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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.
| 43651. |
Making use of the result obtained in the foregoing problem, find the conditions defining the angular position of maxima of the first, the second, and the third order. |
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Answer» Solution :SINCE `I(alpha)` is `+ve` and vanishes for `b sin varphi = k LAMBDA` i.e for `alpha = k lambda`, we expect maxima of `I(alpha)` between `alpha =+ pi & alpha =+ 2pi`, etc. We can get these values by. `(d)/(d alpha) (I(alpha)) = I_(0)2(sin alpha)/(alpha)(d sin alpha)/(alpha) = 0` `(alpha cos alpha- sin alpha)/(alpha^(2)) = 0` or `tan alpha = alpha` SOLUTIONS of this transcendental equation can be obtained graphically. The first three solutions are `alpha_(1) = 1.43pi, alpha_(2) = 2.46, alpha_(3), = 3.47 pi` on the `+ve` side. (On the negative side the solution are `-alpha_(1), -alpha_(2),- alpha_(3), .......)` THUS `b sin varphi_(1) = 1.43 lambda` `b sin varphi_(2) = 2.46 lambda` `b sin varphi_(3) = 3.47 lambda` Asymptotocally the solutions are `b sin varphi_(m) ~= (M+(1)/(2)) lambda` |
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| 43652. |
One electric generator produces power at 220 V potential difference. Its internal resistance is r = 10 Omega . Find power wasted in external resistance R = 100 Omega |
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Answer» 484 W R.=R+r =100+10 =110 `Omega` `THEREFORE` Current in the circuit , `I=V/(R.)=220/110`=2A `therefore` Power wasted in external resistance , `P=I^2R=(2)^2 xx 100` =400 W |
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| 43653. |
Monochromatic light of frequency 6.0xx10^(14) Hz is produced by a laser. The power emitted is 2.0 xx10^(-3) W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on an average, are emitted by the source? |
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Answer» SOLUTION :(a) Each photon has an energy `E=hv=(6.63xx10^(-34)Js)(6.0xx10^(14Hz))` `=3.98xx10^(-19)J` If N is the number of PHOTONS emitted by the source per second, the POWER P transmitted in the beam EQUALS N times the energy per photon E, so that P = N E. Then `N=(P)/(E)=(2.0xx10^(-3)W)/(3.98xx10^(-19)J)` `=5.0xx10^(15)" photons per second."` |
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| 43654. |
If electric force between point charged varies inversely as the cube of the distance, will Gauss.s law be valid? |
| Answer» SOLUTION :No. Gauss.s LAW is only TRUE if inverse square law of ELECTRIC FORCE is true. | |
| 43655. |
In an X-rays tube, if the electrons are accelerated through 140 KV, then anode current obtained is 30 mA. If the whole energy of electrons is converted into heat, then the rate of production of heat at anode will be : |
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Answer» <P>968 CALORIE `P=140xx10^(3)xx30xx10^(-3)` `=4200` watt=4200 Joule/sec or `P=(4200)/(4*2)=1000` calorie. |
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| 43656. |
A point source of light is 60 cm from a screen and is kept at the focus of a concave mirror whichreflects light on the screen. The focal length of the mirror is 20 cm. The ratio of average intensities of the illumination on the screen when the mirror is present and when the mirror is removed is: |
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Answer» `36:1` |
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| 43657. |
The plates of a parallel plate capacitor are, not exactly parallel. The surface charge density therefore : |
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Answer» HIGHER at the closer end Being a conducotr, each plate has the same potential at each point. But because `E=-(DeltaV//Deltax)`, hence `E` is HIGHEST where the plates are closed to each other. Hence, surface charge density is also higher at the cloer end. |
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| 43658. |
Determine the current in each brance of the network showin in |
