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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 a Lyod's mirror set up for interference of waves of frequency 6xx10^(14)s^(-1) emitted from point source S. distance SO, OA & AC are 1mm, 50 cm and 50 cm respectively, then te number of dark and bright fringes obtained on the scren will be |
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Answer» Dark=4,BRIGHT =4 F.W. `=(Dlamda)/(2D)=0.25mm` |
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| 2. |
An electron having momentum 2.4 xx 10^(-23)" kg m s"^(-1) enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30^(@) with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be |
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Answer» 2mm `r=(mv_(_|_))/(eB)=("mv sin "theta)/(eB)=((2.4xx10^(-23)"kg m s"^(-1))xxsin 30^(@))/((1.6xx10^(-19)C)xx(0.15T))` `=5xx10^(-4)m=0.5xx10^(-3)m=0.5mm` |
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| 3. |
A primitive diving bell consists of a cylindrical tank with one end open and one end closed. The tank is lowered into a freshwater lake, open end downward. Water rises into the tank, compressing the trapped air, whose temperature remains constant during the descent. The tank is brought to a halt when the distance between the surface of the water in the tank and the surface of the lake is 40.0 m. Atmospheric pressure at the surface of the lake is 1.01 xx10^5 Pa. Find the fraction of the tank's volume that is filled with air. |
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Answer» `0.205` |
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| 4. |
Two gases A and B having same temperature T, same pressure P and same volume V are mixed. If the mixture is at same temperature T and occupies a volume V then the pressure of the mixture is : |
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Answer» 2P PV +PV=P.V `rArrP.=2P` THUS correct CHOICE is (a). |
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| 5. |
In the ciruit shown, current throught the resistance 2Omega is i_(1) and current through the resistance 30Omega is i_(2). Fing the ratio (i_(1))/(i_(2)). |
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Answer» Solution :POTENTIALS are indicated in figure Current in `2Omega=(10-(-5))/(2)=(15)/(2)=7.5A` , leftwards Current in `30Omega=(10-(-15))/(30)=(25)/(30)=(5)/(6)A` , downwards `(i_(1))/(i_(2))=9` |
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| 6. |
While passing through a prism of angle 6^(@) a ray of light undergoes a devation of 3^(@). The refractive index of the material of the prism is ____________. |
| Answer» SOLUTION :HINT : `because delta=(n-1)A`, hence `3^(@)=(n-1)xx6^(@)RARR n=1.5` | |
| 7. |
Name the material that is used as control rods in a nuclear reactor. |
| Answer» SOLUTION :BORON or CADMIUM | |
| 8. |
उपापचय पदार्थों का शरीर से निष्कासन क्या कहलाता है |
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Answer» श्वसन |
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| 9. |
स्त्री में अण्डे का निषेचन होता हैं- |
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Answer» गर्भाशय में, |
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| 10. |
Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured.Slit is illuminated by the light of wavelength 6000 Å. If slit illuminated by light of another wavelength, angular width decreased by 30%. Wavelength of light used is : |
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Answer» `3500 Å` For first diffraction, n = 1 `therefore d sin theta = lambda` or `d theta = lambda` (`because sin theta = theta` for small angle) For ANGULAR width, `theta = (lambda)/(d)` For full angular width, `OMEGA = 2 theta = (2 lambda)/(d)` Again `omenag. = (2 lambda.)/(d)` `therefore` DIVIDING we get. `(omega.)/(omega) = (lambda.)/(lambda) ` or `lambda. = (lambda. omega.)/(omega)` `therefore lambda. = 4200 Å`. |
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| 11. |
Read the followingpassage and then answer questions (a) - (e) on the basis of your understand- ing of the passage and the related studied concepts.A semiconductor diode is basically a p-n junction and is thus a two terminal device, when an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal, it is forward biased. The direction of the applied voltage is opposite to the built in barrier potential. As a result, the depletion layer width decreases and the barrier height is reduced. If the applied voltage is increased, it may overcome the barrier potential altogether and a large current flows across the junction. When an external voltage (V) is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens. So current flowing across the junction decreases enormously (practically becomes zero) as compared to the diode under forward bias. If an alternating voltage is applied across a diode, the current flows only in that part of the cycle when the diode is forward biased. Why is a p-n junction called semiconductor diode ? |
