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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. |
Sodium light of wavelength 5890 Å is incident normally on a diffraction grating of 5000 rulings per cm. Find the angle of diffraction for the second order. |
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Answer» |
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| 2. |
The magnetic field vec(dB) due to a small current element vec(dl) carrying a current I at a distance vecr is given as |
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Answer» `vec(DB) = (mu_0)/(4pi) I[(vec(DI) XX vec(R))/(r)]` |
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| 3. |
A heating element of piston resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and crosssection A as shown in diagram. The cylinder contains one mole of an ideal diatomic gas. The current flows through the element such that the temperature rises with time t as DeltaT =alpah t+1/2beta t^2 ? (alphaand betaare constants), while pressure remains constant. The atmospheric pressure above the piston is P_0. Then |
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Answer» the RATE of increase in internal energy is `5/2 R ( alpha + betat )` ` therefore` Rateof increase in INTERNALENERGY ` (dU)/( d t)= n C_V (d T)/(d t)` here ` n=1, dT=alpha t+1/2betat^2therefore(dT)/( d t)= alpha+ betat` fordiatomicgas ` C_V=5/2R ,C_P= 7/2R ` `therefore(dU )/( d t)= 1 xx5/2R xx( alpha+ betat)= 5/2R ( alpha+ betat )` heatsuppliedby theheatingelement to thegas atconstantpressure ` d Q= n C_(P )d T ` Rateof hetasupplied ` ( dQ)/(d t)= nC_P( d t)/( d t) = 1 xx (7)/(2)R xx ( alpha+ beta t )` ( using (i )) ` =7/2R ( alpha+ beta t )` LetI becurrentflowingin theelement ` thereforeI^2r =7/2R( alpha+ betat )` ` I^2 = 7/( 2 r)R (alpha + betay)impliesI= sqrt( 7/(2 r)R (alpha+ betat))` accordingto idealgas equation`PV = nRT ` At contain pressure`PdVn RT ` ` PdV= nR( alphat + 1/2beta t^2 )` workdone` w =pdV ` ` =n R( alpha t + 1/2beta t^2 )` LetF beforceexertedbythegasonthe pistonof areaA ` thereforeP = ( F)/( A)orF = pA ` letthepistonmovedupwardsby adistancex . ` thereforeW = F xorx =W /F= (nR)/(pA )( alphat +1/2beta t ^2)`( using (ii ) ) veloity` v=(dx )/ (dt)=(nR)/( PA)( alpha+ betat )` Acceleration` a = ( dv )/( dt ) = ( nR) /(PA )beta ` |
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| 4. |
A convex lens of focal length 24 cm in air is surrounded by different mediums as shown in the fig. A point object O is placed along the principle axis at a distance 30 cm from the lens. Find the number and position of the images formed. |
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Answer» 1 image, at infinity and 4 CM |
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| 5. |
A hollow cylinder is rolling on an inclined plane, inclined at an angle of 30° to the horizontal. Its speed after travelling a distance of 10 m will be? |
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Answer» 49m/sec |
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| 6. |
In a negative feedback amplifier, the gain without feedback is 100, feed back ratio is 1..25 and input voltage is 50mA. Calculate (i) gain with feedback (ii) feedback factor (iii) output voltage (iv) feedback voltage (v) new input voltage so that output voltage with feedback equals the output voltage without feedback |
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Answer» Solution :i) Gain with FEEDBACK `A_(f)=(A)/(1+beta)=(100)/(1+1//25xx100)=20` ii) `beta=(1)/(25)` iii) OUTPUT voltage `V_(0)=A_(f)V_(s)=20xx50mV="1 volt"` iv) Feedback voltage `betaV_(0)=(1)/(25)xx1="0.01 volt"` v) New increased INPUT voltage `V_(i)=V_(i)(1+beta)` `=50(1+(1)/(25)XX100)=250mV` |
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| 7. |
Two point charges 4 muC and 9 muC are separated by 30 cm. Find the point where the strength of the field is zero. |
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Answer» Solution :The distance x of the null POINT `4 mu C` charge and between the two charge is `4/(x^2) = 9/((30 - x)^2) cdot (or) x = 30/(sqrt(9/4) + 1)= (30 xx 2)/(5) = 12 cm`. |
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| 8. |
15 gram of nitrogen is enclosed in a vessel at a temperature of 300 K. |
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Answer» Work done is 0 J `(1)/(2)mv_(rms)^(2)=(3)/(2)KT` (where K is Boltzman.s constant) or `T=propv_(rms)^(2)` So if `v_(rms)` is doubled, the new temperature (T.) will be four times the initial temperature, i.e., `T.=4T`. Change in internal energy of nitrogen gas is given by `DELTAU=nC_(V)=(T.-T)` where n is the number of moles of the gas and `C_(V)` is the molar specific heat at constant volume. `THEREFORE n=(m)/(M)=(15)/(28)` [M(molecular weight of nitrogen) = 28] Also, `C_(V)=(5)/(2)R` [Nitrogen is diatomic] where R is the UNIVERSAL gas constant. `therefore DeltaU=((15)/(28))xx(5)/(2)R(4T-T)""[because T.=4T]` `=(15)/(28)xx(5)/(2)xx(8.31)((300xx4)-300)[because T=300K]` `=1.0016xx10^(4)J` Now, from first law of thermodynamics, we have `Q=DeltaU+W` Here, the gas is enclosed in a vessel `therefore DeltaV=0` Work done, `W=PxxDeltaV=0` `therefore Q=1.0016xx10^(4)+0=1.0016xx10^(4)J` |
