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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. |
The charging current for a capacitor is 0.2A. What is the displacement current? |
| Answer» SOLUTION :REMAIN same `I_C=I_D` | |
| 2. |
An infinite, uniformly charged sheet with surface density sigma cuts through a spherical Gausian surface of radius R at a distance x from its center, as shown in the figure. The electric flux Phi through the Gaussian surface is (kpi(R^(2)-x^(2))sigma)/(epsilon_(0)) find the vlaue of k? |
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Answer» |
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| 3. |
The concept of stationary (non-radiating) orbits was proposed by |
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Answer» J.J. THOMSON |
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| 4. |
Carbon and silicon both have four valence electrons each. How then are they distinguished ? |
| Answer» SOLUTION :The four vaJence electrons in carbon are in the SECOND orbit but in silicon these are in third orbit. Hence, energy required to TAKE out an electron from carbon is `E_(8) = 5.4` eV but for silicon `E_(8)` = 1.1 eV. As a result carbon is an insulator but silicon is a semiconductor. | |
| 5. |
Even though the drift velocity of the electrons is very small,a bulb burns almost at the same instant when the switch is ON.Why? |
| Answer» Solution :As soon as the switch is ON, electric field is set up throughout the CONDUCTOR ALMOST instantly with the speed of LIGHT and the electrons in every PART of the conductor begin to DRIFT. | |
| 6. |
A long resistance wire is divided into 2n parts. Then n parts are connected in series and other .n. parts in parallel separately. Both combinations are connected to identical supplies. Then the ratio of heat produced in series to parallel combinations will be |
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Answer» `1:1` |
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| 7. |
A narrow beam of monoenergetic electrons falls normally on the surface of a Ni single crystal. The reflection maximum of fourth order is observed in the direction forming an anle theta=55^(@) with the normal to the surface at the energy of the electrons equal to T= 180eV. Calcualte the corresponding value of the interplanar distance. |
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Answer» Solution :Path difference is `d+d cos theta= 2d "cos"^(2)(theta)/(2)` Thus for reflection MAXIMUM of the `K^(TH)` order `2 d "cos"^(2)(theta)/(2)k lambda=k(2pi ħ)/(sqrt(2mT))` Hence `d=(kpi ħ)/(sqrt(2mT))"sec"^(2)(theta)/(2)` Subsatitution with `k=4` gives `d= 0.232nm` 
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| 8. |
Two short bar magnets of equal dipole moments 'M' each are fastened perpendicular at their centers as shown in figure. The magnitude of the magnetic field at 'P' at a distance d from their common centre as shown in figure is |
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Answer» `(mu_(0))/(4pi)(M)/d^(3)` |
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| 9. |
A mixture of two diatomic gases exists in a closed eyj ic volumes and velocities of sound in the two gases areV_(1), V_(2),c_(1) and c_(2) , respectively. Determine ihe velocity of sound in the gaseous mixture. (Pressure of gas remains constant)^ |
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Answer» `c_(1)c_(2) SQRT((V_(1) + V_(2))/(V_(1)c_(2)^(2) + V_(2)c_(1)^(2)))` |
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| 10. |
What is the area of the plates of a 2 F parallel plate capacitor given that the separation between the plates is 0.5 cm ? |
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Answer» Solution :Here C = 2F d = 0.5 cm = `5XX10^(-3)` m `in_(0)= 8.85 xx10^(-12) C^(2)N^(-1) m^(-2)` `C = (in_(0)A)/(d)` `:. A = (Cd)/(in_(0)) = (2xx5xx10^(-3))/(8.85xx10^(-12))= 1130xx10^(6) m^(2)` `:. A = 1130 (KM)^(2)` |
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| 11. |
Identify the correctorder in whichthe valueof normal reaction increasesobjectis placedon roughhorizontal surfaces (A ) the object is pushedwiththe forceFat ananglethetawith horizontal (B)the objectis pulledwiththe force Fat an angle theta with horizontal (C )The object is pusheddownwith theforce F normally (D ) The objectis pushed upwith the forceFnormally |
| Answer» Answer :B | |
| 12. |
Let N_(betabe the number ofbeta particles emitted by 1 gram of Na^(24)radioactive nuclei (half ife = 15 hrs) in 7.5 hours, N_betais close to (Avogadro number = 6.023xx10^(23)/g "mole" ) : |
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Answer» `1.75 xx10^(22)` |
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| 13. |
Two infinitely long wires carry current as shown in Figure. Find the magnetic field inten sity at the points P, P and P* |
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Answer» <P> Solution :`At P,B_p = (mu_0I)/(2PI)[1/x + (1)/(r-x)]``At P.B_(p) = (mu_0I)/(2pi)[1/x - (1)/(r+x)]` and at `P..,B_(P^**)=(mu_0I)/(2pi) [1/x-(1)/(r+x)]`(with the same CURRENT ) 
