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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. |
From the initial point A(1,2,0), to the final point D(3,2,0), a charge of 2C is moved through the points B (2,2,0) and C (3, 1, 0) along the path ABCD. The electric field in that region is 4 bar(i) N/C. The work done is given by |
| Answer» ANSWER :C | |
| 2. |
(A) : Ampere's law holds for steady currents in straight conductor which do not vary with time. (R) : Magnetic field lines always form closed loops. |
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Answer» Both 'A' and 'R' are true and 'R' is the correct EXPLANATION of 'A'. |
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| 3. |
Which of the two, namely voltmeter and potentiometer is preferable to measure the emf of a battery? Why? |
| Answer» SOLUTION :Potentiometer is preferable to MEASURE the e.m.f of a battery. While measuring e.m.f. voltmeter draws current from battery. Whereas potentiometer draws no current when it is BALANCED. HENCE the reading of p.d. between two points is more accurate. | |
| 4. |
A 10 watt electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and the container rises by 3^@ K in 15 minutes. The container is then emptied, dried and filled with 2 kg of oil. The same heater now raises the temperature of container-oil system by 2^@ K in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water as 4200 J kg^(-1)K^(-1)the specific heat of oil in the same unit is equal to |
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Answer» `1.50 xx 10^3 ` Massofwater`m_w= 0.5Kg` Massof oil`m_0=2kg ` specificheatof water`s_w= 4200J Kg^(-1) K^(-1)` incase CONTAINER watersystem energysuppliedbythe heaterto thesystemin 15min `= 10 xx 15 xx 60= 90 00 J` ` thereforem_w s_w DeltaT_w+m _c S_cDeltaT_c= 9000 ` ( wheresubscriptsw andc refereto watereand continerrespectively ). `0.5 xx 4200 xx 3 xx m_cxxs_cxx 3=9000` ` 6300+ 3m_cs_c=9000or3m_C s_c2700` `m_c s_c = 900` in caseof container-oilsystem energysupplied by thesame heaterto the systemin 20min `=10 xx 20 xx 60 = 1200 J` ` thereforem_o S_0DeltaT_o+ m_cS_cDeltaT_c= 12000` ( wheresubjscripto referto oil ) ` 2 xx s _0xx2+ 900xx 2= 1200 ` ` 2 xx S_0xx 2+ 900xx 2= 1200 ` ` 4s_0=10200` ` s_0 =2550j Kg^(-1)K^(-1) =2.55 xx 10^3J kg^(-1) K^(-1)` hencethe specificheatof OILIS `2.55 xx 10^3Jkg^(-1)K^(-1)` |
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| 5. |
A wire has mass m=0.3+-0*003g, radius r=0*5+-0*005mm and length l=6+-0*06cm. The maximum percentage error in the measurement of its density is : |
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Answer» `1` `(Deltarho)/(rho)xx100=((DELTAM)/(m)xx100)+2((Deltar)/(r)xx100)` `+((Deltal)/(l)xx100)` PUTTING the various values `(Deltarho)/(rho)xx100[(0*003)/(0*3)xx100]+2[(0*005)/(0*5)xx100]` `+((0*06)/(6)xx100)` `=1+2+1=4%`. Hence correct CHOICE is `(d)`. |
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| 6. |
Apoint P lies on the perpendicular bisector of a short electric dipole, of dipole moment p' at a distance r from the centre of dipole. The electric field at point P is proportional to |
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Answer» `(P)/(r^(2))` |
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| 7. |
A series RLC circuit , driven with E_(rms)=120 V at frequency f_d=60.0 Hz, contains a resistance R=200 Omega, and inductance with inductive reactance X_L=80.0 Omega, and a capacitance with capacitive reactance X_C=150Omega. (a) What are the power factor cos phi and phase constant phi of the circuit ? |
