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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 projectile is thrown at an angle theta with the horizontal and its range is R_(1) It is then thrown at an angle 0 with vertical and the range is R_(2) then : |
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Answer» `R_(1)=4R_(2)` |
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| 2. |
Statement I : The liquids rises higher in a capillary tube of smaller radius than in a tube of bigger radius. Statement II : The height of a liquid in a capillary tube is inversely proportional to diameter of capillary tube. |
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Answer» STATEMENT-I is true, Statement-II is true and Statement-II is correct explanation for Statement-I. |
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| 3. |
As the electrons in Bohr orbit of hydrogen atom passes from state n=2 to n=1, the kinetic energy K and potential energy U change as |
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Answer» K two-fold, U four-fold |
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| 4. |
A car accelerates from rest at constant rest a alpha t for same time after which it deaccelerates at constant rate beta to come to rest. If total time taken is t then distance covered is given by : |
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Answer» `X=t^(2)((alpha BETA)/(alpha+beta))` Let V be themaximum velocity REACHED. `:. X=(v^(2))/(2alpha)+(v^(2))/(2beta)=v^(2)((2beta+2alpha)/(2alpha.2beta))` Since `v=((alphabeta)/(alpha+beta))t:. x=((alphabeta)/(alpha+beta))^(2).t^(2).((alpha+beta))/(2alphabeta)` `x=(1)/(2)((alphabeta)/(alpha+beta)).t^(2)` |
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| 5. |
In the helium-neon laser of laser action occurs between two excited states of the neon atom. However, in many lasers, laser action (lasing) occurs between the ground state and an excited state, as suggested in fig. (a) Consider such a laser that emits at wavelength lambda=550nm. If a population inversion is not generated what is the ratio of the population of atoms in state E_(x) to the populationin the ground state E_(0), with the atoms at room temperature? |
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Answer» SOLUTION :The naturally occurring population ratio `N_(x)//N_(0)` of the two states is due to thermal agitation of the gas atoms `N_(x)//N_(0)=e^(-(E_(x)-E_(0))//kT)` (39-22) To find `N_(x)//N_(0)` with we need to find the energy separation `E_(x)-E_(0)` between the two states. (2) We can obtain `E_(x)-E_(0)` from the given wavelength of 550 nm for the LASING between those two states. The lasing wavelength gives us `E_(x)-E_(0)=hf=(hc)/(lambda)` `=((6.63xx10^(-34)J.s)(3.00xx10^(8)m//s))/((550xx10^(-9)m)(1.60xx10^(-19)J//eV))` `=2.26eV`. We also need the mean energy of thermal agitation kT for an atom at room temperature (assumed to be 300 K), which is `kT=(8.62xx10^(-5)eV//K)(300K)=0.0259eV`, in which k is Boltzmann.s constant. SUBSTITUTING the last two results into give us the population ratio at room temperature: `N_(x)//N_(0)=e^(-(2.26eV)(0.0259eV))` `~~1.3xx10^(-38)` This is an extremely small number. It is not unreasonable, however. Atoms with a mean thermal agitation energy of only 0.0259 eV will not often impart an energy of 2.26 eV to another atom in a collision. (b) For the conditions of (a), what temperature would the ratio `N_(x)//N_(0)` be 1/2? Calculation: Now we want the temperature T such that thermal agiation has bumped ENOUGH neon atoms up to the higher-energy state to give `N_(x)//N_(0)=1//2`. Substituting that ratio into taking the natural logarithm of both sides, and solving for T yield `T=(E_(x)-E_(0))/(k(ln2))=(2.26eV)/((8.62xx10^(-5)eV//K)(ln2))` `=38 000K`. This is much hotter than the surface of the SUN. Thus, it is clear that if we are to invert the population of these two levels, some specific mechanism for bringing this about is needed-that is, we must ..pump.. the atoms. No temperature, however high, will naturally generate a population inversion by thermal agitation. |
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| 6. |
एक कण सरल रेखा y=3x+5 के अनुदेश गतिशील है कोन से निर्देशाक के परिवर्तन की दर अधिक होगी! |
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Answer» X- निर्देशाक |
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| 7. |
a. The small ozone layer on the top of the stratosphere is crucial for human survival. Why ?b. Where is the position of UV radiation in the electromagnetic spectrum ?c.Can you convert UV to visible radiation ? Explain.d.Which radiation is adjacent to UV to the higher frequency side of the electromagnetic spectrum ? |
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Answer» Solution :a. OZONE layer absorbs UV radiation which are HARMFUL for humans. b.Between X - rays and visible LIGHT. c.Yes. Inmercury TUBE light, UV is coverted to visible radiation. d. X-rays. |
