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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 magnetic materials having negative magnetic susceptibility are called ___and those having positive magnetic susceptibility are called __________ . |
| Answer» SOLUTION :DIAMAGNETIC , paramagentic or FERROMAGNETIC | |
| 2. |
A triode valve has an amplification factor 20and its plate potential is 300 V . What should be the grid potential to reduce the plate current zero ? |
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Answer» `+15` VOLT |
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| 3. |
A parallel-plate capacitor is clamped to a stand such that its plates are vertical. It remains connected to a cell through a centre-zero galvanometer. A metal sheet which is parallel to the plate of the capacitor and whose thickness is slightly less than the seperation between these plates is now allowed to fall under gravity though the plates, wihtout touching them. Which of the following correctly described the motion of the metal plate as it passes through the plates of the capacitors ? |
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Answer» It will MOVE with a constant acceleration equla to g the acceleration due to gravity. |
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| 4. |
How do you justify that the rest mass of photons in zero ? |
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Answer» Solution :The MASS of a body moving with speed v m =`m_0/sqrt(1-v^2/C^2)` Rest mass, `m_0 = m Sqrt(1 - c^2/v^2)` For a photon, v = c, THEREFORE `m_0 = m sqrt(1 - C^2/v^2)` = ZERO. |
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| 5. |
A parallel-plate capacitor is clamped to a stand such that its plates are vertical. It remains connected to a cell through a centre-zero galvanometer. A metal sheet which is parallel to the plate of the capacitor and whose thickness is slightly less than the seperation between these plates is now allowed to fall under gravity though the plates, wihtout touching them. In the entire process of the metal sheet passing through the plates, which of the following statements will not be correct ? |
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Answer» Its GAIN in kinetic ENERGY will be less than its loss in gravitational potential energy. |
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| 6. |
If the angular velocity of a body is halved,then the ratio of its angular momentum to the new angular momentumwill be |
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Answer» 2:1 |
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| 7. |
A ball is dropped from a spacecraft revolving around the earth at a height of 120 km. What will happen to the ball ? |
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Answer» it will fall down to the EARTH gradually |
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| 8. |
(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step up transformer. (ii) A step up transformer converts a low input voltage into a high output voltage. Does it violate law of conservation of energy? Explain. (iii) Write any two sources of energy loss in a transformer. |
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Answer» Solution :(i) N/A (ii) A step up TRANSFORMER does not violate the LAW of conservation of energy. ALTHOUGH it enhances the output voltage but correspondingly reduces, the circuit current so that POWER output of even an ideal transformer is same as the power input. (iii) N/A |
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| 9. |
A capacitor of capacitance C has reactance X. If capacitance and frequency become double then reactance will be |
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Answer» 4X |
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| 10. |
A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval.Show the variation of its acceleration with time.(Take acceleration in the backward direction as positive) |
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Answer»
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| 11. |
What is the force of repulsion between two charges of 1C each, kept 1m apart in vacuum ? |
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Answer» `11xx10^(9) N` |
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| 12. |
Discuss the variation of photocurrent with collector plate potential for different intensity of radiations. |
