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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. |
In YDSE, the two slits are separated by 0.1 mm and they are 0.5m from the screen. The wavelength of light used is 5000 Å. Find the distance between 7th maxima and 11th minima on the screen. |
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Answer» SOLUTION :GIVEN, d= 0.1mm, D=0.5m and `lamda = 5000Å = 5.0 xx 10^(-7) m` `Delta y = (y_11)_(dark) - (y_7)_("BRIGHT") =((2 xx 11 - 1)lamda D )/(2d) - (7 lamda D)/(d)` `= 8.75 xx 10^(-3) m = 8.75 mm` |
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| 2. |
The graph between wave number (1) and angular frequency (0) is |
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| 3. |
A particle moves in a straight line with retardation proportional to its displacement. It's loss of kinetic energy for any displacement x is proportionalto |
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Answer» `X^2` |
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| 4. |
What are the limitations of ohm's law ? |
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Answer» Solution :Ohm.s law fails if V -I GRAPHIS non-linear . Ohm.s law fails if the relation between V and I DEPENDSON the signof V .ex: - semi conductor diodes . Ohm.s law fails if the relation between V and I is not - uniqueex:-GAAS . |
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| 5. |
An oscillating LC circuit consists of a 75.0 mH inductor and a 3.60 muF capacitor. If the maximum charge on the capacitor is 5.00 muC, what are (a) the total energy in the circuit, (b) the maximum current, and (c) the period of the oscillations? |
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| 6. |
The current through a wire varies wilh time asI = I_(0)+ alpha t where I_(0)= 20 A and alpha= 3 A s^(-1) . Find the charge that flows across a cross - section of the wire in first 10 seconds. |
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Answer» Solution :350 C Q= `int_(0)^(10) I dt = int_(0)^(10) ( I_(0) + alpha t) dt = int_(0)^(10) ( 20 + 3t) dt ` `thereforeQ = [ 20 t + (3t^(2))/(2)]_(0)^(10)` ` therefore Q = [ 20 XX 10 + (3 xx 100)/(2) ] ` `therefore Q = [ 200 + 150]` `therefore Q = 350 C ` |
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| 7. |
Suppose a hydrogen molecule passed to the first vibration-rotational energy level. What spectral line will be observed when it returns to the ground state? |
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Answer» `epsi_(1)^(vib-rot)=epsi_(0)^(vib)+epsi_(1)^(rot)=(homega)/2+h^(2)/J` The transition to the ZERO level results in the radiation of a PHOTON of energy `epsi_(ph)=epsi_(1)^(rot)=h^(2)//J`. Expressing the photon energy in terms of its wavelength, `epsi_(ph)=2pihc//lamda`, we obtain `lamda=2piJc//h` |
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| 8. |
The refractive index material of an equilateral prism is sqrt3. Then angle of minimum deviation of the prism is: |
| Answer» Answer :B | |
| 9. |
Assertion : The magnetic poles of earth do not coincide with the geographic poles. Reason: The discrepancy between the orientation of a compass and true north-south direction is known as magnetic declination. |
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| 10. |
Three charges 4muC each are placed at the vertices of an equilateral triangle of side 9 cm. Find the electric potential at the centroid of that triangle. |
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| 11. |
Two waves A and B of frequencies 2MHz and 3MHz respectively are beamed in the same direction for communication via sky wave. Which one of these likely to travel longer distance in the ionosphere before suffering total internal reflection ? |
| Answer» Solution :With increases in frequency, refractive index of a medium also increases. For higher frequency waves, angle of REFRACTION is LESS (i.e., bending of wave inmedium BECOMES less). So, condition for total internal REFLECTION is attained after travelling greater distance by higher frequency 3MHZ) | |
| 12. |
Out of the many input signals of different frequencies, a series resonant circuit will accept one which has |
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Answer» the HIGHEST FREQUENCY |
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| 13. |
Blot-Savart law indicates that moving electrons (velocity v) produce a magnetic field B such that. |
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Answer» `Bbotv` |
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| 14. |
Who has first used the properties of showing the direction of the magnet and why ? |
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Answer» Solution :A thin long piece of a magnet when suspended freely, pointed in the north-south DIRECTION. When it was placed on a piece of cork which was then allowed to float in still water. It is shows also north-south direction. The name lodestone (or loadstone) given to a naturally occurring ore of IRON magnetite means leading STONE. The first use of magnet was made by the Chinese people to determine the direction while traveling in the sea. Caravans crossing the Gobi desert also employed magnetic NEEDLES. A Chinese legend narrates the tale of the victory of the emperor Huang-ti about four thousand years ago, which he owed to his craftsmen (engineers).  These engineers built a chariot on which they placed a magnetic figure with arms outstretched. The figure swiveled around so that the finger ofthe statuette on it always pointed south. With this chariot, Huang-ti.s troops were able to attack the enemy from the REAR in thick fog and to defeat them. |
