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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. |
Equal volumes of two liquids A and B are heated to the same temperature and left in the atmosphere in identical vessels. |
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Answer» both will COOL at the same rate |
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| 2. |
An experimenter measures the length of a rod. In the caese listed, all motions are with respect to the lab and parallel to the length will be minimum? |
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Answer» The ROD and the EXPERIMENTER move with the same speed v in the same DIRECTION . |
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| 3. |
Calculate the drift speed of electrons when 1A of current flows in a copper wire of cross section 2 "min"^2 . The number of free electrons in 1 cm^3 of copper is 8.5 xx 10^(22). |
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Answer» SOLUTION :From TE current and drift velocity RELATION I = Ane `v_d`, `v_d = I/("Ane") =(1)/(2XX10^(-6) xx 8.5 xx 10^(28) xx 1.6xx10^(-19))` `=3.7xx10^(-5) ` m/s |
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| 4. |
The ratio of radii of the first three orbits is: |
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Answer» `3/2:1/2:1/3` |
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| 5. |
The time taken by the sunlight to reach the bottom of a tank of depth 4.5m filled completely with water is ....... ns. The refractive index of water is 4/3 |
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Answer» 2 `THEREFORE t =(nd)/(c)` `=4/3xx(4.5)/(3xx10^8)=(18)/(9)xx10^(-8)=2xx10^(-8)` second `therefore t=20xx10^(-9)` second = 20 MS |
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| 6. |
The resultant capacitance of given circuit between point P and Q is |
| Answer» ANSWER :A | |
| 7. |
Name the particle emitted spontaneously in the following nuclear reaction. ""_(15)^(32)Prarr""_(16)^(32)S+vecv+........... |
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Answer» |
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| 8. |
Speed of light wave of 5000 Å wavelength produced from a star is 1.5 xx 10^(6) m//s, then change in wavelength when it reaches to Earth is ...... |
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Answer» `25Å` `=5000xx(1.5xx10^(6))/(3XX10^(8))=25Å` |
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| 9. |
According to Bohr, the difference between the energies of the electron in two orbits is equal to: |
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Answer» HV |
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| 10. |
A mass m_(1)=1kg connected to a horizontal spring performs S.H.M. with amplitude A. While mass m_(1) is passing through mean position another mass m_(2)=3kg is placed on it so that both the masses move together with amplitude A_(1). The ratio of (A_(1))/(A) is ((p)/(q))^(1//2), where p and q are the smallest integers. Then what is the value of p+q? |
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Answer» |
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| 11. |
A particle moves in a region having a uniform magnetic field and a parallel, uniform electic field. At some instnat, the velocity of the particle is perpendicular to the field direction. The path of the particle will be |
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Answer» a STRAIGHT line |
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| 12. |
Assertion:In the phenomenon of mutual induction, self- induction of each of the coil persists. Reason: Self-induction arises when strength of current in one coil changes. In mutual induction current is changing in both the individuals coils. |
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Answer» If both assertion and reason are true and reason is the correct explanation of assertion |
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| 13. |
In each question, there is a table having 3 columns and 4 rows. Based on the table, there are 3 questions. Each question has 4 options (a), (b), (c) and (d), ONLY ONE of these four options is correct. In the given table Column I lists the elements in the circuit, Column II shows their combination and Column III shows the circuit diagram. What are the conditions for 1//L = 1//L_(1) + 1//L_(2)? |