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Answer» Solution :The power factor cos `phi` can be found from the resistance R and impedance Z (cos `phi` =R/Z). Calculations : To calculate Z, we use `Z=sqrt(R^2+(X_L-X_C)^2)` `=sqrt((200OMEGA)^2+(80.0 Omega- 150Omega)^2)` =211.90 `Omega` `cos phi = R/Z = (200 Omega)/(211.90 Omega)=0.9438 APPROX 0.944` Taking the inverse cosine then yields `phi=cos^(-1) 0.944=pm 19.3^@` The inverse cosine on a calculator givens only the positive answerhere , but both `+19.3^@` and `-19.3^@` have a cosine of 0.944. To determine which sign is correct , we must consider whether the current leads or lags the driving emf . Because `X_C gt X_L` , this circuit is mainly capacitive, with the current leading the emf. Thus, `phi` must be negative . `phi=-19.3^@` We could , instead, have found `phi`. A calculator would then have given us the answer with the minus sign. (b) What is the average rate `P_(avg)` at which energy is dissipated in the resistance ? There are TWO ways and two ideas to use : (1) Because the circuit is assumed to be in steady -state operation, the rate at which energy is dissipated in the resistance is equal to the rate at which energy is supplied to the circuit . `(P_(avg)=E_(rms) I_(rms) cos phi)`. (2) The rate at which energy is dissipated in a resistance R DEPENDS on the square of the rms current `I_(rms)` through it, `(P_(avg)=I_(rms)^2 R)`. First way : We are given the rms driving emf `E_(rms)` and we already know cos `phi` from part (a). The rms current `I_(rms)` is determined by the rms value of the driving emf and the circuit impedance Z (which we know ). `I_(rms)=E_(rms)/Z` . `P_(avg)=E_(rms)I_(rms)cos phi=E_(rms)^2/Z cos phi` `=(120V)^2/(211.90Omega)(0.9438)`=64.1 W Second way : Instead, we can write `P_(avg)=I_(rms)^2 R=E_(rms)^2/Z^2 R` `=(120V)^2/(211.90Omega)^2 (200Omega)`=64.1W (c) What new capacitance `C_"new"` is needed to maximize `P_(avg)` if the other parameters of the circuit are not changed? (1) The average rate `P_(avg)` at which energy is supplied and dissipated is maximized if the circuit is brought into resonance with the driving emf. (2) Resonance occurs when `X_C=X_L`. Calculations : From the given data, we have `X_C gt X_L` . Thus , we must decrease `X_C` to reach resonance .From `(X_C=1//omega_d C)`, we see that this means we must increase C to the new value `C_"new"`. we can write the resonance CONDITION `X_C=X_L` as `1/(omega_d C_"new")=X_L`. Substituting `2pif_d` for `omega_d` (because we are given `f_d` and not `omega_d` ) and then solving for `C_"new"`, we find `C_"new" = 1/(2pi f_d X_L)=1/((2pi)(60 Hz)(80.0 Omega))` `=3.32xx10^(-5)F=33.2 muF` Following the procedure of part (b), you can show that with `C_"new"`, the average power of energy dissipation `P_(avg)` would then be at its maximum value of `P_(avg. max)`=72.0 W |
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| 8. |
A block of mass 2 kg is on a horizontal surface. The co-efficient of static & kinetic frictions are 0.6 & 0.2 The minimum horizontal force required to start the motion is applied and if it is continued, the velocity acquired by the body at the end of the 2nd second is (g=10 ms^(-2)) |
| Answer» Answer :A | |
| 9. |
A flood light is covered with a fitter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave E_(x) = 36 sin (1.20 xx 10^(7) z – 3.6 xx 10^(15) t) V // m The average intensity of beam in "watt" // "(metre)"^(2) will be : |
| Answer» Solution :`I_("aver.")=(1)/(2)epsilon_(0)E_(0)^(2) xx C=1.719 W // m^(2)` | |
| 10. |
यदि समीकरण 10x -9y=12 का हल (2k-1,k) हो तो, k का मान |
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Answer» 1 |
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| 11. |
Suggest a method to shield certain region of space from magnetic fields . |
| Answer» SOLUTION :The SPACE has to be enclosed by rings of soft iron or some other ferromagnetic material . The magnetic lines of force would pass through the material and KEEP the space inside FREE from magnetic FIELD . | |
| 13. |
Magnetic susceptibility of a paramagnetic material is x at a termperature of 27^@ C . At what temperature will its susceptibility be 1.5 x? |
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Answer» `13.5^@ C` |
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| 14. |
The reading of an ideal ammeter, in the circuit shown here, equals (i) I when key K_1is closed but key K_2 is open , (ii) I/2when both keys K_1 and K_2are closed. Find the expression for the resistance of X in terms of the resistance R and S. |
