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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A silk piece, when rubbed with glass rod, acquires a charge q=-4.8 xx10 ^(-14)C. Charge developed on glass rod will be. ______________ |
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| 2. |
In India electicity is supplied for domestic use at 220V. It is supplied at 110V in USA. If the reistance of a 60W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be |
| Answer» ANSWER :C | |
| 3. |
As the beam enters the medium, it will |
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Answer» DIVERGE |
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| 4. |
A H-atom moving with speed v makes a head on collisioon with a H-atom at rest . Both atoms are in ground state. The minimum value of velocity v for which one of the atom may excite is (1.25 nxx10^(4)) m/s Find value of x. |
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| 5. |
consider two thin unifromly charged concentric shells of radii r and 2r charges Q and -Qrespectively , as shown. Three points A,BandC are marked at distances r/2 , (3r)/2and (5r)/2 respectively from their common centre. IfE_(A) E_(B) and E_(C) are magniudes of the electric fields at points A, Band C respectivly ,then |
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Answer» `E_(A)gtE_(B)gtE_(C)` |
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| 6. |
If the system is released from rest, determine thespeeds of both masses after B has moved 1 m. Neglect friction, the masses of pulleys andinextensiblestring. |
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| 7. |
Twocapacitors 2 muF and 4 muF are connected in parallel. A third capacitor of 6 muF capacity is connected in series. The combination is connected across a 12 V battery. The voltage across a 2 muF capacitor is |
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Answer» 2 V |
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| 8. |
Write two examples for a diamagnetic material. Explain how the property relates with temperature? |
| Answer» Solution :BISMUTH, Copper, Lead, Nitrogen, WATER. DIAMAGNETISM is INDEPENDENT of temperature. | |
| 9. |
"The susceptibility of a material is small" what do you mean by this statement. |
| Answer» SOLUTION :DIAMAGNETISM developed by external magnetic FIELD is very small. | |
| 10. |
The focal length of the objective of an astronomical telescope is 1 m and it is in normal adjustment. Initially the telescope is focussed to a heavenly body. If the same telescope is to be focussed to an object at a distance of 21 m from the objective, then identify the correct choice |
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Answer» eye piece should be displaced by 2 CM away from the objective `u_(2)=21M, 1/F=1/v_(2)-1/u_(2)implies` find`v_(2)` |
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| 11. |
An Electron oscillating with a frequency of 3xx10^6Hz would generate- |
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Answer» X rays |
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| 12. |
If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty ? |
| Answer» Solution :SLIGHTLY LESS, since bismuth is a diamagnetic material in nature. | |
| 13. |
An a.c. source of 220 V 50 Hz is connected in series with a 50 Omega resistance 150 mu F capacitors and 0.5 H inductor in series. Calculate the current through the combination. |
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Answer» Solution :`Z = SQRT(R^(2) + (X_(L) - X_(C ))^(2))` `X_(L) = 2pifL = 157 Omega` `X_(C ) = (1)/(2pi fC) = 21.23 Omega` To find `Z = 144.68 Omega` `I = (V)/(Z) = (220)/(144.68) = 1.52 A` |
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| 14. |
A 100 W bulb B_(1) and two 60 W bulbs B_(2) and B_(3) are connected to a 250 V source, as shown in the figure. Now W_(1), W_(2) and W_(3) are the output powers of the bulbs B_(1)B_(2) and B_(3) respectively. Then |
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Answer» SOLUTION :A bulb is essentially a resistance `R=(V^(2))/(P)` where P denotes the POWER of the bulb. `therefore"Resistance of "B_(1)(R_(1))=V^(2)//100` `"Resistance of "B_(2)(R_(2))=V^(2)//60` `"Resistance of "B_(3)(R_(3))=V^(2)//60` `therefore I_(1)="CURRENT in "B_(1)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))` `I_(2)="Current in "B_(2)=(250)/((R_(1)+R_(2)))=(250xx300)/(8V^(2))` `I_(3)="Current in "B_(3)=(250)/(R_(3))=(250xx60)/(V^(2))` `therefore""W_(1)" OUTPUT power of "B_(1)=I_(1)^(2)R_(1)` `therefore W_(1)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(100)` `W_(2)=I_(2)^(2)R_(2)"or"W_(2)=((250xx300)/(8V^(2)))^(2)xx(V^(2))/(60)` `W_(3)=I_(2)^(2)R_(3)"or"W_(3)=((250xx60)/(V^(2)))^(2)xx(V^(2))/(60)` `therefore W_(1):W_(2):W_(3)=15:25:64"or"W_(1)LT W_(2) lt W_(3)` |