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| 15. |
A current passes throgugh a copper voltameter and water voltameter I series. How much hydrogen at 27^(@)C and pressure of 100cm of mercury wil be liberated during the time it takes for 17.5g of copper to be deposited? (Atomic weight of copper=63 and density of hydrogen at S.T.P.=0.09 kg m^(-3). |
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| 16. |
The specific resistance for the insulators is in the range of _________ |
| Answer» SOLUTION :`10^(8-)10^(14)OMEGA m` | |
| 17. |
The electric field due to an infinitely long straight uniformly charged wire at a distance r is directly proportional to |
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Answer» R |
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| 18. |
A stream of glass beads, each with a mass of 15 gram, comes out of a horizontal tube at a rate of 100 per second. The beads fals a distance of 5m to a balance pan and bounce back to their original height. How much mass (in kg) must be placed in the other pan of the balance to keep in pointer at zero? |
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Answer» `=2msqrt(2gh)=2xx15xx10^(-3)(10)=0.3kg ms^(-1)` Then i `/_\p_(100)` ( in 1 second) `=30kgms^(-1)` FORCE EXERTED `=30N` So weight requried `=30N`. So mass `=3kg` |
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| 19. |
Internal resistance of a cell is r = 5Omega & is connected in potentiometer (wire length = 1m) circuit arrangements as shown in figure, which shows two circular conducting rings R_(1) & R_(2)each having radii 20 cm cross each other at conducting joints A & B. Section AB subtends 120^(@) at the center of each ring. Resistance per unit length of R_(1) & R_(2) are (3)/(4)Omega//cm & (6)/(pi)Omega//cm respectively. Cell is connected across point C & D of rings lying on perpendicular bisector of AB. Now when switch Sw remains closed, balancing length at null point is 48 cm. Find the balancing length (in cm.) when switch Sw is opened. |
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Answer» SOLUTION :`R_(AC)=R_(CB)=(2pixx20)/3xx3/pi=40 Omega` & `R_(AD)=R_(BD)=(2pixx20)/3xx6/pi=80Omega` `rArr` Balanced W.S.B `rArr R_(CD)=120/2=60 Omega` `epsilonxx(60/(60+5))=48x` & `epsilon=x l` `rArr lxx60/55=48 rArr l=52 cm` |
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| 20. |
In which country people enjoy better physical health? |
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Answer» DEVELOPING countries |
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| 21. |
The current in the arm CD of the circuit will be: |
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Answer» `i_1 + i_2` |
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| 22. |
Define binding energy of a nucleus. |
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| 23. |
A Capacitor of capacitance C=5muF is connected to source of emf varepsilon_(2)=200V with the switch S in the position 1 (Figure 6.2). Subsequently the switch is pushed to the position2. Find the amount of heat generated in R_(1)=500 Omega if R_(2)=300Omega. |
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| 24. |
Under the influence of a uniform magnetic field a charged particle moves with constant speed V in a circle of radius R. The time period of rotation of the particle. |
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Answer» Depends on R and not on V |
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| 25. |
A positively charged glass rod attracts a suspended object. Can we conclude that the object is negatively charged ? |
| Answer» Solution :No, there may be attraction between the rod and SUSPENDED neutral OBJECT DUE to induction. | |
| 26. |
What is the difference between an -type and a p-type intrinsic semiconductor? |
| Answer» Solution :In a n-type SEMICONDUCTOR DOPANT is pentavalent and electrons are majority charge carriers. How- EVER, a trivalent material is used as a dopant to OBTAIN p-type semiconductor and holes are the majority charge carriers. | |
| 27. |
Two heat engines A and B have their sources at 327^(@)C and 227^(@)C and sinks at 127^(@)C and 27^(@)C the ratio of their efficiencies is |
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Answer» `6//5` `=(600-400)/(600)xx(500)/(500-300)=(2)/(6)xx(5)/(2)=(5)/(6)`. |
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| 28. |
जीव विज्ञान का जनक माना जाता है |
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Answer» चियोफ्रेस्टस |
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| 29. |
Electric current is ..... . |
