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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. |
Thedifference between phase and frequency modulation |
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Answer» practically they are same but theoretically they differ |
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| 2. |
In accordance withthe Bohr's model, find the quantum numberthat characterisethe earth'srevolutionaroundthe Sunin an orbitof radius1.5 xx 10^(11)m withorbital speed 3 xx 10^(4) m//s . |
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Answer» SOLUTION :Here , orbital radiusof earth `r = 1.5 xx 10^(11)` m , orbital speed of earth ` v = 3 xx 10^(4) m//s`and mass of earth`m=6.0 xx 10^(24) kg` As per Bohr.smodel, ANGULAR momentum `mv_(n)r_(n) = (n H)/(2pi)` `rArr` Quantum number ` n = (2 pi m v_(n) .r_(n))/(h) = 2 xx 3.142 xx 6.0 xx 10^(24) xx 3 xx 10^(4) xx3 xx 10^(4) xx 1.5 xx (10^(11))/(6.63 xx 10^(-34)) =2.6 xx 10^(74)`. |
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| 3. |
A vessel contains some water. It is moved towards right on a horizontal plane with a constant acceleration 'a'. Which of the following figure represents correct water surface ? |
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Answer»
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| 4. |
One quarter sector is cut from a uniform circular disc ofradius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is |
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Answer» `1/2MR^(2)` Moment of INERTIA of that disc = `1/2(4M)R^(2)` Moment of inertia of sector=`1/2(4M)R^(2)xx1/4=1/2MR^(2)` |
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| 5. |
Define conductance and conductivity. Or. Define electrical conductivity of a conductor and give its S.I. unit. |
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Answer» SOLUTION :Conductance. The reciprocal of resistance is called conductance. It is denoted by G and its unit in S.I. is `OHM^(-1)(Omega^(-1))` or mho or siemen (S). And `G=(1)/(R)` CONDUCTIVITY. The conductivity is reciprocal of resistivity and it measure the ease with which the current FLOWS through a conductor. it is denoted by `SIGMA`. And `sigma=(1)/(rho)` The unit of `sigma` is `Omega^(-1) or Sm^(-1)`. |
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| 6. |
A manwishes tocrossariverflowingwith velocityu joumpsat ananglethetawiththeriverflow. Findoutthenetvelocity of themanwithrespectto groundif hecanswim withspeedv. alsofindhowfarhow fatfromthe pointboatactuallymove. if thewidthof theriveris d. |
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Answer» Solution :velocityof man`= V_M= sqrt( U^2+ v^2+ 2vu cos theta )` `tanphi= ( vsintheta)/(u+ vcostheta )` `( VSIN theta ) t =d impliest =(d)/(v SINTHETA)` `X= (u + v costheta) t= (u + vcostheta) (d)/( v sintheta )` 
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| 7. |
The blue colour of the sky is due to the phenomenon of |
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Answer» scattering |
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| 8. |
Find the change on the capacity C= 1muFin the circuit shown in the figure . |
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Answer» `10MUC` |
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| 10. |
What is meant by persistence of vision? |
| Answer» Solution :After an OBJECT has been removed, its image persists on the retina for short time. This is called persistence of vision. Because of this, a bright point on the CIRCUMFERENCE of a wheel appears as a COMPLETE circle if the wheel is rotated SUFFICIENTLY fast. | |
| 11. |
A wire of identical length and cross-section is used for fixed vibration between two rigid supports slightly more than the length of string. The wire has Young's Modulus =2Y and mass per unit length =mu//2 and v is the fundamental frequency. Find frequency in the 2^(nd) mode of this wire. |
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Answer» V |
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| 12. |
While measuring viscosity of caster oil using terminal velocity concept the following observation table has been taken by a student. Which one is the first connect reading which he should consider for the computation of terminal velocity |
| Answer» ANSWER :C | |
| 13. |
The electric potential V at a point on the axis of an electric dipole depends on the distance 'r of the point from the dipole as |
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Answer» `V prop 1/R` |
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| 14. |
Assertion: The whole charge of a conductor cannot be transferred to another isolated conductor Reason:The total transfer of charge from one to another is not possible |
