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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. |
Nucleus of an atom consists of protons and neutrons packed into a small volume. How will you relate the binding energy and the stability of a nucleus? |
| Answer» SOLUTION :GREATER the BINDING energy PER nucleon, greater will be the STABILITY of that nucleus. | |
| 2. |
The wavelength of spectrumlines of the ""_(1)H^(2) atom varies slightly from the wavelength of the spectrum lines of the hydrogen atom because ..... |
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Answer» the MASS of both NUCLEUS is DIFFERENT. |
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| 3. |
To get a magnification from a compound microscope , |
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Answer» the FOCAL length of the objective should be large while thefocal length of the EYEPIECE should be SMALL . |
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| 4. |
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is |
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Answer» `-5/4` |
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| 5. |
In an electron microscope, if the potential is increased from 20 kV to 80 kV, the resolving power R of the microscope will become : |
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Answer» R here `V_(1)=20 KV` `V_(2)=20 KV` `:.(V_(2))/(V_(1))=4` `:.` So resolving power becomes 2R. |
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| 6. |
Two satellites A and B revolve round a planet in coplanar orbits in the same direction, rheir periods are 1 hour and 8 hours respectively. The orbital radius of A is 10^(4) km the speed of B relative to A when they are closest is : |
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Answer» `10^(4)pi KM//h` `therefore v_(1)=(2pi r_(1))/(T_(1))=(2pi XX 10^(4))/(1)km//hr` As`r_(1)=10^(4)km`  By Kepler.s LAW `(r_(2))/(r_(1))=((T_(2))/(T_(1)))^(2//3)=((8)/(1))^(2//3)` or `r_(2)=4xx10^(4)km` `v_(2)=(2pi r_(2))/(T_(2))=(2pi xx4xx10^(4))/(8)` `=pi xx 10^(4) km//hr`. Correct choice is (a). |
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| 7. |
When two resistances are connected in series. The effective resistance is found to be 48Omega . Their resistances if their values are in the ratio 3:1. |
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Answer» `24Omegaand24 OMEGA ` |
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| 8. |
A light ray incident along vector 2hati+4hatj+sqrt(5)hatk strikes in the x-z from medium of refractive index sqrt(3) and enters into medium II of refractive index is mu_(2). The value of mu_(2) for which the value of angle of refraction becomes 90^(@) is |
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Answer» `(4sqrt(3))/5` `1-cos^(2)theta_(C)=((mu_(2))/(mu_(1)))^(2)` `1-(hatn.hatp)^(2)=((mu_(2))/(mu_(1)))^(2)` `1-[4/(sqrt(25))]^(2)=((mu_(2))/(mu_(1)))^(2)IMPLIES(mu_(2))/(mu_(1))=3/5` `implies mu_(2)=(3sqrt(3))/5` |
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| 9. |
You have ""_6^11C, ""_6^12C and ""_6^13C. a. What are these nuclides ? ""b. How do they differ? |
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Answer» Solution :a. They are isotopes. B. In their MASS NUMBER with DIFFERENCE in number of the neutrons. |
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| 10. |
Focal length of the objective of a terrestrial telescope is 80 cm and, it is adjusted for parallel rays, then its magnifying power is 20. If the focal length. Of erecting lensis 20 cm, then full length of telescope will be (in cm) |
