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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. |
A sample consisting of Hydrogen atoms in the ground state is excited by monochromatic radiation of energy 12.75 eV. If we were to observe the emission spectrum of this sample, then the number of spectral lines observed, will be |
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Answer» 3 |
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| 2. |
An electric charge 10^-3muC is placed at the origin (0, 0) of xy coordinate system. Two points A and B are situated at (sqrt2,sqrt2) and (2, 0) respectively. The potential difference between points A and B will be: |
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Answer» 4.5 V |
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| 3. |
What is the relation between path difference and phase difference? |
| Answer» SOLUTION :Phase difference `sigma=frac{2pi}{LAMBDA}XX` PATH difference. | |
| 4. |
A solid non conducting sphere has charge density pCm^(-3) Electric field at distance x from its centre is ______ [x lt R] |
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Answer» `(PX)/(3epsilon_(0))` |
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| 5. |
M^-1L^3T^-2 is dimensional representation for |
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Answer» Power |
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| 6. |
Heat is flowing through two cylindrical rods of the same material. The fiameters of the rods are in the ratio 1:2 and the lengths in the ratio 2:1. If the temperature difference between the ends be the same, then the ratio of the rate of flow of heat through them will be : |
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Answer» `1:8` `(H_(1))/(H_(2))=(A_(1))/(A_(2))xx(l_(2))/(l_(1))=1/4xx1/2=1/8` Hence the correct choice is (b). |
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| 7. |
Two small identical electrical dipoles AB and CD each of dipole moment 'p' are kept at an angle of120 ^(@)as shown in What is the resultant dipole moment of this combination ? If this system is subjected to electric field. oversetto E directed along +X direction ,what will be the magnitude and direction of the torque acting on this? (##U_LIK_SP_PHY_XII_C01_E09_029_Q01.png" width="80%"> |
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Answer» Solution :As shown in let two identical electric dipoles AB and CD are kept at an ANGLE ` theta= 120 ^(@) ` ,where their dipole moment have magnitude `""| oversetto (p_A) |=|(oversetto p_C)| =p` The resultant dipole moment subtend an angle ` (theta)/(2)= 60 ^(@)` from EITHER of two dipoles ` oversetto (p_A)or oversetto (p_B) ` Therefore `oversetto (p_R) ` subtend an angle `30^(@) `from +X direction If the system is SUBJECTED to electric field `oversetto E ` directed along +X direction, the torque acting on the system is ` oversetto tau = oversetto (p_R) xx oversetto E ` Thus ,the magnitude of torque is ` | oversetto (tau)| =pE sin 30^(@) =(1)/(2)pE `and the torque is directed into the plane of paper i.e., the torquetends to align the system along the direction of electric field `oversetto E ` ` (##U_LIK_SP_PHY_XII_C01_E09_029_S01.png" width="80%"> |
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| 8. |
Why did Anju cover her mouth? |
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Answer» Because she did not want Mini to SEE her |
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| 9. |
What happens if a bar magnet is cut into two pieces: (i) transverse to its length (i) along its length ? |
| Answer» SOLUTION :On cutting a bar magnet into two pieces EITHER transverse to its LENGTH or along its length, each part behaves as a COMPLETE magnet having both N and S poles. | |
| 10. |
The body in shown figure is pivoted at O, and two forces act on its as shown. Find an expression for the net torque on the body about the pivot. |
| Answer» Solution : Take a troque that TENDS to CAUSE a counter CLOCKWISE rotation to be POSITIVE and a torque that tends to cause a clockwise rotation to be negative. Thus a positive torque of magnitude `r_1 F_1sin theta_1` is associated with`vecF_1`and a negative torque of magnitude `r_2 F_2 sin theta_2`is associated with `vecF_2`. Both of these are about O. The net torque about O is`TAU = r_1 F_2 sin theta_2` | |
| 11. |
What is Seebackeffect ? |
| Answer» Solution :SEEBECK DISCOVERED that in a CLOSED circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an EMF (POTENTIAL difference)is developed. | |
| 12. |
What is modulation ? Briefly explain three reasons which make modulation an essential step in communication system. |
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Answer» Solution :Modulation. The process by which some characteristics of a relatively high frequency wave ( called carrier wave ) is varied in accordance with the instantaneous value of a low frequency wave ( called MODULATING wave ) is called modulation. Reasons for modulation `(i)` For transmission of signal, the height of transmitting ANTENNA should be comparable to wavelength of signal. For audio frequency having range of frequency from `20HZ` to `20,000Hz,` the height of antenna should be from `15km` to `15000km` which is not possible, hence modulation of signal is required. `(ii)` The effective power radiated by an antenna is inversely proportional to square of length of antenna height. So for more power to be radiated, the height of the antenna should be small, hence modulation is needed. `(iii)` If large number of signals are to be transmitted in the audio range `(20Hz` to `20kHz)` these signals GET mixed and it becomes almost impossible to distinguish them. To remove this difficulty we communicate at higher frequencies and allotting a band of frequencies to each user. Hence modulation of the signal of low frequency to high frequency is required. |
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| 13. |
The potential difference between point A and B adjoining figure is(##VMC_NEET_XII_PHY_MOD_04_C13_E05_001_Q01.png" width="80%"> |
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Answer» `(2)/(3)V` |
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| 14. |
One weber is equal to |
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Answer» `10^(2)` maxwell |
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| 15. |
An electron and a proton moving parallel to each other in the same direction with equal momentra, enter into a uniform magnetic field which is at right angles to their velocities. Trace their trajectories in the magnetic field. |
Answer» Solution :Both electron and PROTON (having same MOMENTA) describe circular path of EQUAL as shown in FIG. 
