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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 thin magnetic wire of length and moment is bent at its midpoint at an angle of 60^(@). The new magneticinoment after bending will be |
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Answer» `M/2` |
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| 2. |
A soap bubble (surface tension= 30 xx 10^(-3) N.m^(-1)) has radius 2 cm. The work done in doubling the radius is : |
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Answer» 0 For soap bubble increase in area `=2[4pi(2r)^(2)-4pir^(2)]=24pir^(2)` `therefore` Work done = T`xx24pir^(2)` `=30xx10^(-3)xx24xx22/7xx(2xx10^(-2))^(2)` `=4.043xx10^(-4)J` Hence correct CHOICE is (d). |
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| 3. |
Two animals belong to the same kingdom but different classes. They may belong to the same |
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Answer» phylum |
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| 4. |
A battery of e.m.f 6 V and internal resistance 1 Omega gives a p.d of 5.8 V when connected to a resistance . Find the external resistance . |
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Answer» Solution :e.m.f of battery E= 6V Internal resistance of battery R = `1 OMEGA` p.d across the resistance V= 5.8 V Let the external resistance be .R. E= V+ IR E`=V+V/R r` `r/R = (E-V)/(V)` `R= (Vr)/(E-V) = (5.8xx1)/(6-5.8) =(5.8)/(0.2) = 29 Omega` |
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| 5. |
A : For an object in rolling motion rotational kinetic energy is always equal to translational kinetic energy.R : For an object in rolling motion magnitude of linear speed and angular speed are equal. |
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Answer» If both ASSERTION & Reason are TRUE and the reason is the correct explanation of the assertion, |
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| 6. |
A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why? |
| Answer» SOLUTION :A HIGH tension SUPPLY should have high internal resistance. OTHERWISE, if the supply isaccidentally short circuited, the circuit current will be exceedingly high (exceeding the safety limit) which may damage the appliances ETC. | |
| 7. |
In a silicontransistor, a change of 7.89 mA in the emitter current produces a change of 7.89 mA in the collector current. What change in the base current will bring the same change in the collector current ? |
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Answer» |
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| 8. |
Plot the Carnot cycle in the T-S coordinates and calculate its efficiency. |
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Answer» Solution :Plot the GRAPH of the CARNOT CYCLE in the T-S variables (FIG.) `ETA = (Q_1 -Q_2)/(Q_1) = (T_1 (S_2 - S_1) - T_2(S_2-S_1))/(T_1(S_2 - S_1)) = (T_1 - T_2)/(T_1)` . 
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| 9. |
The sources are said to _____ if they have a constant or zero phase difference between them. |
| Answer» SOLUTION :COHERENT | |
| 10. |
The temperature of filament of bulb is 500 K and amount of heat radiated is E the filament is supposed to be a black body, the amount of heat radiated when the temperature of filament is 1000 K is : |
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Answer» 16 E `E_(2)=E((1000)/(500))^(2)""E_(2)=16E` Thus correct choice is (a). |
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| 11. |
How far above the horizon is the moon when its image reflected to calm water is completely polarized (mu_("water") = 1.33) ? |
| Answer» SOLUTION :`30^@` above the HORIZON | |
| 12. |
Name the process responsible for energy production in the sun. |
| Answer» Solution :FUSION of four HYDROGEN NUCLEI into helium NUCLEUS. | |
| 13. |
Monochromatic light of wavelength 589 nm is incident from air on a water surface. If refractive index for water is 1.33, find the wavelength, frequency and speed of the refracted light. |
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Answer» Solution :Here wavelength of incident light in air `lambda = 589 nm = 589 xx 10^(-9)` m, and speed of light in air `c = 3 xx 10^(8) ms^(-1)` `therefore` FREQUENCY of light `v=c/lambda = (3 xx 10^(8))/(589 xx 10^(-9)) = 5.09 xx 10^(14) Hz` `therefore` REFRACTIVE INDEX of water n=1.33 `therefore` Wavelength of REFRACTED light `v. = v = 5.09 xx 10^(14)` Hz and speed of refracted light `v=c/n =(3 xx 10^(8))/1.33 = 2.26 xx 10^(8) ms^(-1)` |
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| 14. |
