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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.
| 6651. |
An n - p -n transistor circuit is arranged as shown in the figure , it is |
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Answer» a COMMON - base AMPLIFIER CIRCUIT |
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| 6652. |
Tow identical short barmagnets each having magnetic moment m are placed a distance of 2dapart with axes perpendicualr to each other a horizontal plane the magnetic inlduciton at point midway between them is |
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Answer» `(mu_(0))/(4pi)sqrt(2)(M)/(d^(3))` `rarr B_(net)=(mu_(0))/(4pi).(sqrt(5)M)/(d^(3))` 
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| 6653. |
It is now belived that protons and neutrons (whichconstitude nuclei of ordinary matter)are themselves built out of more elementaryunitscalled quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so-called 'up'quark (denoted by u) of charges +(2)/(3) e,and the 'down' quark (denoted by d )of charges(-(1)/(3)) e,togetherwith electronsbuild up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter. ) Suggest a possible quark composition of a proton and neutron. |
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Answer» Solution :Possible quark COMPOSITION of a proton is ` 2u+1d i.e. ,u,u,d,` so that TOTAL charges on it ` =+(2)/(3) e+(2)/(3) e-(1)/(3) e=+e.` Possible quark composition of a neutron is `1mu +2d i.e. , u, d,d, `so that total charge on it. `=+(2)/(3) e, -(1)/(3) e-(1)/(3) e=0` . |
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| 6654. |
Calculate the de Broglie wavelength associated with the following : (i) electrons moving with a speed of 10^(5)ms^(-1) (ii) protons moving with a speed of 10^(5)ms^(-1) (iii) a proton of momentum 2.26 xx 10^(-23) kg ms^(-1)(iv) a thermal neutron of energy 0.025 eV. |
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Answer» `3.97 xx 10^(-12) m` `0.29 xx 10^(-10) m` `1.807 Å` `0.1690 Å` `0.01276 Å` |
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| 6655. |
The number of electric lines of force originating from a charge of 1C is |
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Answer» `1.129 XX 10^(11)` |
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| 6656. |
In the circuit shown in fig. power developed across 1 Omega, 2 Omega and 3 Omega resistances are in the ratio of |
| Answer» Answer :D | |
| 6657. |
Inthediagram,grapharedrawnbetweenstopping potentialV_ 0andfrequencyvfortheelementsKandCa.Accordingthistodiagram |
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Answer» theworkfunctionsofKandCaareequal |
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| 6658. |
Alnico and ticonal materials are commonly used for preparing __________ . |
| Answer» SOLUTION :PERMANENT MAGNETS | |
| 6659. |
A parallel beam of light consisting of two wavelenths lambda_1 = 4000 Å and lambda_2 = 8000 Å is incident perpendicular to plane of both slits in a typical Young’s double slit experiment. The seperation between both slits is d = 2mm and the distance between slits and screen is D = 1 meter. In each situation of column-I a point P on screen is specified by its distance ‘l’ from central bright on screen. Match the proper entries from column-2 to column-1 using the codes given below the columns {:("Column-I",,"Column-II"),((P) "At P such that l=0",,"(1) intensity is maximum for " lambda_(1)=4000Å),((Q) "At P such that l=0.1mm",,"(2) intensity is maximum for " lambda_(1)=4000Å),((R) "At P such that l=0.2mm",,"(1) intensity is maximum for " lambda_(2)=8000Å),((S) "At P such that l=0.4",,"(4) intensity is maximum for " lambda_(1)=8000Å):} |
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Answer» <P>`{:(,P,Q,R,S),((A),3,2,4,3):}` |
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| 6660. |
Explain the existence of sharply defined K, and B, characteristic X-rays. |
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Answer» Solution :X-rays are produced when fast ELECTRONS IMPINGE on a metal target. Some of these electrons interact with the nuclei of the target atoms and are decelerated to different degrees. The loss in their kinetic energies CHARACTERISTIC X-rays. that is CONVERTED to X-rays give rise to the continuous X-ray spectrum. When the incident electrons have sufficient energy, some of them knock out tightly-bound inner-shell electrons of the target atoms. Then, an electron from an outer shell or a free electron from outside the atom occupies the energy state of the knocked out electron. The energy difference appears as an X-ray photon. Such photons have discrete energies characteristic of the target metal atom and give rise to the characteristic X-rays. When a K-shell (n =1) electron is knocked out, the transition of an L-shell electron (n=2) to that n=1 state results in the `K_(ALPHA)` line, while the transition of an M-shell electron (n=3) to the n =1 state gives rise to the `K_(p)` line, and so on. |
