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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. |
Two large charged plane sheets of charge densities sigma and - 2sigma C//m^(2)are arranged vertically a separation of d between them. Deduce expressions for the electric field at pointsto the left of the first sheet , (ii) To right of the second sheet, between the two sheets. |
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Answer» Solution :Consider TWO large charged plane sheets of surface charges DENSITIES `+sigma and -2 sigma ` arranged vertically with a separation .d. between them. At any point `P_1` left of first sheet ,electric field due to the two sheets are `E_1 =(sigma)/( 2 in _0)and E_2 (2sigma )/(2 in _0)=(sigma )/( in_0) ` and their DIRECTIONS are MUTUALLY opposite as shown here . ` therefore ` Net electric fieldat ` P_1 , E=E_2-E_1=(sigma )/(in_0)-(sigma )/( 2 in _0)=(sigma )/( 2 in _0) ` towards the first sheet. (ii) At any point ` P_2` right of second sheet `E_1 =(sigma )/( 2 in _0)and E_2 =(sigma )/( in_0)` and their directions are again mutually opposite as shown ` therefore ` Net electric field at `P_2 ,E =E_2 -E_1 =(sigma )/( in_0)-(sigma)/( 2 in _0)=(sigma)/ (2 in _0)` towards the second sheet. (iii) At a point `P_3` between the two sheets both ` E_1and E_2` are DIRECTED in same directions . Hence net electric field at ` P_3 , E= E_1+E_2 =(sigma)/( 2 in _0)+(sigma)/(in_0)=(3 sigma )/( 2 in _0)` towards the second sheet. 
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| 2. |
Can d.c. ammeter use for measurement of alternating current? |
| Answer» SOLUTION :No, it is based on the principle of torque. When ac is PASSING through it (of FREQ. 50 Hz). It will not respond to frequent change in direction due to inertia hence WOULD show zero DEFLECTION. | |
| 3. |
A uniform circular disc of mass M and radius R is pivoted at distance x above the centre of mass of the disc, such that the time period of the disc in the vertical plane is infinite. What is the distance between the pivoted point and centre of mass of the disc ? |
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| 4. |
In Young's experiment with sodium light the slits are 0.586 m aprt. Whatis the angular width of the fourth maximum ? Given that lambda = 589 nm : |
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Answer» `sin^(-1)(3 xx 10^(-6)` `or (x_(n))/(D) = (n lambda)/(d)` `therefore sin theta = (n lambda)/(d) = 3 xx 10^(-6)` `theta = sin^(-1) (3 xx 10^(-6))`. |
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| 5. |
Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric potential at a point due to dipole as compared to that due to a single charge. |
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Answer» SOLUTION :For expression for electric POTENTIAL, see Short Answer Question NUMBER 5 just above. Electric potential at any point distant r from the centre of a dipole varies as `V prop 1/r^2`, whereas potential DUE to a point charge varies as `V prop 1/r`. |
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| 6. |
भारत की सबसे ऊंची चोटी कौन सी है? |
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Answer» माउंट एवरेस्ट |
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| 7. |
The diffraction pattern due to a straight edge contains |
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Answer» alternate bright and DARK bands of same width |
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| 8. |
What is resonance? Give its example. |
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Answer» Solution :A phenomenon in which an external FORCE or a vibrating system forces another system around it to vibrate with greater amplitude at a specified frequency of operation called RESONANCE. Example `:` Letphenomenon is a child on a swing. The swing has a natural freuqency for swinging BACK and FORTH. If the child pulls on the ROPE at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging the amplitude of the swinging will be large. A swing cannot stop. This is called resonance. |
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| 9. |
In a biprism, 21 fringes are distinctly seen on screen at a distance of 1 m, when the sources are 0.5 mm apart, what is the coherent length and coherent time of the set up .(lambda = 6000 Å): |