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Answer» (I) (IV) (L) |
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| 14. |
In the circuit below , A and B represent two inputs and C represents the output . The circuit represents |
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Answer» OR GATE `= W(1+Y)+XW+XY` `=W.1 +XW +XY ""( because 1+Y=1)` `=W(1+W)+XY` `=W +XY ""(because 1 +X=1)` |
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| 15. |
A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If alpha and betabe the base angles and thetabe the angle of projection, prove that tan theta = tan alpha+ tan beta |
Answer» Solution :The SITUATION is SHOWN in figure. From figure, we have `tan alpha + tan beta = y/x + y/(R-x), tan alpha + tan beta = (yR)/(x(R-x))…(i)` But equation of trajectory is `Y = x tan THETA [1-x/R]` `tan theta =(yR)/(x(R-x))...(ii)` From Eqs. (i) and (ii), `tan theta = tan alpha + tan beta` |
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| 16. |
A carnot engine whose sink is at 300^@ K has an efficiency of 40%. By how much should the temperature of the source be increased so as to increase its efficiency by 50% of the original efficiency |
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Answer» 225 K |
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| 17. |
In each question, there is a table having 3 columns and 4 rows. Based on the table, there are 3 questions. Each question has 4 options (a), (b), (c) and (d), ONLY ONE of these four options is correct. In the given table Column I lists the elements in the circuit, Column II shows their combination and Column III shows the circuit diagram. What are the conditions for M =Ksqrt(L_(1)L_(2))? |
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Answer» (III) (i) (K) |
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| 18. |
The image of a candle formed by a concave mirro is cast on a screen. If some parts of the mirror are covered then how will the image be affected ? |
| Answer» Solution :The parts of the mirror that are covered do not take part in image formation. Therefore, the BRIGHTNESS of the image is DIMINISHED. Still, the FULL image will be FORMED - no part of the image will be MISSING. | |
| 20. |
The induced emf across the secondary coil depends upon (i) the number of turns n_(1) and n_(2) in primary and secondary coil respectively. (ii) the permeability of the medium between the two coils (iii) area of cross section of two coils (iv) resistance of wire of two coils Which of the above is/are true? |
| Answer» Answer :B | |
| 21. |
Two identical point charges ,q each , are kept 2 m apartin air. A third pint charges Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. |
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Answer» Solution :Let two CHARGES ,q each , are kept at points A and B respectively separated by a distance 2m. Let a third point charges Q be placed at point C SITUATED at a distance x m from A . For equilibrium of system the net force on any charges must be zero. Considering net force on charge Q, we have ` ""oversetto (F_(CA)) +oversetto (F_(CB)) =oversetto 0or |oversetto (F_(CA))|=oversetto (|F_(CB)|) ` ` therefore ""(1)/(4 pi in _0) (qQ)/(x^(2))=(1)/(4pi in _0).(qQ)/((2-x)^(2)) ` `rArr "" x= 2 -x or x=1 m ` Again considering net force on charge q situated at point A, we have ` ""oversetto (F_(AC)) +oversetto (F_(AB)) =oversetto 0 ` `rArr ""(1)/(4 pi in _0).(qQ)/((1)^(2)) +(1)/(4 PIIN _0) .(q^(2))/((2)^(2)) =0 ` `rArr "" Q +(q)/(4) = 0 or Q= -(q)/(4)` 
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| 22. |
A copper wire of length 5 cm carries a current of density j = 1 A (mm)^(-2). The density and molar mass of copper are 9000 Kg m^(-3) and 63 gmol^(-1). Each copper atom contributes one free electron. The temperature of the wire is 27^(@)C. Estimate the (average) distance travelled by a free electron during the time it moves from one end of the copper wire to the other end. Assume that thermal motion of electrons are similar to that of molecules of an ideal gas. |
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Answer» |
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| 23. |
