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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.
| 41701. |
The drift velocity acquired per unit electric field is the ………………….. . |
| Answer» SOLUTION :mobililty | |
| 41702. |
A magnified image of real object is to be obtained on a large screen 1 m from it. This can be achieved by |
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Answer» using a convex mirror of focal length less than 0.25 m |
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| 41703. |
A ray of light is refracted from medium 1 to medium 2. Show that the ratio of the sine of the angle of incidence and the sine of the angle of refraction is equal to the ratio of speed of light in medium 1 and that in medium 2. |
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Answer» SOLUTION :if the absolute refractive indices of medium 1 and medium 2 are `mu_(1) and mu_(2)` respectively we know, `(sini)/(sinr) = 1^(mu)2 = (mu^(2))/(mu_(1))` Now, `"" mu_(1) = ( C )/(v_(1)) [c = "SPEED of light in vacuum," v_(1) = "speed of light in medium 1"]` `mu_(2) = (c)/(v_(2)) [v_(2) = "speed of light in medium" 2]` `therefore "" (mu_(2))/(mu_(1)) = ((c)/(v_(2)))/((c)/(v_(1))) = (v_(1))/(v_(2))` From (1) and (2) we get, `(sini)/(sinr) = (v_(1))/(v_(2))` |
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| 41704. |
A candle flame 1.6 cm high is imaged ina ballbearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image. |
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Answer» 1.0 mm inside the BALL bearing, 0.08 mm |
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| 41705. |
Magnetic field produced by electron in atom or molecule is due to its ...... |
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Answer» SPIN motion. |
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| 41706. |
Define the S.I unit of magnetic field '' A charge moving at right angles to uniform magnetic field does not undergo change in kinetic energy '' Why ? |
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Answer» Solution :Tesla is the SI unit of magnetic field . The magnetic field at a POINT is one Tesla if a charge fo one coulomb while moving perpendicular to the magnetic field with a velocity of `1 ms^(-1)` experiences a force of 1 NEWTON at that point . The force on a moving charged particle in a magnetic field is perpendicular to its direction of MOTION . So WORK done on the charged particle by the magnetic force is zero . Hence the kinetic energy of charged particle in a magnetic field remains UNCHANGED . |
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| 41707. |
A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster"s angle phi. if n represents the refractive index of glass w.r.t. air, then the angle between the reflected rays will be : |
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Answer» `(90^@ + PHI)` |
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| 41708. |
The motion of a copper plate is damped when it oscillates between the two poles of a horse shoe magnet. The cause of damping is _____. |
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Answer» |
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| 41709. |
Electromagnetic waves are transverse in nature is evident by: |
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Answer» Polarisation |
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| 41710. |
In Lloyd's mirror interference experiment the source slit is at a distance of 2 mm from a plane mirror . The interference fringes , observed on a screen at a distance of 1.5 m from the slit , have a separation of 0.221 mm . Calculate the wavelength of light used . |
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Answer» |
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| 41711. |
The wavelength of K_(alpha) line from an element of atomic number 41 is lamda. Then the wavelength of K_(alpha) line of an element of atomic number 21 is : |
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Answer» `4lamda` |
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| 41712. |
Three photodiodes D_(1), D_(2) and D_(3) are made of semiconductors having band gaps of 2.5 eV,2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å? |
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Answer» Solution :When energy of photon (E ), made INCIDENT on photodiode, is greater than its bnd gap energy `(E_(g))`, no. of minority charge carriers increases due to increase in electron-hole pairs in the region near the junction and so reverse saturation current passing through photodiodeincreases, which can be NOTED by microammeter, connected with photodiode in the external circuit. In the present case, `E=(hc)/(lambda)=(1242.19(eV)(nm))/(6000xx10^(-10)xx10^(9)(nm))` `therefore E = 2.07eV` Among given PHOTODIODES, for `D_(2)`, we have `E_(g)=2eV rArr ` For `D_(2), E gt E_(g)` and for other diodes `E LT E_(g)`. Hence only `D_(2)` can detect given light. |
