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| 6451. |
साम्यावस्था में द्रव के वाष्पन की दर संघनन की दर के ……….. होती है- |
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Answer» समान |
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| 6452. |
What will be the minimum frequency of light source to get photocurrent, from a metal surface having work function 2 eV ? |
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Answer» `4.8xx10^(14)HZ` |
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| 6453. |
The sources of light are ____,____,___ and ____. |
| Answer» Solution :Sun , THERMAL Gas, DISCHARGE TUBE, Luminescent sources. | |
| 6454. |
Two long parallel wires are along z direction at x = 0 and x = d. The magnetic field along x axis has been plotted in the given figure with field (B) positive when it is in positive y direction. The co-ordinate of point R is x = – d. Find co-ordinate of points P and Q shown in figure. |
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Answer» |
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| 6455. |
The current in a conductor of 40 ohm resistance increases linearly from 5 A to 25 A in 10 s. How much heat is liberated in it during this time? Solve the problem using two methods: (a) numerical calculation, (b) integration. |
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Answer» `triangleQ=triangleQ_(1)+triangleQ_(2)+triangleQ_(9)=(i_(1 av)^(2)+i_(2av)^(2)+i_(9av)^(2))R trianglet= 2580 xx 40 x 1 =1.03 xx 10^(5)J` (b) Integration. The CURRENT according to the low i=5+2T. The quantity of heat is `Q=underset(0)overset(10)int i^2 R dt=40 underset(0)overset(10)int (5+2t)^(2) dt-20 underset(0)overset(10)int (5+2t)^(2) d (5+2t)=`  `=(20)/3 (5 2t)^(3) underset(0)overset(10)int =(20)/3 (25^(3)-5^(3))=(20 xx 775 xx 20)/(3)=1.03 xx 10^(5)J` We see that the result of the numerical calculation was ACCURATE. |
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| 6456. |
A copper wire is stretched to make it 0.1 % longer. The percentage change in its resistance is ...... (Assume that the volume of the wire remains constant.) |
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Answer» increase by 0.2 % Suppose the LENGTH of the wire is l and area of cross-section is A. The RESISTANCE of a wire, = R = `(rho l)/(A)` ` therefore R = (rho l^(2))/(Al) = (rho l^(2))/(V) `.... (1) `therefore (dR)/(dl)= (rho)/(V) (2l) ` `therefore dR = (2 rho)/(V) l dl "" ` .... (2) Taking ratio of equations (2) and (1), `(dR)/(R) = ((rho)/(V) 2ldl)/((rho)/(V)l^(2))` `therefore (dR)/(R) = 2 (dl)/(l)` Percentage change = `(dR)/(R) xx 100 %` = 2 `((dl)/(l) ) xx 100 %` = `2 (0.1 %) = 0.2 %` Thus, the resistance of the wire increases by 0.2 % . |
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| 6457. |
if each cm of main scale is divided in q equal parts and p vernier scale divisions coincide with (p-1) main scale divisions then |
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Answer» <P>1 VERNIER scale division `=1/q((p-1)/p) CM ` |
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| 6458. |
A : Duringorbitalmotion of planetaroundthe sunwork doneby the centripetalforce is not zeroat allpoints on the orbit . R : Planet is revolving around the sun in ellipticalorbit . |
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Answer» If both Assertion & Reasonare true . Andthe reasonis the correct explanationof theassertion , then MARK (1) |
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| 6459. |
The equation of a plane progressive wave is y=5sin2pi(8t-5x). Where y and x are in an and t in seconds. Calculate the amplitude, frequency, wavelength and velocity of the wave. |
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| 6460. |
A transmitting antenna of height 20 m and the receiving antenna of height h are separated by a distance of 40km for satisfactory communication in line of sight mode. Then the value of h is |