Calculate the capacitantce of two plates of equal area of 1 m^(2) separated by distance 1 mm |
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Answer» Solution :Capacitance, `C = (in_(0)A)/(d)` `=(8.85xx10^(-12)xx1)/(1xx10^(-3))` `:. C= 8.85 xx10^(-9) F` |
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| 24. |
Potential barier devoloped in a junction diode opposses the flow of. |
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Answer» <P>MINORITY carriers in both regions only |
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| 25. |
The equation of a stationary wave is give by , y=0.05cos(2pix)sin(100pit). Find the frequency of waves. Here x and y are in S.I. units. |
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Answer» 100Hz |
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| 26. |
The measured mass and volume of a body are 53.63 g and 5.8cm^(3)respectively, with possible errors of 0.01 g and 0.1 cm^(3) . The maximum percentage error in density is about |
| Answer» ANSWER :B | |
| 27. |
Radio waves of low frequencies cannot be transmitted to long distances. So thes are superposed on a high frequency carrier signal. This process is known as |
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Answer» amplification |
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| 28. |
Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil. |
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Answer» Solution :Labelled diagrma of AC generator. Alternatively  [Deduct `1//2` MARK, if diagram is not labelled] When the coil is rotated with constant ANGULAR speed `bar(omega)` the angle `theta` betwe the magnetic field and area vector of thecoil at instant, t is by `theta=bar(omega)t`. Therefore , magnetic FLUX `,(phi_(B))` at this instant is `phi_(B)=BA cos bar(omega)t` `:.` induced emf `e=-N(d phi_(B))/(dt)` `e=NBA bar(omega)sin bar(omega) t =e_(0)sin bar(omega)t` where `e_(0)=NBA bar(omega)` |
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| 29. |
(A): An aircraft flies along the meridian, the potential at the ends of its wings will be the same.(R) : Whenever there is change in the magnetic flux e.m.f. induces. |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
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| 30. |
If the tube length of astronomical telescope is 105 cm and magnifying power is 20 for normal setting, then the focal length of the objective is .......... cm. |
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Answer» 10 Tube length `L GE f_0+f_e` …....(1) Magnification m=`(f_0)/(f_e)impliesf_e=(f_0)/(m)`…........(2) From EQN.(1) and (2), `L ge f_0+(f_0)/(m),L=105,m=20,` `105 ge f_0(1+(1)/(20))ge f_0((21)/(20))` `f_0 le (105xx20)/(20) le 100` `implies f_0=100` |
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| 31. |
Hot wire ammeters can be used for measuring |
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Answer» ALTERNATING CURRENT only |
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| 32. |
An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momentum , (b) the energy of the photon and (c) the kinetic energy of electron. |
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Answer» SOLUTION :Here `lambda=1.00nm =1XX10^(-9)` m. (a) Wavelength of electron and photon is same hence according to `p=(h)/(lambda)` their momentum will be equal `therefore` Equal momentum, `p=(h)/(lambda)=(6.63xx10^(-34))/(1xx10^(-9))` `therefore p=6.63xx10^(-25)kg ms^(-1)` (b)Energy of photon, `E=(hc)/(lambda)=(6.63xx10^(-34)xx3xx10^(8))/(1xx10^(-9))` `therefore E=19.89xx10^(-17)J` `therefore E=(19.89xx10^(-17))/(1.6xx10^(-19))EV` `therefore E=12.43xx10^(2)eV` `~~1.243.10^(3)eV` (C )KINETIC energy of electron, `K=(p^(2))/(2m)=((6.63xx10^(-25))^(2))/(2xx9.1xx10^(-31))=(43.9569xx10^(-5))/(18.2xx10^(-31))` `thereforeK=2.4152xx10^(-19)J` `therefore=(2.452xx10^(-19))/(1.6xx10^(-19))eV` `=1.509eV~~1.51 eV` |
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| 33. |
A circular coil of radius 10cm, 500 turns and resistance 2Omega is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180^(@)" in "0.25s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is 3.0 xx 10^(-5)T. |
