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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.
| 40001. |
Justify the output waveform (Y) of the OR gate for the following inputs A and B given in |
Answer» SOLUTION :NOTE the FOLLOWING: 
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| 40002. |
A uniform electric field of 100 V/m is directed at 30^(@) with the positive x-axis as shown in figure Find the potential difference V_(a) - V_(A) if OA = 2 m and OB = 4m. |
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Answer» `100(2-sqrt(3))` VOLT |
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| 40003. |
The current induced in 100Omega coil when the magnetic flux decreases from 1 Wb to 0.1Wb in 0.1s, is |
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Answer» `9A` |
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| 40004. |
A long solid aluminum cylinder of radiusa = 5.0 cm rotates aboutits axis in unidrommagneticfield with induction B = 10 mT. Theangluarvelocityof rotationequlas omega = 45 rad//s withomega uarr uarr B Neglecting the magneticfield of appearingchagres, find theirspcaeand surfaface densities. |
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Answer» Solution :CHOOSE `vec(omega) uarr uarr vec(B)` alongthe z-axisand choose`vec(r)`, as the cylindrical polarradius of a reference point(perpendiculardistance fromthe AXIS). This pointhas the velocity. `vec(v) = vec(omega) XX vec(r)`, and experiences a `(vec(v) xx vec(B))` FORCE, which must becounterbalancedby an electric FIELD, `vec(E) = -(vec(omega) xx vec(r)) xx vec(B) = -(vec(omega). vec(B)) vec(r)`. There must appear a spacecharge density, `rho = epsilon_(0) div vec(E) = -3 epsilon_(0) vec(omega) vec(B) = -8 pC//m^(3)` Since the cylinder, as a wholeis electrically neutralthe surfaceof the cylinder must acquirea positive charge of surface density, `sigma = + (2 epsilon_(0) (vec(omega). vec(B)) pi a^(2))/(2pi a) = epsilon_(0) a vec(omega).vec(B) = +- 2 pC//m^(2)` |
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| 40005. |
In 18.3 and 18.7, 18.8 we have obtained the expression for the potential of the field of a point charge using numerical methods. When you have learned to difierentiate, prove that formula (18.25) leads to the expression for the field intensity of a point charge known from the Coulomb law. |
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Answer» Solution :In this case EQUIPOTENTIAL SURFACES are spheres with a common centre at the source, and radii normal to the surfaces. If the potential is `varphi=q..4pi epsi_(0)r,`the FIELD strength is `E=(dvarphi)/(dr)=-(d)/(dr)((q)/(4pi epsi_(0)r))=(q)/(4pi epsi_(0)r^(2))` and just this was to be proved. |
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| 40006. |
Determine the angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm when light of wavelength 589nm falls normally on it. |
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Answer» |
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| 40007. |
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s_(1) in a uniform horizontal magnetic field of magnitude 3.0 xx 10^(2)T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 102Omega, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? |
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Answer» SOLUTION :Here RADIUS of coil R = 8.0 cm = 0.08 m, hence area of coil `A = piR^(2) =pi xx (0.08)^(-2) = 0.02 m^(2)`, number of turns in the coil N = 20, angular speed of rotation `omega = 50s^(-1)` and magnetic field `B = 3.0 xx 10^(2)T`. `therefere` Maximum induced emf in the coil `lvarepsilon_(MAX) = NBA omega` `= 20 xx 3 xx 10^(-2) xx 0.02 xx 50 = 0.6 V` Average induced emf `vareppsilon_(av)` = average value of `NBA omegasin omegat` for ONE cycle = `N B A omega` [average value of sin t for one cycle] =0. As resistance `R = 10Omega`, hence maximum induced current `I_(max) = (varepsilon_(max))/R = 0.6/10 A = 0.06A` `therefere` Average power loss due to Joule.s heating `P_(av) = (varepsilon_(max).I_(max))/2=(0.6 xx 0.06)/2 =0.018W. Source of power: The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) mustupply torque (and do work) to counter this torque in order to keep the coil rotating uniformly. Thus, the source of the power dissipated as heat in the coil is the external rotor. |
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| 40008. |
When milk is churned, cream gets seperated due to |
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Answer» CENTRIPETAL FORCE |
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| 40009. |