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| 8. |
Value of R required for three, 10W, 20V, 1000mA zener diodes connected as shown is |
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Answer» `2.5Omega` |
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| 9. |
(a) Find the maximum frequency of the X-rays emitted by an X-ray tube operating at 30 kV. (b) An X-ray tube operates at 20 kV. A particular electron loses 5% of its kinetic energy to emit an X-ray photon at the first collision. Find the wavelength corresponding to this photon. (c ) An X-ray tube is operated at 20 kV and the current through the tube is 0.5 mA. Find (i) the number of electrons hitting the target per second, (ii) the energy falling on the target per second as the kinetic anergy of the electrons and (iii) the cut-off wavelength of the X-rays emitted. |
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Answer» Solution :(a) `lambda_(min)=(hc)/(EV), V=30 kV` `lambda_(min)=1242/(E(eV))=1242/(30xx10^(3))=41.4xx10^(-3) nm` `=41.4xx10^(-12)m` `v_(max)=c/lambda(min)=(3xx10^(8))/(41.4xx10^(-12))=7.2xx10^(18) Hz` (B) KINETIC energy of electron `=20 keV` `5%` of this converted to photon `5/100xx20=1 ke V` `lambda=1242/(1xx10^(3)) nm=1.242 nm` (c ) `V=20 kV, i=0.5 mA` (i) `i=n eimpliesn=i/e=(0.5xx10^(-3))/(1.6xx10^(-19))=3.125xx10^(15)` (ii) `P=Vi=20xx0.5==10W` (iii) `lambda_(min)=1242/(20xx10^(3))=0.062 nm` |
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| 10. |
To increase the intensity of image of a distant object formed by a telescope we use a telescope objective of ________. |
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Answer» |
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| 11. |
The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focla length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focla length decreases. For a clear vision of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. from eye-lens. it is about 2.5 cm for a grown-up person A person can theoretically have clear vision of situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most straineed in this position. For an average grown-up person minimum distance of object sould be around 25 cm. A person suffering for eye defects uses spectacles (eys glass). The function of lens of spectacles is to form the image of the object within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina. The number of spectacle-lens used for the remedy of eye defect is decided by the power of hte lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. for example power of lens required is +3D (converging lens of focal length (100)/(3)cm) then number of lens will be +3. For all the calculations required you can use the lens formula and lens maker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necssary for the remedy of this defect will be. |
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Answer» `+1` For lens `u=-inftyv=-100` (1)/(f_(lens))=(1)/(-100)-(1)/(-infty)` |
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| 12. |
In Q. 29, Which os the folllowing graphs best represents the current flowing from X to Y? |
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Answer»
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| 13. |
Define electric flux. Write its SI unit. It a scalar or a vector quantity? |
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Answer» Solution :Electric flux `phi_E` PASSING through a SURFACE area ` oversetto s ` placed in an electric field `oversetto E ` is given by : ` phi__E=oint oversetto E.oversetto (ds) ` SI unit of electric flux is `N m^(2) C^(-1)` and electric flux is a scalar QUANTITY. |
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| 14. |
Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification. |
Answer» Solution :Description : It consists of two convex lenses SEPARATED by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length. The distance of the object can be ADJUSTED by means of a rack and pinion arrangement.  Working : The object OJ is placed outside the principal the principal focus of the objective and the real image is formed on the other side of it. The image `I_(1)G_(1)` is real, inverted and magnified. This image acts as the object for the eyepiece. The position of the eyepieceis so adjusted that the image due to the objectiveis between the optic centre and principal focus to form the final image at the near point. The final image IG is virtual, inverted and magnified. Magnifying Power : It is defined as the ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point. Imagining that the eye is at the optic centre, the angle subtended by the final image is `alpha`. When the object is imagined to be taken at near point it is represented by 1 J' and OJ = I J'. The angle made by I J' at the eye is `beta`. Then the definition of power `m=(alpha)/(beta)~=(tan alpha)/(tan beta)"for small angles"[{:(because Delta" IGO''" rArr Tan alpha=("IG")/("IO''")),(because "IJ'O''" rArrTan beta=("IJ'")/("IO''")):}]` `=("IG/IO''")/("IJ'/IO''")=("IG")/("IJ'")=("IG")/("OJ'")""(because "IJ' = OJ")` DIVIDING and multiplying by `I_(1)G_(1)` on the right side, we get `m=((IG)/(I_(1)G_(1)))((I_(1)G_(1))/(OJ))`. Magnifying power of the objective `(m_(0))=I_(1)G_(1)//OJ`= Height of the image due to the objective/Height of its object. Magnifying power of the eye piece `(m_(e))=IG//I_(1)G_(1)`=Height of the final image / Height of the object for the eyepiece. `THEREFORE m= m_(0)xxm_(e)"...............................