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| 15. |
Two particles of masses m_1 and m_2 in projectile motion have velocities vecv_1 and vecv_2 respectively at time t=0. They collide at time t_0. Their velocities become vecv_1' and vecv_2' at time 2t_0 while still moving in air. The value of |(m_1vecv_1' + m_2vecv_2') - (m_1vecv_1 + m_2vecv_2)| is |
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Answer» Zero At time `t=2t_0 vec(p_2)=m_1v_1+m_2v_2` RATE of change of momentum = FORCE ACTING on the system = `(vec(p_2)-vec(p_1))/(2t_0)=(m_1+m_2)g` Putting the values of `vecp_1` and `vecp_2` `((m_1v_1+m_2v_2)-(m_1v_1+m_2v_2))/(2t_0)=(m_1+m_2)g` `:. (m_1v_1+m_2+v_2_-(m_1v_1+m_2v_2)=2(m_1+m_2)"gt"_0` |
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| 16. |
A light bulb shines light along the x axis and through two parallel ideal polarizing filters one with a fixed polarizing axis and the otherwith an axis that rotates about the x axis and the yz plane . Looking toward the light bulb through the combined filtering system you see |
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Answer» A bulb almost disappearing twice per revolution and reaching a maximum intensity twice per revolution . The maximum brightness is not as bright as looking at an UNFILTERED bulb |
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| 17. |
Derive the expression for the magnetic force experienced by a current carrying conductor . |
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Answer» Solution :Consider a CONDUCTING rod of LENGTH l, and uniform sectional area A in an external magnetic field `vecB`. Let the number densityof the mobilecharge carriers (electrons) be n. `therefore` Number density (n) =`"no. of electrons"/"Volume"` no. of electrons = n x volume no. of electrons = n x Al Ifq is the charge of each charge CARRIER Total charge = q x nAl When a steady CURRENT I flows in this conducting rod , each mobile charge carrier has an average drift velocity`vecV_d` in the presence of an external magnetic field `vecB`. `therefore` Force on these charge carriers is `vecF` =Total Charge `(vecVxxvecB)` `vecF=qnAl(vecV_d xx vecB)` `vecF=Al(vecjxxvecB)` `vecF=I(veclxxvecB)` Magnitude `F=BI l SIN theta` Where `vecl` is a vector of magnitude l and vector sign is transferred from j to `vecl` direction of `vecl` is same as j (Current ) . |
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| 18. |
The quantity having dimensions only in temperature is temperature is |
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Answer» Letent HEAT |
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| 19. |
Astaircase contains three steps each 10 cm high and 20 cm wide as shown in the figure. What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the edge of the lowest plane ? [g=10 m//s^(2)] |
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Answer» SOLUTION :`h=30 cm` `S=60 cm = v sqrt((2h)/(G))` `=v sqrt((2xx30)/(1000))=v sqrt((3)/(50))` `:. V=(60)/(100) sqrt((50)/(3))=(3)/(4)XX sqrt((2)/(3))= sqrt(3) xx sqrt(2)` `=2.45 m//s` 
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| 20. |
A child is standing with folded hands at the center of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is : |
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Answer» 2K `THEREFORE` where `I.=2I` `omega.=(1)/(2)omega` Now K.E. `K.=(1)/(2)I.omega^(.2)` or `K.=(1)/(2)xx2Ixx((omega)/(2))^(2)` `K.=(1)/(2).(Iomega^(2))/(2)=(K)/(2)` |
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| 21. |
Monochromatic lights of wavelengths 420 nm and 540 nm are incident simultaneously and normally on a double slit apparatus with slit separation of 0.0756 mm and screen is at a distance of 1m. The total number of dark fringes due to both wavelenths on the screen is |
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Answer» 360 |
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| 22. |
प्रकाश के अपवर्तन के कितने नियम है? |
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Answer» 3 |
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| 23. |
Define network, junction (branch point), loop. |