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Answer» Solution :Effect of potential on photoelectric current. Light of fixed intensity `I_(1)`and frequency v be made to fall on photosensitive plate. The electrode C is made negative so that the current becomes zero. Let it be `V_(0)`(called stopping potential). Now the potential of C is increased in steps and a graph is plotted. The graph is as SHOWN in the FIGURE.  If the same experiment is repeated with radiation having intensity `I_(2) (I_(2) gt I_(1))` the saturation current will be higher but stopping potential is the same. It MEANS that stopping potential is independent of intensity of incident radiations. The retarding potential, `V_(0)`for which the photoelectric current just becomes zero is called stopping potential or cut-off potential. If mass of electron is m, charge e and maximum velocity is `v_("max")` then : `1/2 mv_("max")^(2) = eV_(0)` And `v_("max") = sqrt((2eV_(0))/m)`. |
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| 13. |
Mention the two characteristic properties of the material suitable for making core of a transformer. |
| Answer» Solution :A MATERIAL suitable for making CORE of a transformer should have (i) a narrow hysteresis curve so as to minimise energy dissipation during each magnetisation cycle, and (ii) a HIGH RESISTIVITY to lower eddy current losses. | |
| 14. |
In the forward bias arrangement of a p-n junction diode |
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Answer» the n-end is connected to positive terminal of the BATTERY. |
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| 15. |
Nanukaka was coming from delhi to- |
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Answer» SEE some people |
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| 16. |
The velocity of a particel is zero at t=0. a) The acceleration at t=0 must be zero. b)The acceleration at t=0 may be zero. c)IF the acceleration is zero from t=0 tot =10s,the speed is also zero in this interval. d)If the speed is zero from t=0 t=10 s the acceleration is also zero in this interval |
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Answer» a,b & d are correct |
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| 17. |
A coil of resistance 20Omegaand self-inductance 5 H connected with battery 100 V. What will be the value of energy stored ? |
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Answer» 31.25 J `=1/2LI^2` `=1/2L[E/R]^2 "" [because I=E/R]` `=1/2xx5xx[100/20]^2=1/2xx5xx25`=62.5 J |
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| 18. |
Define the term ‘Mobility' of charge carriers in a conductor. Write its SI unit. |
| Answer» Solution :Mobility of charge CARRIERS in a CONDUCTOR is the magnitude of DRIFT velocity of the charge carriers per unit electric field. Its SI unit is `m^2 V^(-1) s^(-1)` . | |
| 19. |
A short bar magnet placed with its axis at 30^@ with a uniform magnetic field of 0.16 T experiences a torque of magnitude 0.032 J. The magnetic dipole moment of the bar magnet is |
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Answer» `0.23 J T^(-1)` |
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| 20. |
How can you keep the LC oscillation undamped ? |
| Answer» SOLUTION :By PROVIDING a FEEDBACK CIRCUIT . | |
| 21. |
When sunlight shines on the atmosphere, mass of CO_(2) at an altitude of 75 km undergo natural laser action. The energy levels involved in the action are shown in fig Population inversion occurs between levels E_2 and E_1. The wavelength at which lasing occurs is |
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Answer» 60100Å The `lambda_(La)=(hc)/(E_(2)-E_(1))=100250 Å` |
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| 22. |
An air condenser of 1 miff is immersed in a transformer oil of dielectric constant 3. The capacity of oil capacitor is: |
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Answer» `0.33muF` |
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| 23. |
When sunlight shines on the atmosphere, mass of CO_(2) at an altitude of 75 km undergo natural laser action. The energy levels involved in the action are shown in fig Population inversion occurs between levels E_2 and E_1. 66. The wavelength of sunlight exciting the molecules in laser action is |
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Answer» 25010 Å `(hc)/(lambda_(ex))=E_(2)-E_(0)` `lambda_(ex)="wavelength of exciting radiation"` `(lambda_(ex))/(hc)=(1)/(E_(2)-E_(0))` `rArr lambda_(ex)=(hc)/(E_(2)-E_(0))` `=(6.63 xx 10^(-34) xx 3 xx 10^(8))/((0.289-0) xx 1.6 xx 10^(-19))m` `=43015Å` |