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Answer» 84 |
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| 11. |
A mass M at rest is broken into two pieces having masses m and (M-m). The two masses are then separated by a distance r. The gravitational force between them will be maximum when the ratio of the masses [m : (M-m) of the two parts is |
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Answer» `1:1` Here, `m_(1)=m` and `m_(2)=(M-m) :. F=G(m(M-m))/(r^(2))=0` or `M-2m=0` `RARR m=(M)/(2) :. (m)/((M-m))=(M//2)/((M-M//2))=(1)/(1)` |
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| 13. |
A cord of length 64 m is used to connected a 100 kg astronaut to spaceship whose mass is much larger than that of the astronuat. Estimate the value of the tension in the cord. Assume that the spaceship is orbiting near earth surface. Assume that the spaceship and the astronaut fall on a straight line from the earth centre. the radius of the earth is 6400 km. |
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Answer» `3 xx 10^(-2)N` |
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| 14. |
In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n = 2 to n = 1 transition |
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Answer» Solution :According to formula `E_(n)=-13.6(Z^(2))/(n^(2))` `:.E_(2)=-13.6xx((24)^(2))/((2)^(2)) ( :.` From chromium atom Z=24 here n=2) `:. E_(2)=-1958.4eV` Now `E_(1)=-13.6xx((25)^(2))/((1)^(2))` `:.E_(1)=-7833.6eV` `rArr E_(2) gt E_(1)` `rArr` Energy obtained, `DeltaE=E_(2)-E_(1)` `:.DeltaE=-1958.4-(-7833.6)` `=7833.6-1958.4` `DeltaE=5875.2eV ....(1)` Now, TOTAL energy of electron in n = 4 state in chromium atom, `E_(4)=-13.6eVxx((24)^(2))/((4)^(2))` `:.E_(4)=-489.6eV` `rArr` When above electron is given energy AE = 5875.2 eV, it will firstly USE energy 489.6 eV for its emission and the REMAINING energy will be possessed by it in the form of kinetic energy. Its value would be `K=5875.2-489.6` `:.K=5385.6eV` |
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| 15. |
The rms speed of helium at 24^(@)C and 1 atm pressure is 450ms^(-1). Then the rms speed of the helium molecules at 24^(@)C and 2 atm pressure is |
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Answer» `450ms^(-1)` |
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| 16. |
The magnetic field used to magnetic a material is called the .......... . |
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Answer» |
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| 17. |
Derive the equation for effective focal lengthfor lenses in contact. |
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Answer» Solution :Consider a two lenses of focal length `f_(1) and f_(2)` arrangedcoaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produces deviations `delta_(1) and delta_(2)` respectively and The net deviation `DELTA` is. `delta = delta_(1) + delta_(2)` From Angle of deviation in lens equation, `delta = (h)/(F)` `delta_(1) = (h_(1))/(f_(1)):delta_(2) = (h_(2))/(f_(2)) and delta = (h_(1))/(f)` The equation (1) BECOMES, `(h_(1))/(f) = (h_(1))/(f_(1)) + (h_(2))/(f_(2))` From the geometry. `h_(2)-h_(1)=P_(2)G-P_(2)C=CG` `h_(2) - h_(1) = BG tan delta_(1) = BG delta_(1)` `h_(2) - h_(1) = d(h_(1))/(f_(1))` `h_(2) = h_(1) + d(h_(1))/(f_(1))` Substituting the above equaiton in Equation (3) `(h_(1))/(f)=(h_(1))/(f_(1))+(h_(1))/(f_2)+(h_(1)d)/(f_(1)f_(2))` On further simplification, `(1)/(f) = (1)/(f_(1)) + (1)/(f_(2)) + (d)/(f_(1)f_(2))` The above equation COULD be used to find the EQUIVALENT focal lenght. 