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| 41713. |
The lowest unfilled energy band fonned above the valence band in a crystalline solid is called _________. |
| Answer» SOLUTION :CONDUCTION BAND | |
| 41714. |
Explain the wavefront and its types. |
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Answer» Solution :When we drop a small stone on a calm pool of water, waves spread out from the point of impact. . Every point on the surface starts oscillating with time, hence at any instant, the surface would show circular rings on which the disturbance is maximum. All points on such a circle are oscillating in phase because they are at the same distance from the source. Such a locus of points, which oscillate in phase is called a wavefront. Thus a wavefront is defined as a surface of constant phase. The speed with which the wavefront moves outwards from the source is called the speed of the wave. The energy of the wave TRAVELS in a direction perpendicular to the wavefront. Line perpendicular to the wavefront and indicating the direction of propagation of the wave is called ray. Hence, the wavefront and ray are perpendicular. f a point source EMITTING waves uniformly in all direction, then the locus of points which have the same amplitude and VIBRATE in the same phase are spheres (in THREE dimension) which is known as spherical wave as shown in figure (a) such a waves are diverging.  At a large distance from the source, a small portion of the sphere can be considered as a plane. It is known as a plane wave. It is shown in figure (b).  Wavefronts ORIGINATING from a linear source and propagating in three dimensional homogenous and isotopic medium are cylindrical wavefront. For example : The waves emanating from the tubelight. It is shown in figure (c). |
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| 41715. |
In an interfernce expriment , third bright fringe is obtained on the screenwith a light of 700 nm. Whatshould be thewavelength of the light source in order to obtain fifthbrightfringeat the same point ? |
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Answer» Solution :From ` X = (n lambda D)/d = (n. lambda.D )/d` ` lambda. = ( n lambda)/(n.) = (3 xx 700 nm)/5 = 420 nm ` |
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| 41716. |
Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.5 cm in the same capillary tube. If the density of the mercury is 13.6 gm/cc. and its angle of contact is 135^(@) and density of water is 1 gm/cc. and its angle of contact is 0^(@), then the ratio of surface tension of the two liquids is (cos 135^(@)=0.7). |
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Answer» `1:14` |
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| 41717. |
A light ray is normally incident onone face of equilateral glass prism of refractive index sqrt(2). The deviation of light ray is |
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Answer» `30^(@)` |
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| 41718. |
A planet of mass M and radius a is surrounding by an atmosphere of constant density consisting of a gas of molar mass mu. Find the temperature T of the atmosphere on the surface of the planet if the height of the atmosphere is hlt lta. |
| Answer» Solution :`T= (GMh)/(a^2 R) ` where G= GRAVITATIONAL CONSTANT | |
| 41719. |
The output current of an 80% mondulatingamplitude modulated generator is 1.8A. To what value will the current rise if the generator is additionally modulated by another audiowave of modulationindex 0.6? |
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Answer» Solution :Here, `I_(t)=1.8A A: mu_(1)=(80)/(100)=0.80: mu_(2)=0.6` `I_(t)=I_(c)sqrt(1+mu_(1)^(2)//2)` `I_(c)=(1)/(sqrt(1+mu_(1)^(2)//2))=(1.8)/(sqrt(1+(0.8)^(2)//2))=(1.8)/(sqrt(1.32))` `I_("NET total")=I_(c)sqrt(1+(mu_(1)^(2))/(2)+(mu_(2)^(2))/(2))` `=(1.8)/sqrt(1.32)sqrt(1+((0.8)^(2))/(2)+((0.6)^(2))/(2))=1.91A` |
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| 41720. |
A quantity x is defined by the equation x=3C^(2)B^(2), where C is capacitance in farads and B is magnetic field in tesla, the dimension of x are : |
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Answer» `ML^(-2)` `x=(M^(2)L^(2)T^(-4))/(ML^(2)T^(-2)L^(2)T^(-2))=ML^(-2)` HENCE CORRECT choice is `(a)`. |
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| 41721. |
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60^@ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth's magnetic field at the place. |
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Answer» SOLUTION :As per QUESTION, a magnetic needle free to rotate in a vertical PLANE parallel to the magnetic meridian has its NORTH tip down at `60^@` with the horizontal, hence the angle of dip at given place is `60^@` (i.e. , `delta = 60^@`) . Moreover horizontal component of the earth.s magnetic FIELD at the place `B_H = 0.4 G = 0.4 xx 10^(-4) T ` and `B_H = B_E cos delta` `therefore ` Earth magnetic field at the place `B_E =(B_H)/(cos delta) = (0.4xx10^(-4)T)/(cos 60^@) = 8.0 xx 10^(-5) T ` |