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| 16. |
Two full turns of the circular scale of a screw éauge,cover a distance of Imm on its main scaleThe total number of divisions on the circular scale is 50. Fuerther, it is foun that the screw gauge has a zero error of -0.03 mm. while measuring the diameter ofa thin wire, a student notes the main scale reading of 3mm and the total no.of circular scale divisions in line with the main scale as 35. The diameterof the wire is |
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Answer» 3.32 MM |
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| 17. |
A body of mass 'm' is moving with constant speed v on a track shown in figure. At point A and point B radius ofcurvature is R. N_A,N_B and N_Crepresents normal reactions at A, B and C which of the following option iscorrect ? |
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Answer» `N_A=mg-(mv^2)/R`  `mg-N_A=(mv^2)/R, N_B-mg =(mv^2)/R, N_C=mg` |
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| 18. |
An aeroplane is flying at a height of 2000 m above the ground horizontally with a uniform speed of 720 km/h. At what angle of sight (w.r. to the horizontal) should the pilot drop a bomb so as to hit target on the ground ? |
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Answer» 26°.57 `200ms^(-1)` Height h = 2000 m Now LET X be the horizontal distance, then `h=0+1/2g t^(2)` or, `2000=1/2xx10t^(2)` `t=20sec` x = ut = `200 xx 20` `=1/2` or `theta=tan^(-1)0.5=26^@57.` |
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| 19. |
Name the customs union formed by Prussia to abolish tariff barriers. |
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Answer» Elle |
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| 20. |
There are 10 capacitors. each of capacitance 10 muF. The ratio of maximum to minimum capacitance obtained by their combination is ...... |
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Answer» `5:1` `C_(1) = nC = 10xx10= 100 muF ` In series connection, minimum capacitance `C_(2)` is obtained. `C_(2) = (C)/(n) = (10)/(10)= 1 mu F ` `:. (C_(1))/(C_(2))= (100)/(1) = 100 :1` |
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| 21. |
Following curve shows the variation of intesity of gravitational field (vec(I)) with distance from the centre of solid sphere(r) : |
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Answer»
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| 22. |
Explain how transistor can be used as an oscillator ? |
Answer» Solution :(i) In an oscillator, we get ac output without any external input signal.  (ii) Here L-Ccircuit in emitter-base circuit of transistorwhich is forward biased with battery `V_(B B )` . The collector emitter circuit is reverse biased with battery `V_( C C ) `. (iii) A coil `L_(1)` is inserted in collector emitter circuit . It is coupled with L. Working `:` (i) If we close the key ( K ) , weak collector current start RISING with time due to the inductance `L_(2)`. As a result,INCREASING magnetic flux is linked with `L_(1)` and `l.` brgt (ii) Due to mutual induction, anemf is induced in L which will charge the upper plate ofcapacitor ( C ) , CONSEQUENTLY there will be support to the forward of emitter base circuit. (iii) This results in an increasing in the emitter current and hence an increase in the collector current. (iv) Due to it, more increasing magnetic flux is linkedwith `L_(10` & L . (v) The aboveprocess continues till the collector current becomes maximum ( or ) saturated. (vi) The resonant frequency of tuned circuit at which the oscillator will oscillate. `v =(1)/( 2pi sqrt( LC))` |
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| 23. |
If n is the rigidity modulus, r is the radius , l is the length and C is the moment of the couple, then (2lc)/(pi n r^4) has the dimensions of |
| Answer» ANSWER :A | |
| 24. |
Define the terms input resistance and current amplification factor of a transistorin CE mode. |
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Answer» Solution :Input Resistance :- It is defined as the RATIO of CHANGE in base-emitter voltage `(DELTA V_(BE))`to the change in base current `(Delta I_B)`at constant COLLECTOR-emitter voltage `(V_(CE))` i.e., `r_i = ((Delta V_(BE))/(Delta I_B))_(v_(CE))` Current amplification factor :- It is defined as the ratio of change in collector current `(Delta I_C)`to the change in base current (`Delta I_B)`at constant collector-Emitter `(V_(CE))`when the transistor is in active state. i.e., `beta_(ac) = ((Delta I_C)/(Delta I_B))_(v_(CE))` |
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| 25. |
In fungi, bryophytes and pteridophytes, the fertilisation is |
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Answer» External |
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| 26. |
Total energy possessed by an electron revolving around the nucleus in an orbit of radius r is proportional to: |
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Answer» R |
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| 27. |