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Answer» 40m where `h_(R)` and `h_(T)` are the heights of receiving and transmitting antenna and r is the RADIUS of the EARTH. `40xx10^(3)=sqrt(2xx6400xx10^(3)XXH)+sqrt(2xx6400xx10^(3)xx20)` `40xx10^(3)=sqrt(2xx6400xx10^(3)xxh)+16xx10^(3)` `h=([(40-16)xx10^(3)]^(2))/(2xx6.4xx10^(6))=45m` |
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| 6461. |
A heat engine draws, 800 J of heat from its high temperature source and discards 600 J of exhaust heat into its cold-temperature reservoir each cycle. How much work does this engine perform per cycle, and what is its thermal efficiency? |
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Answer» Solution :The work OUTPUT per cycle is equal to the DIFFERENCE between the heat energy drawn in and the heat energy discarded. `W=Q_(H)-|Q_(C)|=800J-600J=200J`. The efficiency of this ENGINE is `e=(W)/(Q_(H))=(200J)/(800J)=0.25=25%`. |
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| 6462. |
Fluroscent tubes given more light than a filament bulb at same power because |
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Answer» Tube contains gas at LOW temperature |
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| 6463. |
A strong magnetic field is applied on a stationary electron, then ______ |
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Answer» The ELECTRON moves in the direction of the FIELD. |
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| 6464. |
A narrow monochromatic X-ray beam falls on a scattering substance. The wavelengths of radiation scattered at angles theta_(1) = 60^(@) and theta_(2) = 120^(@) differ by a factor eta = 2.0. Assuming the free electrons to be responsible for the scattering, find the incident radiation wave length. |
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Answer» Solution :Let `lambda_(0) =` WAVELENGTH of the INCIDENT radiation. Then wavelength of the radiation scattened at `theta_(1) = 60^(@)` `= lambda_(1) = lambda_(0)+2pi cancel lambda_(c) (1-cos theta_(1))` where `cancel lambda_(c) = (cancelh)/(mc)`. and similarly `lambda_(2)= lambda_(0) + 2pi cancel lambda_(c) (1- cos theta_(2))` From the data `theta_(1) = 60^(@), theta_(2) = 120^(@)` and `lambda_(2) = eta lambda_(1)` Thus `(eta - 1) lambda_(0) = 2pi cancel lambda_(c) [1-cos theta_(2) - eta (1-cos theta_(1))]` `= 2pi cancel lambda_(c) [1-eta +etacos theta_(1) - cos theta_(2)]` Hence `lambda_(0) = 2pi cancel lambda_(c) [(eta cos theta_(1) - cos theta_(2))/(eta - 1)-1]` `= 4pi cancel lambda_(c) [(sin^(2)theta_(2)//2 - eta sin^(2) theta_(1)//2)/(eta - 1)] = 1.21 pm`. The expression `lambda_(0)` given in the bok contains MISPRINTS. |
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| 6465. |
Obtain the condition at resonance draw a plot showing the variation of current with the frequency of a.c. source for two resistances and R_(1)and R_(2)(R_(1) gt R_(2)). Hence define the quality factor Q and write its role in the tuning of the circuit. |
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Answer» Solution :At resonance, `I_(m)` is maximum `rArr X_(L) = X_(C.)` [ALTERNATIVELY : `omega_(0) = (1)/(sqrt(LC))`]  Quality FACTOR of LCR circuit is defined as `(omega_(0))/(2 Delta omega)` `= (omega_(0)L)/(R )` A largr value of quality factor CORRESPONDS to a sharper resonance. |
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| 6466. |
(i) Define ‘activity’ of a radioactive material and write its SI unit. (ii) Plot a graph showing variation of activity of a given radioactive sample with time. (iii) The sequence of stepwise decay of a radioactive nucleus is Doverset(alpha)(to)D_(1)overset(beta^(-))D_(2) If the atomic number and mass number of D_(2) are 71 and 176 respectively, what are their corresponding values for D? |