| Answer» Solution :A p-n junction is based on SEMICONDUCTOR PROPERTY of SILICON and is a two tenninal device, so it is called a semiconductor DIODE. | |
| 12. |
What is the force on a conductor of length l carrying current 'i' when it is situated in a magnetic field of induction B? When is it maximum? |
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Answer» SOLUTION :Force due a magnetic FIELD on a current carrying conductor is GIVEN by` F= Bi l sin theta` Its value is maximum when direction of current and magnetic field are PERPENDICULAR to each other. `thereforeF_("max") = Bil` |
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| 13. |
Inside a uniformly charged spherical shell, electric field is ...... and electrostatic potential is ..... |
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Answer» equal, zero `phi= (Q)/(in_(0))=0` `:. ointvecE.vecda=0` `:. E =0(because phi da ne 0)` Now, potential difference between a point inside charged shell and a point on its surface is `V_(i)-V_(s)=-int_(s)^(i)vecE.vecdl` `:. V_(i)-V_(s)=0 (because E=0)` `:. V_(i)=V_(s) = (KQ)/(R)` EVERYWHERE inside the charged spherical shell, ELECTROSTATIC potential is same, equal to that on the swface. |
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| 14. |
Assertion: At a point in space, the electric field points towards north. In the region, surrounding this point the rate of change of potential will be zero along the east and west. Reason: Electric field due to a charge is the space around the charge. |
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Answer» if both Assertion and Reason are TRUE and the Reason is CORRECT explanation of the Assertion. |
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| 15. |
Explain one similarity and one dissimilarity between nuclear fission and fusion. |
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Answer» SOLUTION :Nuclear FISSION and nuclear fusion are dissimilar in the sense that former involves splitting of a heavy nucleus whereas latter involves fusing of two or more light nuclei. The SIMILARITY between the two is that both involve mass defect `(DELTAM)` and hence nuclear energy is released in both the process as per the relation `E=(Deltam)c^2` |
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| 16. |
What should be the distance between the object in Exercise 30 and the magnifying glass if the virtual image of each square in the figure is to have an area of "6.25 mm"^(2). Would you be able to see the squares distinctly with your eyes very close to the magnifier ? [Note : Exercises 29 to 31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.' |
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Answer» Solution :Here, MAGNIFICATION in area = 6.25 linear magnification `m=sqrt(6.25)=2.5` As `m=(upsilon)/(u) or v=m u=2.5u` As `(1)/(upsilon)-(1)/(u)=(1)/(f)` `(1)/(2.5u)-(1)/(u)=(1)/(10)` `(1-2.5)/(2.5u)=(1)/(10)` `u=-6cm` `y=2.5u=2.5(-6)=-15CM` as the virtual image is at 15 cm, where as distance of DISTINCT vision is 25CM, therefore, the image cannot be SEEN distinctly by the eye. |
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| 17. |
What was the old aunts' mindset for health? |
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Answer» They thought that it is really IMPORTANT to RELIEVE the stress of a distressed person |
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| 18. |
A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is (4)/(3) and fish is 12 cm below the surface, the radius of the circle in cm is : |
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Answer» `36 SQRT(7)` 
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| 19. |
In the following circuit, 5 Omega resistor develops 45J//s due to current flowing through it. The power developed across 12 Omega resistor is |
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Answer» 16 W |
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| 20. |
Can a plane mirror form a real image ? |
Answer» Solution :A plane mirror can form a REAL image of a virtual object As shown in the Fig.1.53 converging rays AB, CD and PO would MEET at p' in the absence of the mirror p' ACTS as the virtual object. But the rays after reflection meet at p and form real image.