A npn transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of 5 m A. The terminal of 10V battery is connected to a collector through a load resistance R_(L) and to the base through a resistance R_(B). The collector emitter voltage V_(CE)=5V, base emitter voltage, V_(BE)=0.5Vand base correct amplification factor beta_(d.c)=100. Calculate the values of R_(L) and R_(B) |
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Answer» Solution :potential difference ACROSS `R_(L)` is `I_(C)R_(L)=10V-V_(CE)=10V-5V=5V` `Or R_(L)=(5)/(I_(c))=(5V)/(5xx10^(-3)A)=10^(3)Omega=1kOmega` Here, `I_(B)=(I_(C))/(beta_(d.c))=(5xx10^(-3))/(100)=5xx10^(-5)A` Potential difference across `R_(B)` is `I_(B)R_(B)=10-V_(BE)=10-0.5=9.5` or `R_(B)=(9.5)/(I_(B))=(9.5V)/(5xx10^(-5)A)=1.9xx10^(5)Omega=190kOmega` |
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| 40010. |
Consider situation in figure. The bottom of the pot is reflecting plane mimor, Fis a small fish and is an object. Refractive index of water is mu. Find the distances from it self the fish can see the image (s) of the object O. |
| Answer» SOLUTION :`H (MU +1//2 ) " above it SELF " H, (mu +3//2 ) " below it self "` | |
| 40011. |
Assertion : Spherical shell a in the shown figure has a charge -q and shell B has charge +q. When these two are connected by a thin conducting wire, then whole of the electrostatic potential energy stored between the shell is lost. Reason : Whole of the inner charge will transfer to outer shell B. |
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Answer» If both ASSERTION and Reason are TRUE and Reason is the correct EXPLANATION of Assertion. |
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| 40012. |
To convert galvanometer into an ammeter we connect |
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Answer» LOW RESISTANCE in series |
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| 40013. |
State Coulomb.s law |
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Answer» Solution :Statement: The electrostatic force between two point charges is directly proportional to the product of magnitude of charges and inversely proportional to the square of the distance between them. The direction of this force is along the line joining the two charges. Explanation: Let `q_(1) and q_2`be the magnitudes of two charges separated by a distance .r..Let F be the force between them, From Coulomb.s LAW, `Falpha(q_(1)q_(2))/r^2` `F = (1)/(4piepsilon_(0))(q_1q_2)/r^2` where `epsilon_0` ISTHE absolute permitivity of free space. 
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| 40014. |
The binding energy of a satellite of mass 1000 kg revolving in circular orbit around the earth when it is' close to the surface, of earth is (G=6.67xx10^-11Nm^2//kg^2), |
| Answer» Answer :B | |
| 40015. |
Consider two sound sources S_1 and S_2 having same frequency 100Hz and the observer O located between them as shown in the fig. All the three are moving with same Velocity in same direction. The beat frequency of the observer is: |
| Answer» ANSWER :C | |
| 40016. |
Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also |
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Answer» SOLUTION :As per question the radius of curvature of CONCAVE mirror R = - 20 CM, hence its focal LENGTH `f = R/2 = -10 cm`, and magnification of real image `m=-2` As, `m=-v/u`, hence, `-v/u =-2` or v= 2u So, from mirror formula `1/v + 1/u = 1/f`, we have and `1/(2u) + 1/u =1/(-10) rArr 3/(2u) =-1/10 rArr u=-15 cm` and `v=2u=2 xx (-15) = -30 cm` Thus, the object is placed in front of mirror at a distance of 15 cm from it and its real, magnified and inverted image is formed at a distance of 30 cm in front of the mirror. |
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| 40017. |
This story is about ? |
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Answer» AUTHOR's Grandmother |
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| 40018. |
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays "1" and "2" are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism. |
Answer» Solution : As per question the REFRACTIVE index of GLASS prism for RAY no. 1 is 1.35, Hence, its critical angle `i_(1) =sin^(-1)(1/1.35) = 47.8^(@)` For ray no. 2 the refractive index of glass prism is 1.45. Hence, its critical angle `i_(2) = sin^(-1) (1/1.45) = 43.6^(@)`. FIG. 9.47 ray no. 1 is refracted from surface AC but ray no. 2 undergoestotal internal REFLECTION at surface AC and comes out from the surface BC. |
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| 40019. |
Two alkenes, X(91% yield) and Y(9% yield) are formed when the following compound is heated The structure of X and Y, respectively are - |
