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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 60 muF capacitor is connected to a 110 V, 60 Hz a.c. supply. Determine the rms value of the current in the circuit. |
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Answer» Solution :Here, C = `60 muF = 60 xx 10^(-6) F, V_(rms) = 110V` and FREQUENCY v = 60 Hz `I_(rms) = V_(rms)/X_(C) = (V_(rms))/(1/C omega) = V_(rms).C.Omega.C.2piv` `=110 xx 60 xx 10^(-6) xx 2 xx 3.14 xx 60 = 2.49` A |
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| 2. |
Can the potential barrier across a p - n junction be measured by simply connecting a voltmeter across the junction ? |
| Answer» SOLUTION :Because there is no FREE CHARGE CARRIER in depletion region. | |
| 3. |
A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index mu. The radius of the sphere is R. Aț.t = 0, the ball is dropped to fall normally on the sphere. For t lt sqrt((2h)/(g))and by a single refraction, what is the speed of image as a function of time ? |
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Answer» `(MU R^2 "gt")/([(mu-1)(H - 1/2 "gt"^2) - R]^2 )` |
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| 4. |
In a pure silicon (n_1=10^16//m^3)crystal at 300 K. 10^21atoms of phosphorus are added per cubic metre. What is the new hole concentration ? |
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Answer» `10^196perm^3` |
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| 5. |
If the specific heat of a gas at constant volume is 3/2R, then value of gamma |
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Answer» `5/2` |
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| 6. |
Galvanometer are of two types____and_____. |
| Answer» SOLUTION :[MCG(MOVING COIL) and MMG (Moving Magnet i.e.TG] | |
| 7. |
How does the resistivity of the following materials vary with the increase in their temperature : (i)metallic conductor and (ii) semiconductor ? |
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Answer» SOLUTION :(i)In metallic conductor , the RESISTIVITY `(RHO)` increases with INCREASE in TEMPERATURE . (ii)In semiconductors , resistivity decreases with increase in temperature . 
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| 8. |
The main scale of vernier callipers reads in mm and its vernier scale is divided into 8 divisions which coincide with 5 divisions of the main scale. When the two jaws touch each other, the zero of the vernier coincides with the zero of main scale. When a rod is tightly placed along its length between the jaws, it is observed that the zero of vernier scale lies just left to 36^(th) division of main scale and fourth division ofvernier scale coincides with one main scale division. The measured value is (in cm) |
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Answer» 3.66 `X + 4V = nM` `X = nM - 4(5/8 M)` `X lt 1` MM for `n = 3` only So, `X=3-5/2=1/2mm` `:.` reading = `35+0.5 mm` ` = 3.55 cm`. |
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| 9. |
How did Pandit Jawaharlal Nehru react to Bhagat Singh's sacrifice? |
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Answer» He made no comment |
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| 10. |
A boy observes a celestial object at a distance of 7 xx 10^(3) km from the Earth. The radius of the object is 4 xx 10^(3) km. Determine the angle subtended by the image on the eye if focal lengths of objective and eyepiece are 5 m and 10 cm, respectively. |
| Answer» SOLUTION :`32.68^(@)` | |
| 11. |
If one microgram of ._92^235U is completely destroyed in an atom bomb, the energy released will be ? |
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Answer» `9xx10^6` J |
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| 12. |
In figure, V_(o) is the potential barrier across a p-n junction, when no battery is connected across the junction. |
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Answer» 1 and3 both correspond to forward bias of junction. When p-n junction is forward BIASED, magnitude of barrier potential (i.e. HEIGHT of potential barrier) becomes smaller than `V_(0)`, which is seen in graph 3. When p-n junction is reverse biased, magnitude of barrier potential becomes greater than `V_(0)`, which is seen in graph 1. Thus, option (B) is CORRECT. |
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| 13. |
The K_(infty) X-ray emission line of tungsten occurs at lambda = 0.021 nm.The energy difference between K and L levels in this atom is about |
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Answer» `0.51`MeV |
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| 14. |
Which animal did the grandmother used to feed in the village? |
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Answer» Dogs |
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| 16. |