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Answer» `2 xx 10^(-14) sec` `therefore Delta_(max) = (x_(max).d)/(D)` For distinct FRINGES, `Delta_(max) = l_(coh)` Here `x_(max) = 10 beta` `l_(coh) = Delta_(max) = (10beta d)/(D)` `therefore l_(coh) = (10d)/(D) .(lambdaD)/(d) = 10 LAMBDA = 6mu m`. `T_(coh) = (l_(coh))/(C) = (6 xx 10^(-6))/(3 xx 10^(8)) = 2 xx 10^(-14)S`. |
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| 10. |
Assertion: A point charge is brought in an electric field. The field at a nearby point will increases whatever be the nature of the charge. Reason : the electric field is independent of the nature of charge. |
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Answer» If both ASSERTION and reason are true and the reason is the CORRECT explanation of the assertion. Explanation : Electric field at the nearby-point will be RESULTANT of existing field and field due to the charge BROUGHT. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought. |
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| 11. |
Assertion: Space waves are used for line -of -sight communication. Reason: Space wave travels in a straight line from transmitting antenna to the receiving antenna. |
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Answer» |
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| 12. |
A ball is projected at t = 0with velocityv_0 at angle thetawith horizontal. It strikes a smooth wall at a distancefrom it and then falls at a distance d from the wall. Coefficient of restitution is 'e'then : |
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Answer» Ball will return to ground at`t = (2v_0 sin THETA)/(G)` So time of flight and maximum height which depend on vertical velocity does not change. `v_(||) = v_0 cos theta` (before collision) `v_(||("after collision")) = ev_0 cos theta` Time taken to reach ground after collision `t' = (2 v_0 sin theta)/(g) "":. "" d = ev_(0) cos theta XX t' = e(R - L)` 
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| 13. |
In an LCR series resonant circuit, the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to |
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Answer» 2L |
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| 14. |
Unstable pions are produced as a beam in a nuclear reaction experiment . The pions leave the target at a speed of 0.995c. The intensity of the beam reduces to half its original value as the beamtravelsa distance of 39 m . Find the half - life of pions (a) in the laboratory frame, (b) in their rest frame. |
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Answer» Solution :(a) The INTENSITY of the pion beam reduces to half its original value in one half - life. The half - life of the pions as measured in the laboratory is ` t_(1/2) = 39 m / 0995c = 39 m / 0.995 xx 3 xx 10 ^(8) m s^(-1)` `= 1.3 xx 10^(-7).`(B) The events -a pion leaving the target and its DECAYING - OCCUR at the same place in the pion - frame . thus, the time measured in the pion - frame is the proper time and is the smallest. It is equal to ` t'_(1/2) = t_(1/2) (sqrt 1 - v^(2) / C^(2)) = (1.3 xx 10^(-7) s ) (sqrt 1- (0.995)^(2)` `= 1.3 xx 10^(-8) s.` |
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| 15. |
Salt P+QtoRoverset(BaCl_(2))towhite ppt (P) is paramagnetic in nature and containsabout 55% K. So (P) is |
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Answer» `KO_(2)` |
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| 16. |
When an electronl beam passes through and electric field they gain kinetic energy. If the same electron beam passes through a magnetic field then their |
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Answer» ENERGY and MOMENTUM both ramain unchanged |
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| 17. |
Two polaroids are in crossed situation and intensity of polarized light is zero. If third polaroid is placed in between such that it makes half angle with optic axis, than angle between two polaroids, then intensity of polarized light will be ....... Where I_(0) is maximum intensity of incident light. |
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Answer» `(I_(0))/(2)` Now, intensity of incident light of secon polaroid is `I_(1)=I_(2)` hence intensity of polarize light from it, `I..=(I_(1))/(2)cos^(2) theta` `=(I_(0))/(4)cos^(2)45^(@)` `=(I_(0))/(4)XX(1)/(sqrt((2)^(2)))=(I_(0))/(8)` |
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| 18. |
Currents flowing in each of the following circuits A and B respectively are |