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| 41722. |
What is the difference between electric lines of force and magnetic lines of induction? |
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Answer» Solution : The ELECTRIC LINES of force always leave or end on the SURFACE of a CHARGED body. Magnetic lines of induction are closed curves starting from north pole and ending on the south pole of the magnet. Within the magnet they run from south to north. |
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| 41723. |
If a wire is stretched under tension fixed at both ends. If the radius of wire is halved and tension is doubled the fundamental frequency becomes |
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Answer» half |
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| 41724. |
What properties em. Waves exhibit ? |
| Answer» SOLUTION :They OBEY PROPERTIES of light, law of reflection and refraction. They exhibit the phenomenon of interference and DIFFRACTION. | |
| 41725. |
The mass of the electron varies with |
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Answer» The SIZE of the CATHODE RAY tube |
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| 41726. |
The electrical potential on the surface of a sphere of radius 'r' due to a charge 3 xx 10^(-6) C is 500 V. The intensity of electric field on the surface of the sphere is [(1)/( 4 pi in_0 ) = 9 xx 10^9Nm^2C^(-2)] ( inNC-1): |
| Answer» Answer :A | |
| 41727. |
Electric field is applied to a semiconductor. In it density of conducting charge is n the drift velocity is v. Now increasing the temperature of semiconductor …….. |
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Answer» v and nincreases INCREASING the temperature more covalent bonds are broken and NUMBER of FREE electrons and holes are increases so n increases. Now increasing the temperatureso n increases. Now increasing the temperature oscillation of charge CONDUCTORS also increasesso drift velocity decreases. |
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| 41728. |
Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed? |
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Answer» SOLUTION :RADIUS of curvature of curved surface R =20 cm Refractive index of medium `n_2 =1.5, ` Refractive index of air `n_1 = 1.0` Object distance U = - 100 cm Equation of refraction by curved surface `(n_2)/(v) - (n_1)/(u) = (n_2 -n_1)/(R)` `(1.5)/(v) - (1)/(-100) = (1.5 - 1.0)/(20)` `therefore (1.5)/(v) = -(1)/(100) + (0.5)/(20)` `therefore(1.5)/(v) = (1.5)/(100)` `therefore v= (1.5 xx 100)/(1.5)` `thereforev = 100 cm` Thus, image will be obtained at 100 cm distance from surface in direction of INCIDENT ray. |
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| 41730. |
A uniform rod of mass m and length l is hinged about one of its end. The hinge is smooth and rod lies on a smooth horizontal surface as shown in Fig. A particle of mass m is coming towards the rod at speed v_(0) perpendicular to the length of rod. (a) Find angular velocity of the rod just after impact. (b) Also find impulse due to hinge on rod during collision. |
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Answer» Solution :(a) (1) When the particle will hit the rod, the hinge will restrict the motion of the end of the rod. We can IMAGINE the rod starting to rotate about hinge and particle continuing to move on same line of motion. The impulse that particle will experience by rod is along the line of motion. (2) We will consider rod and particle as our system. This system will experience external impulse due to hinge. So, the momentum of the system cannot be conserved. But the angular momentum of the system about hinge will remain conserved as TORQUE due to impulse at hinge is zero. Calculation: By conserving angular momentum about hinge `mv_(0)x=mvx+(ml^(2)omega)/3` Here v is the speed of particle after the collision and `omega` is the angular velocity of rod after the collision as shown in Fig. We can also write Newton.s law of collision remembering that it is the velocity of constant points that is required. Thus, `(omegax-v)/(0-v_(0))=-e` Solving Eq. and,we get `omega=(3v_(0)x(1+e))/((l^(2)+3x^(2)))`  (b) (1) Coefficient of restitution is e and perpendicular distance of line of motion from hings is x. Consider e = 1/3, `v_(0)` = 3 m/s, l = 1 m, m = 2 KG and x = 1/3 m. (2) Now to find impulse `J_(1)`, due to the hinge, we will consider rod and particle as our system and write impulse and momentum equations. Calculation: We should remember that impulse between rod and particle during the collision need not be considered since it will be in internal impulse, thus we can write `vecJ_(1)=(vecP_("rod "f)+vecP_("particle "f))-(vecP_("rod "i)+vecP_("particle "i))` `vecP_("particle "i)=6 Nm" "vecP_("rod "i)=0` For `vecP_("particle".f)` we need to find speed of particle from part (a). `omega=(3xx3xx(1//3)(1+1//3))/([1^(2)+3(1//3)^(2)])` Now solving for v by putting value of `omega`, we get `(3xx(1//3)-v)/(0-(3))=-1/3m//s` Solving for v = 0, we get `vecP_("particle".f)=0`. For `vecP_("rod".f)` we need velocity of center of mass of rod as `vecP_("rod".f)=MV_("com")`. Thus, `V_("com")=omegaxxr_(c)=omegaxxl/2=3/2m//s` and `P_("rod "f)=2xx3/2=3kgm//s` Hence, `J_(1)=(3+0)-(0+6)=-3kgm//s` |