The mass defect of ""_(2)^(4)Heis 0.03 u. The binding energy per nucleon of helium (in MeV) is |
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Answer» `27.93` Mass NUMBER = 4 Building ENERGY `= 0.03xx9.31=27.93` Binding energy per nucleon `= (27.93)/(4)=6.9825` |
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| 28. |
The Y axis of the following graph may represent |
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Answer» A current in a circuit CONTAINING a source of constant emf., a PURE RESISTANCE and a prue inductor after a long time, when the source is open at time t = 0 |
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| 29. |
An inductor of 5 H inductance carries a steady current of 2A. How can a 50 V self-induced emf be made to appear in the inductor ? |
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Answer» Solution :L =5H, |e| = 50V. LET US produce the required emf by reducing current to zero. Now, `|e|=L(dI)/(DT)` or `dt=(LdI)/(|e|)=(5 xx 2)/(50)s=(10)/(50)s=(1)/(5)s=0.2s` So, the desired emf can be produced by reducing the GIVEN current to zero in 0.2 second. |
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| 30. |
The charged currents in the outer conducting regions of the earth's core are thought to be responsible for earth's magnetism. What might be the 'battery' (i.e., the source of energy) to sustain these currents ? |
| Answer» Solution :One possibility is the radioactivity in the interior of the EARTH. But NOBODY really knows the REAL SOURCE of ENERGY. | |
| 31. |
Van de Graaf generator is used |
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Answer» In producing nuclear power |
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| 32. |
Nuclear is not possible in laboratory. Why? |
| Answer» SOLUTION :This is because nuclear fusion REQUIRES temperatures as high as `10^6-10^7` K. Such high TEMPERATURE are OFTEN generated in nuclear FISSION. That is why fission precedes fusion. These processes cannot be carried out in laboratory. | |
| 33. |
Which of the following statement (s) is//are correct (may have more than one option correct): |
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Answer» the rest mass of a stable nucleus is less than sum of the rest MASSES of its separated nucleons. |
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| 34. |
Two projectiles A and B are thrown with velcoity v and v/2 respectively. They have the same range. If B is thrown at an angle of 15^(@) to the horizontal. A must have been thrown at an angle. |
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Answer» `SIN^(-1) (1/16)` |
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| 35. |
In an experiment, with the position of the object fixed, a student variesthe position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making anangle of 45^(@) with the x-axis meets the experimental curve at P.The coordinates of P wil be: |
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Answer» `(F/2, f/2)` |
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| 36. |
In magnet, the magnetic attraction is maximum at the ............. Of the magnet |
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Answer» |
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| 37. |
In Q. 133, if the strip is deflected slightly from its equilibrium position in the plane of the figure, the time period of the resulting oscillation is given by : |
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Answer» `(sqrtl)/([g^(2)+((P)/(mc))^(2)]^(1/4))` `T=[(mg)^(2)+((P)/(C))^(2)]^((1)/(2))` or `(T)/(m)=[g^(2)+((P)/(mc))^(2)]^((1)/(2))` This playes the ROLE of EFFECTIVE g. This period of small osicllation is `t=2pi sqrt(((l)/(T))/(m))=2pi (sqrt(l))/([g^(2)+((P)/(mc))^(2)]^((1)/(4)))` |
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| 38. |
An A.C. circuit contains a resistor 'R', an inductor 'L' and a capacitor 'C' connected in series. When it is connected to an A.C. generator of fixed output voltage and variable frequency, the current in the circuit is found to be leading the applied voltage pi//4 rad, when the frequency is f_(1). When the frequency of the generator increased to 'f_(2)' the current is found to be lagging behind the applied voltage by pi//4 rad. The resonant frequency of the circuit is |
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Answer» `(f_1 f_2)/( f_1 +f_2)` |
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| 39. |
The string of a sonometer is divided into two parts with the help of a wedge. The total length of the string is lm and the two parts differ by 2mm. When sounded together chcy produced two beats per second. The frequencies of the notes emitted by the two pints are |
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Answer» 499 & 497 Hz |
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| 40. |
An electron enters into a magnetic field at an angle of 60^@. Its path will be |
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Answer» parabolic |
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| 41. |