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| 21. |
Calculate force between two charges of 1C each separated by 1m in vacuum. |
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Answer» `9 XX 10^(9) N` |
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| 22. |
What is the cause of a small leakage current in reverse bias arrangement ofa p-n junction? |
| Answer» SOLUTION :DRIFT of minority CHARGE carriers across the JUNCTION. | |
| 23. |
Pickthe oddone outfromthe following a. Lyman seriesb.Paschenseriesc .Brackett seriesd.Pfundseriese.Hum[hreyseries |
| Answer» SOLUTION :Lymanseries( it ISIN the UVregion, alltheotherarein the infraredregion) | |
| 24. |
A 1 Kw signal is transmitted using a communication channel which provides attenuation at the rate of -2dB per Km. If the communication channel has total length of 5Km, the power of the signal received is (gain in dB=10logP_0/P_1) |
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Answer» 800W |
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| 25. |
What is responsible for streaks of coloured light from a compact disc ? |
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Answer» Solution :The streaks of coloured light reflected from a compact DISC (CD) resemble the colours that appear when white light passes through a PRISM. However, a CD does not SEPARATE colours on account of refraction. Instead, the light waves undergo interference here. The digital information (alternating pits and smooth reflecting surfaces on the CD) forms closely spaced rows. These rows of data do not reflect nearly as much light as the thin portions of the disc that separate tham. So light reflected from tham undergoes constructive interference in certain directions and destructive interference in certain other directions. The constructive interference would depend upon wavelength of light, orientation of CD and direction of incoming light. Thus when white light is reflected from the CD, each wavelength of light would be SEEN at a particularangle w.r.t. TEH surface of CD. That is how we observestreaks of coloured light from the CD. |
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| 26. |
the electric and the magnetic field, associated with an electromagnetic wave, propagating along X axis can be represented by |
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Answer» `vecE=E_(0)HATJ`and `vecB=B_(0)hatk` |
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| 27. |
Mention two characteristics of electromagnetic waves. |
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Answer» SOLUTION :1.Electromagnetic waves do not require any material MEDIUM for their propagation. They can PROPAGATE in vaccum as well as in a medium. 2.Speed of electromagnetic wave in a medium is GIVEN by `v=1/sqrt(muepsilon)` where `mu` and `epsilon` are respectively the permeability and permitivity of the medium. |
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| 28. |
Read the following passage and then answer questions (a) - (e) on the basis of your understand- ing of the passage and the related studied concepts. India is lucky to receive solar energy for the greater part of the year. Energy received from the sun is about 1.4 kW m^(-2) and it is estimated that during a year India receives energy equivalent to more than 5000 trillion kWh from the sun. Solar energy can be harnessed by the use of solar panels. Each solar panel consists of a number of solar cells which work on photovoltaic effect. With continuous enhancement in technology cost of installing solar power has come down and is now comparable with thermal power stations. As a result, solar power is a fast developing industry in India. During last few years country's installed solar capacity has grown by leaps and bounds and reached 30.1 GW as on 31-07-2019. India aims to have an installed solar power capacity of 100 GW by 2022 and 250 GW by 2030. Name three important criteria for the selection of a material for solar cell fabrication. |
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Answer» Solution :Three important criteria for the selection of a material for solar cell fabrication are : (i) Band gap for semiconducting material must lie between 1.0 eV and 1.8 eV. (II) The material should have rugh OPTICAL ABSORPTION. (iii) Electrical CONDUCTIVITY. |
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| 29. |
The wavelength of light in vacuum is 5000Å when it travels normally through diamond of thickness 1.0 mm find the number of waves of light in 1.0 mm of diamond. (Refractive index of diamond =2.417) |
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Answer» 4834 waves |
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| 30. |
Which of the following will have the dimension of time ? |
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Answer» LC |
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| 31. |
Read the following passage and then answer questions (a) - (e) on the basis of your understand- ing of the passage and the related studied concepts. India is lucky to receive solar energy for the greater part of the year. Energy received from the sun is about 1.4 kW m^(-2) and it is estimated that during a year India receives energy equivalent to more than 5000 trillion kWh from the sun. Solar energy can be harnessed by the use of solar panels. Each solar panel consists of a number of solar cells which work on photovoltaic effect. With continuous enhancement in technology cost of installing solar power has come down and is now comparable with thermal power stations. As a result, solar power is a fast developing industry in India. During last few years country's installed solar capacity has grown by leaps and bounds and reached 30.1 GW as on 31-07-2019. India aims to have an installed solar power capacity of 100 GW by 2022 and 250 GW by 2030. Draw I· V characteristic of a solar cell. |
Answer» Solution :I - V characteristic of a solar CELL [Fig. 14.4] is DRAWN here.