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| 41731. |
Sketch the static characteristics of a common emitter transistor and bring out the essence of input and output characteristics. |
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Answer» Solution :Static Characteristics of Transistor in Common Emitter Mode: The know-how of certain parameters like the input resistance, output resistance, and current gain of a transistor are very important for the effective use of transistors in circuits. The circuit to study the static characteristics of an NPN transistor in the common emitter mode is given in figure. The bias supply voltages `V_(BB)` and `V_(CC)` bias the base-emitter junction and collectoremitter junction respectively. The junction potential at the base-emitter is represented as VBE and the collector-emitter as `V_(CE)`: The rheostats `R_(1)` and `R_(2)` are used to vary the base and collector currents respectively. The static characteristics of the BJT are 1. Input characteristics 2. Output characteristics 3. Transfer characteristics  1. Input Characteristics: Input Characteristics curves give the relationship between the base current `(I_(B))` and base to emitter voltage `(V_(BE))` at constant collector to emitter voltage `(V_(CE))` and are shown in figure. Initially, the collector to emitter voltage `(V_CE))` is set to a particular voltage (above 0.7 V to reverse bias the junction). Then the base-emitter voltage `(V_(BE))` is increased in suitable steps and the corresponding base-current `(I_(B))` is recorded. A graph is plotted with `V_(BE)` along the x-axis and `I_(B)` along the y-axis. The procedure is repeated for different values of `V_(CE)`. The following observations are made from the graph: • The curve looks like the forward characteristics of an ordinary pun junction diode. • There exists a threshold voltage or knee voltage `(V_(K))` below which the base current is very small. The value is 0.7 V for Silicon and 0.3 V for Germanium transistors. Beyond the knee voltage, the base current increases with the increase in base-emitter voltage. • It is also noted that the increase in the collector-emitter voltage decreases a NPN transistor in common emitter the base current. This shifts the curve configuration outward. This is because the increase in collector-emitter voltage increases the width of the depletion region in turn, reduces the effective base width and thereby the base current.  Input resistance: The ratio of the change in base-emitter voltage `(DeltaV_(BE))` to the change in base current `(Delta_(IR))` at a constant collector-emitter voltage `(V_(CE))` is called the input resistance `(R_(i))`.The input resistance is not LINEAR in the lower region of the curve. `R_(i)=((DeltaV_(BE))/(DeltaI_(B)))_(V_(CB))` The input resistance is high for a transistor in common emitter configuration. Output Characteristics: The output characteristics give the relationship between the variation in the collector current `(DEltaI_(C))` with respect to the variation in collector-emitter voltage `(DeltaV_(CE))` at constant input current `(I_(B))` as shown in figure.Initially, the base current `(I_(B))` is set to a particular value. Then collector emitter voltage `(V_(CE))` is increased in suitable steps and the corresponding collector current `(I_(C))` is recorded. A graph is plotted with the `V_(CE)` along the x-axis and `I_(C)` along the y-axis. This procedure is repeated for different values of `I_(B)`.The four importantregions in the output characteristics configuration are:  (i) Saturation region: When `V_(CF)` is increased above 0 V, the `I_(C)` increases rapidly to a saturation value almost independent of `I_(B)` (Ohmic region, OA) called knee voltage. Transistors are always operated above this knee voltage. (ii) Cut-off region: A small collector current `(I_(C))` exists even after the base current `(I_(B))` is reduced to zero. This current is due to the presence of minority CARRIERS across the collector-base junction and the surface leakage current `(I_(CEO))`.This region is called as the cut-off region, because the main collector current is cut-off. (iii) Active region: In this region, the emitter-base junction is forward biased and the