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| 9. |
Magnetic energy stored per unit volume in a magnetic field is _____. |
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Answer» |
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| 10. |
Write the functions of the following in communication systems: (i) Transducer (ii) Repeater |
| Answer» SOLUTION : A repeater, PICKS up the SIGNAL from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier frequency. REPEATERS are used to extend the range of a communication SYSTEM. | |
| 11. |
What are dimesional formulae for magnetic moment |
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Answer» <P> SOLUTION :If `tau` acting on a MAGNET of MAGNETIC moment P when subspended in a magnetic field is :`tau = PB sintheta , P = tau/( Bsin THETA=[M^2L^2T^-2]/([M^1L^0T^-1A^-1} xx [M^0L^0T] therefore [M^0L^2T^0A^1]` |
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| 12. |
The graph between the separation of the slits and fringe width in Young.s double slit experiment is (assume that the distance between the source and the screen and the wavelength of the source are kept constant) |
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Answer» STRAIGHT LINE with NEGATIVE slope |
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| 13. |
A galvanometer has a current sensitivity of 1 mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of 500Omega and cell of internal resistance 1Omega. It gives a deflection of 5 division for shunt of 5 ohm and 20 division for shunt of 25 ohm. The emf of cell is |
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Answer» 47.1 V `I_(g) = E/((R+r) + (GS)/(G+S)) xx S/(G+S)` `therefore I_(g) = (ES)/((R+r)(G+S) + GS)` For `S= 5 "ohm", I_(g) = 5 xx 10^(-3)A` and for `S=25 "ohm", I_(g) = 20 xx 10^(-3)A` Hence, `5 xx 10^(-3) = (E xx 5)/(501(G+5) + 5G)`...........(i) and `20 xx 10^(-3) = (E xx 25)/(501(G+25) + 25G)`.........(ii) Dividing and solving, G= `88.2 Omega` From (i), we get `E=10^(-3)[501 (88.2 + 5) + 5 xx 88.2] = 47.1` volt 
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| 14. |
When a transistor is fully switch on , it is said to be |
| Answer» Solution :Saturated | |
| 15. |
A freely falling from a height h describes (7h)/(16) in the last second of its fall.The height h Is (g=10 ms^(-2)) |
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Answer» 80m |
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| 16. |
For which quantum number of the binding of electron is zero in hydrogen atom ? |
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Answer» 1 BINDING energy `=(13.6)/(N^(2))` `:.0=(13.6)/(n^(2))"":.n^(2)=(13.6)/(0)` `:.n^(2)=oo` `:.n=oo` |
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| 17. |
Graph shows relation between intensity (I),wavelength for X-rays emitted from collidge tube if minimum wavelength is lambda_(mu) and wavelength of K_(prop) line is lambda_(K).By increasing accelerating voltage |
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Answer» `lambda_(K)-lambda_(mu)` increase `therefore lambda_(mu)PROP (1)/(V)` Now INCREASING V,`lambda_(mu)` will decrease HENCE distance between `lambda_(K)` and `lambda_(mu)` will increases. |
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| 18. |
Statement I. Escape velocity of a body is independent of mass of a body. Statement II. Escape velocityof an elephant from the surface of planet is same as that of escape velocity of a mouse. |
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Answer» STATEMENT-I is TRUE, Statement-II is true and Statement-II is correct explanation for Statement-I. So correct choice is a. |
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| 19. |
When light ray travels from ...... then total internal reflection is possible. |
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Answer» from air to WATER |
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| 20. |
Mahatma Gandhi wrote different articles for |
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Answer» The Hindu |
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| 21. |
A capillary of 0.8 mm diameter is immersed in water, and rises 2 cm above the water. To what height will the water rise in it? How can the result be made consistent with the result of the previous problem? |