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Answer» Solution :(i) The activity or the rate of decay (R) of a radioactivesample is defined as the number of disintegrationstaking place per unit TIME in given sample. `THEREFORE` Activity `R=-(dN)/(dt)=lambdaN=lambdaN_(0)e^(-lambdat)` SI unit of activity is `("second")^(-1) or s^(-1)`.  (ii) A graph showing variation of activity of a given radioactive sample with time is given in FIG. 13.07. (iii) The sequence of decays may be expressed as `" "_(72)^(180)D overset(alpha)underset(" "_(2)^(4)He)(to)" "_(70)^(176)D_(1)overset(beta^(-))underset(" "_(-1)^(0)e)(to)" "_(71)^(176)D_(2)` Thus, atomic number and mass number of D are 72 and 180 RESPECTIVELY. |
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| 6467. |
_94U^238 on absorbing a neutron goes over to _92U^239. This nucleus emits an electron to go over to plutonium. The resulting plutonium can be expressed as : |
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Answer» `_94U^239` |
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| 6468. |
When a charged particle enter normal to a uniform magnetic field, it take a circular path. Name the particle accelerator using this principle, |
| Answer» SOLUTION :CYCLOTRON | |
| 6469. |
If radius of " "_(13)^(27)Al nucleus is estimated to be 3.6fm, then the radius of " "_(32)^(125)Te nucleus be nearly |
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Answer» 4fm |
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| 6470. |
Two large conducting planes perpendicular to X-axis are placed at (d.0) and (2d, 0) as shown in the figure. Current per unit width in both the planes is same and current is flowing in the outward direction. The variation of magnetic field as function of x(0 le x le 3d)is best represented by |
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Answer»
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| 6471. |
The process in which the amplitude of the carrier wave is made proportional to the instantaneous amplitude of the signal wave is called |
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Answer» AMPLITUDE modulation |
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| 6472. |
What is the de-Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b)a ball of mass 0.060 kg moving at a speed of 1.0 m/s and (c )a dust particle of mass 1.0xx10^(-9) kg drifting with a speed of 2.2 m/s |
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Answer» Solution :de-Broglie.s WAVELENGTH `lambda=(h)/(mv)` (a) SPEED of bullet, `v_(b)`=1.0 km/s `=1.0xx10^(3)m//s` and mass of bullet `m_(b)`=0.040 kg and `h=6.63xx10^(-34)` Js `THEREFORE lambda_("bullet")=(h)/(m_(b)v_(b))` `=(6.63xx10^(-34))/(0.04xx1xx10^(3))` `=165.75xx10^(-37)` `~~1.7xx10^(-35)`m (b)Speed of ball `v_(g)`=1.0 m/s mass of ball `m_(g)`=0.060 kg `lambda_("ball")=(h)/(m_(g)v_(g))` `=(6.63xx10^(-34))/(0.06xx1)` `=110.5xx10^(-34)` `~~11.1xx10^(-32)` m (c )Speed of dust particle `v_(p)`=2.2 m/s and mass `m_(p)=1.0xx10^(-9)` kg `therefore lambda_("particle")=(h)/(m_(p)v_(p))` `=(6.63xx10^(-34))/(1xx10^(-9)xx22)` `=3.0136xx10^(-25)` `~~3.0xx10^(-25)`m |
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| 6473. |
The angle of incidence is equal to angle of reflection is the statement of law of : |
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Answer» reflection |
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| 6474. |
The charge on 4 mu Fcapacitor in the given circuit is ("in" muC) |
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Answer» 12 |
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| 6475. |
If a charge q is located at a centre of the hypothetical cube, the electric flux through its any face is : |
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Answer» `Q epsilon_0` |
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| 6476. |
A satellite is orbiting round the earth in a circular orbit with speed v. If m is mass of satellite its total energyis: |