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Answer» Solution :Applying KVL `( sum IR = sum EPSILON)` In the CLOSED loop ADCA, we get -4 `(I_(1) - I_(2) ) + 2 (I_(2) + I_(3)- I_(1) )= -10` `therefore 4 (I_(1) - I_(2) ) - 2 (I_(2) + I_(3) - I_(1)) + I_(1) = 10` `therefore 7 I_(1) - 6I_(2) - 2I_(3) = 10 ""` .... (1) Similarly for closed loop ABCDA, ` -4 I_(2) -2 (I_(2) + I_(3)) - I_(1) =- 10 ` `therefore 4I_(2) +I_(2) -2 (I_(2) + I_(3)) + I_(1) = 10` `therefore I_(1) + 6I_(2) + 2I_(3) = 10 ""` ... (2) For closed loop BCDEB, ` - 2 (I_(2) + I_(3)) - 2 (I_(2) + I_(3) - I_(1)) = - 5 ` ` therefore 2 (I_(2) + I_(3)) + 2 (I_(2) + I_(3) - I_(1)) =5 ` `therefore 4 I_(2) + 4I_(3) - 2I_(1) = 5 ` `therefore 2I_(1) - 4I_(2) - 4I_(3) = - 5` `therefore I_(1) - 2I_(2) - 2I_(3) = - 2.5 ""` .... (3) Adding equations (1) and (2), `8I_(1) = 20` `therefore I_(1) = (5)/(2)A ""` .... (4) Adding equations (2) and (3) `2I_(1) + 4I_(2) = 7.5` `therefore2 ((5)/(2)) + 4I_(2) = 7.5 ` `therefore 5 + 4I_(2) = 7.5` `therefore I_(2) = (2.5)/(4) = (25)/(40) = (5)/(8) A "" `... (5) Substituting values of`I_(1) and I_(2)` in EQUATION (3), 2.5 -2 `((5)/(8)) - 2I_(3) =- 2.5 ` `therefore 2.5 - (5)/(4) + 2.5 = 2 I_(3)` `therefore 10 - 5 + 10 = 8I_(3)` `therefore 15 = 8I_(3)` `therefore I_(3) = (15)/(8) A ""` .... (6) |
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| 43659. |
Effective (or rms) current of an a.c. is that steady current which consumes exactly the same power as actual a.c. consumes during one complete cycle when flowing through a given resistor |
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Answer» |
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| 43660. |
What is amplification? With a circuit diagram, explain the working of npn transistor as an amplifier in CE configuration. |
Answer» Solution : The circuit of an n-p-n transistor as an ampliffer in CE-configuration is shown in the above figure. The EB-junction is forward biased with the batter `V _(B B) and CB`-junction is revers biased with the batter `V _(C C).` A suitable load resistor `R _(L)` is connected in the output circuit. The ac SINGNAL to be amplifiedis applied betwen bae and emitter. Thge amplified output is take between collector and emitter. The circuit of an n-p-n transistor an an amplifier in CE-configuration is shown in the above figure. The EB-juctin is forward biased witht the BETTER `V _(B B) and CB-`junction is reverse biased with the battery `V _(C C).` A suitable load resistor `R _(L)` is connected in the output circuit. The ac signal to be amplified is applied between bae and emitter. The amplified output is taken between collector and emitter. When the current `I _(C )` flows through the load resistor `R _(L),` the output voltage is given by. `V _(0) = V _(C ) - I _(C)R _(L)` During positive half cycle of input ac signal, the forward bias of EB-junction increases. DUE to this the emitter current increases and hence the collector current ALSO increases. According to the above equation, as the collector current increases output voltage V. decreases. Since the collector is connected to the positive terminal of the battery Vco the decrease in collector voltage means the collector becomes less positive (or NEGATIVE with respect to the initial value). Thus during positive half cycle of input ac voltage, the output signal varies through a negative half cycle. During negative half cycle of input ac signal, the forward bias of EB-junction decreases. Due to this the emitter current decreases and hence the collector current also decreases. According to eqn (2), the collector voltage `V_(C C)` increases and the collector become more positive. This indicates that during the negative half cycle, the output signal varies through positive half cycle. Thus in Common emitter amplifier, the input voltage and output voltages are 180° out of phase as shown in the figure. |
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| 43661. |
A resistance of 2Omegais connected across one gap of a metre bridge (the length of the wire is 100cm) and an unknown resistance greater than 2Omega , is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm neglecting any corrections, the unknown resistance is |
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Answer» `3 OMEGA ` |
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| 43662. |
Following figure shows a network of eight resistors, each equal to 2 Omega , connected to a 3 V battery ofnegligible internal resistance. The current I in the circuit is :- |
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Answer» 0.25 A `therefore ` AB,BC and CD are in series . Total resistance `R_1=6Omega` AE,EF and FD are in series . Total resistance `R_2=6Omega` When they are is parallel , total resistance = `3Omega` `therefore ` CURRENT =` (3V)/(3Omega)`=1.0 A |
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| 43663. |
The work function of a material is 5.3 eV and the stopping potential for a radiation incident on this surface is 4.2V. The energy of a quantum of incident radiation is |
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Answer» 1.1 eV |
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| 43664. |
Given vecr=2hati+hatj+hatk and vecp=2hati-3hatj+hatk. Then angular momentumvecL is : |
