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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. |
Ten dipoles ,each of dipole moment 'p' , are placed inside a hollow sphere . The total electric flux coming out of the sphere will be |
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Answer» `0` |
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| 2. |
How much positive charge should be given to the earth so as to have the same potential as that of a positively charged sphere of1 muC and radius 1 cm . ( Radius of earth = 6400km ) |
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Answer» 600 C |
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| 3. |
A plane wavefront in incident normally on a circular aperture and diffraction pattern is obtained on a screen on another side of aperture. On displacing the screen towards aperture, the number to half period zone ezposed through the aperture |
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Answer» increases |
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| 4. |
(a) Draw a labelled ray diagram of an astronomical telescope to show the image formation of a distant object. Write the main consideration required in selecting the objective and eyepiece lenses in order to have large magnifying power an dhigh resolution of the telescope. (b) A compound microscope has an objective of focal length 1.25cm and eyepiece of focal lenth 5 cm. A small object is kept at 2.5 cm fromthe objective. If the final image formed is at infinity, find the distance between the objective and the eyepiece. |
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Answer» Solution :(a) For diagram Alternatively ,  For large magnifying power `f_(0) `should be large and `f_(E)` should be small. For higehr resolution DIAMETER of the objective should be large. (b) `(1)/( v_(0))- ( 1)/( U _(0)) = ( 1)/( f_(0))` `(1)/( v_(0)) = ( 1)/( f_(0)) +( 1)/( u_(0)) = ( 1)/( 1.25) - ( 1)/( 2.5) = ( 1)/( 2.5)` `v_(0) = 2.5 CM` ` L = | f_(0)| +| f_(e)|``( :. v_(0) = f_(0) )` `= ( 2.5 + 5.0 ) cm = 7.5 cm` |
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| 5. |
The circuit shown in the following figure is in a steady state with the switch open. When the switch is closed, which of the following will change immediately? (All of them eventually change, but three of them stay the same for an instant.) |
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Answer» The potential DIFFERENCE around the capacitor C |
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| 6. |
Does the stopping potential in photo electric emission depends upon The frequency of incident radiation ? |
| Answer» SOLUTION :YES, the STOPPING potential depends on the frequency of incident radiation, above the threshold frequency `V_0 prop v`. | |
| 7. |
Show that displacement current is equal to conduction current during chargeing of a capacitor. |
Answer» Solution :LET at an instant magnitude of charge on the plates of capacitor be Q. Area of each plate is A. Electric field between the plates of capacitor E = `(Q)/(A epsilon_(0))`........ (1) Flux of this passing through the suface between the plaates is ` phi_(E ) = E xx A = (Q)/(A epsilon_(0)) xx A ` `phi_(E ) = (Q)/(epsilon_(0))` ......... (ii) Displacement current `i_(d)` is `i_(d) = epsilon_(0) xx (d phi_(epsilon))/(dt ) = epsilon_(0) (d)/(dt) ((Q)/(epsilon_(0))) = epsilon_(0) xx (1dQ)/(epsilon_(0)dt)` `I_(d) = (DQ)/(dt)` ....... (iii) `(dQ)/(dt)` is the rate at which charge is reaching to POSITIVE plate of capacitor through conducting wire therefore `I_( c) = (dQ)/(dt)` ... (iv) From equations (iii) and (iv) `I_(d) = I_(c )` |
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| 8. |
A U-tube filled with a liquid of volumetric expansion coefficient 10^(-5)//^(@)C lies in a vertical plane. The height of liquid column in the left vertical limb is 100 cm. The liquid in the left vertical limb is maintained at a temperature =^(@)C while the liquid in the right limb is maintained at a temperature =100^(@) C The difference in levels in the two limbs is |
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Answer» `0.1` CM |
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| 9. |
The fusion reaction in the sun is a multi-step process in which the |
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Answer» helium is BURNED into deuterons |
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| 10. |