A cyclist is moveing on a smooth horizontal curved surface with a speed of 5 ms^(-1) If angle of leaning is 30^(@) radius of curvature of Road should be |
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Answer» `5 SQRT3` |
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| 17. |
What was the most enchanting line in the play? |
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Answer» Now that's too much |
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| 18. |
Mention any one application of LEDs. |
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Answer» Solution :1. They are used in T.V remote 2. They are used in burglar ALARMS. 3. They are used in calculators 4. They are used in traffic signals. 5. They are used in optical communication [Any one APPLICATION] |
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| 19. |
The ratio of adiabatic and isothermal bulk modulus of elasticity for a perfect gas is |
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Answer» `GAMMA` |
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| 20. |
State any three feautures of nuclear force |
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Answer» Solution :Nuclear forces are 1. Strongly ATTRACTIVE forces. 2. Strongest forces in nature 3. Extremely short-range forces. 4. Charge independent 5. SPIN dependent 6. Non-central 7. SATURATED 8. Do not obey inverse square law. (Any three) |
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| 21. |
Find the binding energy of the deuterium nucleus and the specific binding energy (binding energy per nucleon). |
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Answer» |
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| 22. |
A stationary pion decays into a muon and a neutrino: pi^(+)tomu^(+)""v_(mu) Find the ratio of the energy of the neutrino to the kinetic energy of the muon. |
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Answer» `Deltaepsi=(m_(pi)-m_(mu))xx931.5=140-106=34MeV` The rest ENERGY of a muon is `epsi_(0mu)=106MeV`, the rest mass of a neutrino is zero. Assuming the dacaying PION to be at rest, we obtain that the muon and the neutrino momenta are equal in magnitude and opposite in direction. We have `p_(V)=epsi_(v)/c,p_(mu)=1/csqrt(K_(mu)(2epsi_(0mu)+K_(mu)))` which yields `epsi_(v)=sqrt(K_(mu)(2epsi_(0mu)+K_(mu)))` But, from the law of conservation of energy `epsi_(v)=Deltaepsi-K_(mu)`, so `Deltaepsi-K_(mu)=sqrt(K_(mu)(2epsi_(0mu)+K_(mu)))`. Hence it FOLLOWS that the kinetic energy of the muon is `K_(mu)=((Deltaepsi)^(2))/(2(epsi_(0mu)+Deltaepsi))` |
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| 23. |
Why cannot we experience the existence of matter waves in our day to day life . |
| Answer» Solution :`lambda = h/(mv)` , if m is large, `lambda` will be SMALL, THUS the WAVELENGTH will be very very small, compared to the SIZE of the particle.hence, it is not POSSIBLE. | |
| 24. |
In the given figure the capacitor of plate area A is charged upto charge q. The ratio of elongations (neglect force of gravity) in springs C and D at equilibrium position is |
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Answer» `(k_1)/(k_2)` |
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| 25. |
A cyclotron is used to obtain 2 Me V protons. If the frequency is 5MHz and potential is 20kV. The magnetic field necessary for resonance is |
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Answer» 2.32T |
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| 26. |
A: The bandwidth of a modulated signal is 2f_(m) in amplitude modulation, where f_(m) is the frequency of the modulating signal. R: In case of amplitude modulation the frequency of a modulated signal is equal to the frequency of the carrier wave. |
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Answer» If both ASSERTION and reason are true and reason isthe correct EXPLANATION of assertion |
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| 27. |
Describe briefly with the help of a circuit diagram, how the flow of current carriers in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased |
Answer» Solution : The HEAVILY doped emitter has a high concentration of majority carriers, which are the HOLES in a p-n-p transistor. These majority carriers enter the base region in large numbers. As the base is thin and LIGHTLY doped, the majority carriers (holes), entering the base region from emitter, SWAMP the small number of electrons there and, as the COLLECTOR is reverse biased, these holes can easily cross the junction and enter the collector. |
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| 28. |
The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why? |
| Answer» Solution :Because MOLTEN iron (which is the PHASE of the iron at the high TEMPERATURES of the core) is not ferromagnetic on ACCOUNT of its temperature being higher than the Curie temperature for iron. | |