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| 24. |
Geometrical shadow is formed due to the phenomenon of |
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Answer» DIFFRACTION of LIGHT |
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| 25. |
A uniform electric field and a uniform magnetic field are produced, pointing in the direction. If an electron is projected with its velocity pointing in the same direction _____ |
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Answer» The electron will TURN to its right.  As shown in figure, the magnetic force acting on an electron moving PARALLEL with the magnetic field will be zero. So, electric force according to `vecF_(e)=-vecE_(e)` will act on electron in the DIRECTION opposite to its motion. As a result of which the velocity of electron will decrease. |
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| 26. |
A pendulum consisting of a small sphere of mass m suspended by aninextensible and massless string of length l is made to swing in a vertical plane. If the breaking strength of the string is 2 mg, then the maximum angular amplitude of the displacement from the vertical can be |
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Answer» `0^(@)` length of pendulum = l Let at any instant the amplitude = `THETA` Then ACCORDING to CONSERVATION of energy K.E. = P.E. `(1)/(2)mv^(2)=mg(l-lcostheta)` `mv^(2)=2mgl(1-costheta)` Equation of motion is `T-mg=(mv^(2))/(l)` or `T=mg+(mv^(2))/(l)` `T=mg+2mg(1-costheta)` or `2mg=mg+2mg-2mgcostheta` or `2mg costheta=mgorcostheta=(1)/(2)` `therefore theta=60^(@)` |
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| 27. |
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E_(2)-E_(1)).n = sigma/epsilon_(0) where hatn is a unit vector normal to the surface at a point and sigmais the surface charge density at that point. (The direction of hatnis from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is sigma hatn //epsilon_(0). (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.] |
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Answer» Solution :Electric field NEAR the surface of a charged plane PLATE, `E = (sigma)/(2in_(0))` If electric field is considered to be vector then the unit vector `hatn` is one side and - `hatn` is other side of plane.  Electric field from side (1) to side (2), `vecE_(2)= (sigma)/(2in_(0)) . hatn` Electric field from s.ide (2) to side (1), `vecE_(1)=-(sigma)/(2 in_(0)). hatn` Electric field to side (1) `(vecE_(2)-vecE_(1))=[(sigma)/(2in_(0))-(-(sigma)/(2in_(0)))]hatn` `= (2sigma)/(2 in_(0))hatn` `= (sigma)/(in_(0)) hatn` where `sigma` is the surface charge density and unit vector `hatn` is from side (1) to side (2), Since `E_(1)` and `E_(2)` are opposite to each othe so electric field has discontinuity on plat and the electric field DISAPPEAR inside the conductor hence `vecE_(1) =0` So, electric field outside the conductor `vecE=vecE_(2) = (sigma)/(in_(0))hatn` |
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| 28. |
Induction furnace is based on the heating effect of ..... |
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Answer» EDDY current |
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| 29. |
Arrange the following electronmagnetic waves in ascending order of their wavelength: Radio waves, Gamman rays, Infrared waves, X-rays |
| Answer» SOLUTION :`gamma`- RAYS `gt x -` rays `gt` Infrared rays `gt` RADIO WAVES | |
| 30. |
It is found that |a + b| = |a|. This necessarily implies. |
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Answer» B=0 |
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| 31. |
A small particle of mass m and charge Q is dropped in uniform horizontal magnetic field B. The maximum vertical displacement of particle is given by h=(nm^2g)/(2Q^2B^2) . Find the value n. |
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Answer» |
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| 32. |
A beam of unpolarised light passes through a tourmaline crystal A and then it passes through a second tourmaline crystal B oriented so that its principal plane is parallel to that of A. The intensity of emergent light is I_(0). Now B is rotated by 45^(@) about the ray. The emergent light will have intensity |
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Answer» `(I_(0))/(2)` |
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| 33. |
A convex lens of focal length 0.5 m and a concave lens of focal length 1 m are combined.The power of the resulting lens is |