The following figure shows the intensity-wavelength relations of x-rays coming from two different Coolidge tubes. The solid curve represents the relation for tube A in which the potential difference between the target and the filament is V_(A) and the atomic number of the target material is Z_(A). These quantities are V_(B)andZ_(B) for the other tube. Then |
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Answer» `V_(A)gtV_(B),V_(A)gtZ_(B)` |
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| 11. |
A narrow beam of identicalions with specific chargeq//m, possessing differentvelocities, enters the region of space, where thereare unifromparallel electric and marneticfields the strengthE and induction B, at the pointO (see Fig). The beam directioncoincides with thex axisat the point O. A plane screenoriented at rightanglesto the x axisis locatedat a distance l fromthe point O. Find theequation of the tracethat theions leaveon the screen. Demonstrate that at x lt lt l it is theequactionof a parabola. |
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Answer» Solution :The equcation of the trajectory is, `x = (v_(0))/(OMEGA) SIN omega t, z = (v_(0))/(omega) (1 - cos omega t), y = (qE)/(2m) t^(2)` as before SEE (3.384) Now on the screen `x = l`, so `sin omega t = (omega l)/(v_(0))` or, `omega t = sin^(-1) (omega l)/(v_(0))` At that moment, `y = (qE)/(2m omega^(2)) ("sin"^(-1)(omega l)/(v_(0)))^(2)` so,`(omega l)/(v_(0)) = sin sqrt((2m omega^(2) y)/(qE)) = sin sqrt((2q B^(2) y)/(Em))` and`z = (v_(0))/(omega) 2 "sin"^(2) (omega t)/(2) = l "tan" (omega t)/(2)` `= l "tan" (1)/(2) ["sin"^(-1) (omega I)/(v_(0))] = l "tan" sqrt((qB^(2) y)/(2 mE))` For SMALL `z, (qB^(2) y)/(2 mE) = ("tan"^(-1) (z)/(l))^(2) = (z^(2))/(l^(2))` or, `y = (2 mE)/(q B^(2) l^(2)). z^(2)` is a parabola. |
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| 12. |
In electromagnetic spectrum ……… the light has lowest wavelength. |
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Answer» Ultraviolet |
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| 13. |
Given below are four logic gate symbols. Those for OR, NOR and AND are respectively |
| Answer» Answer :D | |
| 14. |
Figure below shows tracks of three charged particles in a uniform clectrostatic field. Give the signs of the three charges. Which partiele has the highest charge to mass ratio? |
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Answer» SOLUTION :Charges 1 and 2 are negative, CHARGE 3 is positive. Particle 3 has the HIGHEST charge to mass ratio. |
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| 15. |
An air craft flying horizontally at 360km/hour releases a bomb ( with no velocity relative to air craft) at a stationary tank 200 metre away , what must br the height to the aircraft above the tank, if the bomb is to hit the tank ? |
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Answer» 15.6 m |
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| 16. |
The type of modulation technique using frequency range 88 MHz to 108 MHz |
| Answer» SOLUTION :FM (FREQUENCY MODULATION) | |
| 17. |
A beam of 35.0 keV electrons strikes a molybdenum target, generating the X-rays. What is the cutoff wavelength ? |
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Answer» `35.5 pm` From relation `lambda_(min) = (HC)/(K_(0)) = ((4.14 xx 10^(-5))(3xx10^(8)))/(35.0 xx 10^(3))` `= 3.55 xx 10^(-11) = 35.5 pm`. |
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| 18. |
Define - Efficiency of rectifier. |
| Answer» Solution :EFFICIENCY (N) is the RATIO of the output de power to the AC input power supplied to the circuit. | |
| 19. |
One end of a string of length l is tied to the ceiling ofa lift accelerating upwards with an acceleration g/2. The linear mass density of the string is mu(x)=mu_0x^(1//2)where, x is measured from the bottom. The time taken by a pulse to reach from bottom to top is |
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Answer» `sqrt((3l)/g)` WEIGHT of this element `=[mu(x)dx](g+g/2)` Tension at this element , `T_x=int_0^x 3/2 mu(x) gdx` `rArr T_x=3/2 g int_0^x mu_0 x^(1//2) dx` `=(3gmu_0)/2x^(3//2)/(3//2) = g mu_0x^(3//2)` VELOCITY of TRANSVERSE wave at this element `v=sqrt(T/mu) rArr v=(dx)/(dt) = sqrt((gmu_0x^(3//2))/(mu_0x^(1//2)))=sqrt(gx) rArr dt= (dx)/sqrt(gx)` INTEGRATING both sides , we get `int_0^t dt = int_0^l (dx)/sqrt(gx)rArr t= 1/sqrtg int_0^l x^(-1//2) dx rArr t=2 sqrt(l/g)` 