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Answer» Solution : We know that resistance of an ideal AMMETER is zero. (i)When only key `K_1`is closed, the current flowing through the ammeter is given as : `I. = (epsi)/( R + (SX)/(S + X))` and the current flowing through the ammeter is given as : `I/2 = (S)/(S + X).I.= ( (S)/(S+X)) .(epsi)/(R +(SX)/(S + X)) = (S epsi)/((S + X)R+SX)`...(ii) Multiplying (ii) by 2 and then equating with (i), we GET `(epsi)/(R+X) = (2S epsi)/((S + X)R+ SX)` ` RARR (S+X) R + SX = 2S (R+X)" or" SR + XR + SX = 2SR + 2SX` ` rArr XR - SX = SR`or`X = (RS)/((R - S))` |
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| 15. |
A spring of spring constant 5 xx 10^(3) N/m is stretched intially by 5 cm from the unstretched position. The work required to stretch it further by another 5 cm is : |
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Answer» 6·25 Nm `W=1/2K(x_(2)^(2)-x_(1)^(2))` `=1/2 5xx10^(3)(10^(2)-5^(2))xx10^(-4)` `=18.75Nm`? CORRECT CHOICE is (b). |
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| 16. |
A circular copper disc 10 cm in radius rotates at 20pi rad/s about an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2 T acts perpendicular to the disc. (i) Calculate the potential difference developed between the axis of the disc and the rim. (ii) What is the induced current if the resistance of the disc is 2Omega? |
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Answer» Solution :Here radius `r = 10 cm = 0.1 m, omega = 20pi rad/s and B = 0.2 T` `THEREFORE` Potential difference between the axis of the disc and the rim `V = 1/2 Br^(-2) omega =1/2 2 xx (0.1)^(2) xx 20pi= 6.28 xx 10^(-2)V`. (ii) As RESISTANCE `R = 2OMEGA`, hence induced current `I = V/R = (6.28 xx 10^(-2))/2=3.14 xx10^(-2)A.` |
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| 17. |
A Fresnel's biprism is used to form the interferencefringes. The distance between the source and the biprism is 20 cms and that between the biprism and the screen is 80 cm. If lambda=6563 Å and the separation between the virtual sources is 3.6 mm, then the fringe width is : |
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Answer» 1.82 CM |
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| 18. |
Magnetic susceptibility of diamagnetic substance is : |
| Answer» Answer :A | |
| 19. |
The position x of a particle varies with time t as X=at^(2)-bt^(3).The acceleration of the particle will be zero at time (t) equal to |
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Answer» `(a)/(3B)` |
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| 20. |
On applying a potential of – 1 volt at the grid of a triode, the following relation between plate voltage V_(p) (volt) and plate current I_(p) (in mA) is found I_(p) = 0.125 V_(p)-7.5 If on applying – 3 volt potential at grid and 300 V potential at plate, the plate current is found to be 5 mA , then amplification factor of the triode is |
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Answer» 100 |
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| 21. |
In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because |
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Answer» they will break up |
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| 22. |
The figure shows the current I in a single-loop circuit with a battery B and a resistance R (and wires of negligible resistance). (a) Should the emf arrow at B be a drawn pointing leftward or rightward? At points a,b and c rank (b) the magnitude of the current )c the electric potential, and (d) the electric potential energy of the charge carriers greatest first. |
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Answer» |
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| 23. |
A closed circular coil having a diameter of 50 cm and of 200 turns, with a total resistance of 10Omega is placed with its plane at right angles to a magnetic field of strength 10^(-2)T. Calculate the quantity of electric charge that flows through it when the coil is turned through 180^(@) about its diameter. |
| Answer» SOLUTION :`0.078C` | |