(1)"` To find `m_(0)` : In figure OJ O' and `I_(1)G_(1)O'` are similar TRIANGLES. `((I_(1)G_(1))/(OG))=((O'I_(1))/(O'O))` Using sign convention, we find that `O'I_(1)=+v_(0) and O'O=-u` where `v_(0)` is the image distance due to the objective and u is the object distance for the objective or the compound microscope. `I_(1)G_(1)` is negative and OJ is positive. `therefore m_(0)=(v_(0))/(uu).(because (I_(1)G_(1))/(OJ)=m_(0))` To find `m_(e)` : The eyepiece behaves like a simple microscope. So the magnifying power of the eye piece. `therefore""m_(e)=(1+(D)/(f_(e)))` Where `f_(e)` is the focal length of the eyepiece. Substituting `m_(0)` and `m_(e)` in equation (1), `m=+(v_(0))/(u)(1+(D)/(f_(e)))` When the object is very close to the principal focus `F_(0)` of the objective, the image due to the objective becomes very close to the eyepiece. `u~~-f_(0) and v_(0) ~~L` Where L is the length of the microscope. Then `m~~-(L)/(f_(0))(1+(D)/(f_(e)))` |
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| 15. |
Two junction diodes, one of germanium (Ge) and other of silicon (Si) are connected as shown in fig to a battery of 12V and a load resistance 10 k Omega . The germanium diode conducts at 0.3V and silicon diode at 0.7V. When current flows in the circuit, the potential of terminal Y willbe(##AKS_NEO_CAO_PHY_XII_V02_C_C06_SLV_013_Q01.png" width="80%"> |
| Answer» SOLUTION :The Ge DIODE CONDUCTS for a p.d of 0.3V, therefore the current passes through it and the Si diode do not conduct.Hence the potential of terminal Y = 12-0.3 = 11.7V | |
| 16. |
A coil having an area 2m^2is placed in a magnetic field which changes from 1Wb//m^2to 4Wb/m^2in an interval of 2 seconds. The average e.m.f induced in the coil will be |
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Answer» 4V |
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| 17. |
For a transistor, the value of B = 99, then the value of alphais |
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Answer» 99 |
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| 18. |
whether electromagnetic waves are transverse or longitudinal in nature. |
| Answer» SOLUTION :TRANSVERSE | |
| 19. |
Which of the following material has zero magnetic moment of single atom? |
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Answer» Paramagnetic |
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| 20. |
The wavelength of H_(alpha) line in the hydrogen spectrum is found to be 6563Å in the laboratory. If the velocity of the milky way is 1.05xx10^(6)ms^(-1), then the wavelength of H_(alpha) line in the spectrum of milky way will be |
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Answer» `6457Å` |
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| 21. |
Equivalent self inductance of above arrangement is ____ |
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Answer» 1.0 H `L=0.75+(0.5xx0.5)/(0.5+0.5)` `=0.75+0.25/1.0`=1.0 H |
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| 22. |
A fixed mass of a gas expands at constant temperature. Which property of the gas molecules increases? |
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Answer» average NUMBER per unit volume |
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| 23. |
It is possible to achieve positive health in which countries? |
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Answer» HAPPY Countries |
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| 24. |
A beam of 10.6 V photons of intensity 2.0W//m^(2)falls on a platinum surface of area 1.0xx10^(-4)m^(2) and work function 5.6eV. If 0.53% of the incident photons eject photo electrons, then find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). |
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Answer» Solution :Number of PHOTONS incident on the metal per second `=("Total energy incident on the metal per second")/("Energy of each photon")` `=("Intensity x AREA")/("Energy of each photon")` Number of electrons emitted per second `=((0.53)/(100))` Number of photons incident on the metal surface Number of electrons emitted per second `=((0.53)/(100))("Intensity x Area")/("Energy of each photon")` `=((2.0)(1.0xx10^(-4)))/((10.6xx1.6xx10^(-19)))XX(0.53)/(100)=6.25xx10^(11)` Minimum KINETIC energy of PHOTO electrons = 0 Maximum kinetic energy is, `K_(max)=E-W_(0)` `=10.6-5.6=5eV`, |
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| 25. |