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| 20. |
A long straight wire with current i passes (without touching) three square wire loops with edge lengths 2L, 1.5L and L. The loops are widely spaced (so as do not affect one another). Loops 1 and 3 are symmetric about the long wire. Rank the loops according to the size of the current induced in them if current i is (a) constant and (b) increasing. |
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Answer» SOLUTION :(a) No INDUCED CURRENT (b) Current will be induced only in loop 2. |
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| 21. |
Kirchhoff.s first and second laws of electrical circuits are consequences of ......(1)....and (2)….. respectively |
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Answer» CONSERVATION of MOMENTUM,conservation of charge |
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| 22. |
A 15 mu F capacitor is connected to a 220V, 50Hz a.c source Value of capacitance reactance is …………Omega |
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Answer» 106 |
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| 23. |
At which place direction of earth's magnetic field is horizontal ? |
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Answer» MAGNETIC pole. |
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| 24. |
Electrical multimeters, or multitesters, are widely used by technicians when working with electrical and electronic circuits. In this instrument, a single milliammeter connected through different resistances is used to measure currents, potential differences and resistance over different ranges. Current always enters the multimeter at the same terminal, A, and then passes through the milliammeter as well as through other resistances, placed in series with or parallel to it. The choice of the terminal at which current leaves the instrument decides its role (ammeter, voltemeter, etc.) and its range (maximum current or voltage whichit can measure). The milliammeter shown in the circuit has a coil (or internal ) resistance of 0.9 omega and gives the full-scale deflection for a current of 10mA. IfA and D are used as the terminals of the multimeter , i.e., current eneters atA and leaves at D, it will function as a voltmeter of which range? |
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Answer» `1 V` |
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| 25. |
Electrical multimeters, or multitesters, are widely used by technicians when working with electrical and electronic circuits. In this instrument, a single milliammeter connected through different resistances is used to measure currents, potential differences and resistance over different ranges. Current always enters the multimeter at the same terminal, A, and then passes through the milliammeter as well as through other resistances, placed in series with or parallel to it. The choice of the terminal at which current leaves the instrument decides its role (ammeter, voltemeter, etc.) and its range (maximum current or voltage whichit can measure). The milliammeter shown in the circuit has a coil (or internal ) resistance of 0.9 omega and gives the full-scale deflection for a current of 10mA. If A and B are used as the terminals of the multimeter, i.e., current enters at A and leaves at B, it will function as an ammeter of range |
| Answer» ANSWER :B | |
| 26. |
A coil of 40Omega resistance has 100 turns and radius 6 mm is connected to ammeter of resistance of 160 ohms. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, 32 muC charge flows through it. The magnetic induction field strength is |
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Answer» 6.55 T |
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| 27. |
Electrical multimeters, or multitesters, are widely used by technicians when working with electrical and electronic circuits. In this instrument, a single milliammeter connected through different resistances is used to measure currents, potential differences and resistance over different ranges. Current always enters the multimeter at the same terminal, A, and then passes through the milliammeter as well as through other resistances, placed in series with or parallel to it. The choice of the terminal at which current leaves the instrument decides its role (ammeter, voltemeter, etc.) and its range (maximum current or voltage whichit can measure). The milliammeter shown in the circuit has a coil (or internal ) resistance of 0.9 omega and gives the full-scale deflection for a current of 10mA. If A and C are used as the terminals of the multimeter, i.e., current enters at A and leaves at C, it will function as an ammeter of range |