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Answer» 1 A, 2 A The total resistance R is given by, `1/R = 1/4+1/4 = 2Omega` According to ohm.s law, `V = I_(A)R` `implies I_(A) = V/R = 8/2 = 4 A` Circuit-B `V = I_(B)R` `implies I_(B) = V/R = 8/4 = 2 A` |
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| 19. |
A carnot engine has the same efficiency between 100 K and 500 K and T and 900 K. What is the value of the temperature T ? |
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Answer» 130 K |
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| 20. |
A ball is thrown vertically upwards .Which of the following graph/graphs represent velocity-time graph of the ball during its flight (air resistance is neglected ) |
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Answer»
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| 21. |
the wavelengths of two waves are 50 and 51 cm respectively. If the temperature of the room is 20^(@) C, then what will be the number of beats produced per second by these waves, when the speed of sound at 0^(@) C is 332 m/s ? |
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Answer» 14 `v_(0) = 332sqrt((293)/(273)) = 332 xx 1.036 = 344 ms^(-1)` `n_(1) = (344 xx 100)/(50) = 688` HZ. `n_(2) = (344 xx 100)/(51) = 674 ` Hz. `therefore` No. of beats /sec = 688 - 674 = 14. correct choice is (a) . |
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| 22. |
A capacitor of capacitance 50 muF is charged to 10 V. Its electrostatic potential energy is |
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Answer» `2.5 XX 10^(-3)J ` |
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| 23. |
The spectral line for the Lyman series is transist to Pfund series, the number of spectral lines ...... |
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Answer» increases The NUMBER of spectral lines in Lyman series is maximum and in PFUND series it is MINIMUM. |
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| 24. |
An extremely long straight wire, with radius of cross-section a, carries current I. Then ratio of magnetic fields at distances a/2 and 2a from its axis would be ______ . |
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Answer» `1/4` `thereforeB_(i)=(mu_(0)I)/(2pia^(2))(a/2)=(mu_(0)I)/(4pia)""...(1)` For second point, `B_(0)=(mu_(0)I)/(2pia^(2))` `thereforeB_(0)=(mu_(0)I)/(2pi(2A))=(mu_(0)I)/(4pia)""...(2)` From equation (1) and (2), `B_(i)/B_(0)=1` |
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| 25. |
What is the reason for magnetic moment of material ? |
| Answer» Solution :The ORBITAL and spin PROPERTIES of electron in ATOMS of the MATERIAL. | |
| 26. |
A closed pipe emitting its second overtone is in unision with an open pipe emitting its third overtone. The length of the open pipe will be- |
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Answer» `1.6L_em` |
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| 27. |
The arm PQ of the rectangular conductor is moved from x=0, outwards in the uniform magnetic field which extends from r=0 to x=b and is zero for x > b as shown. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance. |
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Answer» Solution :Let us first consider the forward motion from X = 0 to x = 2b The flux `phi_B` . linked with the circuit SPQR is `phi_B = Blx"" 0 lt= x lt b = Blb "" b lt= x lt 2b` The induced emf is,`epsi = - (dphi_B)/(dt) = - BLV "" 0 lt= x lt b` When the induced emf is nonzero, the current I is `I = (Blv)/(r )`(in magnitude)  The force required to keep the ARM PQ in constant motion is IlB. Its direction is to the LEFT. ` F = (B^2 l^2 v)/(r ) 0 lt= x lt b` ` F = 0b lt= x lt 2b` The Joule heating loss is `P_f = I^2 r = (B^2 l^2 v^2)/(r ) 0 lt= x lt b` `P_i = 0 b lt= x lt 2b` One obtains similar expressions for the inward motion from x=2b to x=0. One can appreciate the WHOLE process by examining the sketch of various quantities displayed in Fig. |
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| 28. |
What did the boy decide? |
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Answer» He WOULD not be a cripple |
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| 29. |
Power in A.C. circuits in rms of I and E are related by |
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Answer» Solution :It is GIVEN by `cos phi=R/Z=R SQRT R^2+(OMEGA L- 1/omega C)^2` |
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| 30. |
The velocity of a body of mass 2.5 kg is changed from vecV_i =(3hati-4hatj) m//s to vecV_f = (2hati-9hatj)ms. What is the work done on the body: |