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Answer»
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| 40020. |
A particle is moving in a circle of radius 1 cm and with a constant speed of 8 cm s^(-1) . Centre of the circle lies on principle axis of a converging lens of focal length 50 cm and at a distance of 75 cm from the lens . Plane of the circle is perpendicular to principle axis. The correct statement is |
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Answer» Speed of the image is `4 CM s^(-1)` |
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| 40021. |
A top of mass m=1.0kg and moment of inertia relative to its own axis I=4.0g*m^2 spins with an angular velocity omega=310rad//s. Its point of rest is located on a block which is shifted in a horizontal direction with a constant acceleration w=1.0m//s^2. The distance between the point of rest and the centre of inertia of the top equals l=10cm. Find the magnitude and direction of the angular velocity of precession omega^'. |
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Answer» Solution :The effective `g` is `SQRT(g^2+W^2)` inclined at angle `"tan"^-1w/g` with the vertical. Then with reference to the new "vertical" we proceed as in problem. Thus `omega^'=(mlsqrt(g^2+w^2))/(Iomega)=0*8rad//s`. The vector `overset(rarr')omega` forms an angle `theta="tan"^-1(w)/(g)=6^@` with the NORMAL vertical. |
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| 40022. |
Wavelength of light incident on a photo cell is 3000 019Å if stopping potential is 2.5 volt, then work function of the cathode of photo cell is : |
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Answer» 1.41 eV |
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| 40023. |
If angle of deviations near the first and the second refractory surfaces are delta_1, and delta_2, then ...... |
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Answer» `delta_1=delta_2` |
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| 40024. |
An AM broadcasting has a vertical telescopic transmitting antenna and a receiver has a vertical telescopic antenna. The receiver will respond to |
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Answer» ELECTRIC component of the electromagnetic wave PRODUCED by antenna |
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| 40025. |
A charged particle of mass m = 1 mg and charge q = 1 (mu) C enter along AB at point A in a uniform magnetic field B = 1.2 T existing in the rectangular region of size a xx b, where a = 4 m and b = 3 m. The particle leaves the region exactly at corner point C. What is the speed v (in m s^(-1)) of the particle? |
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Answer» `r=AO = CO = (mv).(QB) also a/r = sin theta, (r-B)/(r) = cos theta` Solve the above equaton to get  `v=qB(a^(2)+b^(2))/(2mb) = 10^(-6) xx 1.2 (4^(2)+3^(2))/(2 xx 10^(-6) xx 3) = 5 m//s`. |
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| 40026. |
Which important property differentiates magnetic field lines and electric field lines? |
| Answer» Solution :MAGNETIC FIELD LINES forms CONTINUOUS closed loops and electric field lines do not form closed loops.: | |
| 40027. |
A particle of mass 2 kg is projected vertically upward with a velocity of 20 m/s. It attains maximum height of 17 m. The loss in mechanical energy due to air drag is ( g = 10 ms^-2) |
| Answer» Answer :d | |
| 40028. |
A series LCR circuit is tuned to resonance. The impedance of the circuit now is |
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Answer» `[R^(2)+(OMEGAL-(1)/(OMEGAC))^(2)]^(1//2)` |
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| 40029. |
In Rutherford experiments on alpha-ray scattering the number of particles scattered at 90^(0)angle be 28 per minute. Then the number of particles scattered per minute by the same foil, but at 60^(0) is |
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Answer» 56 |
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| 40030. |
The ratio of the radius of the earth to that of moon is 10. The ratio of acceleration due to gravity on earth and on moon is 6. The ratio of their escape velocity from earth to that from the moon is : |
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Answer» 10 `=7.74~8` HENCE, the correct CHOICE is (C ). |
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| 40031. |
Three sound waves of equal amplitudes have frequencies (V-1), V. (V+1). They superpose to give beats. The number of beats produced per second will be |
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Answer» 3 |