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Answer» `8hatk` |
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| 43665. |
What is Tangent law ? |
| Answer» SOLUTION :If the magnetic needle is placed under the ACTION of TWO uniform magnetic fields B and H acting at right ANGLES to one onther,it is deflected through an angle `theta` GIVEN by. | |
| 43666. |
Specific heat at constant pressure C_(p) for a diatomic gas at N.T.P. in the units of cal mole k is: |
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Answer» <P>`4 CDOT 95` `=(1)/(2) K.T.` `therefore` The energy of 1 molecule `=5xx(1)/(2) KT` `therefore` 1 mole of gas contians N molecules. `therefore` Total energy of 1 mole of diatomic gas `U=(N cdot 5)/(2) kT. =(5)/(2) RT" "[because k=(R)/(N)]` Since `C_(v) =(dU)/(dT)` `C_(v)=(5)/(2) R` `therefore C_(p) =C_(v) +R =(7)/(2) R=(7)/(2) xx1 cdot 98=6 cdot 93" cal mole"^(-1)k^(-1)` THUS, correct choice is (b). |
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| 43667. |
(a) An a.c. source of voltage V=V_(m) sin omega tis connected across a series combination of an inductor, a capacitor and a resistor. Use the phasor diagram to obtain the expression for (i) impedance of the circuit, and (ii) phase angle between the voltage and the current (b) A capacitor of unknown capacitor, a resistor of 100 Omegaand an inductor of self-inductance L = 4/pi^(2) H are connected in series to an a.c. source of 200 V and 50 Hz. Calculate the value of the capacitance and the current that flows in the circuit when the current is in phase with the voltage. |
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Answer» Solution :(a) N/A (B) Here, `R = 100 Omega, L = 4/pi^(2) H, V_(rms) = 200V` and v= 50 HZ Since current flowing in the circuit is in phase with voltage, hence resonant frequency `v_(r) =v = 50 Hz`, As `v_(r) = 1/(2pi sqrt(LC))` `THEREFORE C = 1/(4pi^(2) Lv_(r)^(2)) = 1/(4pi^(2) xx (4/pi^(2)) xx (50)^(2)) = 25 xx 10^(-6) F = 25 MUF` and current `I_(rms) = V_(rms)/Z = V_(rms)/R = 200/100 A = 2.0 A` |
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| 43668. |
Define electric dipole and dipole moment. What is it's unit ? |
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Answer» Solution :An electric dipole is a pair of equal and OPPOSITE CHARGES SEPARATED by some distance. The electric dipole moment is the product of CHARGE q and VECTOR `vec 2a` drawn from the - ve charges to the + ve charges. `vec P= q xx vec 2a` |
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| 43669. |
The energy of the electron in the ground state of hydrogen is - 13.6 eV. Calculate the energy of the photon that would be emitted if the electron were to make a transition corresponding to the emission of the first line of the (i) Lyman series, (ii) Balmer series of the hydrogen spectrum. |
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Answer» Solution :ENERGY of ELECTRON in ground state (n = 1) of hydrogen E, = - 13.6 eV. (i) Firstline of Lyman series is emitted fortransition n = 2 ton=1 . Hence . `E_(2) = (E_(1))/((2)^(2)) = - (13.6)/(9) = - 3.4 eV` `therefore ` ENERGYOF the emitted photon `E_(1)= E_(3) - E_(2) = - 1.51 -(-3.4) = 1.89 eV` |
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| 43670. |
...... found the phenomena of diffraction. |
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Answer» Young |
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| 43671. |
Assertion : Paramagnetism and ferromagnetism are associated with orbital motion of electrons. Reason : In ferromagnetics, the magnetic effect is increased due to the formation of domains. |
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Answer» Assertion is TRUE, REASON is True, Reason is a CORRECT explanation for Assertion |
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| 43672. |
Answer the following questions: (e) 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 :JUSTIFICATION based on what is EXPLAINED in (d). Typical sizes of APERTURES involved in ordinary optical instruments are much larger than the WAVELENGTH of LIGHT. | |
| 43674. |
Given the wave function (in S.I. units) for a wave to be, Psi (x,t) = 10^(3) sin pi (3 xx 10^(6) x - 9 xx 10^(14) t). The speed of the wave is : |
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Answer» `3 xx 10^(6) m // s` |
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| 43675. |
A person looking through a telescope focuses lens at a point on the edge of the bottom of an empty cylindrical vessel. Next he fills the entire vessel with a liquid of refractive index mu, without disturbing the telescope. Now, he observes themid point of the vessel. Determine the radius to depth ratio of the vessel. |