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Answer» 75 J |
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| 31. |
A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities K_(1) and K_(2), the equivalent conductivity of the combination is : |
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Answer» `2/3K` CORRECT choice is (d). |
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| 32. |
Magnetic lines of force never intersect each other . What is the reason ? |
| Answer» Solution :The DIRECTION of the magnetic FIELD is given by the direction of the tangent DRAWN to the line of force. If two lines of force INTERSECT, at the intersecting point there will be two directions for the field . This is not possible SINCE field is unidirectional. | |
| 33. |
Which of the following is preferred modulation scheme for digital communication ? |
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Answer» PULSE Code Modulation (PCM) |
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| 34. |
The plates of a parallel plate condenser are being moved away with velocity v. If the plate separation at any instant of time is d then the rate of change of capacity with time is proportional to |
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Answer» `1//d` |
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| 35. |
A n object is weighed at the North Pole by a beam balance and a spring balance, giving reading of W_(B) and W_(S) respectively. It is again weighed in the same manner at the equator, giving readings of W_(B)' and W_(S)' respectively. Assume that the acceleration due to gravity is the same everywhere and that the balances are sensitive. |
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Answer» `W_(B) = W_(S)` |
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| 36. |
A particle of mass M and charge q , initially at rest , is accelerated by a uniform electric field E through a distance D and is then allowed to approach a fixed static charge Q of the same sign . The distance of the closest approach of the charge q will then be |
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Answer» `(Q)/(4piepsilon_(0)D)` `threfore` WORK done on particle by electric field. W=qED. `thereforeQED=(1)/(4piepsilon_(0))rArrr_(0)=(Q)/(4piepsilon_(0)ED)` |
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| 37. |
(A) : Infrared waves are often called heat waves. (R) : Infrared waves, invole vibrations of electrons, atoms or molecules of a substance. This increases the internal energy of the substance. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 38. |
In the determination of the internal resistance of acell with apotentiometer, the error in the measurement of the balancing length is pm 1m m . When the cell alone is connected in the circuit, thebalancing length is obtained at 60cm and when the cell is shunted with a resistance of 10 Omega pm 2%, the balancing length is ovtained at 50 cm.The error int he determination ofThe internal resistance ofthe cell is .......... |
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Answer» 2 . 4 % `:. (Deltar)/(r) xx 100 = (Delta R)/(R) xx 100 + (Delta((I_(1))/(I_(2))))/(((I_(1))/(I_(2))))XX100 ` = 4 + 02 = 42% |
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| 39. |
P,Q R and S four resistance wires of resistance 1,2,3 and 4 ohm respectivily. They are connected in form the four arms of Wheatstone bridge circuit. Find out the resistance with whichS meat be shunted in order the bridge may be balanced |
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Answer» Here `P = 3 OmegaQ = 3 Omega , R = 3 Omega ` `:. (3)/(3) = (3)/(S) ORS =3 Omega ` Therefore the resistance of `4 Omega ` must be shunted with a resistance `x`so that the conbination resistance of `4 Omega` and `x` is equal to `3 Omega ` so `(1)/(x )+(1)/(4) + (1)/(3) or (1)/(x) = (1)/(4) - (1)/(12) or x = 12 Omega ` |
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| 40. |
A conductor having cuboid shape has dimensions in ratio 1: 2:5, the ratio of maximum to minimum resistance across two opposite faces is |
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Answer» 1:25 |
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| 41. |
A beam of unpolarised light of intensity I_(0) is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45^(@) relative to that of A. The intensity of the emergent light is : |
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Answer» `(I_(0))/(2)` 
