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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.
| 39851. |
What is a LED? Give the principle of operation with a diagram. |
Answer» Solution :Light Emitting Diode (LED): LED is a p-njunction diode which emits visible or invisible light when it is forward biased. Since electrical energy is converted into light energy, this process is also called electroluminescena The cross-sectional view of a commercial LED is shown in figure (B). It consists of a p-laver n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads, anode and cathode When the p-n junction is forward biased, the conduction band electrons on n-side and VALENCE band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the electrons in the conduction band recombine with holes in the valence band as shown in the figure (C). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hu is emitted. For nonradiative recombination, energy is liberated in the form of heat. The colour of the light is DETERMINED by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AIGP) and red (GaAsP). Now a days, LED which emits white light (GAINN) is also available. |
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| 39852. |
Two point charges q_(1), and q_(2) are kept at a distance of r_(12) in air. Deduce the expression for the electrostatic potential energy of this system. |
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Answer» Solution :Let a point charge 4, be situated at a given point A. Electric potential due to it at another point B, situated at a distance `r_(12)` will be `V= (1)/(4pi in_(0)).(q_(1))/(r_(12))` If a point charge `q_(2)` is brought from infinity to point B, then work done to bring this charge against the field of charge `q_(1)` will be `W= q_(2)V= (1)/(4pi in_(0)). (q_(1)q_(2))/(r_(12))` This work done is stored in the SYSTEM of TWO point charges as their electrostatic potential energy U. Thus `U= (1)/(4piin_(0)).(q_(1)q_(2))/(r_(12))` |
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| 39853. |
Establish the fact that the relative motion between the coil and the magnet inducesan emf in the coil of a closed circuit. |
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Answer» Solution :i. whenever the magnetic flux linked with a CLOSED coil changes, an emf (electromotive force) is induced and hence an electric current FLOWS in the circuit. ii. This current is called an induced current called an induced emf. This phenomenon is known as electromagnetic induction. iii. A bar magnet is PLACED CLOSE to a coil, some of the magnetic lines of forces of the bar magnet pas through the coil i.e., the magnetic flux is linked with the coil. When the bar magnet and coil approach each other, the magnetic flux linked with the coil INCREASES. iv. So this increases in magnetic flux induces an emf and hence a transient electric current flows in the circuit in one direciton. At the same time, when they recede away from one another, the magnetic flux linked with the coil decreases. v. The decrease in magnetic flux again induces an emf in opposite direction and hence a electric current flows in the galvanometer when there is a relative motion between the coil and the magnet. |
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| 39854. |
What is binding energy of a nucleus? Give its expression. |
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Answer» <P> Solution :When Z PROTONS and N neutrons combine to form a nucleus, mass equal to mass defect disappears and the corresponding ENERGY is released. This is called the BINDING energy of thenucleus (BE) and is equal to `(Delta m)c^(2)`.`BE = (Zm_(p) + Nm_(n) -M ) c^(2)`. |
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| 39855. |
The ratio of energy of emitted radiation of a black body at 27^(@)C and 927^(@)C is : |
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Answer» `1:4` `(E_(1))/(E_(2))=((1)/(4))^(4)""RARR(E_(1))/(E_(2))=(1)/(256)` Thus CORRECT CHOICE is (d). |
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| 39856. |
There are two spherical balls A and B of the same material with same surface finish but the radius of A is half than that of B. If A and B are heated to the same temperature and allowed to cool then: |
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Answer» RATE of cooling of both is same |
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| 39857. |