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| 6661. |
To emit a free electron from a metal surface a minimum amount of energy must be supplied. When light of frequency 7.21xx10^14 Hz is incident on a metal surface, the maximum speed of the ejected electrons is 6xx10^5 ms^-1. Calculate the threshold frequency for the metal. |
| Answer» SOLUTION :`hV=hV_0+1/2mv^2 or V_0=V-1/2(mV^2)/H=(7.21xx10^14)-1/2[9.1xx10^-31xx((6xx10^5))^2/(6.626xx10^-34))=4.74xx10^14 HZ` | |
| 6662. |
Two resistors of 5 Omega and 10 Omega are connected in parallel with a cell of emf 3 V and internal resistacne 1 Omega. Calculate the current through each of the resistors. |
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Answer» <P> Solution :`R _(P) = ( 5 xx 10 )/( 5 + 10 ) = (50)/( 15) = (10)/(3) Omega``I = (E)/( R _(P) +r ) = (3)/(( 10)/(3) +1 ) = (9)/( 13) A` `I_(1) = (IR _(2))/( R _(1) + R _(2)) = ((9)/( 13) xx 10 )/( 5 + 10 ) = (6)/(13) A` `I _(2) = I- I _(1) = (3)/(13) A.` |
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| 6663. |
Series combination of 10Omegaresistance and 5 cm long solenoid having 5 mH self inductance are connected to 10 V battery. Current passing through solenoid in steady state is ___ |
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Answer» 5A `I=I_0 (1-e^(-Rt//L))` `=I_0(1-1/e^(Rt//L))` Theoretically , STEADY state is reached when `t to OO` . At this TIME `I to I_0`. But `I_0=V/R=10/10`=1A |
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| 6664. |
A steel girder of length l rests freely on two supports (figure). The moment of inertia of its cross-section is equal to I (see the foregoing problem). Neglecting the mass of the girder and assuming the sagging to be slight, find the deflection lambda due to the force F applied to the middle of the girder. |
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Answer» Solution :ONE can THINK of it as analogous to the previous case but with a BEAM of length `l//2` loaded UPWARD by a force `F//2`. THUS `lambda=(Fl^3)/(48EI)`, On using the last result of the previous problem. 
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| 6665. |
Prove that not more than two electrons can occupy the s-state and not more than six can occupy the p-state. |
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Answer» Solution :The orbital quantum number corresponding to the s-state is l = 0. Therefore the appropriate MAGNETIC quantum number is also m = 0. Hence the ELECTRONS may differ only in their spin projections: s = 1/2 and s = -1/2. Thus there are TWO possible sets of quantum NUMBERS: n, 0, 0, 1/2 and n, 0, 0, -1/2. The orbital quantum number corresponding to the p-state is l = 1. Therefore the magnetic quantum number can assume three values: m = 1, m = 0 and m = –1. Since there are two possible spin projections corresponding to each magnetic number, the possible sets of quantum numbers are six in all: n, 1, 1, 1/2, n, 1, 1, -1/2, n, 1, 0, 1/2, n, 1, 0, -1/2, n, 1, -1, 1/2 and n, 1, -1, -1/2. |
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| 6666. |
By means of plotting find : (a) the path of a ray of light beyound thin converging and diverging lenses (Fig. where OO' is the optical axis, F and F' are the front and rear focal points), (b) the position of a thin lens and its focal points if the position of the optical axis OO' and the positions of the cojugate3 points P, P (see Fig.) are knows , the media on both sides of the lenses are identical, (c ) the path of ray 2 beyond the converging and diverging lenses (Fig. ) if the path of ray 1 and the positions of the lens and of its optical axis OO' are all known, the media on both sides of the lenses are identical. |
Answer» Solution :Clearly the media on the sides are different. The front focus `F` is the position of the object (virtual or real) fro which the imega is formated at infinity. The rear focus `F'` is the position o fthe image (virtual or real )of the object at infinity. (a) FIGURES (a) & (b). This geomertical CONSTRUCTION ensists that the second of the equations is obeyed.  (c) Figure (a) & (b). Clearly, the important CASE is that when the rays (1) & (2) are not symmetric about the principle axis, otherwise the figure can be complated by reflection in the principle axis. Knowing onepath we know the path of all rays connecting the TWO points. For a different object. We proced as shown below, we use the fact that a ray incident at a given height above the topic centre surfcaes a denifite DEVIATION. The concave lens can be disussed similarly.  