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| 40032. |
The shunt required to send 10 % of the main current through a moving coil galvanometer of resistance 99 Omega is |
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Answer» a. `99 Omega` |
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| 40033. |
Match list I with list II for a projectile. |
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Answer» a - F, b - n, C - G, d - e |
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| 40034. |
A point charge +q is placed at the centre of a cube of side L. The electric flux emerging from the cube is |
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Answer» `(Q)/(epsi_(0))` |
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| 40035. |
200 Mev energy is released when one nucleus of .^235U undergoes fission. Find the number of fissions per second required for producing a power of 1 mega watt. |
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Answer» `3.125xx10^14` |
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| 40036. |
Find the thickness of a plate which will produce a change in optical path equal to half the wavelength lambda of the light passing through it normally. The refractive index of the plate is mu. |
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Answer» `(LAMBDA)/(4(MU - 1))` |
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| 40037. |
A galvanometer has a resistance of 100 Omega. A current of 10^(-3) A is permissible through the galvanometer. How can it be converted into a) an ammeter of range 10 A and b) a voltmeter of range 10 V. |
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Answer» <P> Solution :G = 100 `OMEGA p, i_1 = 10^(-3)A `a) `i_2 = 10 A , n = (i_2)/(i_1) = 10^4` `S = (G)/((n-1)) = (100)/((10^4 -1)) = 100/999 Omega` b) `V_1 = i_1 G = 10^(-3) XX 100 = 10^(-1) V` `V_2 = 10 V rArr n = (V_2)/(V_1) = (10)/(10^(-1)) = 100` `therefore R = G (n-1) = 100 (100 -1) = 9900 Omega` |
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| 40038. |
v_(1) is the frequency of the series limit of Lyman series, v_(2) is the frequency of the first line of Lyman series and v_(3) is the frequency of the series limit of the Balmer series. Then |
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Answer» `v_(1)-v_(2)=v_(3)` where n=2,3,4,........ For the series limit of Lyman series `n=oo` `v_(1)=RC[1/1^2-1/oo^(2)]=RC .........(i)` For the first line of Lyman series n=2 `v_(2)=RC [1/1^(2)-1/2^(2)]=3/4RC ..........(II)` For Balmar series `v=RC (1/2^(2)-1/n^(2))` where n=3,4,5.... For the seris limit of Balmer series `n=oo` `v_(3) =RC[1/2^(2)-1/oo^(2)]=(RC)/(4)...........(III)` From EQUATIONS (i), (ii) and (iii), we get `v_(1)=v_(2)+v_(3) rArr v_(1)-v_(2)=v_(3)` |
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| 40039. |
The function sin^(2)(omega t) represents |
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Answer» a PERIODIC, but not simple HARMONIC motion with period `2pi//omega`  `(dy)/(dt) = 2omega sin omega t - omega sin 2omegat` `(d^(2)y)/(dt^(2)) = 2omega^(2) cos2 omega t` For SHM, `(d^(2)y)/(dt^(2))prop -y` HENCE, function is not SHM, but periodic. From the y-t graph, time period, `T = pi/omega` |
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| 40040. |
The resonance frequency in series and parallel resopnance circuits are given by: |
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Answer» `F=2pisqrtLC` |
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| 40041. |
Dielectric constnat of air is 1.005 and velocity of electromagnetic wave propagating in air is a xx 10^(10)cm//s then a = ……. |
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Answer» 3 |
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| 40042. |
The temperature-time graphs obtained in Newton's cooling experiment is |
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Answer» a STRAIGHT LINE with POSITIVE slope |
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| 40043. |
A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 M Omega across a 2V battery (Fig. 8.3). Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after t = 10^(-3) s. (The charge on the capacitor at time t is q (t) = CV [1 – exp (–t// tau)], where the time constant tau is equal to CR.) |