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Answer» `1/2sqrt((1-MU^(2))/(mu^(2)+1))` |
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| 43676. |
A particle of mass 'm' is suspended from the ceiling through a string of length L. The particle moves in a horizontal circle of radius r. The speed of the particle is. |
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Answer» `(rg)/(sqrt(L^2 - r^2))` |
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| 43677. |
Water penetrated into a crack in a rock and froze there. What is the resulting pressure? |
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Answer» |
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| 43678. |
Suppose a pure crystal has 5xx10^(28) atoms m^(-3). It is doped by 1 part per million concentration of pentavalent As . Calculate the number of electrons and holes . Given that n_(i)=1.5xx10^(16)m^(-3). |
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Answer» Solution :Note that thermally generated electrons `(n_(i)~10^(16)m^(-3))` are NEGLIGIBLY small as compared to those produced by DOPING . Therefore, `n_(e)~~N_(D)`. Since `n_(e)n_(H)=n_(i)^(2)`, `n_(e)=(5xx10^(28))/(10^(6))=5xx10^(22)` The number of holes `n_(h)=(2.25xx10^(32))//(5xx10^(22))~4.5xx10^(9)m^(-3)` |
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| 43679. |
An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path. |
| Answer» SOLUTION : B should be in a VERTICALLY DOWNWARD DIRECTION. | |
| 43680. |
A surface, value of electric potential on whichis same at all the points, is called ...... surface. |
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Answer» GAUSSIAN |
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| 43681. |
A : To increase the resolving power of a microscope, .oil immersion objective. canbe used. R : Resolving power of the microscope is given by the reciprocal of the maximum separation of two objects distinctly seen. |
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Answer» Both A and R are true and R is the CORRECT explanation of A |
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| 43682. |
An of the following statements are correct except (for real object): |
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Answer» the magnification produced by a convex mirroris always less than or equal to one |
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| 43683. |
A bullet of mass a moving with velocity b strikes a large stationary block of wood of mass c, and remains embed in it, the final velocity of the system is : |
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Answer» `(a)/(a+b)C` `V=((a)/(a+c))b` |
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| 43684. |
Assertion: Neutrons penetrate matter more readily as compared to protons. Reason: Neutrons are slightly more massive than protons |
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Answer» If both ASSERTION and REASON are TRUE and the reason is the correct EXPLANATION of the assertion |
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| 43685. |
(a) For the circuit shown in the figure, how would the balancing length be affected, if(i) R_1 is decreased, (ii) R_2 is increased, the other factors remaining the same in the circuit ?Justify your answer in each case. (b) Why is a potentiometer preferred over a voltmeter ? Give reason. |
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Answer» Solution : (a) (i) When `R_1` is decreased, the current flowing in the POTENTIOMETER wire AB and consequentlythe POTENTIAL gradient along the potentiometer wire increases. As a result, the balancinglength decreases i.e., the balancing point shifts towards left (towards point A). (ii) When `R_2` is increased, the current flowing through `R_2`decreases and as a result the potentialdifference across the terminals of the CELL, whose internal resistance is to be determined, increases `[V = epsi- Ir]` . As a result, the balancing length increases, i.e., the balancing pointshifts towards right (towards the point B). (b) A potentiometer is preferred over a VOLTMETER for MEASURING potential differences and emf of cells because measurement is done here in balance condition, when no current is being drawn fromthe given cell. |
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| 43686. |
Which scientist experimentally proved, for the first time, the existence of electromagnetic waves? |
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Answer» SIR J.C. Bose |
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| 43687. |