| Answer» ANSWER :C | |
| 28. |
A conducting rod held horizontally along East-West direction is dropped from rest at certain height near Earth’s surface. Why should there be an induced e.m.f. across the ends of the rod? Draw a graph showing the variation of e.m.f. as a function of time from the instant it begins to fall. |
Answer» SOLUTION :`E = B//V and v = "GT"` 
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| 29. |
The frequency range of band cast band (medium) wave is from _____ to _____. |
| Answer» SOLUTION :1550 KHZ, 1600 kHz | |
| 30. |
If a charge is enclosed by the surface of the sphere then total flux emitted from the surface will be: |
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Answer» (the CHARGE ENCLOSED by surface)/`epsi_(0)` |
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| 31. |
From the following equations pick out the pssible nuclear fusion reactions (i) ._(5)C^(13) +._(1)H^(1)rarr._(6)C^(14) + 4.3MeV (ii) ._(6)C^(12) +._(1)H^(1)rarr._(7)C^(13) + 2MeV (iii) ._(7)C^(14) +._(1)H^(1)rarr._(8)C^(15) + 7.3MeV (iv) ._(92)^(235)U +._(0)^(1)n rarr ._(54)^(140)Xe +._(38)^(94)Sr +2._(0)^(1)n + gamma + "energy" |
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Answer» `(i),(II)` |
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| 32. |
An electron does not suffer any deflection while passing through a region. Are you sure that there is no magnetic field ? |
| Answer» SOLUTION :No, because the ELECTRON SUFFERS no force when it moves parallel or antiparallel to MAGNETIC field. So the magnetic field may exist parallel or antiparallel to the motion of electron. | |
| 33. |
A radio active material decreases by simultaneous emissions of two particles with half lifes 1620 and 810 years. The time after which 1/4 of the material remained is |
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Answer» 1080 YEARS |
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| 34. |
Modified Ampere's circuital law state ointvecB.vecdt= _____. |
| Answer» SOLUTION :`I_C + epsilon_0 (dphi_E)/DT` | |
| 35. |
Photons emitted by a gas consisting of excited hydrogen like atoms (A) during a transition from a higher quantum state (quantum no. n) to a lower quantum state (quantum no :m ) are incident on a metallic surface (B) causing the emission of photoelectrons . The fastest photoelectrons pass undeviated through a region consisting of electric field E_0 =3.7 V/cm and magnetic field B_0 =10^3T , oriented in perpecdicular directions and the photoelctrons enter the region perpendicular to both the electric and magnetic field The threshold wavelength for the metal B equal 830 nm. the spectrum of radiations emitted by the excited hydrogen -like atoms (A) consists of 15 different wavelengths . Find the quantum numbers of the states n,m and atomic number (Z) of the element (A) . (Take 1.89//13.6~~1//7) |
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Answer» |
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| 36. |
Mention the types of charges produced on each when (i) glass rod is rubbed with silk and (ii) an ebonite rod is rubbed with wool. |
| Answer» SOLUTION :on EBONITE ROD - ve CHARGE and on WOOL + ve charge. | |
| 37. |
If vecA=3hati+4hatj and vecB=7hati+24hatj find a vector having thesame magnitude as vecB and parallel to vecA. |
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Answer» Solution :Now `vecAxxvecB`=(hati+2hatj+3hatk)XX(2hati-hatj+hatk)=-hatk-hatj-4hatk+2hatj+6hatj+3hati=5hati+5hatj-5hatk` `abs(vecAxxvecB)=Absintheta` we get `SIN theta = abs(vecAxxvecB)//AB` SINCE A=`sqrt(1+4+9)=SQRT14` and B=sqrt(4+1+1)=sqrt6` |
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| 38. |
An EM wave of intensity ‘I’ falls on a surface kept in vacuum. What is the radiation pressure if wave is totally reflected? |
| Answer» SOLUTION :`21/C,c to `SPEED of LIGHT | |
| 39. |
Electromagnetic waves can be propagated throught conductor. Is the above statement true or false ? Explain. |
| Answer» Solution :True. electric and MAGNETIC FIELD VARIATION can be propagated through a CONDUCTOR. | |
| 40. |
name the spectral series of hydrogen which lies in the ultraviolet region of elecrongnetic spectrum . |