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Answer» Solution :The time constant of the CR circuit is `tau = CR = 10^(-3) s`. Then, we have `q(t)=CV [1-"exp"(-t//tau)]` `=2xx10^(-9)[1-"exp "(-t//10^(-3))]` The electric field in between the plates at time t is `E=(q(t))/(epsi_(0)A)=(q)/(pi epsi_(0)), A= pi (1)^(2) m^(2)`= area of the plates. Consider now a circular loop of radius (1/2) m parallel to the plates passing through P. The magnetic field B at all points on the loop is along the loop and of the same value. The flux `Phi_(E)` through this loop is `Phi_(E)=Exx` area of the loop `=E xx pi xx((1)/(2))^(2)=(pi E)/(4) =(q)/(4 epsi_(0))` The displacement CURRENT `i_(d) =epsi_(0)"" (d Phi_(E))/(dt) =(1)/(4)""(dq)/(dt) =0.5 xx10^(-6)" exp "(-1)` at `t=10^(-3)s`. Now, applying Ampere-Maxwell law to the loop, we get `B xx 2 pi xx((1)/(2))= mu_(0) (i_(c)+i_(d))=mu_(0)(0+i_(d))=0.5xx10^(-6) mu_(0)" exp"(-1)` or, `B=0.74xx10^(-13) T` |
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| 40044. |
An observer moves towards a stationary source of sound with speed (1)/(5) th of the speed of sound. The wavelength and frequency of sound emitted are lambda and f respectively. The apparent frequency and wavelength recorded by the observer are respectively : |
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Answer» 1.2 F, 1.2 `lambda` f. `= ((V + U_(0))/(V)) `f `f ((V + (V)/(5))/(V)) ` f f. = 1.2 f . WAVELENGTH is not affected by the MOTION of observer. CORRECT choice is c. |
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| 40045. |
Can the relative permittivity of a medium be less than 1? |
| Answer» Solution :No. Air or vacuum has MINIMUM relative permittivity `(K=epsilon_(0)epsilon_(0)=1)`. The relative permittivity of all other MEDIA is greater than 1. | |
| 40046. |
When a plane electromagnetic wave enters a glass slab, then which of the following will not change? |
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Answer» a) Wavelength |
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| 40047. |
Assertion: A thin aluminium disc, spinning freely about a centre pivot, is quickly brought to rest when . placed between the poles of a strong U-shaped magnet Reason:A current induced in a disc rotating in a magnetic field produces a force which tends to oppose the disc's motion. |
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Answer» both ASSERTION and REASON are both are both wrong |
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| 40048. |
A charge is distributed uniformaly over is ring of radius 'a' .Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge. |
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Answer» Solution :Consider a UNIFORM circular ring of radius .a. carrying a charge Q distributed uniformaly over its surface Let P be a point situated at a distance r from the centre of ring along its axis. Consider an element of length Dl around point A of the ring carrying a charge Dq `=(Q)/(2 pi A) Delta l ` The electrical field at point P due to this element is given as ` ""| oversetto (Delta E) |=(1)/( 4 pi in _0).(Delta q)/((AP)^(2))=(1)/( 4 pi in _0).(Q)/(2 pi a ( a^(2) +r^(2)) ` The ELECTRIC field ` oversetto (Delta E )`, directed along the direction AP, subtends an angle ` theta ` with the axis of ring and can be resolved into TWO components namely (i)`Delta Ecos theta ` along the axis of ring , and (ii)` Delta Esin theta ` normal to the axis of ring. It is clear fromsymmetry that the normal components `Delta E sin theta ` due to mutuallyopposite charges elements at A and B nullify each other. Hence , net electricfield due to whole ring will be ` E= sum Delta E cos theta =sum ((1)/(4 pi in _0) .(QDelta l )/(2 pi a (a^(2) +r^(2)) ). (r)/( (a^(2) +r^(2))^(1//2)) ) ` ` =(1)/(4 pi in_0).(Qr)/( 2pi a (r^(2) +a^(2))^(3//2)) .sum Delta l =(1)/( 4 pi in _0) . (Qr)/( 4 pi a(r^(2)+a^(2) )^(3//2)). 2pi a =(1)/( 4 pi in _0) .(Qr)/( (r^(2) +a^(2)) ^(3//2))` The field `oversetto E ` is directed along the axisOP of the charged ring. If ` gtgt a,`then the above relation may be expressed as ` oversetto E= (1)/( 4 pi in _0) .(Q)/(r^(2)) hat r ` This shows that for far POINTS at long distances from the ring. it behaves lika a point charges. ` (##U_LIK_SP_PHY_XII_C01_E10_001_S01.png" width="80%"> |
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| 40049. |
In AM wave , carrier power is given by |
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Answer» <P>`P_(c) = (2 E_(c)^(2))/(R)` |
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| 40050. |
Consider the fission of ""_(92)^(238)U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ""_(58)^(140)Ce and ""_(44)^(99)Ru. Calculate Q for this fission process. The relevant atomic and particle masses are m(""_(92)^(238)U) =238.05079 u m( ""_(58)^(140)Ce ) =139.90543 u m(""_(44)^(99)Ru ) = 98.90594 u |
| Answer» Solution :`Q=[m(""_(92)^(228)U)+m(n)-m(""_(58)^(140)CE)-m(""_(44)^(99)RU)]c^(2)=231.1MeV` | |