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Answer» <P>scalar quantity  Consider three conductors, meeting at junction p. Here when `I_(1) and I_(2)` CURRENTS COMBINE at point P, they always give `I_(1) + l_(2)= I` Only irrespective of angle `theta`. It MEANS that currentis a scalar quantity. Moreover , I = `(Q)/(t)` where charge Q and time t both are scalar. Hence current is also a scalar quantity. |
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| 30. |
An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R ) respectively, in a uniform magnetic field vecB=B_(0)hati, each with an equal momentum of magnitude p = eBR. Under what conditions on the direction of momentum will the orbits be non-intersecting circles ? |
Answer» Solution :1. If RADIUS of the circle made from combination of electron and positron pair then if this is more than 2R, than these two circle does not intersect with each other.  2. Magnetic field `vecB` is in the direction of x, so momentum of circular system is in plane yz. 3. Both particles are opposite to each other from orbital radius R. 4. Let `P_(1)andP_(2)`, are momentum of electron and positron. 5. Suppose `P_(1)` form `theta` angle with y-axis and `P_(2)` also form same angle. 6. Centre of RESPECTIVE circle are at origin position and perpendicular to R distance. 7. Centre point of path of electron-positron circle as shown by `C_(E)andC_(p)`. 8. Coordinate of `C_(e)(0,-Rsintheta,Rcostheta)` co-ordinate of `C_(p)(0,-Rsintheta,3/2R-Rcostheta)` 9. If distance between centre point to radius for circle is GREATER than 2R than circle doesn.t inter sect with each other. 10. Suppose, distance between `C_(p)andC_(e)` be d. `therefored^(2)=(2Rsintheta)^(2)+(3/2R-2Rcos0^(@))^(2)` `d^(2)=4R^(2)sin^(2)theta+9/4R^(2)-6R^(2)costheta+4R^(2)cos^(2)theta` = `4R^(2)+9/4R^(2)-6R^(2)costheta` 11. If distance of d be more that 2R, `dgt2R` `d^(2)gt4R^(2)` `therefore4R^(2)+9/4R^(2)-6R^(2)costhetagt4R^(2)` `therefore9/4R^(2)gt6R^(2)costheta` `therefore9/4gt6costheta` `thereforecosthetalt3/8` So, this is necessary condition. |
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| 31. |
If the magnetic dipole moment of an atom of diamagnetic material paramagnetic material andferromagnetic material are denoted by m_(d), m_(p) and m_(f) respectively then |
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Answer» `m_(d) = 0 and m_(p) pm 0` `m_(d) = 0. m_(p) ne 0.` |
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| 32. |
A light bulb is rated at 200W for a 230 V supply. Findthe rms current through the bulb. |
| Answer» SOLUTION :`I APPROX` 0.87 A | |
| 33. |
Mona and Anushka are friends, both studying in class 12. Mona is in Science stream and Anushka in in Arts stream. Both of them go to market to purchase sunglasses. Anushkafeels that any coloured glasses with fancy look are good enough. Mona tells her to look for UV protection glasses, polaroid glasses and photo sensitive glasses. Read the above passage and answer the following questions : (i) What are UV protection glasses, polaroid glasses and photo sensitive glasses ? (ii) What values are displayed by Mona ? |
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Answer» Solution :(i) `UV` protaction glasses are those which filter ultra-violet rays that are HARMFUL to our eyes. Polaroid glasses help in reducing the glare. Photo-sensitive glasses GET DARKER in strong day light. They PROTECT our eyes from strong sunlight especially at noon. Mona has displayed concern for her friend. She has put to use the knowledge she acquired in her science classes. Mugging up things for examination is of no use. What we are taught in class room must be used in practice. |
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| 34. |
Electric current density is a_______ quantity whereas electric current is a ______ quantity. |
| Answer» SOLUTION :VECTOR , SCALAR | |
| 35. |
A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth.s magnetic field B. (i) Write the expression for the instantaneous value of the emf induced in the wire. (ii) What is the direction of the emf ? (iii) Which end of the wire is at the higher potential ? |
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Answer» SOLUTION :(i) `epsilon= BLv` (II) WEST to east (III) East |
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| 36. |
In the electric field of a point charge Q placed at the centre of a circle, as shown in fingure, a certain test charge q_0 is carried from point A to B, C B and D. Then the work done |
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Answer» is LEAST along the path AB. |
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| 37. |
A student finds the balancing length to be 'l' with a cell of constant emf in the secondary circuit. Another student connects the same cell in the secondary circuit of potentiometer of double the length but with a cell of half the emf in the primary circuit. The balancing length will be [cell in primary is ideal and no series resistance is present in primary circuit.] |
| Answer» ANSWER :A | |
| 38. |
When the electromagnetic radiation of maximum wavelength of 2480 nm is incident on semiconductor. The band gap energy of semiconductor is ……eV. [6.6xx10^(-34)JS, C=3xx10^(8)m//s, 1eV=1.6xx10^(-19)J] |