The microspores, as they are formed, are arranged in a cluster of four cells-the microspore tetrad. As the anthers mature and dehydrate, the microspores dissociate from each other and develop |
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Answer» POLLEN grains |
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| 15. |
For which 2 reasons the discussion of LC oscillations is not realistic ? |
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Answer» Solution :(i) Every inductor has some resistance. The effect of this resistance is to introduce a damping effect on the charge and current in the circuit and the OSCILLATIONS finally die AWAY. (ii) Even if the resistance were ZERO, the total energy of the system would not remain constant. It is radiated away from the system in the form of electromagnetic wave. In face, radio and TV TRANSMITTERS depend on this radiation. |
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| 16. |
B along the axis of a long solenoid is given by |
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Answer»
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| 17. |
What is Lenz force ? |
| Answer» Solution :When rod is moved perpendicular to MAGNETIC FIELD, current is induced in it and DUE to it force BIL is induced in the opposite to velocity. Now to move rod with constant velocity we need to apply same force which is called LENZ force. | |
| 18. |
The first overtone of a open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the open pipe. (Speed of sound v = 330 m/s) |
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Answer» `l_0 = 3.8937 m` |
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| 19. |
Force on a 1 kg mass on earth of radius R is 10 N. Then the force on a satellite revolving around the earth in the meanorbital radius 3R//2 will be (mass of satellite is 100 kg): |
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Answer» `4.44xx10^(2)N` When radius of satellite`r=(3)/(2)R`. Then `F.=(GMm.)/(r^(2))=(GMxx100)/(((3)/(2)R)^(2))=(GM)/(R^(2)).(100)/(9)xx4=10xx(400)/(9)` `=4.4xx10^(2)N`. THUS correct choice is (a). |
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| 20. |
A rocket is receding away from earth with velocity 0.2C. The rocket emits signal of frequency 4 xx 10^7 Hz . The apparent frequency of the signal produced by the rocket observed by the observer on earth will be |
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Answer» `3 XX 10^6Hz` |
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| 21. |
A flywheel rotating about a fixed axis has a K.E. of 360 J, when its angular speed is 30 rad s^(-1). The M.I. of the flywheel about the axis of rotation is : |
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Answer» `0.15kgm^(2)` implies `I=(720)/(900)=0.8kgm^(2)` |
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| 22. |
As the speed of the fan increases, current: |
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Answer» increases |
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| 23. |
Harmonic vibrations comply to the law s=0.20cos(300t+2) Find the amplitude, the frequency, the period and the initial phase of the vibrations. |
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Answer» |
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| 24. |
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing continuously. |
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Answer» Solution :(i) CLOCKWISE in 1 (II) Anticlockwise in 2 |
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| 25. |
Natural light falls at the Brewster angle on the surfcae of glass. Using the Fresnel equations, find (a) the reflection coefficient, (b) the degree of polarization of refracted light. |
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Answer» Solution :From Fresnel's equations `{:(I'_(bot)=I_(bot)(sin^(2)(theta_(1)-theta_(2)))/(sin^(2)(theta_(1)+theta_(2)))),(I'_(||)=0):}}` at Brewste's angle `I'_(||) = 0` `I'_(bot) = I_(bot) sin^(2)(theta_(1) - theta_(2))` `= (1)/(2)I (sin theta_(1) COS theta_(2) - cos theta_(1) sintheta_(2))^(2)` Now `tan theta_(1) = N, sin theta_(1) = (n)/(sqrt(n^(2) + 1))` `cos theta_(1) = (1)/(sqrt(n^(2) + 1)), sintheta_(2) = cos theta_(1)` `cos theta_(2) = sin theta_(1)` `I'_(bot) = (1)/(2)I ((n^(2) - 1)/(n^(2) + 1))^(2)` Thus reflection coefficient `= RHO = (I'_(bot))/(I)` `= (1)/(2) ((n^(2) - 1)/(n^(2) + 1))^(2)0.074` on putting `n= 1.5` (b) For the REFRACTED light `I''_(bot) = I_(bot) - I'_(bot) = (1)/(2)I {1-((n^(2) - 1)/(n^(2) + 1))^(2)}` `= (1)/(2)I (4n^(2))/((n^(2) + 1)^(2))` `I'_(||) = (1)/(2)I` at the Brewster's angle. Thus the degree of polarization of the refraced light is `P = (I''_(||) - I''_(bot))/(I''_(||) + I''_(bot)) = ((n^(2) + 1)^(2)-4n^(2))/((n^(2) + 1)^(2) + 4n^(2))` `= ((n^(2) - 1)^(2))/(2(n^(2)+1)^(2) -(n^(2) -1)^(2)) = (rho)/(1-rho)` On putting `rho = 0.074` we get `P = 0.080`. |