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| 18. |
What is packing efficiency of Simple Cubic |
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Answer» 0.74 |
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| 19. |
The semicircular non - conducting ring of radius R containing +ve and negative charge as shwon are joined in tw perpendicular planes. The magnitude of linear charge densities is unifrom and is lambda. The electric field at O will be : |
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Answer» `(lambda)/(2PI epsilon_(0)R)` |
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| 20. |
A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is manitained in a direction normal to the common axis of the coils. A narrwo beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostaic field. If the beam remains undeflected when the electrostatic field is 9.0 xx 10^(-5) V m^(-1), make a simple guess as to what the beam contains. Why is the answer not unique? |
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Answer» Solution :Let charged be having a charge q, mass m and velocity ACQUIRED by them, when accelerated through a potential DIFFERENCE V, be v, then `v = sqrt((2 q V)/(m))` For the beam to remain undeflected `vecF_(E) and vecF_B` must be equal in magnitude but mutually opposite in direction. It is possible when `qE = q v B "or" v = E/B` `implies sqrt((2qV)/(m)) = E/B "or" q/m = (E^2)/(2 B^2 V)` In present case `E = 9.0 xx 10^(-5) V m^(-1) , B = 0.75 T and V = 15 kV = 15 xx 10^3 V` `:."" q/m = ((9.0 xx 10^5)^2)/(2 xx (0.75)^(2) xx 15 xx 10^(3)) = 4.8 xx 10^(7) C kg^(-1)` A simple guess suggests that HTE beam may contain deutron particle having mass `m = 3.34 xx 10^(-27) kg`and charge `q = +e = 1.6 xx 10^(-19)C`, thus , having a value of `q/m = (1.6 xx 10^(-19))/(3.34 xx 10^(-27)) = 4.8 xx 10^(7) C kg^(-1)` However , this answer is not UNIQUE because value of the ratio `q/m` may be `4.8 xx 10^(7) C kg^(-1)` for other ions like `Li^(+++), He^(++)` too. |
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| 21. |
Two blocks A and B of masses M_(A) and M_(B) respectively, are located 1.0 m apart on a horizontal surface. The coefficient of static friction mu_(s) between the block and the surface is 0.50. Block A is secured to the surface and cannot move, What is the minimum mass of Block A that provides enough gravitation attraction to move Block B ? Th universal gravitation constant is 6.67xx 10^(-11) Nm^(2)/kg^(2) . |
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Answer» `7.5 XX 10^(9) `kg |
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| 22. |
The frequency of oscillation of the electric field vector of a certain electromagnetic wave is 5xx10^(14) Hz. What is the frequency of oscillation of the corresponding magnetic field vector and to which part of the electromagnetic spectrum does it belong ? |
| Answer» Solution :Wavelegnth `lambda=(C )/(V)=(3XX10^(8))/(5xx10^(14))=6xx10^(-7)m=600nm`. It BELONGS to visible light region of the electromagnetic spectrum. | |
| 23. |
What fraction of the energy drawn from the charging battery is stored in a capacitor ? |
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Answer» 2 |
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| 24. |
A laser beam of pulse power 1012 watt is focussed on an object are 10–4cm . The energyflux in watt|cm at the point of focus is |
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Answer» `10^(-)` |
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| 25. |
The lifetime of a neutral pion is 8.0xx10^(-17)s. What is the accuracy with which its mass can be determined? |
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Answer» `delta=(Deltam)/m=(Deltaepsi)/epsi~~h/(Deltatau)` |
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| 26. |
A column of soldiers, marching at 100 paces per minute, keep in step with the beat of a drummer at the head of the column. The soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right foot. What is the approximate length of the column? |
| Answer» SOLUTION :`2.1 XX 10^2 m ` | |
| 27. |
A particle executive SHM has maximum speed of 0.5ms^(-1) and maximum acceleration of 1.0ms^(-2) . The angular freqyency of Oscillation is |
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Answer» `2" RAD s"^(-1)` |
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| 28. |
The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B.Which one of the two has higher value of work function ?Justify your answer. |
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Answer» SOLUTION :METAL A Since work function `W = hv_(0)` and `V'_(0) gt v_(0)` so work function of metal A is more Aliter `:` On stopping potential AXIS` - ( W'_(0))/( e ) gt- ( W_(0))/( e ) ` Hence work function `W'_(0) ` of metal A is more. |
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| 29. |
Choose the reverse biased p-n junction : |
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Answer»
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| 30. |
A beam of white light is incident on glass air interface from glass to air such that green light just suffers total internal reflection. The colors of the light which will come out to air are |