collector base junction is reverse biased. The transistor in this region can be used for voltage, current and power amplification. (iv) Breakdown region: If the collector-emitter voltage (VCP) is increased beyond the rated value given by manufacturer, the collector current (1) increases enormously leading to the jung breakdown of the transistor. This avalanche breakdown can damage the transistor. Output Resistance: The ratio of the change in the collector-emitter voltage `(DeltaV_(CE))`to the corresponding change in the collector current `(DeltaI_(C))` at constant base current `(I_(B))` is called output resistance `(R_(O))`. `R_(0)=((DeltaV_(CE))/(DelatI_(C)))_(I_(B))` The output resistance for transistor in common emitter configuration is very low. |
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| 41732. |
Equal currents are flowing in three infinitely long wires along positive x,y and z -directions. The magnetic field at a point (0,0-a) would be (i=current in each wire) |
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Answer» `(mu_0i)/(2pia)(hatj-HATI)` `B=mu_0/(2pi) i/a` |
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| 41733. |
If oint E.ds = 0 over surface, then |
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Answer» the electric field inside the surface and on it is zero |
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| 41734. |
A conducting rod of length 2l is rotating with constant angular speed omega about is perpendicular bisector. A uniform magnetic field vec(B) exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is |
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Answer» `(1)/(2)B omega l^(2)` `e_(1)=(1)/(2)Bl^(2)omega` and between centre and other end `e_(2)=-(1)/(2)Bl^(2)omega`. Therefore between two ENDS `e=e_(1)+e_(2)=0` `therefore` Between two ends of the rod, e.m.f. induced = 0. |
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| 41735. |
The average energy of molecules in a sample of oxygen gas at 300 K are 6.21xx10^(-21)J. The corresponding values at 600 K are |
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Answer» `12.12xx10^(-21)J` `KE_(avg)=(3)/(2)k_(B)Tor((KE_(avg))_(1))/((KE_(avg))_(2))=(T_(1))/(T_(2))` Here, `(KE_(avg))_(1)=6.21xx10^(-21)J,T_(1)=300K`, `T_(2)=600K,(KE_(avg))_(2)=?` `therefore (6.21xx10^(-21))/((KE_(avg))_(2))=(300)/(600)=(1)/(2)or(KE_(avg))_(2)=12.42xx10^(-21)J` |
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| 41736. |
Fig. shows a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I_(0) is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima. |
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Answer» Solution :As is known, RESULTANT amplitude is the sum of amplitudes of either beam in perpendicular and parallel polarization, i.e.,`A = A_("perp".) + A_("parallel")` Now,`A_("perp".) = A_("perp".)^(1) + A_("perp".)^(2)` `= A_("perp".)^(0) sin (kx - OMEGA t) + A_("perp".)^(0) sin (kx - omega t + phi) ` Similarly,`A_("parallel") = A_("parallel")^(1) + A_("parallel")^(2)` `= A_("parallel")^(0)[sin(kx - omega t) + sin (kx - omega t + phi)` `:.`Inetensity `= A_("perp".)^(2) + A_("parallel")^(2) = [A_("perp".)^(0) + A_("parallel")^(0^(2))] [sin^(2)(kx - omega t) + (kx - omega t + phi)]` average `= [ A_("perp".)^(0^(2)) + A_("parallel")^(0^(2)) ]((1)/(2)) 2(1 + COS phi)` As `|A_("perp".)^(0^(2))|"average" = |A_("parallel")^(0^(2))|"average"`, THEREFORE, Without `P`, Intensity `= 2|A_("perp".)^(0^(2))| (1 + cos phi)`...(i) With polariser `P`, suppose `A_("perp")^(2)` is blocked. `:.` Intensity `= {A_("parallel")^(1) + A_("parallel")^(2)}^(2) + (A_("perp".)^(1))^(2) = |A_("perp".)^(0) |^(2) (1 + cos + phi) + | A_("perp".)^(0) |^(2) xx (1)/(2)`...(II) We are given that without polarizer, intensity of principal maximum is `I_(0) = 4 |A_("perp".)^(0)|^(2)`(iii) `:.` From (ii), intensity of principal maximum with polariser would be `I = |A_("perp".)^(0)|^(2) [(1 + 1) + (1)/(2)] = (5)/(2)|A_("perp".)^(0)|^(2)` Using (iii), we get`I = (5)/(2)((I_(0))/(4)) = (5)/(8)I_(0)` Again, intensity at first minima with polsrizer [From (ii)] `I' = |A_("perp".)^(0)|^(2)(1 - 1) + |A_("perp".)^(0)|^(2) xx (1)/(2) = (|A_("perp".)^(0)|^(2))/(2) = (I_(0)//4)/(2) = (I_(0))/(8)` |
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| 41737. |
The nucleus is approximately spherical in shape. Then the surface area of nucleus haviing mass number A varies as. |
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Answer» `A^(2//3)` `4/3pi R^3 prop A , R =R_0 A^(1//3)` So, `4piR^2=R_0A^(2//3) RARR 4piR^2 prop A^(2//3)` SURFACE area is proportional to `"(mass number)"^(2//3)` |
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| 41738. |
After that, another pump with the same chamber volume V, begins to suck in the atmospheric air, also making n double strokes. What will be the pressure inside the vessel? |
