r and R, with Rgtr. what is the magnitude and dirction of the net electrostatic force on the central particle due to the other particles? |
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Answer» Solution :Net FORCE on charge due to outer ring will be zero. Due to INNER ring. The MAGNITUDE of force on `-q` at CENTER is `F_(n)=sqrt(F^(2)+F^(2))=sqrt(2)F=sqrt(2)xx(1)/(4 pi epsilon_(0))xx(2qxxq)/(r^(2))` `=(q^(2))/(sqrt(2)pi epsilon_(0)r^(2))` The direction is at angle `45^(@)` from the x-axis in second quardrant. |
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| 2797. |
How much water should be filled in a container of height 21 cm, so that it appears half filled to the observer when viewed from the top of the container (mu=4//3). |
Answer» Solution : `(REAL depth)/(APPARENT depth)=mu` `x/(21-x)=4/3 IMPLIES x=12 CM` |
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| 2798. |
How electron -hole pairs are created in a semiconductor material ? |
| Answer» SOLUTION :The FREE electrons from electron hole pairs, enable current to flow in the semiconductor when an external voltage is APPLIED. The holes in the valence BAND is applied. The holes in the valence band ALSO allow electron movement within the valence band itself and this also contributes to current flow. This process is called electron-hole pair generation. | |
| 2799. |
ABC is a right angled glass prism of refractive index 1.5. angleA , angleB and angle C are 60^@, 30^@ and 90^@ respectively. A thin layer of liquid is on the face AB. For a ray of light which is incident normally on AC to be totally reflected at AB, the refractive index of the liquid on AB should be |
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Answer» 1.5 |
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| 2800. |
Two charges -q each are separated by distance 2d. A third charge +q is kept at mid-point O. Find potential energy of +q as a function of small distance x from O due to -q charges. Prove that the charge at O is in an unstable equilibrium. |
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