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Answer» Violet, INDIGO, Blue |
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| 31. |
Give the location of magnetic south pole. |
| Answer» SOLUTION :At 79. 74° S, 108.22° E in ANTARCTICA | |
| 32. |
The active medium in a particular laser that generates laser light at a wavelength of 694 nm is 6.00 cm long and 1.00 cm in diameter. (a) Treat the medium as an optical resonance cavity analogous to a closed organ pipe. How many standingwave nodes are there along the laser axis? (b) By what amount Deltaf would the beam frequency have to shift to increase of the travel by one? (c ) Show that Deltaf is just the inverse of the travel time of laser light for one round trip back and forth along the laser axis. (d) What is the corresponding fractional frequency shift Deltaf//f? The appropriate index of refraction of the lasing medium (a ruby crystal) is 1.75. |
| Answer» SOLUTION :(a) `3.03xx10^(5)`, (B) `1.43xx10^(9)HZ`, (d) `3.31xx10^(-6)` | |
| 33. |
The electric field in a region of space is given by E= 5i + 2j N//C. The electric flux due to this field through an area 2m^(2) lying in the Yz plane, in S.I. units, is |
| Answer» ANSWER :A | |
| 34. |
कावेरी नदी का उद्गम स्थल कहां है? |
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Answer» पश्चिमी घाट के ब्रह्मगिरि श्रृंखला से |
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| 35. |
Consider a photon of continuous X-ray and a photon of characteristic X-ray of the same wavelength. Which of the following is different for the two photons |
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Answer» FREQUENCY |
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| 36. |
In a certain region of space electric field vecE and magnetic field vecB are perpendicular to each other and an electron enters in region perpendicular to the direction of vecB and vecE both and move underflected, then velocity of electron is |
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Answer» `(ABS(VECE))/(abs(VECB))` |
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| 37. |
The two lenses of a compound microscope are of foca lengths 2 cm and 5 cm. If an object is placed at a distance of 2.1 cm from the objective of focal length 2 cm the final image forms at the least distance of distinct vision of a normal eye. Find the distance between the objective and eyepiece |
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Answer» `46.17 CM ` |
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| 38. |
Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convexlens of smaller and smaller focal length and achieving greater and greater magnifying power ? |
| Answer» Solution :First, grinding lens of very SMALL focal length is not easy. More imprortant, if you decrease focal length, aberrations (both SPHERICAL and chormatic ) become more PRONOUNCED. So, in practice, you cannot get a magnifyng power of more than 3 or so with a simple convex lens. However, using an ABERRATION corrected lens system, ONE can increase this limit by a factor of 10 or so. | |
| 39. |
What materials should be used to make permanent magnets ? |
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Answer» Solution :The hysteresis curve allows us to select suitable materials for permanent MAGNETS. The material should have high retentivity so that the MAGNET is STRONG. These material have high coercivity, so that the magnetization is not erased by stray MAGNETIC fields, temperature fluctuations or minor mechanical damage. Steel is one favoured choice. It has a SMALLER retentivity than soft iron but has much smaller coercivity of soft iron. Other suitable materials for permanent magnets are alnico, cobalt steel and ticonal. |
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| 40. |
How does the mutual inductance of a pair of coils change when (i) distance between the coils is decreased and (ii) number of turns in the coils is decreased ? |
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Answer» Solution :(i) When the distance between the two coils is INCREASED, MUTUAL inductance of the PAIR of coils is decreased. (ii) If NUMBER of turns in the coils is decreased then mutual inductance of the given pair of coils decreased. |
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| 41. |
Show that the direction of electric field at a given is normal to the equipotential surface passing through that point. |
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Answer» Solution :If the electric field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the DIRECTION of the component of the field work would have to be done. But this is in contradiction to the definition of an equipotential surface because the potential difference between any TWO POINTS on the surface is `DELTAV` is zero so in `:. ` Work W `= qDeltaV , DeltaV=0 :. W =0` and work done in electric field `vecE` with displacement `vecdl` `W = vecE. vecdl` = Edl cos `theta` 0 = Edl cos `theta` ` :. 0 = cos theta [becausevecEne0vecdl ne0]` `:. theta= (pi)/(2)` Hence, the electric field `vecE` is normal to theequipotential surface at that point. |
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| 42. |
Explain theterms (i) Attenuation and (ii)Demodulationusedin CommunicationSystem. |