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Answer» Solution :`rArr` Simple electrical circuit can be analyzed by using Ohm.s law. When number of resistors, capacitors, Inductors and cells connected in COMPLEX way , then Ohm.s law is not sufficient . In this condition Kirchhoff.s LAWS are useful. `rArr` Junction ( Branch point) : In network, point at which TWO or more conductor MEET is called junction or branch point. `rArr` Loop : A closed PATH comprising of component is called Loop. |
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| 24. |
Who changed the perception of the Shehnai? |
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Answer» ZAKIR Hussain |
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| 25. |
A line charge lambdaper unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig.) . A uniform magnetic field extends over a circular region within the rim. It is given by, {:(B=-B_0 K,(r le a , a lt R)),(=0,"(otherwise)"):} |
| Answer» SOLUTION :`-(BPI a^2 LAMBDA)/(MR) HATK` | |
| 26. |
A block of steel heated to 100^(@)C left in a room to cool. Which of the following curves represent the correct behaviour ? |
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Answer» CURVE A `:.` Correct choice is (a). |
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| 27. |
In a car spark coil when current changes from 3A to zero in 15 mus the n induced e.m.f. in the secondary is 30000 volt. Mutual inductance M is: |
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Answer» 150H |
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| 28. |
In an ac circuit with voltage V and current I in phase, dissipated is. |
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Answer» VI |
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| 29. |
The total energy of an electron in the first excited state of H atom is - 3.4 eV, then its potential energy in this state is ...eV. |
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Answer» `+3.4` `:.U=2E :. U=2(-3.4)EV:. U=-6.8eV` |
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| 30. |
Two inductors coils of self inductance 3H and 6H respectively are connected witha resistance 10Omega and a battery 10V as shown is figure. The ratio of total energy stored at steady state in the inductors to that of heat developed in resistance in10second at hte steady state is (neglect mutual inductance between L_(1) and L_(2) |
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Answer» `(1)/(10)` |
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| 31. |
Define inductive and capacitive reactance. Give their units. |
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Answer» Solution :Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source (Figure). The alternating voltage is given by the equation. `v=V_(m)sinomegat""...(1)` ii. The alternting current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by Back emf, `epsi=-(di)/(dt)` By applying Kirchoff's loop to the purely inductive circuit, we get `v+epsi=0` `V_(m)sinomegat=L(di)/(dt)` `di=(V_(m))/(L)sinomegatdt` Integrating both sides, we get `i=(V_(m))/(Lomega)(-cosomegat)+"constant"`  iii. The INTEGRATION constant in the above eqution is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero. `cosomega=sin((pi)/(2)-omegat)` `-sin((pi)/(2)-omegat)=sin(omegat-(pi)/(2))` `i=(V_(m))/(omegaL)sin(omegat-(pi)/(2))` `i=I_(m)sin(omegat-(pi)/(2))""...(2)` iv. where `(V_(m))/(omegaL)=I_(m)` the peak value of the alternating (1) and (2), it is evident that current lags behind the APPLIED voltage by `(pi)/(2)` in an inductive circuit. This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by `90^(@)` (Figure). v. Iductive reactance `X_(L)` The peak value of current `I_(m)` is given by `I_(m)=(V_(m))/(omegaL).` Let us compare this equation with `I_(m)=(V_(m))/(R)` from resistive circuit The quantity `omegaL` PLAYS the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reatance `(X_(l)).` It is measured in ohm. `X_(L)=omegaL` AC circuit containing only a capacitor i. Consider a circuit containing only a capacitor of capacitance C connected across an alternating voltage is given by `v=V_(m)sinomegat`  ii. Let q be the instantaneous charge on the capacitor. The emf across the capacitor at that INSTANT is `(q)/(C)` According to Kirchoff's loop rule, `v=(q)/(C)=0` `q=CV_(m)sinomegat` By the definition of current, `i=(dq)/(dt)=(d)/(dt)(CV_(m)sinomegat)` `=CV_(m)(d)/(dt)sinomegat` `ori=(V_(m))/((1)/(Comega))sin(omegat+(pi)/(2))` Instantaneous value of current, `i=I_(m)sin(omegat+(pi)/(2))""...