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Answer» `1D` |
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| 34. |
A laser beam is used for locating distant object because it |
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Answer» has small angular SPREAD |
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| 35. |
24 ग्राम मे उपस्थित कार्बन के मोल की संख्या |
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Answer» 1mol |
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| 36. |
Statement (A) : NAND gate output is inverse of AND gate. Statement (B): The Boolean expression for a two input NAND gate is Y = bar(A.B) |
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Answer» A is TRUE, B is FALSE |
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| 37. |
Obtain the resonant frequency and Q-factor of series LCR circuit with L=3.0H, C = 27 muF, and R = 10.4 Omega. It is desired to improved the sharpness of the resonance of the circuit by reducing its 'full width at half maximum' by a factor of2. Suggest a sultable way. |
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Answer» Solution :Given, inductance L = 3H Capacitance of capaitor `C=27muF = 27xx10^(-6)F` Ristance `R =7.4 Omega` the resonant FREQUENCY of circuit `omega_(r )=(1)/(sqrt(LC))=(1)/(sqrt(3xx27xx10^(-6)))=(1000)/(9)` = 111.1 rad/s. Q-factor of a series LCR circuit Q-factor `= (omega_(r )L)/(R )=(111.1xx3)/(7.4)=45.04`. To reduce the width at half by factor Q, we have to reduce the value of R as R/2 `(R )/(2)=(7.4)/(2)=3.7 Omega` |
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| 38. |
Statement (A) : Ampere's law states that the line integral of vecB.vecdl along a closed path round the current carrying conductor is equal to μ_(0)i (i is the net current through the surface bounded by the closed path) Statement (B) : Ampere's law can be derived from Biot savart's law |
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Answer» A is TRUE B is false |
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| 39. |
An infinitely thin straight wire has uniform linear charge density lamda . Obtain the expression. For the electric field (E) at a point lying at a distance x from the wire, using Gauss' law. (b) Show graphically the variation of this electric field E as a function of distance x from thewire . |
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Answer» Solution :Consider an infinitely long straight charged wire of linear charge DENSITY `lamda`. To find electric field at a point P situated at a DISTANCE r from the wire by using Gauss. LAW consider a cylinder of length l and radiusr as the Gaussian surface. From symmetry consideration electric field at eac point of its curved surface is `vecE` and is POINTED outwards normally. Therefore , electric flux over the curved surface. `=intvecE.hatnds = E 2pirl` On the side faces 1 and 2 of the cylinder normal drawn on the surface is perpendicular to electric field E and hence these surface do not contribute towards the total electric flux. `:.` Net electric flux over the entire Gaussian surface `phi_E=E.2pirl ""....(i)` By Gauss law electric flux `phi_E=1/in_0` (charge enclosed ) `(lamdal)/in_0 ""...(ii)` Comparing (i) and (ii) , we have `E.2pirl=(lamdal)/in_0` `implies E=lamda/(2piin_0r) and vecE=lamda/(2piin_0r).hatr` (B) The E - r graph is as shown here.  
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| 40. |
Identify the electromagnetic waves whose wavelengths lie in the range (a) 10^(-11)m lt lambda lt 10^(-14)m (b) 10^(-4)m lt lambda lt 10^(-6)m Write one use of each. |
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Answer» Solution :(a) `gamma-` RAYS have their wavelengths within the range `10^(-11)m LT lambda lt 10^(-14)m`. These rays are used for treatment of cancer and the TECHNIQUE is KNOWN as, .radio therepy.. (b) Infrared waves have their wavelengths within the range `10^(-4)m lt lambda lt 10^(-6)m`. These rays are widely used in the remote switches of household electronic gadgets such as TV sets, video RECORDERS. Wi-Fi systems etc. |
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| 41. |
Give the location of magnetic north pole. |
| Answer» Solution :At a LATITUDE of `79,74^@` N and a LONGITUDE of 71.8° WIN north CANADA | |
| 42. |
Two rain drops falling through air have radii in the ratio 1:2. They will have terminal velocity in the ratio |
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Answer» `4:1` |
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| 43. |