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| 6667. |
Which is the incorrect match for the given compounds ? |
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Answer» Napthalene`":"`Homocyclic `&` Aromatic |
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| 6668. |
Define current gain in common base and common emitter amplifier and find the relation between them. |
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Answer» Solution :Current gain `alpha` is defined as the RATIO of collector current to the emitter current at constant collector VOLTAGE. Current gain `BETA` is defined as the ratio of the collector current to the base current at constant collector voltage. i.e. `beta=((I_(c))/(I_(B)))_(E_(ie))` Relation between `alpha` and `beta`. For n-p-n and p-n-p transistor, we have `I_(e)=I_(b)+I_(c)` or `(I_(e))/(I_(c))=(I_(e))/(I_(c))+1`......`(i)` Since current gain of common emitter amlifier, `alpha=(I_(c))/(I_(e))` And current gain of common base amplifier, `beta=(I_(c))/(I_(b))` So Eq. `(i)` becomes `(1)/(alpha)=(1)/(beta)+1` or `(1)/(beta)=(1)/(alpha)-1` or `beta=(alpha)/(1-alpha)` |
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| 6669. |
The rad is the correct unit used to report the measurement of : |
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Answer» The ability of a beam of `gamma`- rays to produce IONS in a target. Hence CORRECT choice is `(b)`. |
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| 6670. |
Using Huygens principle , draw a diagram to show the refraction of plane wave front incident obliquely on a surface separating two media. |
Answer» SOLUTION :
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| 6671. |
Two identical magnetic magnetic dipoles of magnetic moment 2Am^2 are placed at a separation of 2 m with their axis perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is |
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Answer» `4sqrt5xx10^(-5)T` |
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| 6672. |
Maxwell’s equationintvecE.vecdl=(-dphi_(m))/(dt) is astatement of |
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Answer» AMPERE’s LAW |
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| 6673. |
Name the laws associated with the following equation. ointvecE.vecdt = m_0(I_C+epsilon_0 (dphi_E)/dt |
| Answer» SOLUTION :`ointvecE.vecdt` = 0. This EQUATION is BASED on the fact that magnetic monopoles do not exist. | |
| 6674. |
The Quantisation condition in Bohr model of atom is given by mvr = ? |
| Answer» SOLUTION :`NH/(2PI)` | |
| 6675. |
Define the term drift velocity |
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Answer» SOLUTION :The average velocity acquired by the free electrons of a conductor in a DIRECTION opposite to the externally applied electric FIELD is called drift velocity. Drift Velocity. `v_d=(-eEtau)/m` Where, e = charge on electron E = EXTERNAL electric field `tau`= relaxation TIME m = mass of electron |
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| 6676. |
The unit of expression mu_(0)in_(0) are : |
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Answer» `m//s` |
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| 6677. |
In the circuit of CE amplifier, a silicon transistor is used. The value of V_("CC")=+20V, R_(L)=3kOmega, collector voltage =5v, beta=100. Then the base resistance R_(B) should be ……xx10^(5)Omega (Take input voltage drop across Si transistor = 0.7 V) |
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Answer» |
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| 6678. |
In Davisson-Germer experiment in order to increase velocity of electron emitted from electron gun…… |
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Answer» potential ddifference between ANODE and filament should be increased |
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| 6679. |
In the above problem the displacement after 2 s is: |
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Answer» `20[1+(SQRT(3)-1)i]` |
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| 6680. |
A glass cube of edge 1 cm and mu = 1.5has a spot at the centre. The area of the cube face that must be covered to prevent the spot from being seen is (in cm^2) |
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Answer» `SQRT5 PI` |
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| 6681. |
The principalbehind opticalfibres is |
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Answer» total internal REFLECTION |
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| 6682. |
In an experiment the angle are required to be measured using an instrument 29 divisions of the main scale exactly coinecide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (=0.5^(@)) then the least count of the instrument is |
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Answer» `1` minute `=(1)/(30)xxS.D=(1)/(30)xx0.5^(@)=(1^(@))/(60)=1.` `:.(a)` is the CORRECT choice. |
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| 6683. |
Two cars P and Q start from a point at the same time in a straight line and their position are represented by x_(p)(t)=at+bt^(2) and x_(1)(t)=ft-t^(2).At What time do the cars have the same velocity ? |
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Answer» `(a-f)/(1+b)` |
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| 6684. |
Energy of all molecules of a monatomic gas having a volume V and pressure P is 3//2PV. The total translational kinetic energy of all molecules of a diatomic gas at the same volume and pressure is |
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Answer» `1//2PV` where f = degrees of freedom Montomic or diatomic GASES posses equal degree of freedom for translational MOTION and that is equal to 3, i.e., `f=3` `therefore E=(3)/(2)PV` Although TOTAL energy will be different For monatomic gas, `E_("total")=(3)/(2)PV""("As" f=3)` For diatomic gas, `E_("total")=(5)/(2)PV""("As" f=5)` |
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| 6685. |
Write any five properites of ferromagnetic materials |
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Answer» Solution :1) FERROMAGNETIC MATERIALS are STRONGLY attracted by the magnets. 2) They obey Curie.s law above curie temperature 3) Their magnetization is large and POSITIVE 4) Their susceptibility is high and positive. |
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| 6686. |
In a transistor both emitter base jucntion and colector base junction are reverse biased how will the concentration of chargecarrier across the each junction change |
Answer» SOLUTION :Whenboththe collectorbaseandemitterbasejunctionare REVERSED theemitterwill notsupplychargecarriersand henceno currentwillfowin thetransistor.