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Answer» Solution :INITIAL flux through the coil, `phi_(B_("initial"))=BA cos theta` `=3.0xx10^(-5)xx(pixx10^(-2))xxcos 0^@` `=3pixx10^(-7)Wb` Final flux after the rotation. `phi_(B_("final"))=3.0xx10^(-5)xx(pixx10^(-2))xxcos 180^(@)` `=-3pixx10^(-7)Wb` THEREFORE, estimated value of INDUCTED emf is `epsi=N(DELTAPHI)/(Deltat)=500xx (6pixx10^(-7))//0.25` `=3.8xx10^(-3)V` `I=epsi//R=1.9xx10^(-3)A` Note that the magnitudes of `epsi` and I are the estimated values. |
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| 34. |
When a monochromatic light passes through a slit 0.2 mm wide and falls on a screen 3.5 m away, the first minimum of the diffraction pattern is 9.1 mm from the centre of the central maximum. The wavelength of the light is |
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Answer» `2600 Å` |
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| 35. |
Compare the V_("peak") due to HWR and FWR. |
| Answer» Solution :`(V_(p))_("FWR") = 2(V_(p))_("HWR") and (V_(p))_("HWR") = (V_(0))/(PI)`. | |
| 36. |
Two polaroids are kept corssed ("transmission axesat " 90^(@)) to each other. (i) What will be the intensity of the light coming out from the seond polaroid when an unpolarised light of intensity I falls on the first polaroid? (ii) What will be the intensity of light coming out from the second polaroid if a thrid polaroid is kept at 45^(@) inclination to both of them. |
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Answer» Solution :(i) As the intensity of TH umpolarised light FALLING on the first polaroid is I, theintensity of POLARIZED light emergingfrom it will be `I_(0)=((1)/(2)).` Let I. be the intensity of light emerging from the SECOND polaroid. Malus. law, `I. = I_(0) cos^(2) theta` Here `theta is 90^(@)` as the transmission axes are perpendicular to eachother. Substituting. `1.=((1)/(2))cos^(2)(90^(@))=0"" [thereforecos(90^(@))=0]` No light comes out from the second polaroid. (II) Let the first polaroid be` P_(1)` and the second polaroid `P_(2)`. They are oriental at `90^(@)`. The third polaroid `P_(1)` introduced between them `45^(@)`. Let I. be the intensity of light emerging from `P-(3)`. Angle between `P_(1) and P_(3) is45^(@)`. The intensity of light coming out from `P_(3)is I. = I_(0) cos^(2) theta` Substituting, `I.=((1)/(2))cos^(2)(45^(@))=((1)/(2))((1)/(sqrt(2)))^(2)=(1)/(4):I.(1)/(4)` Angle betweem `P_(3) and P_(2) is45^(@)` . Let I. is the intensityof light coming ut coming `P_(2) I. = I.cos^(2) theta`.  Here, the intensity of polarized light existing between `P_(3)` and `P_(2) " is"(1)/(4)`. Substituting, `1^(n)=((1)/(4))cos^(2)(45^(@))=((1)/(4))((1)/(sqrt(2)))^(2)=(1)/(8)` `1^(n) = (1)/(8)` |
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| 37. |
Which medium is needed for propagation of electromagnetic waves? |
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Answer» Air |
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| 38. |
Masses of three wires are in the ratio 1:3:5. Their lengths are in the ratio 5:3:1. When they are connected in series to an external source, the amounts of heats produced in them are in the ratio |
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Answer» `125: 15:1` |
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| 39. |
In amplitude modulation, the modulation index m, is kept less than or equal to 1 because |
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Answer» `m gt 1`, will RESULT in interference between CARRIER frequency and message frequency, RESULTING into distortion |
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| 40. |
Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero ? |
| Answer» Solution :(e) Yes, it is quite POSSIBLE that a system with ZERO net CHARGE can have non-zero magnetic moment. For example (i) atom of a PARAMAGNETIC material (ii) DOMAIN of a ferromagnetic material and (iii) permanent magnet have zero net charge and yet they have non-zero magnetic moment. | |
| 41. |
Object of 100 g mass moves with velocity of 36 km/hr .de-Broglie wavelength associated with it will be of ………order (h=6.626xx10^(-34)Js) |
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Answer» `10^(-14)` `=(36xx1000)/(3600)(m)/(s)=10(m)/(s)` de-Broglie WAVELENGTH , `lambda=(h)/(p)=(h)/(mv)=(6.626xx10^(-34))/(0.1xx10)` `lambda=6.626xx10^(-34)m` `therefore 10^(-34)m` ORDER |