If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material be denoted by mu_d, mu_p and murespectively, then |
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Answer» `mu_d NE0 and mu_f =0` |
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| 39858. |
The average emf induced in a coil in which the current changes from 2 A to 6 A in 0.05 s is 8 V. Self-inductance of the coil is |
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Answer» 0.1 H |
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| 39859. |
A coil of inductance 0.50 H, and resistance 100Omega is connected to a 240V, 50Hz ac supply What is the maximum current in the coil? |
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Answer» Solution :For an LR circuit, if `V=V_(0)sin omega t` `I=(V_(0))/(SQRT(R^(2)+omega^(2)L^(2))) sin (omega t -phi)`, where `tan phi=(omega L//R)`. (a) `I_(0)=1.82A` (b) V is maximum at t = 0, I is maximum at `t=(phi//omega)`. Now, `tan phi=(2pi vL)/(R ) = 1.571 or phi ~~57.5^(@)` Therefore, time lag `=(57.5pi)/(180)xx(1)/(2pi xx 50)=3.2 ms` |
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| 39860. |
When does (i) a plane mirror and (ii) a convex mirror produce real image of objects |
| Answer» SOLUTION :Plane and convex mirror produce REAL image when the object is virtual that is rays convering to a point behind the mirror are reflected to a point on a SCREEN. | |
| 39861. |
(A): Thermistors are used in temperature con trol unit of industry. (R): Thermistor is a heat sensitive device. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A' |
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| 39862. |
What does Indian cinema need? |
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Answer» a STYLE, an IDIOM a SORT of ICONOGRAPHY of cinema |
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| 39863. |
Canal effect. Figureshows an anchored barge that extends across a canal by distance d = 30 m and into the water by distance b = 12 m. The canal has a width D = 55 m, a water depth H = 14 m, and a uniform water-flow speed v_(i) = 1.2 m/s. Assume that the flow around the barge is uniform. As the water . passes the bow, the water level undergoes a dramatic dip known as the canal effect. If the dip has depth h = 0.80 m, what is the water speed alongside the boat through the vertical cross sections at (a) point a and (b) point b? The erosion due to the speed increase is a common concern to hydraulic engineers. |
| Answer» SOLUTION :(a) 2.37 m/s ~~ 2.4` m/s (B) 2.3 m/s | |
| 39864. |
Two co-axisal rings, each of radius R, are separated by a small distance l(ltltR) and carry charges q and -q.Find the electric potential and field at a point on the axis at a distance x from the midpoint between the rings, |
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Answer» |
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| 39865. |
The current I in the adjoining circuit is |
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Answer» `1/45 A` `1/R = 1/30+ (1)/((30 + 30)) = 1/30 + 1/30 RARR R = 20Omega` ` therefore ` Circuit current `I= V/R = (2V)/(20Omega ) = 1/10A` |
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| 39866. |
The barrier potentials for silicon and Germanium diodes are about |
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Answer» 0.7V , 0.3V |
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| 39867. |
The satellite of mass m is orbiting around the earth in a circular orbit with a velocity v.What will be its total energy ? |
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Answer» `(3//4)MV^(2)` `K.E.=(|P.E.|)/(2), K.E. =(1)/(2)mv^(2)` `:.` Total energy `=(-1)/(2)mv^(2)`. |
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| 39868. |
A telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 15 cm is focused on a distant object in such a way that parallel rays comes out from the eye lens. If the object subtends an angle 2^(0) at the objective, the angular width of the image is |
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Answer» `10^(0)` |
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| 39869. |
In the circuit shown in Fig, the transistor used has a current gain beta=100. If the bias resistance R_(B)=axx10^(5)Omega, then V_(CE)=5 V. Neglecting V_(BE), find the integer value of a? |
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Answer» `-V_(C C)+I_(C)xx10^(3)+V_(CE)=0` or `V_(C C)=I_(C)xx10^(3)+V_(CE)` 