Two cells are connected in series and e.m.f. of combination is found to balance against length of 450 cm ofpotentiometer wire. Whentthetwo cells are connected in parallel emf of combination balances against50 cm length of wire. The ratio of e.m.f. of two cells is |
| Answer» Solution :`(E_(1))/(E_(2))=(l_(1)+l_(2))/(l_(1)-l_(2))=(450+50)/(450-50)=5/4=1.25` | |
| 43688. |
A monochromatic light passes through a glass slab (mu = 1.5) of thickness 9 cm in time t_1/ If it takes a time t_2to travel the same distance through water (mu = 4//3). The value of (t_1 - t_2)is |
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Answer» `5 xx 10^(-11) SEC ` |
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| 43689. |
Draw the block diagram of generalised communication system. |
Answer» SOLUTION :
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| 43690. |
Soil formed by intense leaching is: |
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Answer» ALLUVIAL soil |
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| 43691. |
(A): The mutual induction between the two coils infinitely apart is zero. (R): If the mutual induction between the two coils is zero, it means that their self inductances are also zero. |
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Answer» Both A and R are TRUE and R is the correct EXPLANATION |
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| 43692. |
How does declination at a place helps in navigation? |
| Answer» Solution :Declination at a PLACE is the angle between the geographic MERIDIAN and magnetic meridian. With the help of knowledge of declination at a place, we can STEER the ship/boat in the desired direction to REACH a destination POINT. | |
| 43693. |
A bus is travelling at 20 m/s on a circular road of radius 400 m. its radial acceleration is |
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Answer» `1 m/s^2` |
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| 43694. |
A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4thth of the angle of prism. Calculate the speed of light in the prism. |
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Answer» Solution :For equilateral triangular prism A = 60° and the angle of incidence for minimum deviation STATE `i=3/4 A = 45^(@)`. As in minimum deviation position i = e `therefore D_(m) = i+e-A=2i -A = 2 xx 45^(@) - 60^(@) = 30^(@)` `therefore n=(sin(A+D_(m))/2)/(sin(A/2)) =(sin 45^(@))/(sin 30^(@)) =(1//sqrt(2))/(1//2) = sqrt(2)` But, `n=c/v`, hence, SPEED of light in the PRIM `v=c/n = (3 xx 10^(8))/sqrt(2) =2.12 xx 10^(8) ms^(-1)` |
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| 43695. |
A body of mass 2 kg is acted upon by two perpendicular forces 4 N along the X-axis and 3 N along the Y-axis. What is the magnitude of the acceleration of the body ? |
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Answer» `1.5 ms^(-2)` :.acceleration `a = F//m=(5)/(2)=2.5m//s^(2)` Hence correct CHOICE is (C). |
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| 43696. |
(a) Write two characteristic features of nuclear force. (b) Draw a plot of potential energy of a pair of nucleons as a function of their separation |
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Answer» Solution :(a) (i) NUCLEAR forces are INDEPENDENT of charge (ii) They are the STRONGEST forces in nature (B) 
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| 43697. |
A cylindrical bar magnet is rotated about its axis as in figure. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then |
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Answer» a direct current FLOWS in the ammeter A. |
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| 43698. |
What are electromagnetic waves? State their characteritics/properties. |
| Answer» SOLUTION :The speed of electromagnetic WAVE in free SPACE is `C = 1/(sqrtmu_0epsilon_0)` | |
| 43699. |
What is the change in the energy of a charged particle when it enters in a magnetic field and experiences a magnetic force ? |
| Answer» Solution :There is no change in the ENERGY of a chargedparticle because when a CHARGED particle enters in a magnetic field, it experiences a FORCE at RIGHT angles to its velocity. As the particle is moving in a circle due to centripetal force, therefore, SPEED of the charged particle is constant and hence , its energy remains constant. | |
| 43700. |
Three thin disc A, B, C of the same material are coated with carbon black radii of discs are 2cm, 4cm, & 6cm respectively. Wavelength corresponding to maximum intensity are 300nm, 500nm, & 600 nm respectively. Power radiated by the discs are Q_(A), Q_(B) & Q_(C) |
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Answer» `Q_(A)` is MAXIMUM |
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