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Answer» `(1)/(2)MV^(2)` THUS CORRECT choice is (c ). |
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| 6477. |
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be |
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Answer» 36 CM towards the mirror  `(1)/(v_1)=1/f-(1)/(u_1)=(u_1-f)/(u_1f)` `therefore v_1=(u_1f)/(u_1-f)=((-40)(-15))/(-40-(-15))` `therefore v_1=(600)/(-25)=-24` cm Now `u_2`=-20 cm ,`v_2=?`  `(1)/(v_2)=(1)/(f.)-(1)/(u_2)=(u_2-f)/(u_2f)` `therefore v_2=(u_2f)/(u_2-f)=((-20)(-15))/(-20-(-15))` `therefore v_2=(300)/(-5)=-60` cm Distance between images = `v_2-v_1` =-60+24 =-36 cm As distance is POSITIVE hence 36 cm away from mirror. |
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| 6478. |
The velocity time relation of an electron starting from rest is given by v=kt where k=2m/s^2 . The distance traversed in 3 second is |
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Answer» 9m |
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| 6479. |
Define the terms (i) drift velocity, (ii) relaxation time. A conductor of length L is connected to a d.c. source of emf epsi. If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L, how will the drift velocity change ? |
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Answer» Solution :Drift velocity : The AVERAGE velocity with which the free electrons drift under the INFLUENCE of an external field. Relaxationk time: Average time interval between two succesive collisions of an electron with the ions/atoms of the conductor. The drift velocity will be inversely proportional to (or `V _(d) prop (1)/(I))` and HENCE it will become and third of its initial value. |
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| 6480. |
The time period of the magnetic in an oscillation pmagnetometer in the earth magnetic field is 2s. A short bar magnet is placed to the north of the magnetometer, at a separation 10cm from the oscillation magnet, with its north pole pointing towards north. The time period beceomes half. Calculate the magnetic moment of this short magnet. |
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Answer» Solution :Time period T=`=2pi sqrt((I)/(MB)) Rightarrow T alpha (1)/(sqrtH)B alpha (1)/(T^(2))` Let , is magnet moment DUE to short magnet and B' is a MAGNETIC field due to short magnet, along SOUTH to NORTH Given , `T_(2)=2s, B_(h)=12muT` `"" T_(2)=1s, B_(h)=B+B'=12+B'` ` "" (B+B')/(B)=(T_(1)^(2))/(T_(2)^(2))` ` "" (12+B')/(12)=((2)/(1))=4` `""B'=36muT` `""B'=(mu_(0))/(4pi)(2M)/(r^(3))` `Rightarrow "" 36xx10^(-6)=10^(-7)xx(2M)/((0.10)^(3))` `"" M=0.18Am^(2)` |
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| 6481. |
A: The trajectory of a charge when it is projected perpendicular to an electric fiedl is a parabola. R: A moving charge entering parallel to the magnetic field lines moves in a circular path. |
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Answer» If both Assertion & Reason are true and the reason is the CORRECT explanation of the assertion then mark (1) |
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| 6482. |
Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequences: AtoBtoC Here, B is an intermediate nucleus which is also radioactive. Considering that are N_(0) atoms ofA initially, plot the graph showing the variation of number of atoms of A and B versus time. |
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Answer» Solution :The given sequence of radioactive decay is `AtoBtoC` C is STABLE. At t=0, `N_(A)=N_(0) and N_(B)=0`. As time passes on, t increases.`N_(A)` FALLS of exponentially to zero at `t=oo`. The number of atoms of B goes on increasing with time, becomes maximum and finally decays to zero (at `t=oo`) following exponential decay law. The variation of the number of atoms of A and B versus time is SHOWN in FIG. 