A non magnetic material is that which is ..... |
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Answer» not ATTRACTED by a magnet |
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| 42. |
A beam of light consisting of two wavelengths 800nm and 600nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1.4m away. If the two slits are separated by 0.28mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. |
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Answer» Solution :The two bright fringes will coincide when `mlambda_(1)=(m+1)lambda_(2)` `mxx800xx10^(-9)=(m+1)xx600xx10^(-9)` `:.m=3` `X_(m)=(mDlambda_(1))/(d)` `=((3xx1.4xx800xx10^(-9)))/((0.28xx10^(-3)))m` `=12XX10^(-3)m` |
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| 43. |
A narrow slit of width 1 mm is illuminated by monochromatic light of wave length 600 nm .The distance between the first minima on either side on a screen at a distance of 2m is |
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Answer» `1.2 `CM |
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| 44. |
एक वैद्युत द्विध्रुव को 2 xx10^5 N / C तीव्रता के वैद्युत क्षेत्र से 30^(@) कोण पर रखने के उस पर 4 N-m का बल-आघूर्ण लगता है। यदि द्विध्रुव की लम्बाई 2 सेमी हो, तो उस पर आवेश होगा: |
| Answer» ANSWER :C | |
| 45. |
Light waves producing interference have their amplitudes in the ratio 3 : 2. The intensity ratio of maximum and minimum of interference fringes is |
| Answer» Solution :`(I_("MAX"))/(I_("MIN"))= (A_("max")^2)/(A_("min")^2)= 25""(THEREFORE I propto A^2)`. | |
| 46. |
Which of the following is wrong for interference fringes? |
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Answer» FRINGES are due to LIMITED portion of wave front |
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| 47. |
A very ftexible chain of length L and mass M is vertically suspended with its lower end just touching the table. If it is is released so that each link strikes the table and comes to rest, what force the chain will exert on table at the moment 'y' part of length falls on table ? |
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Answer» Solution :Since chain is uniform, the mass of y part of chain will be `((M)/(L)y)`. When this part reaches the table, its total FORCE exerted must be equal to weight of y part RESTING on table + Force due to MOMENTUM imparted `= (M)/(L)YG +(((M)/(L)dy)sqrt(2gy))/(dt)=(Mg)/(L)y+(M)/(L)y+(M)/(L)v sqrt(2gy)` `(as(dy)/(dt)=v)=(Mg)/(L)y +(M)/(L)sqrt(2gy).sqrt(2gy)=3(My)/(L)g` |
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| 48. |
Describe Young's double slit experiment. |
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Answer» Solution :Young has kept slit S on screen A and two parallel slits having `SS_(1)` = `SS_(2)` on screen B parallel to the slit S and screen A and B PLACED parallel. Now placing screen C at far from screen B so that straight line fringes are PRODUCED by the SUPER imposing of the CYLINDRICAL wavefronts emanating from the slits `S_(1)S_(2)`.  WIDTH="80%"> The intensity distribution curve for interference of Young.s double slit experiment is in figure.  From the graph, it can be said that all interference fringes are the same width and the same intensity. Points on which the estructive interference occur have zero intensity. The distance between two successive bright fringes or two successive dark fringes is same and the intensity of fringe does not depend on the order of fringes. |
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| 49. |
A point object is placed infront of a plane mirror as shown in figure. X_(OM) rArr x - co-ordinate of object relative to mirrorX_(IM) rArr x -x-co-ordinate of image relative to mirrorX_(IM) = -X_(OM)differentiating V_(IM) = 0-V_(OM)V_I - V_M = - (V_0 - V_M) Velocity if image relative to mirror = velocity of object relative to mirror. basing on this information answer the questions The reflection surface of a plane mirror is vertical. A particle is projected in a vertical plane which is also perpendicular to the mirror. The initial velocity of the particle is 10 m/s and the angle of projection is 60^@. The point of projection is at a distance 5 m from the mirror. The particle moves towards the mirror. Just before the particle touches the mirror, the magnitude of relative velocity of approach of the particle and its image is |
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Answer» 10m/s |
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| 50. |
A point object is placed infront of a plane mirror as shown in figure. X_(OM) rArr x - co-ordinate of object relative to mirrorX_(IM) rArr x -x-co-ordinate of image relative to mirrorX_(IM) = -X_(OM)differentiating V_(IM) = 0-V_(OM)V_I - V_M = - (V_0 - V_M) Velocity if image relative to mirror = velocity of object relative to mirror. basing on this information answer the questions A point object is moving with a speed v before an arrangement of two mirrors as shown in figure. The magnitude of velocity of image in mirror M_1 with respected to image in mirror M_2 is |
| Answer» Answer :B | |