(2)` iii. where `(V_(m))/((1)/(Comega))=I_(m)` the peak value of the alternating current. From equations (1) and (2), it is clear that curret leads the applied voltage by `(pi)/(2)` in a capacitive circuit. This is shown pictorially in Figure. The wave diagram for a capacitive circuit also shows that the current leads the applied voltage by `90^(@).` Capacitive reactance `X_(C)` The peak value of current `I_(m)` is given by `I_(m)=(V_(m))/((1)/(Comega)).` Let us compare this equation with `I_(m)=(V_(m))/(R)` from resistive circuit. The quantity `(1)/(Comega)` plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance `(X_(C)).` It measured in ohm. `X_(C)=(1)/(Comega).` |
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| 32. |
How many significant figures are there in sum 18.5+0.4235 |
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Answer» 2 |
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| 33. |
भारतीय भ्रूण विज्ञान के जनक है |
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Answer» सर जगदीश चंद्र बोस |
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| 35. |
Heat is kinetic energy of molecule |
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Answer» KINETIC ENERGY of molecule |
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| 36. |
f A and B persons are moving with V_(A) and V_(B). velocities in opposite directions. Magnitude of relative velocity of B w.r.t. A is x and magni-tude of relative velocity of A w.r.t B is y. Then |
| Answer» ANSWER :B | |
| 37. |
A polythenepiecerubbedwithwool is found to have a negativechargeof 3xx10^(-7)c (a) estimate the numberof electronstransferred (from which towhich ) (b) is therea transfer of mass fromwoool to polythene |
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Answer» Solution :(a) `2XX10^(12)` form woolto polythene (b) YES but of a negligible AMOUNT (`=2xx10^(-18)` KG in the EXAMPLE ) |
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| 38. |
Assertion In Young's double slit experiment, ratio I_(max)/I_(min) is infinite Reason If width of any one of the slits is slightly increased, then this ratio will decrease. |
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Answer» `:. I_(max)/I_(min)="infinite"` If width of one slit is slightly increased `I_(min) GT0` therefore this ratio will be LESS than infinite |
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| 39. |
Inside an infinitely long circular cylinder charged uniformly with volume density P ther is a circular cylindrical cavity. The distance between the axis of the cylinder and the cavity is equal to a. Find the electric field strenght E inside the cavity. The permittivity us assumed to be equal to unity. |
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Answer» `E =(arho)/(epsilon_(0))BAR(a)` is DIRECTED toword the AXIS of cavity |
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| 40. |
Three point charges Q_1, Q_2, and Q_3 in that order are placed equally spaced along a straight line. Q_2 and Q_3 are equal in magnitude but opposite in sign. If the net force on Q_3 is zero, the value of Q_1 is |
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Answer» `Q_1 = |Q_3|` |
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| 41. |
A film projector magnifies a 100 cm^(2)film strip on a screen. If the linear magnification is 4, the area of magnified film on the screen is |
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Answer» `1600 cm` |
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| 42. |
A body absorbs 5 xx 10^(29) photons of frequency 10^(20) Hz. Which of the following information is correct? [Assume, all the energy of photons is transformed into mass] |
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Answer» MASS of the BODY REMAINS unchanged |
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| 43. |
At a place, the horizontal component of earth's magnetic field is B and angle of dip is 60^@. What is the value of horizontal component of the earth's magnetic field at equator? |
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Answer» SOLUTION :The value of horizontal COMPONENT of the earth.s magnetic field at equator is equal to the TOTAL magnetic field `.B_C.` of earth i.e., `(B_H)` equator =` B_E` As at a given place, horizontal component `B_H = B_E cos delta`, where `delta ` is the angle of dip. Hence , we have `B= B_E cos 60^@ = 1/2 B_E impliesB_E = 2B` `implies(B_H)_("equator") = B_E = 2B` |