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| 39870. |
A water tank is partially fill upto 2 m height. As shown in figure, 0.1 m thick glass slab is placed on water surface. If refractive indices of water and glass are 1.3 and 1.5 respectively, what will be virtual depth of object placed at bottom when it is viewed from top ? |
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Answer» 3.2 m |
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| 39871. |
When an electric cell drives current through load resistance, its Back emf, |
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Answer» Supports the original emf |
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| 39872. |
A bomb explodes into two pieccs of masses 4 kg and 8 kg. The velocity of 8 kg mass is 6 m/s. The KE of the other mass |
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Answer» 48 J |
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| 39873. |
In the water a of a lakea blastoccurs. Thewaves produced in water will be, |
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Answer» TRANSVERSE |
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| 39874. |
A radioactive substance with decay constant of 0.5 s^(-1) is being produced at a constant rateof 50 nuclei per second. If there are no nuclei present initially, the time in second) after which 25 nuclei will be present is |
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Answer» 1 `(DN)/(DT)=50-lamdbaN rArr (dN)/(50-lambdaN)=dt` Integrate both sides, we get `int_(0)^(N)(dN)/(50-lambdaN) =int_(0)^(t) dt rArr -(1)/(lambda)[ln(50-lambdaN)]_(0)^(N)=t` `ln((50-lambdaN)/(50))=-lambdat` `(50-lambdaN)/(50)=e^(-lambdat), 1-(lambdaN)/(50)=e^(-lambdaN), N=(50)/(lambda)(1-e^(-lambdat))` As, `N=25 and lambda=0.5s^(-1)` `25=(50)/(0.5)(1-e^(-0.5t)) or e^(-0.5t)=(3)/(4) therefore t=2ln((4)/(3))` |
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| 39875. |
In, all the capacitors are in steady state initially. i. What is the charge flowing through the switch when it is closed? ii. What is the charge flowing section AB? iii. What is the work done by the battery? iv. What is the heat produced when (S) is closed? . |
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Answer» Solution :Before closing the SWITCH, `4x + (x-6) 2+ (x-6)2=0` or `x=4 V` `|Q_(4muF)|=16 muC,|Q_(2muF)|=4muC` `|Q_(6muF)|=12 muC` When switch `S_(w)` is closed, the `4 muF` capacuit is short-circuited, and the potential difference across it BECOMES zero. The circuit after closing the switch is shown in   .  . shown the charges coming and going out of the capacitors. So the charge through section `AB` is `32 muC`, and the charge through the switch is `48 muC`. When the switch is closed, `16 muC` charge from the left plate goes the switch is charge PASSING through th battery is `32 muC`. so the work done by the battery is `W_("battery")=DeltaqV=(32 muC)(6V)=192 mu J` Heat produced is `H=W_("battery")-DeltaU=W_("battery")-(U_(f)U_(i))` `U_(f)=1/2(2)xx6^(2)+1/2(6)(6)^(2)=144 mu J` and `U_(i)=1/(2)4(4)^(2)+1/2(8)2^(2)=48 muJ` `H=[192-(144-48)]mu J=96 mu J`. |
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| 39876. |
Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction. |
| Answer» Solution : The TWO processes are :(i) Diffusion (ii) Drift Diffusion : Holes DIFFUSE from p-side to n-side ( p `to `n) and electrons diffuse from n-side to p-side ( n`to`p). Drift: The motion of charge carriers, DUE to the applied ELECTRIC field (`vecE`) which results in drifting of holes along `vecE` and of electrons OPPOSITE to that of electric field (`vecE`). | |
| 39877. |
When a metallic rod falls freely, keeping its length horizontal and parallel to north-south direction, emf induced across it is ______ |
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Answer» increases with time |
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| 39878. |
A disc of mass 100 g slides down from rest on an inclined plane of 30^(@) and come to rest after travelling a distance of 1m along the horizontal plane. If the coefficient of friction is 0.2 for both inclined and horizontal planes, then the work done by the frictional force over the whole journey, approximately, is (Acceleration due to gravity, g = 10 ms^(-1) ) |