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| 6483. |
Describe briefly Davisson-Germer experiment which demonstrated the wave nature of electrons. |
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Answer» Solution :Davisson -Germer experiment `:` (i) De Brogli hypothesis of matter waves was experimentally CONFIRMED by Clinton Davisson and Lester Germer. (ii) They demonstrated that electron beams are diffracted when they fall on crystalline solids. Since crystal can act as a three -dimensional diffraction grating formatter waves, the electron waves incident on CRYSTALS are diffracted off in certain specific directions. (iii) The filament F is heated by a low tension ( I.T. ) battery. Electrons are emitted from the hot filament by thermionic emission. (iv) They are then accelerated due to the potential difference between the filament and the anode aluminium cylinder by a high tension ( H.T. ) battery . (v) Electron beam is collimoted by USING two thin aluminium diaphragms and is allowed to strike a single crystal of Nickel. (vi) The electrons scattered by Ni atoms in different directions are received by the electron detector which measures the intensity of scattered electron beam. (viii) The detector is rotatable in the plane of the paper so that the angle `phi` between the incident beam and the scattered beam can be changed at our will. The intensity of the scattered electron beam is measured as a function of the angle `theta `.  (viii) FIGE shows the variation of intensity of the scatter electrons with the angle `theta` for the accelerating voltage of 54V. For a given accelerating voltage V, the scattered wave shows a peak or maximum at an angle of `50^(@)` to the incident electron beam. (ix) This peak in intesity is attributed to the constructive interference of electrons diffracted from various atomic layers of the target material. (x) From the known value of the interplanar spacing of Nickel, the wavelength of the electron wave has been experimentally calculated as `1.65 Å`. (xi ) The wavelength can also be calculated from de-Broglie relation from `V =54V` from equation as `lambda = ( 12.27 )/( sqrt( V ))Å= ( 12.27 )/( sqrt( 54))` `lambda = 1.67 Å` (xii) This value agrees well with the experimentally observed wavelength of `1.65Å` .Thus this experiment directly VERIFIES de Broglie's hypothesis of the wave nature of moving paritcles. |
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| 6484. |
The dimensional formula of the pnysical quantity whose S.I. unit is farad is |
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Answer» `ML^(2) T^(-3) I^(-2)` |
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| 6485. |
The coil of a sensitive M.C.G. swings too far on both sides. This movement can be quickly stopped by |
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Answer» holding a magnet near the coil |
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| 6486. |
At the time of the Chinese traveller It-Sing's visit there were ………………………. big rooms |
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Answer» 300 |
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| 6487. |
In the Young.s double slit experimentset up , source S of wavelength 500 nm illuminates two slitS_(1) and S_(2)which set as two coherent sources . The source S oscillates about its own position according to theequation y = 0.5 sin pi t where y is in mm and t in seconds . the minimum value of time t for which the intensity at point p on the screen exactly infrontof the upper slit becomes minimum iswhichthe intensity at point P on teh screen exactlyinforntof the upper slit becomesminimum is |
| Answer» ANSWER :A | |
| 6488. |
In a singleslit diffraction experiment, when a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why ? State two points of difference the interference pattern obtained in Young's double slit experiment and the diffraction pattern due to a single slit. |
Answer» Solution :The bright spot is DUE to constructive INTERFERENCE of waves DIFFRACTED from the edge of the CIRCULAR obstacle 
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| 6489. |
The magnetic induction at point 'O' in the following fig. will be |
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Answer» `(mu_0 I)/( 4r)[(3)/(2) -(1)/(PI )o.` |
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| 6490. |
Explain internal reflection and total internal reflection. |