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| 6687. |
Let f_(1) be the frequency |
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Answer» `f_(1) -f_(2) = f_(3)` `f_(2)` relates `rArr n_(1) =1, n_(2) =2` `f_(3)` relates `rArr n_(1) =2, n_(2) = alpha` . |
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| 6688. |
(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision. (b) The total magnification produced by a compound microscope is 20. The magnification produced by the piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14cm. If least distance of distinct vision is 20cm, calculate the focal length of the objective and the eye piece. |
Answer» SOLUTION :
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| 6689. |
In the circuit shown, the potential difference between x and y will be |
| Answer» Answer :B | |
| 6690. |
A ball is projected at an angle 30^(@) with the horizontal. What is the component of accepleration along the velocity of projection : |
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Answer» G |
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| 6691. |
A body projected up with a velocity of 39.2 m/s reaches a point in its path after t_1 and t_2 sec from start. Then (t_1+t_2) is |
| Answer» Answer :B | |
| 6692. |
You are standing upright in a room in front of a vertical mirror.In this mirror, you can see from your position only the upper two - third of your body.You wish to see the entire lenght of your body reflected in the mirror.Which combination of the follwoing three courses of action will achieve this? (I) Move away from the mirror (II) Move toward the mirror (III) Use a mirror whose height will allow you to see your whole image. |
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Answer» (I) Only |
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| 6693. |
Absolute refractive index of a medium is X. Refractive index of same medium w.r.t to air is Y and absolute refractive index to air is Z. Then the relation between them is |
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Answer» X=Y/Z |
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| 6694. |
A car starting from rest, accelerates at a constant rate of 5 ms for some time. It then retards at a constant rate of 10 ms and finally comes to rest. If the total time taken is 6 s, what is the maximum speed attained by the car: |
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Answer» `10 ms^(-1)` |
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| 6695. |
In a hydrogen atom, the radius of n^(th) bohr orbit is r_n. The graph between log (r_n//r_1) and logn will be |
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Answer»
So, LOG `(r_n//r_1)=21`OGN Hence, the GRAPH between log `(r_n//r_1)` and logn will be a straight line passing through origin. The POSITIVE SLOPE is given by `tantheta=2`. |
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| 6696. |
The work function of caesium is 2.14 eV. The threshold frequency is_____ |
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Answer» `5.16xx10^19 HZ` |
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| 6697. |
S is the surface of a lump of magnetic material. |
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Answer» Lines of `overset(to)(B)` are necessarily continuous across s. Magnetic intensity outside magnet `(H) = (B)/( mu_0)` and for inside the magnet `H= (B)/(mu_(0) mu_(r) )` (where `mu_r` is the RELATIVE permeability of material). Magnetic intensity H forms continuous magnetic field lines hence field lines of B need to pass from S. Magnetic intensity varies for inside and outside of lump. So lines of H can not all be continuous across S. |
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| 6698. |
If v_(K_(alpha)),v_(K_(beta)) and v_(L_(alpha)) are the frequencies of K_(alpha),K_(beta) and L_(alpha) lines, then |
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Answer» `v_(K_(ALPHA))+v_(K_(beta))=v_(L_(alpha))` |
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| 6699. |
Five equal capacitors connected in series have a resultant capacitance 4uF. What is the ratio of energy stored when the capacitors are connected in series and then parallel and connected to the same source of emf in both the cases is |
| Answer» ANSWER :C | |
| 6700. |
A long straight horizontal cable carries a current of 2.5 A in the direction 10^@ south of west to 10^@ north of east. The magnetie meridian of the place happens to be 10^@ west of the geographie meridian. The earth's magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral potnts, magnetic field due to a current-carying cable is equal and opposite to the horizontal component of earth's magnetic field.) |
Answer» Solution : As shown in the diagram, for all the points along the horizontal line at perpendicular distance R below the given horizontal cable carrying current from a to b, if `B.=B_h` then all these points (like `P_1, P_2, ….)` would be NULL points. HENCE, `B.= B_h` `therefore (mu_(0 ) I)/( 2pi r) = B cos phi` (Where `B=` magnetic FILED of Earth and `phi =` dip angle ) `therefore (mu_(0 ) I)/( 2pi r) = B ( because "Here", phi =0^@` ) `therefore r= (mu_(0) I)/( 2pi B)` `therefore ((4pi xx 10^(-7) )(2.5) )/( (2pi ) (0.33 xx 10^(-4) ) )` `therefore r= 1.5 xx 10^(-2) m` |
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