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Answer» 0.106 J From third equation of the motion, velocity of disc when it leaves the inclined plane,  `V^(2) - u^(2)- 2as` or `"" mu = sqrt(2as)""(because v = 0)` `therefore` Acceleration of a block on a horizontal, a = `mu` G `rArr"" a = 0.2 xx 10 = 2m//s^(2)"" ("given", mu = 0.2 )` After putting the value of a in Eq. (i), we get `therefore "" u = sqrt(2 xx 2 xx 1 )` = 2 m/s `"" ( because s = 1 ` m, given ) From above diagram, The FRICTIONAL force applied on the disc inclined plane r is F = `mu` N = `mu` mg cos `30^(@)` and the net acceleration force down the inclinedplane, mg sin `30^(@)` -f = mg sin `30^(@) - mu mg cos 30^(@) = ma_(1)` or `a_(1) = g(sin 30^(@) = mu cos 30^(@)) = 5( 1 - sqrt(3) mu )`... (1) HENCE, from the third equation of motion, `v_(1)^(2)= mu_(1)^(2) + 2a_(1) `s When s is distance travelled by the disc, `v_(1) = u = 2 m//s m_(1) = 0` `therefore "" (2)^(2) = 2a_(1) s " or " s = (2)/(a_(1)) ` Putting the value of `a_(1)` form Eqs (i) we get `rarr "" s = (2)/(5 (1 - sqrt(3) mu ) ) ` = `(2)/( 5 (1 - sqrt(3) xx 0.2)) = 0.612 `m Hence, the work done by the frictional force, W = work done on inclined plane + work done on horizontal plane. = `( mu mg cos 30^(@) ) s + (1)/(2) mu^(2)` Putting the given values , we get ` = 0.2xx (100)/(1000) xx 10 xx cos 30^(@) xx 0.612 + (1)/(2) xx (100)/(1000) xx (2)^(2)` = 0.306 J |
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| 39879. |
As per de-Broglie explanation of Bohr's quantum condition, for an electron revolving in nth circular orbit the total circumference of the orbit 2pi r, is equal to______. |
| Answer» SOLUTION :`n lambda ` , where `lambda` is the de- BROGLIE WAVELENGHTOF ELECTRON wave. | |
| 39880. |
When a conducting wire is connected in the right gap and known resistance in the left gap, the balancing length is 60cm. The balancing length becomes 42.4 cm when the wire is stretched so that its length increases by |
| Answer» ANSWER :B | |
| 39881. |
A particle is executing simple harmonic motion along a straight line PQ. At three points A, B and C on the line PQ, lying on one side of the mean position, the velocities of the particles are 8ms^(-1), 7ms^(-1) " and "4ms^(-1), respectively. If AB = BC = 1m, the velocity of the particle at mean position is |
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Answer» `9ms^(-1)` velocity of particle A, `v_(A)=8ms^(-1)` velocity of particle B, `v_(B)=7ms^(-1)` velocity of particle C, `v_(C)=4 ms^(-1)` velocity of a particle in harmonic motion (SHM), `RARR ""V=omegasqrt(a^(2)-x^(2))` Square on the both sides, we get `""v^(2)=a^(2)omega^(2)-x^(2)omega^(2)` For particle A, `""v_(A)^(2)=a^(2)omega^(2)-x^(2)omega^(2)` `rArr ""64=a^(2)omega^(2)-x^(2)omega^(2) "...(i)"` For particle B, `""v_(B)^(2)=a^(2)omega^(2)-(x+1)^(2)omega^(2)` `rArr ""49=a^(2)omega^(2)-(x+1)^(2)omega^(2) "...(ii)"` For particle C, `""v_(C)^(2)=a^(2)omega^(2)-(x+2)^(2)omega^(2)` `rArr ""16=a^(2)omega^(2)-(x+2)^(2)omega^(2) "...(iii)"` By solving Eqs. (i), (ii) and (iii), we get `""x=1/3, a^(2)omega^(2)=65 ""(because v=aomega)` HENCE, velocity of particle at the MEAN position, `v=aomega=sqrt(65)ms^(-1)`. |
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| 39882. |
A particle moves in a circle with a uniform speed. When it goes from a point A to a diametrically opposite point B, the momentum of the particle changes by vecP_(A)-vecP_(B)=2 kg m/s (hatj) and the centripetal force acting on it changes by vecF_(A)-vecF_(B)=8N(hati) where hati,hatj are unit vectors along X and Y axes respectively. The angular velocity of the particle is |
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Answer» `"8 RAD s"^(-1)` |
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| 39883. |
0.7040 मे कितने सार्थक अंक है - |
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Answer» 1 |
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| 39884. |
Faraday's law on electromagnetic induction gives ___ |
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Answer» DIRECTION of induced emf |
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| 39885. |
Represent the variation of PEI v/s anode potential. |
Answer» SOLUTION :
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| 39886. |
A siren placed at a railwary platform is emitting sound of frequency 5 k Hz. A passenger sitting in a moving train approaches the siren. During his return journey in a different train B, he records a frequency of 6.0 K Hz. While approaching the same siren. The ratio of velocity of train B to the velocity of train A is : |