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| 26. |
Sate the law gives the polarity of the induced emf. |
| Answer» SOLUTION :Lenz's law the polarity of the induced emf is such that it TENDS to produce CURRENT which oppose the change in MAGNETIC flux that PRODUCES it. | |
| 27. |
When metal surface is incident with photon after how much time photoelectron are emitted? |
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Answer» `10^(-1)S` |
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| 28. |
The rays of the sun are focused on a piece of ice through a lens of diameter 5 cm , as a result of which 10g ice melts in 10 minutes. The amount of heat received from the sun per unit area per minute is |
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Answer» `4 "CAL cm"^-2min.^(-1)` |
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| 29. |
In Young's double slit experiment , describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit . Hence obtain the expression for the fringe width. OR Describe Young's double slit experiment to produce interference pattern due to a monochromatic source of light. Deduce the expression or the fringe width. (b) The ratio of the intensities at the minima to the maxima in the Young's double slit experiment is 9 : 25 . Find the ratio of the widths of the two slits. |
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Answer» Solution :(a) (b) We have `(I_(max))/(I_(MIN))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=(25)/(9)` `:.(a_(1)+a_(2))/(a_(1)-a_(2))=(5)/(3)implies(a_(1))/(a_(2))=(4)/(1)` Then ratio of width `omega_(1)` & `omega_(2)` of two SLITS is given by `:.(omega_(1))/(omega_(2))=(I_(1))/(I_(2))=((a_(1))^(2))/((a_(2))^(2))=(16)/(1)` |
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| 30. |
If hati denotes a unit vector along an incident ray hatr a unit vector along the refracted ray into a medium of refractive index mu and hatn a unit vector normal to the boundary of the media directed towards the incident medium, then the law of refraction can be written as |
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Answer» `hati.hatn=MU(hatr.hatn)` |
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| 31. |
Charge density of the given surface is s. Then electric field stranght at the centre (of quarter sphere) is |
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Answer» `(SIGMA)/(4sqrt(2epsilon_(0)))` |
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| 32. |
Given figure shows the distance-time graph of the motion of a car. It follows from the,| graph that the car is |
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Answer» at rest |
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| 33. |
f(x)=(3-x^3)^1/3 द्वारा परिभाषित फलन F:RrarrRहै, तब (fof)X है |
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Answer» `X^1/3` |
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| 34. |
In condition of stopping potential value of photo-electric current will be…… |
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Answer» Zero |
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| 35. |
A rock is dropped off a cliff that's 80 m high. If it strikes the ground with an impact velocity of 40 m//s, what acceleration did it experience during its descent ? |
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Answer» SOLUTION :If something is dropped, then that MEANS it has no initial velocity : `v_(0)= 0`. So, we're given `v_(0), DELTA s` and v, and we're asked for a. Since t is missing, we USE Big Five #5. `v^(2)= v_(0)^(2)+2a Delta s implies v^(2)= 2a Delta s ("since " v_(0)=0)` `a = (v^(2))/(2 Delta s)=((40 m//s)^(2))/(2(80 m))= 10 m//s^(2)`. |
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| 36. |
In hydrogen atom, the electron is moving round the nucleus with velocity 2.18 xx 10^(6) m//s is an orbit of radius of 0.528 Å. The acceleartion of the electron is |
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Answer» `9 xx 10^(18)m//s^(2)` Given `v=2.18 xx 10^(6) m//s`, `r=0.528 Å=0.528 xx 10^(-10)m` `a=((2.18 xx 10^(6))^(2))/(0.528 xx 10^(-10))=9 xx 10^(22) m//s^(2)` |
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| 37. |
A long straight conductor carrying a current of 2 A is in parallel to another conductor of length 5 cm and carrying a current 3A. They are separated by a distance of 10 cm. Calculate a) B due to first conductor at second b)the force on the short conductor. |