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| 32. |
Read the following passage and then answer questions (a) - (e) on the basis of your understand- ing of the passage and the related studied concepts. India is lucky to receive solar energy for the greater part of the year. Energy received from the sun is about 1.4 kW m^(-2) and it is estimated that during a year India receives energy equivalent to more than 5000 trillion kWh from the sun. Solar energy can be harnessed by the use of solar panels. Each solar panel consists of a number of solar cells which work on photovoltaic effect. With continuous enhancement in technology cost of installing solar power has come down and is now comparable with thermal power stations. As a result, solar power is a fast developing industry in India. During last few years country's installed solar capacity has grown by leaps and bounds and reached 30.1 GW as on 31-07-2019. India aims to have an installed solar power capacity of 100 GW by 2022 and 250 GW by 2030. What type of bias is employed for a solar cell and why ? |
| Answer» Solution :No external bias is APPLIED on a solar CELL because it is a device which GENERATES an emf when solar radiation are incident on it. | |
| 33. |
The distance between two successive collisions is called |
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Answer» WAVELENGTH |
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| 34. |
find the potential difference between th point A and B and between the point B and C of figure in steady state. |
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Answer» Solution :`C_eq = [(3muf P 3muf ) S ( 1 mu fP 1 mu f ) (P mu f) ` ` [ S = SERIES , P= parallel ] ` ` = [(3 + 3 ) mu f S (2 mu f)] P 1 muf ` ` = 3/2 + 1 = 5/2 mu f ` ` V = 100 V ` `Q = CV = 5/2 xx 100 = 250 mu C . ` Charge stored across 1 mu fcapacitors ` = 100 mu c ` Charge flowing from A to B ` = 150 mu c = Q ` ` Ceq betweena and b is 6 mu f = C ` ` potential drop across AB = V ` ` Q/C = 25 V ` ` Potential drop across BC = 75 V. ` |
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| 35. |
Energy is being emitted from the surface of a black body at 127^(@)C temperature at 1xx10^(6)" J s"^(-1)"m"^(-2). Temperature of the black body at which the rate of emission is 16xx10^(6)"J s"^(-1)"m"^(-2) will be : |
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Answer» `254^(@)C` `(16xx10^(6))/(1xx10^(6))=((T_(2))/(400))` `2=((T_(2))/(400))rArrT_(2)=800K=527^(@)C` Thus CORRECT choice is (c ). |
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| 36. |
What provides centripetal force in Electron revolving around nucleus? |
| Answer» SOLUTION :ELECTROSTATIC FORCE between ELECTRON and the NUCLEUS. | |
| 37. |
A battery of emf 100V and a resistor of resistance 10k(Omega)are joind in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100(Omega), the current through it is constant up to two significant digits.Find its value, This is the basic principle of a constant-current source. |
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Answer» Solution :`E = 100 volt, R' = ` 100 K Omega = 100000 Omega`, ` R=1 UPTO 100 ` ` when no other register is added or r =0 ` ` i= E/R = 100/10000 = 0.001 Amp.` ` when r = 1` ` i= (100/100000+1) = 100/100001` ` = 0.000999A` up tor = 100, the CURRENT does not up to two SIGNIFICANT digits. Thus it is proved. |
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| 38. |
In Young's experiment ratio of maximum and minimum intensities of fringes is 9: 1. Ratio of amplitude of superposing waves = ..... |
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Answer» Solution :`(I_(max))/(I_(min))= :.(9)/(1)=((E_(1)+E_(2))/(E_(1)-E_(2)))^(2)` `:. (E_(1)+E_(2))/(E_(1)-E_(2))=(3)/(1) [ :.` Taking SQUARE ROOT] Now, Taking componendo- dividendo, `(2E_(1))/(2E_(2))=(3+1)/(3-1) :. (E_(1))/(E_(2))=(2)/(1) =2:1` |
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| 39. |
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k_(1),k_(2) and k_(3) as shown. If a single dielectric naterial is to be used have the same capacitance C in this capacitor, then its dielectric constant k is given by |
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Answer» `(1)/(k)=(1)/(k_(2))+(1)/(k_(2))+(1)/(2k_(3))` |
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| 40. |
When electric current in a coil steadily changes from +2 A to -2 A is 0.05 s, an induced emf of 0.8 V is generated in it. Then the self Inductance of the coil is ....... H. |
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Answer» Solution :`epsilon=-L (DI)/(dt)` `therefore 8=(-L xx (-4))/0.05` `therefore L=(8xx0.05)/4` = 0.1 H |
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| 41. |
How many elements are there in a row between those whose wavelength of k_(alpha) lines are equal to 250 and 179 p m? |
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Answer» Solution :We calculate the `Ƶ` values CORRESPONDING to the given wavelength using Moseley's law.See problem (134). Substitution GIVES that `Ƶ=27` corresponding to `lambda=179 p m` There are thus three elements in a row between those whose wavelengths of `K_(ALPHA)` LINE are equal to `250 p m` and `179p m)`. |
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| 42. |
The potential difference across the 3 resistor shown in figure is |
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Answer» |
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| 43. |