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| 14. |
Let B_(P),B_(Q)andB_(R) be the magnetic field produced by the three infinite long wires P, Q and R respectively. The three wires are placed symmetrically inside an equilateral triangular loop as shown in figure. Current in 3 wires are shown in figure If int_(A)^(B)vec(B_(P)).dvec1=4mu_(0)T-mandint_(A)^(B)vec(B_(P)).dvec1=-15mu_(0)T-m, then the value of i is |
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Answer» 15 A |
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| 15. |
why there are two propellers in a helicopter ? |
| Answer» SOLUTION :It there were only one propeller in the helicopter the helicopter itself would have TURNED in opposite direction due to CONSERVATION of angular MOMENTUM. | |
| 16. |
A copper wire of length 3m and diameter 1 mm is stretched to increase it's length by0.3 cm. What is the lateral contraction produced in the wire ?poison ratio is 0.25. |
| Answer» Solution :`(DELTA D)/D = delta (delta L)/L = (0.25 xx 0.3 xx 10^-2)/3` = `025 xx 10^-3` there for `delta D = 25 xx 10^-5 xx 1 xx 10^-3 = 2.5 xx 10^-7` | |
| 17. |
An electron, an alpha-particle and a proton have the same kinetic energy. Which one of these particles has the largest de-Broglie wavelength ? |
| Answer» SOLUTION :As `lamda=(h)/(sqrt((2mK)))` and KINETIC enerrgy is CONSTANT, hence `lamdaprop(1)/(sqrtm)` i.e., ELECTRON particle, being lightest, will have the LARGEST de-Broglie wavelength. | |
| 18. |
A body with area A at maintained temperature T and emissivity e=0.6 is kept inside a spherical black body. What will be the maximum energy radiated per second ? |
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Answer» `0*60sigma"AT"^(4)` CORRECT CHOICE is (a). |
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| 19. |
A shell of mass 2 kg and moving at a rate of 4 m/s suddenly explodes into two equal fragments. Thefragments go in directions inclined with the original line of motion with equal velocities. If the explosion imparts 48 J of translational kinetic energy to the fragments divided equally, the velocity (magnitude and direction both) of each fragment :- |
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Answer» 2 m/s , `45^@` 2 kg x 4m/s = 1kg x v cos `theta` m/s+1 kg x v cos `theta` m/s  `8 = 2v cos theta` `v cos theta` = 4 m/s ….(i) The initial kinetic energy of the shell `=1/2xx2kgxx(4m//s)^2` =16 J An amount of 48 J is imparted by EXPLOSION. Thus, the total energy of the fragments is 64 J, i.e. each fragment has 32 J kinetic energy `=1/2xx 1kg xx v^2 `=32 J `v^2`=64 v=8m/s and `v cos theta `=4 `cos theta =1/2 RARR theta =60^@` |
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| 20. |
Write the definition of valence band and conduction band. |
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Answer» Solution :Some solids have a crystalline STRUCTURE means there is a systematic arrangement of atoms or molecules in it. When atoms are arranged close to each other the ATOM gets interact with neighbouring atoms and other atoms as a RESULT energy of electrons of atoms are changed. The energy level of inner shells electrons are not affected, but the energy levels of the outer shells electrons (valence electron) are changed, since these electrons are SHARED, by more than one atom in the crystal. The electrons of theatoms in the crystal has closely spaced different energy levels instead of widely separated energy level of electrons of isolated atom. Such a band of close energy levels is called energy band. The valence electrons (in last counter orbit) are called as valence electrons in incomplete orbit. The energy band whichincludesthe energy levels of the valence electrons is called the valence band. The energy band above the valence band is called the conduction band. Usually a valence band has a valence electronwhile no electron in conduction band. There is some gap between the conduction band and the valence band. The energy difference of this empty space is called band gap energy `(E_(g))`. When the minimum energy level of the conduction band is lower than the higher energy level of the valence band, the electrons in the valence band can easily MOVE to the conduction band. This happens in metals (conductors). If there is some gap between the conduction band and the valence band, electrons in the valence band all remain bound and no free electrons are available in the conduction band. This makes the material an insulator. If the distance between the conduction band and the valence band is relatively small and if some of the electrons in the valence band get external energy crosses band gap energy and goes into the conduction band then there is a possibility of flow in conduction band as well as the electron going from the valence band in the conduction band there is a possibilityof conduction due to empty spaces in the valence band. This is only possible in the semiconductor. 