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| 44. |
SSI integrated chip contains |
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Answer» LESS than 10 gates |
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| 45. |
Depict the direction of the magnetic field lines due to a circular current carrying loop. |
| Answer» SOLUTION :MAGNETIC FIELD lines have been shown in FIG. | |
| 46. |
In a series LCR circuit connected to an ac source of variable frequency and voltage V=V_(M) Vsinomega t,draw a plot showing the variation of current (l) with angular frequency (omega) for two different values of resistance R_(1) and R_(2)(R_(1) gt R_(2)).Write the condition under which the phenomenon of resonance occurs.For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance. |
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Answer» Solution :Figure shows the VARIATION of `i_(m)` with `omega` in a `LCR` series circuit for two values of resistance `R_(1)` and `R_(2)(R_(1) gt R_(2))` The CONDITION for RESONANCE in the `LCR` circuit is. `omega_(0)=1/sqrt(LC)` We see that the current amplitude is maximum at the resonant FREQUENCY `omega`.Since `i_(m)=V_(m)//R` at resonance. the current amplitude of case `R_(2)` is shaper to that for case `R_(1)`. Quality factor or simply the `Q`-factor of a resonant `LCR` circuit is defined as the ratio of voltage drop across the resistance at resonance. `Q=V_(L)/V_(R)=(omegaL)/R` Thus, finally, `Q=1/Rsqrt(L/C)` The `Q` factor determines the sharpness at resonance as for higher value of `Q` factor the tuning of the circuit and it sensivity to accept resonating frequency signals will be much higher. 
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| 47. |
How did the grandmother spend her time in the city? |
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Answer» FEEDINGS dogs |
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| 48. |
There is one long coaxial cable which consists of two concentric cylinders of radii a and b. Central conductor carries steady current I and outer conductor acts as return path for the current. (i) Calculate energy stored in the magnetic field of length l of such a cable. (ii) Calculate self-inductance for l length of this cable. |
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Answer» Solution :(i) Magnetic field intensity at a point distance r from the centre of the cable is given by `B=(mu_(0)I)/(2pir)` Energy density in terms of distance r can be written as follows: `u=(B^(2))/(2mu_(0))` `implies u=(((mu_(0)I)/(2pir))^(2))/(2mu_(0))` `implies u=(mu_(0)I^(2))/(8pi^(2)r^(2))`  Consider a volume segment dV which is in the form of a cylindrical shell of radii r and r + dr as shown in the figure. Length of this cylindrical shell is l which is perpendicular to the plane shown in figure. Energy density can be assumed to be constant THROUGHOUT this segment and hence energy stored in this segment can be written as follows: Energy dU = udV `dU=(mu_(0)I^(2))/(8pi^(2)r^(2))2pirldr` `dU=(mu_(0)I^(2)l)/(4pi)(dr)/(r)` Total magnetic energy can be calculated by integrating above relation. `U=(mu_(0)I^(2)l)/(4pi) int_(a)^(b)(dr)/(r)` `U=(mu_(0)I^(2)l)/(4pi) int_(a)^(b)(dr)/(r)` `U=(mu_(0)I^(2)l)/(4pi)"In"(b)/(a)` (ii) We can use above calculated VALUE of energy stored to calculate self-inductance of this cable. We know that energy stored in an inductor is given as follows: `U=(1)/(2)LI^(2)` We can equate it with value calculated in part (i) `(1)/(2)LI^(2)=(mu_(0)I^(2)l)/(4pi)"In"(b)/(a)` `L=(mu_(0)l)/(2pi)"In"(b)/(a)` |
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| 49. |
The time period ofa seconds pendulum is mea- sured repeatedly for three times by two stop watches A, B. If the readings are as follows S.NO"AB" 1."2.01 sec2.56 sec " 2."2.10 sec2.55 sec " 3. "1.98 sec2.57 sec " |
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Answer» A is more ACCURATE but B is more PRECISE |
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| 50. |
Two identical loudspeakers are placed on a line at distance 4m and 6m from a microphone lying on the line joining them. The output of each speaker at the microphone is 2mW. If the two speakers are connected in series and excited by the same oscillator and they vibrate in phase with each other, the resultant ouput of the microphone will be (frequency of oscillator =660 Hz and speed of sound =330 m//s) |
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Answer» zero |
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