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Answer» Solution :Given `i_1 = 2 A , i_2 =3 A , r =10 cm = 10 xx 10^(-2) m , l_2 = 5 cm ` a) `B = (mu_0 i_1)/(2pi r) = 2 xx 10^(-7) xx (2)/(10 xx 10^(-2)) = 4 xx 10^(-6)` TESLA `b) F = (mu_0 i_1 i_2)/(2pir) xx l_2 = 2 xx 10^(-7) xx (2 xx 3)/(10 xx 10^(-2)) xx 5 xx 10^(-2)` `= 6 xx 10^(-7)` N |
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| 38. |
A tritrium gas target is bombared with a beam of monoenergetic protons of kineticenergy K_(1) =3 MeV The KE of the neutron emiited at 30^(@) to the inicdent beam is K_(2) ? Find the value of K_(1)//K_(2) (approximately in whole number). Atomic masses are H^(1) =1.007276 amu, n^(1)=1.008665 amu, ._1 H^3=3.016050 amu, ._2He^3=3.016030 amu. |
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Answer» `Q=-1.2745 MEV` Using CONSERVATION of linear momnetum, we get `K_(2) =1.44 MeV, K_(1) =3 MeV` . |
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| 39. |
In an npn transistor circuit , the collector current is 20 mA . If 90%of the electrons emitted reach the collector |
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Answer» the EMITTER current will be9mA |
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| 40. |
For the circuit of Fig. 30-29, assume that ~ = 10.0 V, R = 112 Omega, and L = 5.50 H. The ideal battery is connected at time t = 0. (a) How much energy is delivered by the battery during the first 2.00 s? (b) How much of this energy is stored in the magnetic field of the inductor? (c) How much of this energy is dissipated in the resistor? |
| Answer» SOLUTION :(a) 1.74 J, (B) `2.19 XX 10^(-2)` J , (C) 1.72 J | |
| 41. |
Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II. |
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Answer» |
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| 42. |
In a car a convex mirror of focal length 1 cm is being used as rear view mirror. A car of height 5 m and width 2.5 m behind this car, (i) what will be the size and position of the image of the second car on the viewfinder of the first car? (ii) If the second car tries to overtake the first car with a relative velocity 20 m* s^(-1) what will be the magnitude and direction of the velocity of the image? |
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Answer» |
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| 43. |
A body in linear S.H.M. has maximum velocity, when it crosses the mean position. What is its velocity when it is midway between the mean position and the extreme position ? |
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Answer» SOLUTION :`v=omegasqrt(a^2-x^2) = omegasqrt(a^2-(a/2)^2)` `=omegasqrt(a^2-a^2/4)=omegasqrt((3a^2)/4)=sqrt3/4a OMEGA` |
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| 44. |
What we call the spectra produced by substances in atomic state ? |
| Answer» SOLUTION :LINE SPECTRA | |
| 45. |
Making use of the formulation and the soulution of the foregoingproblem, findthemagnitudefo the force exertedby the charges bound on the surfaceof the dielectricon the pointcharge q. |
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Answer» Solution :The force on the POINT CHAGRE `q` is due to thebound chagres. This can be calculated from the field at thischarge after extracing out the SELF field. This image field is `E_("image") = (epsilon -1)/(epsilon +1) (q)/(4pi epsilon_(0) (2L)^(2))` Thus, `F = (epsilon - 1)/(epsilon +1) (q^(2))/(16pi epsilon_(0) l^(2))` |
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| 46. |
The applied a.c power to a half-wave rectifier is 200W. The d.c power output obtained is 50W. The rectification efficiency is |
| Answer» ANSWER :B | |
| 47. |
A balloon is rising vertically up with a velocity of 29 ms^(-1). A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it is |
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Answer» 400 m Here `h= -ut+(1)/(2)g t^(2)` `=-29xx10+(1)/(2)xx9.8xx100` `=-290+490=200 m` |
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| 48. |
A stretched string is taken and stretched such that elongated by 1% then the fundamental frequency decreased by x xx10^(-1) %what is the value of x ? |
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Answer» |
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| 49. |
The velocity of a particle performing S.H.M. at mean position is, |
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Answer» Maximum |
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| 50. |
Alpha ray...........led of the discovery of.......... . |
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Answer» |
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