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| 21. |
A rod AB of length L and mass M is free to move on a frictionless horizontal surface.It is moving with a velocity v, as shown in figure. End B of rod AB strikes the end of the wall. Assuming elastic impact, the angular velocity of the rod AB, just after impact, is |
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Answer» `(V)/(2L)` |
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| 22. |
A ball of 200g is at one end of a stringof length 20 cm. it is revolved in a horizontal circle at an angular frequency of 6 rpm. Find (i) the angular velocity (ii) the linear velocity, (iii) the centripetal acceleration. |
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Answer» SOLUTION :(i) The angular velocity, ` omega=(2piN)/(t)=(2pi XX 6)/(60)=(pi)/(5)=0.6284 rad s^(-1)` (ii) the linear velocity `V= r omega` `=0.20xx0.6284=0.1257 m s^(-1)` (iii) The CENTRIPETAL ACCELERATION `a_(c)=r omega^(2)` `=0.20 xx (0.6284)^(2)=0.0970 m s^(-2)` |
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| 23. |
The two capacitors, shown in the circuit, are initially uncharged and the cell is ideal.The switch S is closed at t = 0. Which of the following functions represents the current i(t), through the cell as a function of time?Here i_(0), i_(1), i_(2) are constants. |
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Answer» `i(t) = i_(0) + i_(1)e^(-t/tau), tau = 3C xx (R)/(3)` |
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| 24. |
A block of mass 2 kg is placed on the floor The coefficient of static friction is 0.4. If a force of 2.8 N is applied on the block parallel to the floor, the force of friction between the block and the floor (take g =10ms^-2) is |
| Answer» Answer :A | |
| 25. |
The total energy of a hydrogen atom in its ground state is -13.6 eV. If the potential energy in the first excite state is taken as zero then the total energy in the ground state will be |
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Answer» `-3.4eV` |
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| 26. |
Assertion (A) : Hydrogen atom consists of only one electron but its emission spectrum has many lines. Reason (R) : Only Lyman series is found in the absorption spectrum of hydrogen atom where as in the emission spectrum all the spectral series all present. |
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Answer» If both ASSERTION and reason are true and the reason is the correct explanation of the assertion. |
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| 27. |
Photons of wavelength 660 nm are emitted from a 60 watt lamp. What is the number of photons emitted per second ? |
| Answer» Answer :C | |
| 28. |
The dimensions of Rydberg constant are: |
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Answer» `[M^0L^-4T^0]` |
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| 29. |
In deriving the single-slit diffraction pattern, it was states that the intensity is zero at angles (nlamda)/(a). Justify this by suitable dividing the slit to bring out the cancellation. |
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Answer» Solution :Let us consider a single-slit of width a. for diffraction pattern due this slit nth principal MINIMA will be obtained at an angle `theta` such that `asintheta=nlamda or sintheta=(nlamda)/(a)`, and for small angle `sintheta=theta=(nlamda)/(a)`. To justify this, let us consider that the given slit is divided into 2n smaller slits, each of EQUAL width `a.=(a)/(2n)`. then in the direction of diffraction angle `theta`, the path DIFFERENCE between wavelets from two adjacent slits will be `a.sintheta=((a)/(2n))sintheta=(a)/(2n)*(nlamda)/(a)=(lamda)/(2)`. it means that light from each odd NUMBERED slit is exactly nullified by the light from SUBSEQUENT even numbered slit and resultant intensity is zero. |
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| 30. |
A 2muFcapacitor is charged to a p.d. of 60V. The charging battery is disconnected and the capacitor is connected in series with a coil of inductance 10mH, so that LC oscillations occur. What is the maximum current in the coil? Assume that the circuit contains no resistance. |
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Answer» SOLUTION :`C = 2 xx 10^(-6) F , L = 10 xx 10^(-3) H, V = 60 V` ` (Li_m^2) /(2) = (Q^2)/(2C)` ` i_m = sqrt( (Q^2)/(LC) )` But Q = CV `therefore i_m = sqrt( (C^2V^2)/(LC) ) = sqrt( (CV^2)/(L) )` ` therefore= i_m = sqrt( (C^2 V^2)/(LC) ) = sqrt( (CV^2)/(L) ) = sqrt( (2 xx 10^(-6) xx 60 xx 60 )/(10 xx 10^(-3) ))=6 sqrt(2 xx 10^(-2) ) = 0.848 A = 848 mA` |