In the above question the instantaneous power developed by the gravitational force at time t will be : |
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Answer» `(1)/(2)Mg t^(2)` `P=Mgxxv` But `v=0+at` `therefore P=Mgxxat""(because a=2G//3)` `=Mgxx(2g)/(3)xxt` `P=2//3Mg^(2)t` |
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| 26. |
A given wire is bent into a rectangular loop of size 15cm xx 5cm and placed perpendicular to a magnetic field of 1.0 Tesla. Within 0.5sec, the loop is changed into a 10cm square and the field increases to 1.4 Tesla.Calculate the value of e.m.f. induced in the loop? |
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Answer» |
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| 27. |
In metre bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is .t.. If the temperature of water is increased, the new balancing length is |
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Answer» Solution :According to the question, `(R_("unknown"))/(R_("standard"))=(1)/((1-1))` If temperature increases than RESISTANCE increases. Hence the new balancing is GREATER than 1. |
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| 28. |
(A): In equation F=q(E+v xx B) when v = 0, any force on the charge must arise from the electric field term E alone (R): To explain, the existence of induced emf or induced current in static conductor kept in time - varying magnetic field, we must assume that a time - varying magnetic field generates an electric field |
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Answer» Both A and R are true and R is the CORRECT EXPLANATION |
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| 30. |
Consider two adjacent transparent media. The speed of light in medium 1 is v_(1), and the speed of light in medium 2 is v_(2). If v_(1) lt v_(2), then total internal reflection will occur at the interface between these media if a beam of light is |
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Answer» INCIDENT in MEDIUM 1 and strikes the INTERFACE at an angle of INCIDENCE GREATER than `sin^(-1)(v_(1)//v_(2))`. |
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| 31. |
Magnetic field and magnetic potenitial are related |
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Answer» `B = (-DV)/(DX)` |
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| 32. |
State Brewster's law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason. |
| Answer» Solution :For Brewster.s law it is clear that VALUE of polarising ANGLE DEPENDS on the REFRACTIVE index of the MEDIUM. As refractive index of a medium is different for light of different colours, hence it is obvious that value of polarising angle is different for light of different colours. | |
| 33. |
Find the Q value of the reaction. ""_(1)H^(2)+""_(3)Li^(6) to ""_(1)H^(1)+""_(3)Li^(7) The rest mass of ""_(1)H^(2)=2.01410 a.m.u. ""_(3)Li^(6)=6.01513 a.m.u. ""_(3)Li^(7)=7.01601 a.m.u. ""_(1)H^(1)=1.00783 a.m.u. |
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Answer» Solution :`""_(1)H^(2)+""_(3)Li^(6)=""_(3)Li^(6)+""_(1)H^(1)+Q` Then, 8.02933=8.02384+Q Q=0.00549 a.m.u. `=0.00549 XX 931 MeV=5.1MeV` |
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| 34. |
The temperature at which the reading of a Fahrenheit thermometer will be double that of centigrade thermometer is |
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Answer» a)`160^@C` |
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| 35. |
Variation of magnification"("m")" produced by a thin convex lens versus distance "("v")" of image from pole of the lens is shown in the graph. Which of the following statements is not correct ? |
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Answer» FOCAL length of the lens is equal to intercept on v-axis |
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| 36. |
The diagram shows the energy level for an electron in a hydrogen atom. Which transition, shown here, represents the emissions of a photon of maximum energy? |
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Answer» A |
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| 37. |
Which of the following is not a vector quantity ? |
| Answer» Solution :Potential energy | |
| 38. |
In a series resonant circuit, the A.C. voltages across resistance R, inductor L and capacitor C are 5V, 10V and 10V respectively. The A.C. voltage applied to the circuit will be ....... |
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Answer» 25 V `=sqrt((5)^2 + (10-10)^2)` `therefore` V=5 V |
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| 39. |
What is the radius of the path of an electron (mass 9xx10^(-31)kg and charge 1.6xx10^(-19)C) moving at a speed of 3xx10^(7)m//s in a magnetic field of 6xx10^(-4)T perpendicular to it? What is its frequency ? Calculate its energy in keV. (1eV=1.6xx10^(-19)J) |