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Answer» Solution : When light travels from an optically denser medium to a rarer medium at the interface, it is partly REFLECTED back into the same medium and partly refracted to the second medium. This reflection is called the internal reflection. When a ray of light enters from a denser medium to a rarer medium it bends away from the normal.  The ray `AO_1B` in FIGURE the incident ray `AO_1`, is partially reflected `(O_1C)` and partially transmitted `(O_1B)` or refracted. The angle of refraction (r) being larger than the angle of incidence (i). As the angle of incidence increases, so does the angle of refraction, till for the ray `AO_3`, the angle of refraction is `pi/2` The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray `AO_3D`. The angle of incidence for which the refraction angle becomes `pi/2`, that incidence angle is called critical angle `.i_C.`. The deviated ray obtained after incidence at critical angle is called critical ray. If the angle of incidence is increased still further e.g. the ray `AO_4`, refraction is not possible and the incident ray is TOTALLY reflected. This is called total internal reflection, When light GETS reflected by a surface, normal some fraction of it gets transmitted. The reflected ray, therefore, is always less inten: than the incident ray, howsoever smooth th reflecting surface may be. In total intern reflection on the other hand, no transmission light takes place. According to Snell.s law at point `O_3`, `nisin i_c = n_2 sin r` `therefore (n_1)/(n_2)sin i_c = sin" "r` `therefore n_(12) sin i_c = sin 90^@" " [because r = 90^@]` `therefore n_(12)sin i_c = 1` `therefore _(12) = (1)/(sin i_c)`(where `n_(12)` = refractive inde of denser medium-1 w.r.t. rarer medium-2) If `n_1 = n` and `n_2` = 1 (air) then, sin `i_c` = 1 `therefore n=(1)/(sin i_c)` |
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| 6491. |
Light year is |
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Answer» Light EMITTED by sun in ONE year |
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| 6492. |
The wing span of an aeroplane is 20 metre. It is flying in a field, where the vertical component of magnetic field of earth is 5 xx 10^(-5) tesla, with velocity 360 km/h. The potential difference produced between the blades will be |
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Answer» 0.10V |
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| 6493. |
The M.I. of a ring about a transverse axis through its centre is MR^2. What is the M.I. about a tangent in its plane. |
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Answer» `MR^2` |
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| 6494. |
The mass of proton is 1.0073 u and that of neutron is 1.0087 u ( u=atomic mass unit ). The binding energy of ._2^4He , if mass of ._2^4Heis 4.0015 u is |
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Answer» 0.0305 erg =2 x 1.0073 + 2 x 1.0087 - 4.0015 =0.0305 u Binding ENERGY = 0.0305 x 931 MeV = 28.4 MeV |
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| 6495. |
4000 मे कितने सार्थक अंक है - |
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Answer» 1 |
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| 6496. |
In an oscillating circuit shown in figure, the coil inductance is equal to L=2.5 mu F. The capacitor have capacitances C_(1)=2.0 mu F and C_(2)=3.0 mu F. The capacitors were charged to a voltage V=180 V, and then the switch S w was closed. Find : (a) the naturaloscillation frequency, (b) the peak value of the currentflowing through the coil. |
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Answer» Solution :The equations of the `L-C` circuit are `L(d)/(dt)(I_(1)+I)=(C_(1) V-int I_(1)dt )/( C_(1))=(C_(2)V-int I_(2) dt)/( C_(2))` Differentiating again `L(I_(1)+I_(2))=-(1)/(C_(2))I_(1)=-(1)/(C_(2))I_(2)` Then ` I_(1)=(C_(1))/(C_(1)-C_(2))I, I_(2)=(C_(2))/(C_(1)+C_(2))I`, `I=I_(1)+I_(2)` so `L(C_(1)+C_(2))I+I=0` or `I=I_(0) sin ( omega_(0)t + ALPHA)` where `omega_(0)^(2)=(1)/( L(C_(1)+ C_(2))) (` PART `a)` `(` Hence `T=(2PI)/( omega_(0))=0.7 ms )` At ` t=0 , I=0` so `alpha=0` `I=I_(0)sin omega_(0)t` The peak value of the CURRENT if `I_(0)` and it is RELATED to the voltage `V` by the first equation `LI=V-intI dt//(C_(1)+C_(2))` or `+ L omega_(0) I_(0) cos omega_(0)t = V-(1)/( C_(1)+ C_(2))int_(0)^(t)I_(0) sin omega_(0) t dt ` `(` The `P.D.` across the inductance is `V` at `t=0)` `=V+(1)/(C_(1)+C_(2)). (I_(0))/(omega_(0))( cos omega_(0)t -1)` Hence `I_(0)=(C_(1)+C_(2))omega_(0) V=Vsqrt((C_(1)+ C_(2))/( L))=8.05A.` 
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| 6497. |
10 % of the total energy of a 100 W bulb is converted into visible light. Calculate the average intensity out a spherical surface which is at a distance of 1m from the bulb, consider the bulb to be a point source and let the medium be isotropic. |