Assertion :- At any instant , if the current through an inductor is zero, then the induced emf may not be zero Reason :- An inductor tends to keep the flux (i.e. current ) constant |
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Answer» If the ASSERTION & REASON are TRUE& the Reason is a correct explanation of the Assertion . |
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| 44. |
Explain with reason if there can be an acceleration in the motion of a body when the velocity of a body is zero. |
| Answer» SOLUTION :YES, This happens when the body CHANGES its direction of motion under acceleration. For example in S.H.M, when the body is in the position of maximum DISPLACEMENT, it's VELOCITY is zero but it's acceleration is maximum. | |
| 45. |
Input characteristics are shown for CE configuration of n-p-n transistor for different output voltage. Here. |
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Answer» `V_("CE"_(1) ) GT V_("CE"_(2) )` |
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| 46. |
A crate, in the form of a cube with edge lengths of 1.2 m, contains a piece of machinery, the center of mass of the crate and its contents is located 0.30 m above the crate's geometrical center. The crate rests on a ramp that makes an angle theta with the horizontal. As thetais increased from zero, an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefficient of static friction between ramp and crate is 0.30, (a) does the crate tip or slide and (b) at what angle theta does this occur? If mu_(s) = 0.70, (c) does the crate tip or slide and (d) at what angle theta does this occur? (Hint: At the onset of tipping, where is the normal force located?) |
| Answer» Solution :As`THETA` is increased from zero the CRATE slides before it TIPS, (b) It STARTS to slide when `theta= 17^(@)`, (c) The crate begins to slide when `theta = 35.0^(@)` and begins to TIP when `theta = 33.7^(@)`, thus, it tips first as the angle is increased, (d) Tipping begins at `theta= 33.7^(@)~~ 34^(@)` | |
| 47. |
Two wires A and B of same material and mass have their lengths in the ratio 1 : 2. On connecting them separately to the same source, the rate of heat dissipation in B is found to be 5W. The rate of heat dissipation in A is |
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Answer» 10W |
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| 48. |
Suppose that the electric field part of an electromagnetic wave in vacuum is vecE={(3.1N//C)cos[(1.8"rad/m")y+(5.4xx10^(6)"rad/s")t]}hati. (a) What is the direction of propagation? (b) What is the wavelength lambda? (c ) What is the frequency v? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. |
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Answer» Solution :Here electric field part is of the form `vecE=E_(0)cos(ky+omegat)hati` where `E_(0)=3.1N//, K=1.8" rad/m and "omega=5.4xx10^(6)"rad/s"` (a) The equation for `vecE` clearly shows that dreiction of PROPAGATION is along negative DIRECTION of y - axis. (a) The equation for `vecE` clearly shows that direction of propagation is along negative direction of y - axis. (b) `lambda=(2pi)/(k)=(2xx3.14)/(1.8)=3.5m` (c ) Frequency `v=(omega)/(2pi)=(5.4xx10^(6))/(2xx3.14)=8.6xx10^(5)Hz or 0.86MHz` (d) Amplitude of the MAGNETIC fieldpart of the wave `B_(0)=(E_(0))/(c )=(3.1)/(3xx10^(8))=1.033xx10^(-8)T=10nT.` (e) As `vecE` is along - ve y - axis, hence `vecB` must be along - ve - z axis and, therefore, expression for the magnetic field part of the wave should be `vecB=B_(0)cos(ky+omegat)hatk` `rArr""vecB=(10nT)cos[(1.8"rad/m")y+(5.4xx10^(6)"rad/s")t]hatk`. |
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| 49. |
A : Lenses of large aperture suffer from spherical aberration. R : The curvature of the lens at central and peripheral regions is different. |
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Answer» If both ASSERTION & Reason are true and the reason is the correct explanation of the assertion, then MARK (1). |
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| 50. |
A ring of mass m, radius r with charge per unit length lambda. Encloses a magnetic field such that B=-B_(0)hatk, when r le a, B=0 when r gt a when the magnetic field is switched off. The rings starts to rotate due to induced electric field with varying flux. Find angular velocity (in 10^(-2)rad//s) with which ring rotates after the magnetic field has been compeletely turned off. B_(0)=1 Tesla, a=1cm, r=2cm, m=0.5kg, lambda=(4)/(pi)C//m. |
| Answer» ANSWER :B | |