| 24. |
when electrons drift in a metall from lower to higher potential does it mean that all the free electrons of the metal are moving in the same direction? |
| Answer» Solution :By no MEANS. The drift velocity is superposed over the large RANDOM VELOCITES of electrons | |
| 25. |
A solenoidal coil has 50 turns per centimetre along its length and a cross-sectional area of 4 xx 10^(-4) m^(2). 200 turns of another wire is wound round the first solenoid co-axially. The two coils are electrically insulated from each other. Calculate the mutual inductance between the two coils. |
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Answer» Solution :`n_(1)= 50` turns per CM =5000 turns per metre `n_(2)L= 200, A= 4 xx 10^(-4)m^(2)` `M= mu_(0)n_(1) (n_(2)l)A` `=4pi xx 10^(-7) xx 5000 xx 200 xx 4 xx 10^(-4)H` `=5.03 xx 10^(-4)H` |
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| 26. |
Determine the rms value of a semi-circular current wave which has a maximum value of a. |
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Answer» |
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| 27. |
The refracting angle of a prism is A and the refractive index is cot (A//2). The angle of minimum deviation is : |
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Answer» `180^(@) - 2A` Given `"" mu = cot (A)/(2)` `therefore cot""(A)/(2)=sin(A+(delta_(m))/(2))/(sin((A)/(2)))` `rArr "" COS((A)/(2))=sin((A+delta_(m))/(2))` `rArrsin(90^(@)-(A)/(2))=sin""((A+delta_(m))/(2))` `rArr "" 90^(@)-(A)/(2)=(A+delta_(m))/(2)` `sin = 180^(@) - 2A`. |
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| 28. |
Current and voltage in AC are I = underset(o)(I) sin (omegat + pi/4), Then |
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Answer» `underset(L)(X)` GT `underset( C)(X)` |
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| 29. |
A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in : |
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Answer»
or `N=(mv^(2))/(r)-mgorNprop(1)/(r)` This means N will be MAXIMUM for the track having minimum radius of curvature. |
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| 30. |
At the given point of time, the earth receives energy from sun at 4 cal cm^(-2) mi n^(-1) .Determine the number of photons received on the surface of the Earth per cm^(2) per minute. (Given : Mean wavelength of sun light = 5500 Å ) |
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Answer» Solution :Numberof photons received on the SURFACEOF the earth, from `E = NHV |{:(E = "4 calorie"),("= 4 xx 4.184 J),(LAMBDA = 5500 Å):}` `n = (E lambda)/(hc)` ` = (4 xx 4.184 xx 5500 xx 10^(-10))/(6.6 xx 10^(-34) xx 3xx 10^(8)) = (9.2048 xx 10^(-10))/(19.8 xx 10^(-26)) = 4648xx 10^(16)` ` = 4.648 xx 10^(19)` ` n = 4.65 xx 10^(19)` |
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| 31. |
Why cannot we experience the existence of matter waves in our day life. |
| Answer» SOLUTION :`lambda = h/(mv)`, if m is large, `lambda` will be small, thus the WAVELENGTH will be very very small COMPARED to the size of the particle. HENCE , it is not observable. | |
| 32. |
Figure shows graph of stopping potential v frequency in experiment of photo-electric effect work function of metal will be……e=change of electron. |
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Answer» `Ovxxe(in eV)` `V_(0)=((h)/(e ))V-(phi_(0))/(e )` Comparing with y=mx+c Intercept of `V_(0)` axis =`-(phi_(0))/(e )` `thereforeOB=(phi_(0))/(e )impliesphi_(0)Obxxe` (in electron volt) Neglecting negative SIGN. |
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| 33. |
Explain the design, working, uses and benefits of light emitting diode. |