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Answer» 0.9 `E_(g)=(hc)/(lambda)` `E_(g)=(6.6xx10^(-34)xx3xx10^(8))/(248xx10^(-8)xx1.6xx10^(-19))` `=(19.8xx10^(-26))/(396.8xx10^(-27))=0.049899xx10^(1)~~0.5eV` |
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| 39. |
Give any two applications of X-rays. |
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Answer» Solution :(i) X-rays are used as DIAGNOSTIC TOOL in medicine. (ii) To TREAT CERTAIN forms of CANCER. |
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| 40. |
Fraunhofer lines of the solar system is an example of |
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Answer» emission LINES spectrum |
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| 41. |
Explain how, using a Polaroid and a quarter-wave plate made of positive uniaxial crystal (n_(e) gt n_(0)), yo distinguish (a) light with left-hand circualr polarization from that with right-hand polarization, (b) natural light from light with circualr polarization and from the compoitive of natural light and that with circular polarization. |
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Answer» Solution :The light with right circular polarization (viewed against the oncoming light, this means that the light vector is moving clock WISE) becomes plane polarized on passing through a quarter-wave plate. In this case the direction of oscillations of the ELECTRIC vector of the electromagnetic wave forms an angle of `+45^(@)` with the axis of the CRYSTAL `OO'` (see Fig. `(b)` below). In the cae of left hand circular polarizations, this angle will be `- 45^(@)` (Fig.) (b) If for any position of the plate the rotation of the polaroid (located brhind the palte) does nor bring about any variation in the intensity of the TRANSMITTED light, the INCIDENT light is unpolarized (i.e. natural). If the intensity of the transmitted light can drop to zero on rotating the analyzer polarid for some position of the quarter wave plate, the incident light is circulary polarized. If it varies but does not drop to zero, it must be a maxture of natural and circualry polarized light. 
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| 42. |
In the given Boolean expression, Y = A.barB+ B.barA if A=1, B=1 then Y will be |
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Answer» 0 |
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| 43. |
If a satellite performes elliptical path, then |
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Answer» `v_(h) LT v_(c)` |
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| 44. |
An inductor (L=0.03H) and a resistor (R=0.15kOmega) are connected in series to a battery of 15 V emf in a circuit shown below. The key K_1 has been kept closed for a long time. Then at t=0 , K_1 is opened and key K_2 is closed simulatenously. At t=1 ms, the current in the circuit will be (e^5cong150) |
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Answer» 100 mA `I_0=e/R=15/(0.15times10^3)=0.1A` Here, t=1 ms =`10^-3` s Time CONSTANT of the LR circuit, `t_0=L/R=0.03/(0.15times10^3)=2times10^-4s` `thereforet/t_0=10^-3/(2times10^-4)=5` `thereforeI=I_0e^(-t//t_0)=0.1e^-5=0.1/150=0.67times10^-3A` =0.67 mA |
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| 45. |
Arrange the following communication frequency bands in the increasing order of frequencies a. AM broadcast b. Cellular mobile radio c. FM broadcast d. Television UHF e Satellite communication |
| Answer» Answer :D | |
| 46. |
A hollow copper tube of 5 m length has got external diameter equal to 10 cm and its walls are 5 mm thick. The specific resistance of copper is 1.7 xx 10^(-8) Omega m. The resistance of the copper tube, approximately |
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Answer» `5.4 xx 10^(-3)Omega` |
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| 47. |
Surface charge density of soap bubble of radius ‘r’ and surface tension T is sigma. If P is excess pressure, the value of sigma is |
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Answer» <P>`in_0[(4T)/R - P]^(3//2)` |
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| 48. |
An electric dipole of moment p is lying along a uniform electric field vec(E). The workdone in rotating the dipole by 90^(@) is |
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| 49. |
Give main characteristics of intrinsic semiconductors. |
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Answer» Solution :CHARACTERISTICS of intrinsic semiconductor `(i)` It is a pure semiconductor without any ADDITION of impurity ATOMS. `(ii)` Outer shell is completely filled at `0K`. `(iii)` At room temperature, electrical conductivity is low. `(iv)` Its electrical conductivity increases with rise in temperature. `(v)` Number of ELECTRONS (in. C.B.) is equal to number of holes (in V.B.). |
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| 50. |
Minimum deviation of prism having refractive index mu and small angle of prism A is shown by ..... |
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Answer» `delta_m=(mu-1)A` For small ANGLE of PRISM and `delta_m` `sin((A+delta_m)/(2))~~(A+delta_m)/(2)` and `sin(A/2)~~A/2` `THEREFORE mu=(A+delta_m)/(2)` `therefore muA=A+delta_m``therefore delta_m=(mu-1)A` |
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