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Answer» Solution :1. When a PARTICLE of mass m and charge q moves in a uniform magnetic field B with speed v radius of its circular path is GIVEN by, `r=(mvsintheta)/(Bq)` (where `theta` = angle between `vecvandvecB`) `thereforer=0.2812m=28.12cm` 2. We have, `v=romega` `thereforev=r(2pif)` `thereforef=v/(2pir)` = `(3xx10^(7))/((2)(3.14)(0.2812))` `thereforef=1.699xx10^(7)Hz~~17MHz` 3. Now, kinetic energy of electron is, `K=1/2mv^(2)` = `1/2xx9xx10^(-31)xx(3xx10^(7))^(2)` `thereforeK=4.05xx10^(-16)J` `thereforeK=(4.05xx10^(-16))/(1.6xx10^(-19))EV` `thereforeK=2.531xx10^(3)eV` |
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| 40. |
Considerthe chargesq,q and -q placed at the verticesof an equilateraltrianglewhat is the force on each charge |
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Answer» Solution :The forces actingon CHARGE q at a due to charges q at B and -q at c are `F_(12)`along BA and `F_(13)` along AC respectivelyq at a is GIVEN by the force of attaraction or repulson for each pari of charges has THESAME magnitude `F=(q^(2))/(4piepsilon_(0)l^(2))` The totalforce `f_(2)` on chargeq at b is thus `f_(2)` =f `vecr_(2)`where `vecr_(2)` is a unit vectoralong AC `F_(1)+F_(2)+F_(3)=0` The result is not at all surprisingit follows straightfrom the fact that colulomb law is consistent with newton THRID law the proof is left to you as an exercise |
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| 41. |
What is the radius of the path of an electron (mass 9xx10^(-31)kg and charge 1.6xx10^(-19)C) moving at a speed of 3xx10^(6)m//s in a magnetic field of 6T perpendicular to it? What is its frequency ? Calculate its energy in keV. (1eV=1.6xx10^(-19)J) |
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Answer» |
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| 42. |
An electron in a hydrogen atom is in a state with l=5. What is the minimum possible value of the semiclassical angle between vecLandL_(z)? |
| Answer» SOLUTION :`24.1^(@)` | |
| 43. |
Figure shows the orientation of two vectors u and v in the XY-plane If u = ahati + bhatj and v =phati + qhatj which of the following is correct ? |
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Answer» <P>a and p are POSITIVE while B and Q are NEGATIVE |
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| 44. |
Define a reversible process. What is an irreversible process ? |
| Answer» Solution :A reversible process is one which is performed in such a way that, at the end of the process, both the system and its local surroundings can be RESTORED to their INITIAL states, without producing any change in the rest of the universe. A process may be reversible if it takes PLACE quasistatically and without dissipative EFFECTS. A process which does not fulfill the rquirements of reversiblity is said to be an IRREVERSIBLE process. In this case, the system and the local surroundings cannot be restored to their initial states without affecting the rest of the unierse. All natural processes are irreversible. | |
| 45. |
What will be the potential at the middle spherical shell in the previous example? |
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Answer» Solution :Note that the MIDDLE spherical shell is outside the spherical shell of radius a, and inside the spherical shell of radius c. For outside points potential is same as that of point charge as if it were placed at the centre of the shell, but for points inside potential is same as that on the surface. `V=(1)/(4PI epsilon_(0)) (q_(1))/(b)+(1)/(4pi epsilon_(0)) (q_(2))/(b)+(1)/(4pi epsilon_(0)) (q_(3))/(c )` |
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| 46. |
To reduce the resonant frequency in an LCR series circuit with a generator |
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Answer» the GENERATOR frequency should be REDUCED |
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| 48. |
A proton and an alpha-particle are accelerated through the same potential, which one of the two has (i) greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy. Give reasons to support your answer. |
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Answer» SOLUTION :We know de-Broglie wavelength associated with a particle of CHARGE q, mass m, accelerated through a potential V is given by `lamda=(h)/(sqrt(2qmV))`. (i) Here accelerating potential V is CONSTANT for proton as well as `alpha`-particle. Hence `lamdaprop(1)/(sqrt(QM))`. As value of .qm. is less for proton, hence de-broglie wavelength for proton is more. (ii) The kinetic energy of a charged particle K=qV. as charged of proton is less than that of `alpha`-particle, the kinetic energy of proton is less. |
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| 49. |
A radioactive nucleus has specific binding energy E_1. It emits an alpha -particle. The resulting nucleus has specific binding energy E_2. Then |
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Answer» `E_(2) =0` |
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