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Answer» Solution :It is a heavily doped p-n junction which under forward bias emits spontaneous radiation, hence DIODE have to provide forward bias to operate. The diode is encapsulated with transparent COVER so that emitted light can come out. The outline of the LED is shown in figure.  When the diode is forwardbiased, electrons are sent from `n to p` where they are minority carriers and hole are sent from `p to n` where they are minority carriers. At the junction boundary the concentration of minority crriers increases when there no bias. Thus at the junction boundaryon either side of the junction, excess minority carriers are there which recombine with majority carriers near the junction as a result, the energy is RELEASED in the form of photons. Photons with energy equal to or slightly less than the band gap are emitted. When the forward current of the diode is small, the intensity of light emitted is small. As the forward current increases, intensity of light increases and reaches a maximum. Further increase in the forward current results in DECREASE of light intensity. LED are biased such that the light emitting efficiency is maximum. The V-I characteristics of a LED is similar to that of a Si junction diode. But the threshold voltages are much HIGHER and slightly different for each colour. The reverse breakdown voltages of LED are very low, typically around 5V. So care should be taken that high reverse voltages do not appear across them. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap `E_(g)` of 1.8eV (spectral range of visible light is from about 0.4 `mu`m to 0.7`mu`m, that is from about 3 eV to 1.8eV). Hence `E_(g_(1))=(hc)/(lambda_(1)) and E_(g_(2)) = (hc)/(lambda_(2))` `=(6.625xx10^(-34)xx3xx10^(8))/(4xx10^(-7)xx1.6xx10^(-19))eV` `=3.10eV` `=3eV` and `E_(g_(2))=(hc)/(lambda_(2))=(6.625xx10^(-34)xx3xx10^(8))/(7xx10^(-7)xx 1.6xx10^(-19))eV` `=1.774 eV` `=1.8eV` The compound semiconductor Gallium Arsenidephosphide `[GaAs_(1-x)P_(x)]`is used for making LEDs of different colours. `GaAs_(0.6)P_(0.4)[E_(g)=1.9eV]` is used for red LED. GaAs `[E_(g)=1.4eV]` is used for making infrared LED. The uses of LED are in remote controls, burglar alarm systems (For prevent stealing) optical communication. Extensiveresearch is being done for developingwhite LEDs which can replace incandecent lamps. LEDs have following advantagesover conventionalincandescent low powerlamps. (1) Low operational voltageand less power. (2) Fast action and no warm-up time required. (3) The bandwidth of emitted light is 100 Å to 500 Å or in other words it is nearly monochromatic. (4) Long life and ruggedness. (5) Fast ON/OFF switching capability. |
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| 34. |
A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is : |
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Answer» `F/(2l)` `=1/2F/Axxl/LxxAL` `=1/2Fl` THUS correct CHOICE is (b). |
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| 35. |
A semiconductor has equal electron and hole concentration of 6 xx 10^(6) m^(-3) . On doping with certain impurity, electron concentration increases to 9 xx 10^(12)m^(-3). (i) Identify the new semiconductorobtained after doping. (ii) Calculate the new hole concentration. |
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Answer» SOLUTION :Here `n_(i) = 6 XX 10^(8) m^(-3)` and on doping `n_(e) = 9 xx 10^(12) m^(-3)` . (i) As on doping electron concentration has increased, hence the doped SEMICONDUCTOR should behave as n-type semiconductor. (i) We KNOW that for a doped semiconductor `n_(e).n_(h) = n_(i)^(2)` `RARR n_(h) = (n_(i)^(2))/(n_(e)) = ((6 xx 10^(8))^(2))/(9 xx 10^(12)) = 4 xx 10^(4) m^(-3)` . |
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| 36. |
Name the process involved in the formation of p-n junction diode. |
| Answer» SOLUTION :DRIFT and DIFFUSION | |
| 37. |
Draw a plot showing the variation of (i) electric field (E ) and (ii) electricpotential (V) with distance r due to a point charge Q. |
Answer» SOLUTION :
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| 38. |
In single slit experiment, the width of the slit is reduced. Then, the linear width of the principal maxima |
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Answer» DECREASES but becomes less bright |
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| 40. |
(a) write three characteristic properties of nuclear force. (b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph. |