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Answer» 242/252 `v_(1) = 0.1` v In second case, 6 = `((V + v_(2))/(v) ) 5 rArr v_(2) = 0.2 `v `therefore (v_(2))/(v_(1)) = 2 `. CORRECT choice is (c). |
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| 39887. |
How old was margie? |
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Answer» 12 |
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| 39888. |
In the following problems, assume that the resistance of each diode is zero in forward bias and infinity in reverse bias . (a) Find the current supplied by the battery in the following cases. (b) Find the currents through the resistance in the circuits shown in the figures. (c ) Find the current through the battery in each of the circuits shown in the figures. (d) Find the equivalent resistance of the network shown in the figure between the points A and B. |
Answer» Solution :(a)  `D_(1)` is in forward BIAS, `i_(1)=(5)/(10)=0.5A` `D_(2)` is in reverse bias, `i_(2)=0` `i=i_(1)+i_(2)=0.5A` (ii)  `D_(1)` is in reversible bias, `i_(1)=0``D_(2)` is forward bias, `i_(2)=(5)/(20)=0.25A` `i=i_(1)+i_(2)=0.25A` (b) (i)  Both diodes are in forward bias, `i=(2)/(2)=1A` (ii)  `D_(1)` is in forward bias, `D_(2)` is in reverse bias `i=0` (iii)  Both diodes are in forward bias `i=(2)/(2)=1A` (iv)  `D_(1)` is in forward bias, `D_(2)` is in reverse bias `i=(2)/(2)=1A` (c ) (i)  Both diodes are in forward bias, `i_(1)=5//10=0.5A,i_(2)=(5)/(10)=0.5A` `i=i_(1)+i_(2)=1A` (ii)  Upper diode is in forward bias `i_(1)=(5)/(10)=0.5A` Lower diode is in reverse bias `i_(2)=0` `i=i_(1)+i_(2)=0.5A` (d)  If`V_(A)gtV_(B)` Diode is in forward bias, it offers zero RESISTANCE. `R_(eq)=(10)/(2)=5OMEGA` If `V_(A) lt V_(B)`, Diode is in reverse bias, it offers infinite resistance. `R_(eq)=10OMEGA` |
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| 39889. |
The separation between two plates of a parallel plate condenser is filled with two dielectric media as shownin figure . The ratio of its capacities , with and without dielectric , is |
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Answer» `K_(1)(K_(1)+K_(2))` Without dielectric , `C_(0)=(epsilon_(0)A)/(d)` The two capacitors formed with `K_(1)andK_(2)` are in series `C_(1)=(K_(1)epsilon_(0)A)/(d//2)=2K_(1)*(epsilon_(0)A)/(d)=2K_(1)C_(0)` `C_(2)=(K_(2)epsilon_(0)A)/(d//2)=2K_(2)*(epsilon_(0)A)/(d)=2K_(2)C_(0)` In series COMBINATION, `C_(eq)=(C_(1)C_(2))/((C_(1)+C_(2)))=((2K_(1)C_(0))(2K_(2)C_(0)))/(2K_(1)C_(0)+2K_(2)C_(0))` `C_(eq)=(4K_(1)K_(2)C_(0)^(2))/(2C_(0)(K_(1)+K_(2)))=(2K_(1)K_(2)*C_(0))/((K_(1)+K_(2)))` `(C_(eq))/(C_(0))=(2K_(1)K_(2))/((K_(1)+K_(2)))` |
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| 39890. |
A radioactive substance has half-life T_(1/2) and number of radioactive nuclides present initially was N_(0). What is the number of radioactive nuclides at time t ? |
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Answer» Solution :We know that number of radioactive nuclides left intact after a time `T_(1/2)` is `N = N_(0)/2` , Thus, after ANOTHER time equal to `T_(1/2)` (i.e., after 2 half-lives) VALUE `N =1/2(N_(0)/2)=N_(0)(1/2)^(2)`. It FOLLOWS that after a time equal to n half-lives, the number of radioactive nuclides left intact will be `N=N_(0)(1/2)^(n)` For time t, `n=t/T_(1/2)` and hence, `N=N_(0)(1/2)^(t/T_(1/2)`. |
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| 39891. |
Which of the following is used as a moderator in nuclear reactors? |
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Answer» Plutonium |
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| 39892. |
A charge is distributeduniformly overa ringof radius'a'Obtain anexpressionfor theelectricintensityE at apointon theaxisof thering . Henceshow thatforpointa lergedistancefrom theringitbehaves likea pointcharge . |