| 29. |
A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is |
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Answer» virtual and at a distance of 16 cm from the mirror {(1/v) + (1/30) }= 1/15 rArr v = 30 cm` so, object for place mirror is 20 cm from it towards right. `rArr`Image formed = 20 cm left from plane mirror. This behaves as virtual object for lens at a distance of 10 cm left.  using lens formula, (1/v.) + (1/u) = 1/f `rArr (1//v.) + (1//-10) = 1//15 rArr v. = 6 cm` So, image formed is real and at a distance 16 cm from mirror. `mu = A + (B)/(lambda^(2)) + (C )/(lambda^(4)) + ...` `f_(R) ge f_(v)`. |
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| 30. |
WHAT WAS DHOBI BRANDISHING? |
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Answer» CHEQUE |
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| 31. |
Electromagnetic radiation used to sterilise milk is |
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Answer» X-ray |
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| 32. |
A sensitive magnetic instrument can be shielded very effectively from outside magnetic fields by placing it inside a box of |
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Answer» TEAK WOOD |
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| 33. |
If potential (in volts)in regions is expressed as V (x,y,z) = 6xy -y + 2yz, the electric field (in N//C) at point (1,1,0) is |
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Answer» `-(6 hat(i) + 9 hat(j) + hat(k))` |
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| 34. |
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1//2) QE, where Q is the charge on the capacitor , and E is the magnitude of electric field between the plates . Explain the origin of the factor 1//2. |
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Answer» Solution :If F is the force on each plates of parallel plate CAPACITOR, then work done in increasing the seperation between the plates by `Delta x = f Delta x` This must be the increase in potential energy of the capacitor Now the increase the volume of capacitor is `= A Delta x` If U = energy DENSITY = energy stroed/volume then the increase in potential energy `= U.A Delta X` `:. f Delta X = U. A Delta X` `f = U.A = ((1)/(2) epsilon_(0) epsilon^(2)) A = (1)/(2) (epsilon_(0) A epsilon) epsilon` `F = (1)/(2) (epsilon_(0) A (V)/(d)) epsilon [ :' C = (epsilon_(0) A)/(d)]` `F = (1)/(2) (CV) E = (1)/(2) QE` The origin of factor 1/2 in force can be explained by the fact that inside the conductor field is zero and outside the conductor, the field is E. Therefore the AVERAGE value of the field (i.e E/2) contributes to the force. |
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| 35. |
The polariser and analyser are inclined to each other at 60^(@). If I/2 is the intensity of the polarised light emergent from analyser. Then the intensity of the unpolarised light incident on the polariser is |
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Answer» 8I |
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| 36. |
What is the difference between terrestrial and astronomical telescope? |
| Answer» SOLUTION :The IMAGE in a terrestrial TELESCOPE will be erect while the image in the ASTRONOMICAL telescope is inverted. | |
| 37. |
Fibre optic communication is gaining popularity among the various transmission media-justify. |
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Answer» Solution :(i) The method of transmitting information from one place to another in terms of light pulses. Through an optical FIBRE is called Fibre optic communication. (ii) It is in the process of replacing wire transmission in communication systems. (iii) As fibres are not electrically conductive, it is preferred in places where MULTIPLE channels are to be laid and isolation is required from ELECTRICAL and electromagnet INTERFERENCE. (iv) Fibre cables are very thin and weigh lesser than copper cables. (v) This system has much larger band width . This means that its information carrying CAPACITY is larger. (vi) Fibre optic system is immune to electrical interferences. (vii)Fibre optic cables are cheaper than copper cables. |
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| 38. |
A particle is dropped along the axis from a height f/2 on a concave mirror of focal length f as shownin the figure. The acceleration due to gravity is g. Then the maximum speed of the image is given by 3/4 sqrt(x fg)where x = ..... |
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Answer» |
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| 39. |
A beam of light consisting of two wavelength 800 nm and 600 nm is used to obtain the interference fringes in YDSE on a screen held 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from thecentral bright maximum, where the bright fringes of the two waveength coincide. |