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| 31. |
Two mass particles of masses 1 kg and 2kg have coordinates (2,-3) and (-3,2) respectively, the coordinates of their centre of mass will be |
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Answer» (1.95, 1.95) |
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| 32. |
(a) explainbriefly with the help of necessary diagrams the forward and the reversecharqcteristic cuves inthe two cases (b) A semicionductor has equal electron and holw concentration of 6xx10^(8) on doping with increases to 9xx10^(12)//m^(3) (i) identifty the new semiciondcutor obtained after doping (ii) calculate the new hole concentrati9no |
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Answer» <P> SOLUTION :`(b) (i)n-type`(II)`n_(C ) n_(p)= n_(1)^(2)` `n_(p)= (n_(2)^(2)) /(n_(e))= (6xx 10^(8))/(9xx 10^(12))^(2)` `=(36 XX 10^(16))/( 9xx 10^(12))` `=4 xx 10^(4) //m^(8)` |
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| 33. |
Two nuclei P, Q have equal no.of atoms at t= 0. Their half-life are 3 hours, 9 hours. Compare their rates of disintegration after 18 hrs from the start. |
| Answer» ANSWER :A | |
| 34. |
Give any three differences between a compound microscope and a telescope. |
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Answer» Solution :The FOCAL length and the size of the objective will be less than the eye piece in the case of a compound microscope whereas in the case of a TELESCOPE, the focal length of the objective and its size will be LARGER than in the microscope. The MAGNIFICATION in the case of a compound microscope is `m=(L)/(-f_(0))((D)/(f_(e)))`, whereas in the case of a telescope `m=(f_(0))/(f_(e))` for image at INFINITY and .L. is the length of the tube of the microscope. The length of the telescope `=L=f_(0)+f_(e)` and the length of compound microscopeis `v_(0)+f_(e)` where `v_(0)` is the image distance obtained in the case of refraction through the object lens. |
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| 35. |
Total heat required to convert 50 kg of water at 10^(@)C to steam at 100^(@)C (L.H =2.25 xx 10^(6)" J kg"^(-1)): |
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Answer» `3.13xx10^(6) cal` `1000 xx50 xx(100-10)+(50xx2.25xx10^(6))/(4.2)` `=4.5xx10^(6)+26.8xx10^(6)=31.3xx10^(6)` cals. `therefore` Correct CHOICE is (b). |
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| 36. |
What is meant by self sustained nuclear reaction? |
| Answer» SOLUTION :If the NEUTRONS released in nuclear fission can be USED to promote further fission, then the reaction is a self sustained nuclear reaction. | |
| 37. |
(A): Electric field outside the wire due to a steady current through it is zero. (R): Net charge present on current carrying wire is zero. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A' |
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| 38. |
A chemical cell of emf E has negligible internal resistance. It is connected to a variable resistance (R) which changes linearly from 20 Omega to 40 Omega in 20 minute and thereafter becomes constant. It was found that the cell lost 10 % of its total chemical energy in first 20 minute after the switch was closed. How long will the energy in the cell last? |
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Answer» |
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| 39. |
A ball is dropped from the window of a moving train on horizontal rails. What is the path followed by the ball on reaching the ground ? |
| Answer» SOLUTION :A PARABOLIC PATH | |
| 40. |
Explain why TV transmission towers made high. |
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Answer» Solution :(i) The TV signals are frequency modulated the transmission of TV signals cannot be obtained the ground wave propagation, as such signal get absorbed by ground due to their high frequency. (II) T.V signals cannot be TRANSMITTED via SKY wave propagation as ionosphere is unable to reflect radio waves of frequencies greater than 40 MHz. (iii) Therefor, the only way out for the transmission of TV signals is that receiving antenna should DIRECTLY INTERCEPT the signal from the transmitting antenna. |
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| 41. |
If energy E, velocity v and time T are chosen as the fundamental quantities. Find dimensional formula of surface tension ? |
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Answer» `MLT^(-1)A^(-1)` |
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| 42. |