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Answer» SOLUTION :Water will start rising and will reach the upper end of the capillary. Here the RADIUS of curvature will decrease until the PRESSURE of the curved surface becomes equal to the hydrostatic pressure of the water column. Then the water will stop rising. The condition for EQUILIBRIUM is `Deltap=(2 SIGMA cos theta)/(r)= rho gh` For the contact angle we obtain `"cos" theta= (r rho gh)/(2sigma)=0.544, theta=57^(@)` |
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| 22. |
Complete the equation for the following fission process _92U^236 + _@n^1 to…… _38Kr^90 + …… : |
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Answer» `_50Xe^143 + 3_@n^1` |
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| 23. |
State the principle which a transformer works. Describe the woeking of a transformer with necessary theory. |
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Answer» Solution :Transformer is a device to convert a low alternating current of high voltageinto high alternating current of low voltage and VICE versa. Principle : It works on the principle of mutual induction between two coils. Working : When an alternating emf is applied across the PRIMARY COIL, the input voltage changes with time. HENCE the magnetic flux through the primary also changes with time. This changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.  Theory : Let `N_(1)` and `N_(2)` be the NUMBER of turns in the primary and secondary. Let `V_(P)` and `V_(S)` be the emf's across the primary and secondary. `(V_(S))/(V_(P))=("Out put emf")/("In put emf")=(-N_(2)(d phi)/(dt))/(-N_(1)(d phi)/(dt))=(N_(2))/(N_(1))` `therefore (V_(S))/(V_(P))=(N_(2))/(N_(1))` = Transformer ratio. Efficiency of transformer : It is the ratio of output power to the input power. `eta = ("Output power")/("Input power")xx100` |
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| 24. |
Two large on conducting plates having surface densities +sigma and -sigma, respectively, are fixed .d. distance apart. A small test charge q of mass m is attached to two non - conducting identical springs of spring constant k as shown in the adjacent fig. The charge q is now released from rest with springs in natural lenght. The q will (neglect gravity) |
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Answer» perform SHM with angular frequency `SQRT((2k)/(m))` |
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| 25. |
Assertion: In a transistor the base is made thin. Reason: A thin base makes the transistor stable. |
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Answer» If both the assertion and reason are true and reason is a true explantion of the assertion. |
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| 26. |
number of solutions for x between 3 and 15 if int_(0)^(x)[t]dt=int_(0)^([x]) tdt where [.] represents greatest integer function is - |
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Answer» 11 |
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| 27. |
If the emf of an AC source is given by 6 sin omegat + 4 sin2omegat, V then rms values of the emf is |
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Answer» `SQRT(10)`V E = 6 sin `omega t + 4 sin 2 omega t ` Now, `E^(2) = (6 sinomega t + 4 sin 2 omega t )^(2)` `therefore E^(2) = 36 sin^(2) 2 omega t +16 sin^(2) 2 omega t + 48 sin omega t . Sin 2 omega t ` `= 36 sin^(2) omega t+ 16 sin^(2) 2 omega t + 48 ( cos"" (omega t)/(2) - cos ""(3 omega t)/(2) )` Now, THEAVERAGE value of `E^(2)`, `therefore ""E_("avg") = (1)/(T) int_(0)^(T) [ E^(2) dt ] ` ` = (1)/(T) int_(0)^(T) [ 36 sin^(2) "" omega t + 16 sin^(2) 2 omega t + 48 ( (cos omega t)/(2) - (cos 3 omega t)/(2) ) ] ` ` = (1)/(T) [ 36. (T)/(2)+ 16 xx (T)/(2) + 48 (0-0) ] ` `therefore "" E_("avg") = 18 + 8 = 26 `V `therefore "" E_("rms") = sqrt( E_("avg")) = sqrt(26) V `. |
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| 28. |
An a.c. voltage given by e = 100 sin (3.14t + 60^@) is applied to a machine having a power rating of 500 W. The r.m.s. value of the current in the circuit is : |
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Answer» `10 A` |
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| 29. |
The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation ofMagnetic flux versus the current |
Answer» SOLUTION : MAGNETIC FLUX VERSUS the CURRENT 
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| 30. |
A diode rectifier |
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Answer» CONVERTS A.C. into D.C. |
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| 31. |
Assertion:At neutral point, a compass needle may point out in any arbitrary direction. Reason:Magnetic field of earth is balced by fied due to manetic at neutral point. |
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Answer» If both Assertain and Reason are TRUE and Reason is the correct explanation of Assertain |
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| 32. |
A cyclotron adjusted to give proton beam, magnetic induction is 0.15wbm^(-2) and the extreme radius is 1.5m. The energy of emergent proton in Me V will be |