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Answer» Solution :`x=N lamda_1D/d=(n+1)lamda_2 D/d` `n times 800=(n+1) lamda_2 D/d` `n=3` `x=nlamda_1 D/d=3 times 800 times (10^-9 times 1.4)/(0.28 times 10^-3)=12 mm` |
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| 40. |
What is current density ? Derive Ohm's law in form of current density? |
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Answer» Solution :`RARR` Current density : The electric current density at any point is defined as amount of electric current flowing per unit cross section PERPENDICULAR to the current at that point. Current density is VECTOR quantity, `vec(J) = (I)/(vec(A))` Unit = `(A)/(m^(2)) = Am^(-2)` Dimensional formula = `M^(0) L^(-2) T^(0) A^(1)` Ohm.s law V = IR `"" `....(1) Consider conductor of length l electric field be E then potential DIFFERENCE between ends of conductor be V then, `E = (V)/(l)` V = El resistance R= `(rho l)/(A)` V = IR El = `I ((rho l)/(A))` `(I)/(A) = J` E = J`rho ` `J = SIGMA ` E `vec(J) = sigma vec(E)` which also represent Ohm.s law (in vector form). |
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| 41. |
For a transistor, the current gain of common- base configuration is 0.8. If the transistor is in common-emitter configuration and the base current changes by 5mA, then the change in collector current is |
| Answer» Answer :B | |
| 42. |
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A. |
| Answer» SOLUTION :`2 xx 10^(-5) N`, ATTRACTIVE FORCE normal to A towards B | |
| 43. |
In modern diodes the anode is often brought very close to the cathode so Dial their areas are approximately equal. Assuming the electrons to leave the cathode with zero velocity find the force with which they act on the anode. The current in the tube is i_("sat")=500mA, the anonded voltage is varphi_(a)=600V. |
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Answer» `F=(imv)/(e)=(i)/(e)SQRT(2evarphim)` |
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| 44. |
A ray of light is to travel from point A to point B in figure in the shortest possible time after reflecting from P. Then OP is ..... cm. |
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Answer» |
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| 45. |
The time period of a thin bar magnet is T. It is cut into 'n' equal parts by cutting it normal to its length. What will be the time period of each of them in the same field? |
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Answer» Solution :The expression for time period of a magnet `T= 2pi SQRT((I)/(MB))` `:. T prop sqrt((I)/(M))`........... (1) [`:.` B is same] But `I=m((l^(2) + b^(2))/(12))`, where m is mass For a THIN BAR magnet breadth `b lt lt l` , So neglect `b^(2)` in the above expression . ` :. ~~ (ml^(2))/(12) or I prop ml^(2)` .......... (2) From eq.s (1) and (2) , `T prop sqrt((ml^(2))/(M))` `:. (T_2)/(T_1) = (l_2)/(l_1) sqrt((m_2)/(m_1) xx (M_1)/(M_2))` But `l_2 = (l_1)/(n) , m_2 = (m_(1))/(n) and M_2 = (M_1)/(n)` `:. (T_2)/(T_1) = (1)/(n)=(1)/(n) sqrt((1)/(n) xx n) = (1)/(n) implies T_2 = (T)/(n)` |
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| 46. |
4^(3/5)⋅4^(7/5) निम्नलिखित में किसके बराबर है? |
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Answer» `(16〗^(4/5)` |
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| 47. |
The value of electric potential at any point due to any electric dipole is (k = (1)/(4pi epsi_(0))) |
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Answer» `k.(VEC(p) XX vec(R ))/(r^(2))` |
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| 48. |
A cell is balanced on 125 cm length of a potentiometer wire. Now the cell is short circuited by aresistance of 2 Omegaand the balance is obtained at 100 cm. The internal resistance of the cell is____. |
| Answer» SOLUTION :`0.5Omega` | |
| 49. |
To a man walking at the rate of 3 km/h the rain appears to fall vertically. When he increases his speed 106 km/h it appears to meet him ar an angle of 450 with vertical. Find the angle made by the velocity of rain with the vertical and its value. |
Answer» Solution : From the diagram `Tan 45^(@)=(3)/(4) ....(1) and Tan theta=(3)/(4) ......(2)` From (1) and (2)`theta =45^(@)` and `:. SIN5^(@)=(3)/(V_(R)),(1)/(sqrt(2))=(3)/(V_(R)) "" V_(R)= 3sqrt(3)` KMPH |
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| 50. |
Assertion: Lenz's law is about violation of energy conservation in electromagnetic induction . Reason: When magnetic flux linked with a coil changes then emf is induced in such a manner that it opposes the cause that has induced it. |
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Answer» If both ASSERTION and REASON are CORRECT and reason is correct EXPLANATION of the assertion |
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