Figure 10-80 is an overhead view of a rod with length 2L and negligible mass and which lies on a frictionless surface. Two bullets, each with mass m and speed v and traveling parallel to the x axis, hit the ends of the rod simultaneously and are buried in the rod. After the collision, what are (a) the angular speed of the system and (b) the speed of the system's center of mass? (c ) What is the change in the total kinetic energy because of the collisions? |
| Answer» SOLUTION :(a) `omega=v//L`, (B) Zero, ( C) Zero, hence, the kinetic energy of the system is conserved. | |
| 44. |
Explain the function of moderator, control rods andcoolant in a nuclear reactor. |
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Answer» Construction. It consists of thick blocks of graphite in which long cylindrical holes have been drilled . Fissionable material is inserted through these holes. Fuel. It consists of certain mass of fissionable material taken in the form of rod of few centimetres in diameter. It may consist of enriched uranium which is 3.5 % uranium -235 and 96.5% of uranium-238. Moderator. To slow down the neutrons, they must suffer collision against atoms of material of moderator. Suitable material used as moderator are heavy water and graphite.  Controlling rods. Cadmium rods are used as controlling rods. The rods are inserted in the reactor and the chain reaction is controlled by careful adjustment of their length inside the carbon blocks. The movement of a cadmium rod by a little (say one cm distance) causes a significant difference in the performance of the reactor. Working. Neutrons produced by the action of `alpha`- particle on polonium or beryllium are slowed down and are used initially to bring about fission of say `U^(235)` nuclei. The emitted neutrons are slowed down by the passage through the moderator to split further `U^(235)` nucleus. The chain reaction is said to be steady so that effective multiplication factor `K_(e)` is 1. `K_(e)=("Rate of emission of neutrons")/("Rate of loss of neutrons")` Some lost neutrons are absorbed by `""_(92)U^(238)` present in enriched uranium which gives rise to `""_(94)U^(239)` as shown below: `""_(0)N^(1)+""_(92)U^(238)RARR ""_(92)U^(239)rarr""_(-1)beta^(0)+""_(93)Np^(239)` `""_(93)Np^(239) rarr ""_(-1)beta^(0)+""_(94)Pu^(239)` `""_(94)Pu^(239)` is produced as a bye-product and can be used for fission purpose. India used `""_(94)Pu^(239)` in nuclear device which was exploded on 18TH May 1974 and later on May 11 and 13, 1998 at Pokhran. Uses. A nuclear reactor is used for generation of electricity, preparation of radio-isotopes. Most of energy SET free is in the form of heat which through heat exchange is made to furnish steam to operate conventional turbined and electricgenerators. 
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| 45. |
In photoelectric effect threshold wavelength of sodium is 5000Å .Find its work function. [h=6.6xx10^(-34)Js,c=3xx10^(8)m//s,1eV=1.6xx10^(-19)Js] |
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Answer» 2.5 eV |
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| 46. |
A rectifier which rectifies half of an a.c. input supply is called_____and which rectifies both the halves of each input cycle is called_____. |
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Answer» |
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| 47. |
The twocoherent sources of equal intensity produce maximum intersity of 100 units at a point.Ifthe intensityof one of the sources is reduced by 50 % by reducing its widththen the intensity of light at the same point will be |
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Answer» 90 |
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| 48. |
The potential difference across the terminals of a battery is 10 V when there is a currentof 3A in the battery from the negative to the positive terminal. When the current is 2A in the reverse direction, the potential difference becomes 15 V. The internal resistance of the battery is |
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Answer» Solution :`E-3r=10"………(i)"` `E+2r=15"……….(II)"` SOLVING these TWO EQUATIONS, we get, `r=1Omega` |
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| 49. |
Statement-1: Sky wave signals are used for long distance radio communication. These signals are in general less stable than ground wave signals Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season |
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Answer» S-1 is FALSE, S-2 is true |
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| 50. |
Explain the term 'drift velocity' of electrons in a conductor. Hence obtainthe expression for the current through a conductor in terms of drift - Temperature T velocity. |
Answer» Solution : Consider a CONDUCTOR of uniform cross-section AREA A, carrying a current I. Consider a SMALL section KL of the conductor having a length `Delta x `or having a VOLUME A. `Delta x` , then number of free electrons present in this section =` n A. Delta x `, where n = Number density of free electrons. Total charge carried by these electrons while CROSSING the given section `DeltaQ = n A: Deltax.e` and total time taken by the electrons to cross this section is `Delta t = (Delta x)/(v_d)`where `v_d = `drift velocity of electrons ` therefore I= (Delta Q)/(Delta t) = (nA Delta x.e)/((Deltax)/(v_d)) = nAev_d`. which is the requisite relation. |
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