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| 42. |
If f is focal length and u is object distance for spherical mirror, then lateral magnification m = ...... |
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Answer» `(f)/(u-f)`<BR>`(f)/(u+f)` `1/f=1/u+1/v implies 1/f-1/u=1/v`< br>`therefore (u-f)/(fu)=1/v` `therefore(u-f)/(f)=u/v` `-v/u=-((f)/(u-f))therefore m=(f)/(f-u)` |
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| 43. |
An object is thrown horizontally with an initial speed of 10 m//s. How far will it drop in 4 seconds ? |
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Answer» SOLUTION :The first step is to decide whether this is a horizontal question or a vertical question, since you must consider these motions separately. The question " How far will it drop ? " is a vertical question, so we'll USE the set of equations listed above under vertical motion. NEXT, " how far..? " implies that we will use the first of the vertical-motion equation, the one that gives vertical displacement, `DELTA y`. Since the object is thrown horizontally, there is no vertical component to its initial velocity vector `v_(0)`, that is , `v_(0_(y))=0`. So `Delta y= v_(0_(y)) t+(1)/(2)(-g)t^(2) to Delta y =(1)/(2)(-g)t^(2) ("because " v_(0_(y))=0)` `=(1)/(2)(-10)(4^(2))` `= -80 m`  The fact that `Delta y` is negative means that the displacement is down. Also, notice that the information given about `v_(0_(x))` is irrelevant to the question. |
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| 44. |
In a series LCR circuit, obtain the conditions under which (i) the impedance of the circuit is minimum , and (ii) wattless current flows in the circuit. |
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Answer» Solution :Impedance of series LCR circuit is given by `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2)` or for Z to be minimum, `X_(L) = X_(C)` (or `omega = (1)/(sqrt(LC))`) For WATTLESS CURRENT to flow CIRCUITS should not have any ohmic resistance i.e., R = 0 Alternatively : Power `= V_(rms)I_(rms) cos phi` for `phi = 90^(@) = pi//2` Power = 0 `:.` Wattless current flows when the impedance of the circuit is purely inductive / capacitive of the combination of the two. |
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| 45. |
A star emiting light of wavelength 4500Å is moving towards the eart with a speed of 3600 km/sec. Determine the wavelength shift and apparent wavelength due to Doppler effect. The speed of light is 3xx10^(8)m//s |
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Answer» SOLUTION :DATA supplied `v=3.6xx10^(6)m//s, "" lamda=4500xx10^(-10)m""c=3xx10^(8)m//s` `DELTA lamda=(v lamda)/c` and `Delta lamda=-lamda-lamda.` or `lamda.-lamda` `:.Delta lamda=(3.6xx10^(6)xx4500xx10^(-10))/(3xx10^(8))=54.00xx10^(-10)m=54Å` The apparent WAVELENGTH `lamda.= Delta -Delta lamda=4446Å` |
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| 46. |
State with reason where the balance point will be shifted when resistance S is increased keeping R constant. |
| Answer» Solution :Balanacing POINT shifted towards B when R is INCREASED current through the potentiometer wire DECREASES so much length is needed to balance the emf of the SECONDARY cell Q. | |
| 47. |
A ball of mass m and density (rho)/2 is completely immersed in a liquid of density rho, contained in an accelerating vessel as shown.Select the correct answer using the codes given below the columns. |
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Answer» <P>`{:(,P,Q,R,S),((A),1,3,2,4):}`  
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| 48. |
Assertion : The wave nature of electrons was first experimentally verified by Davisson and Germer Experiment. Reason : From the electron diffraction measurements, the wavelength of meter waves was found to 0.165 nm. |
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Answer» If both assertion and REASON are TRUE and reason is the correct explanation of assertion. |
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| 49. |
Commonly used materials for fabricating solar cells are ______ and ______ . |
| Answer» SOLUTION :SILICON (SI), gallium-arsenide (GA As) | |
| 50. |
SI unit of the ratio of Electric flux and Magnetic flux is |
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Answer» `m` but `C= AS` `therefore (phi_(E))/( phi_(m) ) = (mA)/( AS) = (m)/(S)` `therefore` Unit of `(phi_E)/( phi_m)` is the unit of velocity of light `=ms^(-1)`. |
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