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Answer» 3.42 |
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| 33. |
The current flowing through an inductor of self inductance L is continuously increasing. Plot a graph showing the variation of Induced emf versus dI/dt |
Answer» SOLUTION : INDUCED EMF VERSUS dllcli 
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| 34. |
In a Fresnel’s diffraction arrangement, the screen is at a distance of 2 meter from a circular aperture. It is found that for light of wavelengths lambda_(1)and lambda_(2)the radius of 4 thzone for lambda_(1) coincides with the radius of 5th zone for lambda_(2). Then the ratio lambda_(1) : lambda_(2) is |
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Answer» `SQRT(4//5)` |
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| 35. |
A short bar magnet is dropped through a coil of wire of similar length. Which one of the graphs below shows best how the current through the coil varies with time? |
| Answer» Solution :D. Because GRAPH D shows best how the current flows through the coil varies with TIME. It depends on the nature of CAUSE that produces it. | |
| 36. |
To change de-Broglie wavelength from 0.5xx10^(10)m to 1xx10^(-10)m its energy should be…… |
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Answer» <P>half of INITIAL KINETIC ENERGY but p =`sqrt(2mK)` `therefore` In `lambda=(h)/(sqrt(2mK)),(h)/(sqrt(2m))` same `therefore lambda prop (1)/(sqrt(K))` `therefore K prop (1)/(lambda^(2))` `therefore (K_(2))/(K_(1))=((lambda_(1))/(lambda_(2)))^(2)` `=((0.5xx10^(-10))/(1xx10^(-10)))=((1)/(2))^(2)=(1)/(4)` ` therefore K_(2)=(K_(1))/(4)` |
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| 37. |
Calculate the half-life period of a radioactive substance, if its activity drops to 16^(th) of its initial value in 30 years. |
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Answer» Solution :`N= (N_(0)) /(16)`, where, `N_(0) = 30` years `N= N_(0) ((1)/(2) )..` `(N)/( N_0) = ((1)/(2))^(4) ""[because "No. of half lives" =4]` `4= ("Time of DISINTEGRATION")/("Half LIFE period")` `RARR (30 " years") /( 4)` = half life period `therefore` Half-life period `=7.5` years |
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| 38. |
What kind of end does the poet talks about? |
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Answer» DISMAL end |
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| 39. |
A cobalt (Co) plate is placed at a distance of 1 m from a point source of power 1 W. Assume a circular area of the plate of radius, r= 1 A is exposed to the radiation and ejects photo electrons. The light energy is considered to be spread uniformly and the work function of cobalt is 5 eV. The minimum time the target should be exposed to the light source to eject a photoelectron (Assuming no reflection losses) is |
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Answer» 320 s Source power , P = I W Intensity of light source at a distance of 1m is given by `I = (P)/(4pir^2) = (1)/(4pil^2) = (1)/(4pi) W//m^2` Power absorbed by the circular plates of RADIUS 1Å is , ` P = I xx ` area of PLATE ` = (1)/(4pi) xx pi r^2 ""[ becauser = 1 Å = 10^(-10) m ]` ` P = 1/4 (10^(-10) )^2 (10^(-20) )/(4)` Work function of cobalt `W = 5 eV = 5 xx 1.6 xx 10^(-91) J` `P = W/t` `t = W/P` ` THEREFORE t = (5 xx 1.6 xx 10^(-19) )/( (10^(-20 ) )/(4) )= 320 s ` |
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| 40. |
A diatomic gas which has initial volume of 10 liter is isothermally compressed to 1//15^(th) of its original volume where initial pressure is 10^(5) Pascal. If temperature is 27^(@)C then find the work done by gas |
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Answer» `-2.71xx10^(3)J` `w=P_(0)V_(0)ln ((v_(2))/(v_(1)))` `w=10^(5)xx10xx10^(-3)ln ((1)/(15))` `w=- 2.70 xx10^(3)J` |
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| 41. |
How does self inductance of a coil depend on number of turns in the coil? |
| Answer» SOLUTION :`L PROP N^2`. | |
| 42. |
Find the magnetic induction due to a long straight conductor using Ampere's circuital law. |
Answer» Solution :Consider a straight conductor of infinite length carrying current I and the DIRECTION of magnetic field lines is shown in Figure. Since the WIRE is geometricallycylindrical in shape the wire is geometrically cylindrical in shapeand symmetrical about its axis, we construct an Amperian loop in the form of a circular shape at a distance r from the centre of the conductor as shown in Figure. From the AMPERE's law , we get `underset(C)ointvecB.vec(dl)= mu_(0)I` where `vec(dl) ` is the line element along the amperian loop ( tangent to the circular loop ). Hence, the angle between magnetic field vector and line element is zero . Therefore, `underset(C)ointBdl = mu_(0) I ` where I is the current enclosed by the Amperianloop. Due to the symmetry, the magnitude of the magnetic field is uniform over the Amperian loop, we can take B out of the INTEGRATION. `Bunderset(C)ointvec(dl) = mu_(0)I ` For a circular loop, the circumference is `2 pi r`, which implies, `Bunderset(0)overset(2 pi r)ointdl = mu_(0)I ` `vecB.2pir = mu_(0)I` `B = (mu_(0)I)/(2 pi r) ` In vector form, the magnetic field is `vecB = (mu_(0)I)/(2 pi r) hatn`where `hatn` is the unit vector along the tangent to the Amperian loop . |