In a single slit diffraction experiment, the width of the slit is increased. How will the (i) size and (ii) intensity of central bright band be affected ? Justify your answer. |
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Answer» Solution :We know that angular WIDTH of central BRIGHT band in a single slit diffraction pattern is given as `alpha=2 theta= (2lambda)/(a)` where a = slit width and size of central bright band x = `D alpha=(2lambdaD)/(a)` where D is the distance of screen from the slit. (i) If slit width .a. is INCREASED then obviously a as well as x will decrease i.e., central bright band will become NARROWER. (ii) On increasing the size of slit more light PASSES through it and the size of central bright band is decreased. Therefore, it is very much clear that intensity of central bright band will definitely increase. |
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| 43. |
Doubly ionised helium atoms and hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of the helium and the hydrogen ion is |
| Answer» ANSWER :C | |
| 44. |
Two metallic sphenrs of radil R_(1) and R_(2) are charged. Now they are brought into contact with each other with conducting wire and then separated. If the electric fields on their surfaces are E_(1) and E_(2) respectively (E_(1))/(E_(2)) = ....... . |
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Answer» `(R_(2))/(R_(1))` `:. V_(1) = V_(2)` `:. (kQ)/(R_(1))=(kQ)/(R_(2))` `:. (kQ)/(R_(1)^(2))R_(1)=(kQ)/(R_(2)^(2))R_(2)` `:. E_(1)R_(1)= E_(2)R_(2)` `:. (E_(1))/(E_(2))=(R_(2))/(R_(1))` |
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| 45. |
Find the equivalent inductance of two inductors having inductances L_1 and L_2 connected in parallel with the help of appropriate DC circuit. |
Answer» Solution :Let two inductor `L_1` and `L_2` are connected in parallel across a battery whose emf can be varied continuously and having no INTERNAL resistance.  Let currents passing through `L_1` and `L_2` are `I_1` and `I_2` at time t and the rates of change of currents through them are `(dI_1)/(DT)` and `(dI_2)/(dt)` respectively. `therefore` p.d. between their two ENDS are respectively `-L_1 (dI_1)/(dt)` and `-L_2(dI_2)/(dt)` Current passing through the main circuit is I at time t then, `epsilon =-L (dI)/(dt)=-L ((dI_1)/(dt)+(dI_2)/(dt))`...(1) But, `L_1(dI_1)/(dt)=-epsilon` and `L_2(dI_2)/(dt)=-epsilon` `therefore (dI_1)/(dt)=-epsilon/L_1` and `(dI_2)/(dt)=-epsilon/L_2` Putting these values in eqn. (i) `epsilon=-L (-epsilon/L_1-epsilon/L_2)` `therefore 1/L=1/L_1+1/L_2` |
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| 46. |
Assertion: A metal rod is moved perpendicular to the length of rod and uniform magnetic field exits perpendicular to the plane of motion of rod. Electric field is generated along the length of the rod. Reason: When metallic object moves inside magnetic field then free electrons experience force and are moved in the direction of force. Due to redistribution of electrons, electric field is created in the metal. |
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Answer» If both ASSERTION and REASON are CORRECT and reason is correct EXPLANATION of the assertion |
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| 47. |
Consider the venturi tube of Sample Problem 14.21 and Fig.without the manometer. Let A equal 5a. Suppose the pressure p_(1) at A is 3.0 atm. Compute the values of (a) the speed V at A and (b) the speed vat a that make the pressure p_(2) , at a equal to zero. (c) Compute the corresponding volume flow rate if the diameter at A is 5.0 cm. The phenomenon that occurs at a when p, falls to nearly zero is known as cavitation. The water vaporizes into small bubbles. |
| Answer» SOLUTION :(a) 5.0 m/s (b) 25 m/s (C) `2.0 XX 10^(-3) m^(3)//s` | |
| 48. |
Light is an electromagnetic wave. Its peed in vacuum is given by the expression |
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Answer» `sqrt((mu_(0).in_(0))` |
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| 49. |
Assertion : The effective resistance of the network between P and Q is (4)/(5)r Reason : Symmetry can be applied to the networkwith respect to center |
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Answer» bothassertionandreason are true and the reason is Assertion and Reason EXPLANATION of the Assertion  `(1)/(R_(PQ)) = (1)/(R ) + (1)/(2r) + (3)/(8R) = (15)/(8r)` or `R_(PQ) = (8r)/(15)` |
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| 50. |
Refractive index of cladding of optical fibre is ...... that of core of optical fibre. |
| Answer» Solution :less than | |