Answer» Solution : Suppose that theringis placed withits planeperpendicularto thex-axisas shown in Fig. Consider asmallelementdl of the ring .  Asthe totalchargeq isuniformlydistributedthe charge dq on the elementdl isdq `=(Q)/(2pia) .dl` `:.` Themagnitudeof thefieldd `bar(E)` produced bytheelementdlat thefieldpoint P is `bar(DE) =k .(dq)/(r^(2)) =(kq)/(2pia) .(dl)/(r^(2))` Thefield `dbar(E)` has twocomponents (a) the axialcomponent dEcos `theta` , and (b)the perpendicularcomponentdEsin `theta` Sincethe perpendicularcomponentsof any twodiametricallyoppositeelements are equalandoppositethey allcancelout inpairs.Onlythe axialcomponents will addup toproducethe resultantfield `bar(E)` at pointP whichis givenby `E = overset(2pia)underset(0)(int) dE " cos" theta""[:' "onlythe axialcomponentscontributetowards" E]` `=overset(2pia)underset(0)(int) (kq)/(2pia) ,(dl)/(r^(2)).(X)/(r ) =(kpx)/(2pia) .(1)/(r^(3)) overset(2pia)underset(0)(int) dl` `=(kpx)/(2pia).(1)/(r^(3)) [l]_(0)^(2pia) =(kpx)/(2pia).(1)/((x^(2)+a^(2))^(3//2)).2pia ""[:' r^(2) =x^(2) +a^(2)]` `"or"""E =(kpx)/((x^(2) +a^(2))^(3//2))=(1)/(4piepsilon_(0)).(qx)/((x^(2) +a^(2))^(3//2))` LTBGT `" if" ""x gt gt a` `x^(2) +a^(2) ~~x^(2)` `E =(1)/(4pi epsilon_(0)) .(qx)/(x^(3)) rArr E=(1)/(4piepsilon_(0)).(q)/(x^(2))` i.e., Whenthe pointis FARAWAY from thecentrecharged ringacts likea pointcharge . |
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| 39893. |
A point source of light is placed 4 m below the surface of a transparent liquid of refractive index 5/3. The minimum diameter of a disc which should be placed over the surface of the liquid to cut off all light coming out of the liquid is : |
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Answer» 3 m |
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| 39894. |
In Pascal's law , the pressure P in terms of force and area is given by |
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Answer» P=Fa |
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| 39895. |
If the electric field components due to electric chargeincubeshowninfigureare E_x = 600x^(1/2) and E_y = 0 and E_z = 0, then charge within the cube………… |
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Answer» |
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| 39896. |
A uniform rod of mass M and length L is free to rotate about a frictionless pivot located L/3 from one end. The rod is released from rest incrementally away from being perfectly vertical, resulting in the rod rotating clockwise about the pivot. when the rod is horzontal what is the magnitude of the tangential acceleration of its center of mass? |
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Answer» g/6  Torque about P `MG((l)/(6))=(Ml^(2))/(9)alpha` `IMPLIES=(3)/(2)(g)/(l)` `impliesa_(t)=alpha(l)/(6)=(g)/(4)` |
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| 39897. |
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? |
| Answer» SOLUTION :DIVERGING, 60 CM | |
| 39898. |
Two cells of same emf epsibut internal resistances r_1 and r_2are connected in series to an externalresistor R. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero. |
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Answer» Solution : Here net emf of combination = `2epsi`and net resistance of the circuit = `r_1 + r_2 + R` CURRENT flowing in the circuit `I = (2 EPSI)/(r_1 + r_2 + R)` As potential difference across FIRST CELL `V_1 = epsi - Ir_1= 0, ` hence `epsi - (2epsi)/((r_1 + r_2 + R)) r_1 = 0 ` or`epsi[r_1 + r_2 +R - 2r_1]= 0` ` rArr R = r_1 - R_2 ` |
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| 39899. |
(A): Electron capture occurs more often than positron emission in heavy elements. (R) : Heavy elements exhibits radioactivity. |
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Answer» Both .A. and .R. are true and .R. is the CORRECT explanation of .A. |
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| 39900. |
A charged metallic sphereAis suspendedby a nylon thread. Another chargedis brought close to A such that the distancebetweentheir centers is 10cm as shownin Fig. 1(a).12(a). The resultingrepulsion of Ais noted(for example,byshininga beamof lightand measuringthe deflectionof its shadowon acalibratedscreen). Spheres A and B are touchedby uncharged spheresC and Drespectively, asshownin Fig. 1(a). 12b. C and D are thenremovedand B is broughtcloserto A to a distance of 5.0 cmbetweentheir centers as shown in Fig. 1(a). 12(c). What is the expected repulsionof A on and spheres B and D haveidenteical sizes. Ignorebetweenthier centers. |
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Answer» Solution :Let the origanalrepulsiveforce between A and B be `F = (k q_(1) q_(2))/(r^(2))` As A and C havesame size, their charges are also sharedequally. As charges on A and B are halvedand DISTANCEBETWEEN them is alsohalved from 10 cm to 5 cm, therefore, `F = k((q_(1)//2) (q_(2)//2))/((r//2)^(2)) = (k q_(1) q_(2))/(r^(2)) = F` |
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