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Answer» SOLUTION :(i) Lossof strength ofa signalduringpropagation throughcommunication CHANNEL. (II) Demodulationis thephenomemon ofretrieval OFINFORMATION signalfromthe MODULATED weve atreceiver. |
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| 43. |
Two identical cells of emf 1.5V and internal resistance 1Omega are joined in series and the combination is connected in parallel with a third cell of the same emf and internal resistance. Calculalte the terminal voltage of the cells. |
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Answer» |
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| 44. |
Statement - (A): Resistivity of insulators is about 10^22 times the resistivity of metallic conductors. Statement - (B): Metals like silver, copper and aluminium have very high values of resistivity. |
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Answer» A and B are TRUE |
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| 45. |
If a point source is placed at a distance of 18 cm from the pole of a concave mirror, its image is formed at a distance of 9 cm from the mirror. A glass slab of thickness 6 cm is placed between the point source and the mirror such that the parallel faces of the glass slab remains perpendicular to the principal axis of the mirror. If the refractive index of glass is 1.5 what will be the displacement of the image? |
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Answer» Solution :Let P be the position of the object and in the absence of the glass slab, Q be the position of image formed by the concave MIRROR [Fig. 2.23]. Here, `mu = 18` `OQ = v = 9 cm` `therefore " " "According to" (1)/(v) + (1)/(u) = (1)/(f) "we get"` `(1)/(-9) + (1)/(-18) = (1)/(f) or, f = - 6cm` If the slab of thickness 6 cm is PLACED between the point source and the concave mirror, apparent displacement of the point source will take place towards the mirror. The RAYS coming from P appear to come from P. after refraction. Apparent displacement of P. `PP. = d(1-(1)/(mu)) = 6(1-(1)/(1.5)) = 2cm` So in the second case, object distanceu = -(18 - 2) = - 16cm, f = - 6 cm , v = ? `"We know" , (1)/(v) + (1)/(u) = (1)/(f) or, (1)/(v) + (1)/(-16) = (1)/(-6)` `or, "" (1)/(v) = (1)/(-6) + (1)/(16) = (-10)/(96) or, v = -(96)/(10) = - 9.6cm` `therefore "" |OQ.| = 9.6` `therefore " " "Displacement of the image".` `Q Q. = OQ. - OQ = 9.6 - 9 = 0.6cm` |
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| 46. |
What excuse does Beinkensopp give in the opening stanza? |
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Answer» GRANDMA's death |
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| 47. |
On the basis of your understanding of the following paragraphand the related studied concepts. An equipotential surface is a surface with a constant value of electric potential at all points on the surface. The component of electric field parallel to an equipotential surface is zero, as the potential does not change in this direction. Thus, the electric field is perpendicular to the equipotential surface at each point of the surface. Further, say the direction of electric field electrical potential gradually fall with distance. A uniform electric field vecE = 1000 hati V m^(-1) exists in space at a given place. What is the shapeof equipotential surfaces there ? |
| Answer» SOLUTION :Since electric FIELD `vecE` is a uniform field ALONG the DIRECTION of + x-axis, the equipotential surfaces MUST be planer surfaces in Y - Z plane. | |
| 48. |
Figure shows a snaps hot of a sinusoidal travelling wave taken at t = 0.3s. The wavelength is 7.5 cm and the amplitude is 2 cm. If the crest P was at x = 0 at t=0), write the equation of travelling wave |
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Answer» Solution :The wave has travelled a distance of 1.2 CM in 0.3s Hence SPEED of the wave, `v=(1.2)/(0.3)=4 cm//s` `:.` Angular frequency `omega=(v)(k)=3.36` rad /s Since the wave is travelling ALONG positive x-direction and crest (maximum displacement) is at x=0 at t=0 we can write the wave equation as, `y=A sin (kx-omegat)+(pi)/(2) (or) y(x,t)=a cos(kx-omegat)` THEREFORE, the desired equation is, `y(x,t)=(2)cos[(0.84)x-(3.36)t]cm` |
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| 49. |
On the basis of your understanding of the following paragraphand the related studied concepts. An equipotential surface is a surface with a constant value of electric potential at all points on the surface. The component of electric field parallel to an equipotential surface is zero, as the potential does not change in this direction. Thus, the electric field is perpendicular to the equipotential surface at each point of the surface. Further, say the direction of electric field electrical potential gradually fall with distance. Draw equipotential surfaces around an isolated positive point charge + q. Mark direction of electric field vecE also. |
| Answer» Solution :Equipotential SURFACES are drawn below and direction of ELECTRIC FIELD `vec E ` us ALSO marked. | |
| 50. |
लीची का सबसे अधिक उत्पादन होता है? |
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Answer» बिहार |
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