| Answer» SOLUTION : LYMAN SEREIS. | |
| 41. |
The density of diamond, a form of carbon, is 3520 kg/m^3 If the atomic mass of carbon is 12.011 u, how many carbon atoms are there in a solid diamond sphere with a radius of 0.033 m? |
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Answer» `5.16 XX 10^(26) ` ATOMS |
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| 42. |
See the electrical circuit shown in Fig. Which of thefollowing equations is a correct equation for it ? |
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Answer» `epsi_1 - (i_1 + i_2 )R - i_1 r_1 = 0` |
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| 43. |
Spherical aberration in a thin lens can be reduced by |
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Answer» using a MONOCHROMATIC light |
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| 44. |
If vecP+vecQ=vecR and vecP-vecQ= vecS then R^(2)+S^(2) is equal to |
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Answer» `P^(2)+Q^(2)` |
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| 45. |
In YDSE ten maxima of two slit pattern (Interference) if formed in central maxima of single slit pattern diffraction. If separation in between two slits is 1mm, then what is the width of each slit ? |
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Answer» 0.2 mm |
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| 46. |
A slit or an aperture diffracts light. Even then we say light travels in a straight lineand ray optics is valid. Comment. |
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Answer» Solution :We have studied that a slit of WIDTH a diffracts a beam of light. Half angular width a central maximum is `theta ~~ sin theta = (lambda)/(a)` This is the half angular size of BRIGHT central maximum. In travelling a distance `Z`, the diffracted beam, therefore, acquires a width `Z lambda//a` just due to diffraction. This width will become more than width of slit, i.e. `(Z lambda)/(a)ge a`, when `Z ge (a^(2))/(lambda)` We call `(a^(2))/(lambda) = Z_(F)`, the Fresnel's distance , For example, for human EYE, aperture `a = 3 mm = 3 XX 10^(-3) m and lambda = 5000 Å = 5 xx 10^(-7)m` (for visible light) `Z_(F) = (a^(2))/(lambda) = (3 xx 10^(-3))^(2)/(5 xx 10^(-7)) = 18m` This means width of beam due to diffraction does not become more than `3 mm` unless distance travelled by the beam is more than `18m`. Therefore, we can negelct broadening of beam by diffraction upto distance as large as a few meters. Hence ray optics is valid in all common situations we comeacross. |
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| 47. |
What is her work known for? |
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Answer» for its ORIGINALITY, VERSATILITY and FLAVOUR of the soil |
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| 48. |
Find the capacitance of a system of two identical metal balls of radius . a . if the distance between their centres in equal to .b., with to b gt gtaThe system is located in a uniform dielectric with relative permittivity E. |
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Answer» `C=piepsilon_(0)epsilona` |
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| 49. |
Which of the following diodes is operatedin reverse bias mode ? |
| Answer» Answer :A | |
| 50. |
Define self-inductance of a coil and hence write the definition of 'Henry'. |
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Answer» Solution :Self induction is the phenomenon on production of induced emf in a coil when changing current passes through it. The self-inductance of a coil is said to be one Henry if an induced emf of one volt is SET up in it when the current in it changes at the rate of one amphere per second. 1. Number of turns : Larger the number of turns in the solenoid, largerf is its self- inductance. 2. Area of cross-section : Larger the area of cross-section of the solenoid. larger is self -inductance. (ii) The mutal inductance of two solenoids depends on their geometry and the magnetic permeability of the CORE material. 1. Number of turns : Larger the number of turns in the two solenoids, larger will be their inductance. 2. Relative sepatation : Larger the distance between two solenoids, SMALLER will be the magnetic flux LINKED with the secondary coil due to current in the primary coil. Hence, smallerf will be the value of M. |
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