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| 43. |
n alpha particles per second are emitted from N atoms of a radioactive element. Then find the half life of radioactive element ? |
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Answer» Solution :`(DN)/(DT)=-lambdaN rArr n=-LAMBDA N rArr lambda = (-n)/N` Half LIFE = `0.693/lambda=(0.693N)/n` |
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| 44. |
When a resistance of 2 Omega is placed across the terminals of a battery, the current is 0.5 A. When the resistance across the terminals of the battery is 5 Omega, the current is 0.25 A. Calculate the emf of the battery and also its internal resistance. |
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| 45. |
Two cells of 6 V and 4 V having internal resistance of 3Omega and 2Omegarespectively are connected in parallel so as to send a current through an external resistance 8Omegain the same direction. Find the current through the cells and the current through the external resistance. |
Answer» SOLUTION : From Kirchoff.s LOOP rule, To the loop ABEFA, we GET `(I_1 + I_2) 8 + 3I_1 = 6` `8I_1 + 8I_2 + 3I_1 = 6` `11I_1 + 8I_2 = 6`……(1) The loop BEDCB, we get `(I_1 + I_2) 8 + 2I_1 = 4` `8I_1 + 8I_2 + 2I_2 = 4` `8I_1 + 10I_2 = 4`.....(2)  `I_1 = 36/44` `I_1 = 0.818 A` From (2)` 8(0.818) + 10I_2 = 4` `6.544 + 10I_2 = 4` `10I_2 = 4 - 6.544` `10I_2 = -2.544` `I_2 = -0.2544A` The current through the EXTERNAL resistance `I = I_1 + I_2 = 0.818 - 0.2544` `I = 0.5636 A` |
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| 46. |
Purpose of oscillator in the AM transmitter |
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Answer» to produce MODULATING signal |
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| 47. |
Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of J (current density) changes in an exact manner, while the current I remainunaffected. The agent that is essentially responsible for is |
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Answer» 0.13 A `I_(DC)=(2I_(m))/(PI)` |
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| 48. |
The force F acting on a particle plotted against time t is shown in figure given. Its velocity v is plotted against t in the following figure. Which of these represents the resulting curve best ? |
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Answer»
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| 49. |
A radioactive sample decays with an average life of 20 ms. A capacitance 100muFis charged to some potential and then the plates are connected through a resistance R. What should be the value of R so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time? |
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| 50. |
Fiberoptic communication is gaining popularity among the various transmission media -justify. |
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Answer» Solution :Fiberoptic communication is gaining popularity among the various transmission media -justify. The method of transmitting information from one place to another in terms of light.pulses through an optical fiber is called fiber optic communication. It is in the process of replacing wire transmission in communication systems. Light has very high frequency (400 THz - 790 THz) than microwave radio systems. The fibers are made up of silica GLASS or silicon dioxide which is highly abundant on Earth Now it has been replaced with materials such as chalcogenide glasses, fluoroaluminate crystalline materials because they provide larger infrared wavelength and better transmission capability. As fibers are not electrically conductive, it is preferred in places where multiple channels are to be laid and ISOLATION is required from electrical and electromagnetic interference. Applications Optical fiber system has a number of applications namely, international communication, inter-city communication, data links, plant and traffic control and defense applications. Merits (i) Fiber cables are very thin and WEIGHT lesser than copper cables. (ii) This system has MUCH larger bandwidth. This means that its information carrying capacity is larger (iii) Fiber optic system is immune to electrical interferences. < (iv) Fiber optic cables are CHEAPER than copper cables. Demerits (i)Fiber optic cables are more fragile when compared to copper wires. (ii) It is an expensive technology |
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