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| 42. |
A galvanometer coil has a resistance of 12Omega and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V ? |
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Answer» Solution :1. Here, current capacity of galvanometer is, `I_(g)=3xx10^(-3)A` 2. Series resistance REQUIRED to CONVERT above galvanometer into voltmeter of voltage capacity V = 18 VOLT is, `R_(S)=V/I_(g)-G` `thereforeR_(S)=18/(3xx10^(-3))-12=5988Omega` THUS, we should connect `5988Omega` resistance in series with above galvanometer and then we should CALIBRATE its scale from 0 to 18 volt in order to prepare required voltmeter. |
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| 43. |
Explain the working of p-n junction diode as a full wave rectifier with circuit diagram. Give input and output waveforms. |
Answer» Solution : The circuit connections are made for full wave RECTIFIER as shown in circuit diagram. The circuit contains ac source, step down centre tapped transformer(T), two diodes (`D_(1)` and `D_(2)`) and a load resistance `(R_(L))`. During positive half cycle of input ac, the point A is possitive with respect to the point B. Then `D_(1)` is FORWARD biased and `D_(2)` is REVERSE biased. Thus `D_(1)` allows current `I_(1)` through the load `R_(L)` from M to N. During NEGATIVE half cycle of input ac, the point A is negative with respect to the point B. Then `D_(1)` is reverse biased and `D_(2)` is forward biased. Thus `D_(2)` allows current `I_(1)` through `R_(L)` from A to B. Thus the diodes allows both positive and negative half cycles of input ac through `R_(L)` in the same direction (from M to N). Hence the diode acts as full wave rectifier. Input and output wave forms as below 
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| 44. |
In an A.C. series circuit of L-R voltage applied is 220 V, if the potential difference across two ends on inductor is 176 V then the potential difference across two ends of the resistance will be …… |
Answer» Solution : USING Kirchoff.s second law for CLOSED loop, `V^2=V_R^2+V_L^2` `therefore V_R^2=V^2-V_L^2` `=(220)^2 - (176)^2` =48400-30976 = 17424 `therefore V_R`= 132 V |
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| 45. |
There is a plane mirror at the bottom of tank filled with water having refractive index 4//3. A point object is placed at a height 10 cm above the bottom. What will be the apparent distance between point object and its image in plane mirror as observed by an observer outside the liquid in air ? Line joining observer and object is normal to the surface of water. |
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Answer» 10 cm `("ACTUAL distance")/(MU)=(20)/(4//3)=15 cm` Hence, option (b) is correct. |
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| 46. |
In a p-n junction, there is no appreciable current if |
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Answer» p-section is made positive and n-section negative. |
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| 48. |
Mukesh wants to learn to become a motor mechanic by: |
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Answer» FINDING a tutor |
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| 49. |
A straight line conductor of length 0.4 m is moved with a speed of 7 m/s perpendicular to a magnetic field of intensity 0.9 Wb/m^2. The induced emf across the conductor is ____ |
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Answer» 1.26 V `=0.9xx7xx0.4xxsin90^@` =2.52 V [`because sin 90^@`=1] |
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| 50. |
Find the work done in the bringing a charge q from perpendicular distance from the line of charge . A device X is connected an ac source of voltage V= V_0sin omega t . The current through X is given asI=I_ssin (omegat+(pi)/(2)) |
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Answer» SOLUTION :work DONE in moving charge .q. Through small displacement .qr. dW =`oversetto E d oversetto r ` ` dW =qoversetto Ed oversetto r` ` =qE dcos theta` `dW =qxx (lambda )/( 2pi in _0r) dr ` work done in moving the GIVEN CHARGES from ` r_1to r_2(r_2gt r_1) ` ` W=int r_1^(r_2_) dwint int _r-1^(r_2) (lambda qdr)/( 2piin_0r) ` W = `(lambda q)/( 2pi in _0) [log (r_2)/( r_1) ]` |
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