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Answer» Solution :Electrical energy used per second in bulb, U = P of `10%=100xx0.1` `THEREFORE U=10 J` Considering bulb as centre, area of circular SURFACE `A=4pi d^(2)=4xx3.14xx(1)^(2)=12.56m^(2)` Average INTENSITY on circular surface `I=(U)/(A)=(10)/(12.56)=0.796 Wm^(-2)` `therefore I=0.8 Wm^(-2)` (Approximate). |
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| 6498. |
Explain the magnetic hysteresis loop ? What are its uses ? |
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Answer» Solution :(a) Hysteresis. The lagging of intensity of magnetisation behind the magnetising FIELD is called hysteresis. To explain term hysteresis, suppose a bar of iron is magnetised slowly. The intensity of magnetisation M increases with increase in the STRENGTH of magnetising field along OA as shown in the figure. At A, any further increase in H does not PRODUCE any increase in M and therefore, corresponding to point A, the bar has acquired a state of magnetic saturation. If the magnetising field is now decreased slowly, the intensity of magnetisation also decreases but corresponding to point B when H becomes zero the intensity of magnetisation does not become zero. The value of intensity of magnetisation retained by the magnetic material, even when the magnetising field is reduced to zero is called its retentivity or remanent magnetism or residual magnetism. Thus, OB represents the retentivity of the material under study. If now the direction of magnetising field is reversed, the intensity of magnetisation reduces along BC till it becomes zero at C. Thus to reduce the residual magnetism to zero, magnetising FILED equal to OC has to be applied in reverse direction. The value of the reverse magnetising field which is to be applied to the magnetic material so as to reduce the residual magnetism to zero is called its coercivity.  When the magnetising field is further increased in reverse direction, the intensity of magnetisation increases along CD till corresponding to point B, it again acquires saturation value symmetrical to point A. On increasing the field again, the intensity of magnetisation follows the path DEFA, and the closed curve ABCDEFA is obtained for complete cycle of magnetisation. This closed curve is known as hysteresis loop. On repeating the process, the same closed curve is obtained again and again but portion OA is never obtained. Corresponding to point B, H is zero but M has still some finite value and becomes zero after increasing H in reverse direction. Therefore, intensity of magnetisation does not become zero on making magnetising field zero but does so a little and this effect is called hysteresis. (b) Uses. When a ferromagnetic substance is taken over a complete cycle of magnetisation, the energy spent per unit volume is numerically equal to the area of the hysteresis loop.  1. The shape of the hysteresis loop is characteristic of the magnetic materials. For example, for soft iron, the hysteresis loop is narrow and large in height, while that for steel, it is quite wide and small in height, while that for steel, it is quite wide and small in height, [as shown in the fig.] The area of hysteresis loop for soft iron is found to be much smaller than for steel. Due to this, the loss of energy in case of soft iron bar will be very small as compared to that in case of steel, when both are taken over complete cycle of magnetisation. On account of this, soft iron is used to make the core of transformers and generators. Silicon, iron and mumetal (76% nickel, 17 % iron and small percentage of copper and chromium) also possess narrow hysteresis loop and can be used to make the core of a transformer. 2. Due to high value of coercivity and fairly large value of retentivity, steel is used to make permanent magnets. The area of hysteresis loop is large for steel but it is of no consideration as a permanent magnet has never to be taken through a cycle of magnetisation. Materials suitable for making permanent magnets are cobalt steel (containing cobalt, tungsten and CARBON) chromium, steel, tungsten steel. The alloy alnico (54% iron, 18% nickel, 10% aluminium, 12% cobalt, and 6% copper) is also very suitable for making permanent magnets. The only disadvantage is that alnico is brittle. 3. For soft iron, coercivity is very small and area of hysteresis loop is very small. Because of these characteristics soft iron is an ideal material for making electromagnets. |
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| 6500. |
In Rutherford scattering experiment ,What will be the scattering angle for an impact parameter zero? |
| Answer» ANSWER :D | |