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Answer» Solution :(a) Characteristic properties of nuclear forces are : (i) The nuclear forces are neither gravitational nor electrostatic as the former is too weak to hold nucleus together while the LATTER will blow it APART. (ii) They become negligible for distances greater than gf( 1 fermi = `10^(-15)` m (III) They do not follow inverse SQUARE law. They are short range, strong ATTRACTIVE forces. (b) N/A |
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| 41. |
The activity of a radioactive substance reduces to (1)/(32) of its initial value in 22.5 xx 10^9 yrs. Find its disintegration constant : |
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Answer» `1.54 xx 10^(-10)//yr` `A/A_(0)=[1/2]^(n) rArr 1/(32)=(1/2)^(n) rArr (1/2)^(5)=(1/2)^(n)` n=5 ALSO, `n=t/T rArr T=t/n` i.e. `T=(22.5 xx 10^(9))/(5)=4.5 xx 10^(9) yrs` |
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| 42. |
Metal wire is connected in the left gap, semi conductor is connected in the right gap of meter bridge and balancing point is found. Both are heated so that change of resistances in them are same. Then the balancing point |
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Answer» will not shift |
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| 43. |
Explain the.Conduction band. .Valence band. and .Energy gap. in semiconductors. |
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Answer» Solution :Energy GAP is finite but small. It is less than 3eV. Because of small energy gap, at room TEMPERATURE some electrons from valance band can acquire enough energy to cross the energy gap and enter the conduction band. These electrons (small in number) can MOVE in the conduction band. |
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| 44. |
Distance between two points on paper is They are observed such that view line observer is perpendicular to them. Diameter lens of an eye of an observer is 2 mm. Fo which minimum value of d both points will h observed ? |
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Answer» `1.25 cm` where d = diameter of lens of eye If DISTANCE between two points is Y and distance from eye is D. then `theta=(y)/(D)` `(y)/(D)=(lambda)/(d) rArr Y=(lambdaD)/(y) ""...(1)` `Y=(5xx10^(-7)xx50)/(2XX10^(-3))=12.5xx10^(-3)m=12.5 cm` |
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| 45. |
What is double refraction? |
| Answer» Solution :When a ray of unpolarised LIGHT is INCIDENT on a calcite crystal, two refracted rays are produced. HENCE, two IMAGES of a single object are FORMED. This phenomenon is called double refraction. | |
| 46. |
A river is flowing from the west to the east at 5m/min. A swimmer on the southerm bank can swim at 10 m/min in still water. In what direction should he swim if (i) He wishes to cross the river in minimum time (ii) He wishes to cross te river through shortest route |
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Answer» (II) `30^(@)` west of north |
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| 47. |
10g of ice at 0^@C is mixed thoroughly with 50g of water at 10^@C. The temperature of mixture will be |
| Answer» Answer :B | |
| 48. |
The primary use of a zener diode is |
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Answer» RECTIFIER |
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| 49. |
The effective resistance between A and B is the given circuit is |
| Answer» Answer :A | |
| 50. |
The binding energy of the valance electron in a Li atom in the states 2S and 2P is equal to 5.39 and 3.54 eV respectively. Find the Rydberg corrections for s and p terms of the atom. |
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Answer» <P> SOLUTION :From the RYDBERG formula we write`E_(n)= -( ħR)/((n+alpha_(l)^(2))` we use ` ħR= 13.6eV`. Then for n= 2 state `5.39= -(13.6)/((2+alpha_(0))^(2)), l=0(S)` state `alpha_(0)~~ -0.41` for `p` state `3.54= -(13.6)/((2+alpha_(1))^(2))` `alpha_(1)=-0.039`. |
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