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Answer» Solution :In the figure we can see that when the piston comes out, the valve CONNECTING CHAMBER to the vessel will close and another valve will open to atmosphere. Thus, air from the atmosphere will come into the pump.s chamber. Now when the piston goes in, the valve connecting chamber to the vessel will open and the other valve to atmosphere will close. So all the air inside the chamber will go to the vessel. We can say that the total volume gone in is nV, as number of moles entering the chamber will remain constant in every stroke. Also, during delivery, the pumpateachdoublestrokesucksin airwitha constantpressure`p_0 ` Calculations: TOTALNUMBEROF molesthatwillbepresentin thevesselare ` ((p.V+P_0nV_0))/(RT ) ` IF finalpressureispthenwe canwrite ` ((P.V+ P_0n V_0 ))/( RT ) = (pV ) /(RT ) ` solvingweget ` P=p.+(p_0nV _0 ) /(V )=P_0{((V )/(V +V_0 ))^N+(nV_0 )/( V )}` here `p gt p .`for anyn,sinceduringdelivery, thepumpat eachdoublestokesucksin airwitha pressure`p.`andduringevancuationvolume`V_0`of theairis beingpumpedoutat pressurebelow`p_0` |
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| 41739. |
A gun of mass 10Kg fires four bullets per second. The mass of each bullet is 20g and the velocity of the bullet when it leaves the gun is 300 ms. The force required to hold the gun while firing is |
| Answer» Answer :C | |
| 41740. |
गंगा की सहायक नदियों में इनमें से कौन सी नदी है जो प्रायद्वीपीय पठार से निकलकर गंगा में मिलती है? |
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Answer» चंबल |
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| 41741. |
What is an ideal inductor? |
| Answer» Solution :An IDEAL INDUCTOR is ONE that has ZERO RESISTANCE | |
| 41742. |
A nonrelativistic electron enters a uniform magnetic field of 50 mT with a speed of 10^(6) m//s at right angles to it and emerges from the field after 0.1 nanosecond. What is the deviation produced the field ? Show that it does not depend on the initial speed of the electron. |
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Answer» |
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| 41743. |
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. (a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m. (b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~10^(-10) W m^(-2)). Take the area of the pupil to be about 0.4 cm2 , and the average frequency of white light to be about 6 xx 10^(14) Hz |
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Answer» `2.51xx10^(31)` |
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| 41744. |
A biconvex thin lens is prepared from glass of refractive index 3/2. The two bounding surfaces have equal radii of 25 cm each. One of the surfaces is silvered from outside to make it reflecting. Where should an object be placed before this lens so that the image coincides with the object. |
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Answer» Solution :Here ` R_1= + 25cmR_2=- 25cm, andmu= 3//2` imagecoincides withobject, hence ` U= v=-X `(say ) `(1)/(F ) =(2)/(F_L) +(1)/(F_C ) =2(3/2-1) (2)/( 25 )+ (2)/(25 ) implies(1)/(F )= (4)/( 25)` By using ` (1)/( v ) - (1)/( u )= (1)/(f) ` `implies - (1)/(x)- (1)/(x )= (4)/(25 )` `x= 12.5cm` Hence, the object should be placed at a distance 12.5 cm in front of the SILVERED LENS. |
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| 41746. |
A bar magnet has coercivity 4xx10^(3) Am^(-1)it is desired to demagnetise it by inserting it inside a solenoid 12 the current that should be sent through the solenoid is |
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Answer» 2a |
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| 41747. |
a. What factors make a fusion reaction difficult to achieve ? b. Why does a fusion reactor produce less radioactive waste than a fission reactor? |
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Answer» Solution :a. Nuclear FISSION occurs at a high temperature of `10^5 - 10^6K`. It is difficult to attain such a temperature. b. Whether t is fission or fusion, the product obtained is radio active. For a certain output ENERGY the number of fusion reaction should be corresponding LARGE. |
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| 41748. |
The difference between the mass of a nucleus and the combined mass of its nucleons is |
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Answer» GREATER than mass of NUCLEUS |
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| 41749. |
If proportion of impurityis less in p-n junction …….. |
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Answer» the WIDTH of depletion region is more. |
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| 41750. |
A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained? |
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Answer» |
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