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Answer» Both ASSERTION and Reason are true and Reason is the correct explanation of Assertion |
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| 15. |
A gas consisting of monatomic molecules (degrees of freedom = 3) was expanded in a polytropic process so that the rate of collisions of the molecules against the vessel's wall did not change. Find the molar heat capacity of the gas in the process. |
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Answer» |
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| 16. |
For a conductor, the given Ogure shows the graph of V rarrI for different temperatures then ..... . |
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Answer» `T_(1) lt T_(2) lt T_(3)`  As temperature increases resistance also increases and slope of graph `V rarr ` I shows resistance . For `T_(1)` temperature `(DELTA V)/(Delta I)` is MINIMUM and for `T_(3)` temperature `(Delta V)/(Delta I)` is maximum. |
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| 17. |
A 50g lead bullet (sp. heat = 0.02) is at 30^(@)C. It is fired vertically upwards with a speed of 840 m s^(-1). On returning to the starting level it strikes the ice cake at 0^(@)C. How much ice is melted ? (L.H. of ice 80 cal/gm): |
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Answer» 52.875 g `=4200` cals. Also heat given by bullet to cool to `0^(@)C mc theta =50xx0.02 xx30=30` cals TOTAL heat =4230 =ML or `M=(4230)/(80)=52.875 g`. `THEREFORE` Correct choice is (a). |
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| 18. |
Singly charged ions He^(+)are accelerated in a cyclotron so that theirmaximumorbitalradiysis r = 60 cm. The frequencyof a cyclotron'soscillator is equal to v = 10.0 MHz, theeffectiveaccelearatingvolatageacross the deos is V = 50 kV. Neglectingthe gap between the does, find: (a) the total timeof acceleration of the ion, (b) the appoximatedistancecovered by the ion in theprocess of its acceleration. |
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Answer» Solution :(a) The total time of acceleration is, `t = (1)/(2v).n`, where `n` is the number of passage of theDoes. But, `T = "neV" = (B^(2) e^(2) r^(2))/(2m)` or, `n = (B^(2) e r^(2))/(2 mV)` So, `t = (pi)/(eB//m) xx (B^(2) er^(2))/(2 mV) = (pi B r^(2))/(2V) = (pi^(2) mv r^(2))/(eV) = 30 MU s` (b) The distance covered is, `s = sum v_(n). (1)/(2v)` But, `v_(n) = SQRT((2eV)/(m)) sqrt(n)` So, `s = sqrt((eV)/(2MV^(2))) sum sqrt(n) = sqrt((eV)/(2 mv^(2))) f sqrt(n) dn = sqrt((eV)/(2mv^(2))) (2)/(3) n^(3//2)` But, `n = (B^(2) e^(2) r^(2))/(2 eV m) = (2 pi^(2) m V^(2) r^(2))/(eV)` THUS, `s = (4pi^(3)n v^(2) mr^(2))/(3eV) = 1.24 km` |
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| 19. |
The horizontal component of earth's magnetic induction at a place is 0.32 xx 10^(4)T. The angle of dip at the pointis60^(@), then find value of vertical component ? |
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Answer» Solution :Vertical component , `B=B_H TAN THETA = 0.32 xx 10^(-4) xx sqrt(3) = 0.55 xx 10^(-4)T ` |
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| 20. |
The resultant of two forces 3P and 2P is R. If the first vector is doubled, the resultant of the vector is also doubled. The angle between the vector is |
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Answer» `30^@` |
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| 21. |
An NPN BJT having reverse saturation currents I_S=10^(-15) A is biased in the forwared active region withV_BE=700mV and the current gain (beta)may vary from 50 to 150 due to manufacturing variation. What is the maximum emitter currents(inmuA). |
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Answer» SOLUTION :`I_S=10^(-15)A` `V_BE=0.7` `V_T=25mV` `beta` RANGE from 50 to 150 `I_C=I_o e^(V_BE//V_T)` `I_E=(beta+1)/(beta)I_C` `I_E=(beta+1)/(beta)I_Se^(V_BE//V_T)` `I_E`be MAXIMUM when `beta` is 50 `=1.02xx10^(-15)xxe^(700xx10^(-3)//25xx10^(-3))` `I_E=1475muA` |
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| 22. |
Two simple pendulum of lengths 16I and I are in phase at the mean position at a certain time.IF T is the time period of shorter pendulum. The maximum time after which they will be again in phase. |
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Answer» `1/3T` `Talpha sqrt(l)` The PENDULUM having time period makes one oscillation LESS than the other. `(T_1-1)/(T_1)=sqrt(l)/(16l)=1/4` `implies (T_1-1)/(T_1)=1/4` `implies1-(1)/T_1=1/4` `T=4//4` Theywill be again in PHASE after`4/3T`, |
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| 23. |
An alternating current in a circuit is given by I=20sin(100 pi t+0.05 pi) A. The r.m.s value and the frequency of current respectively are |
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Answer» 10 A and 100 Hz COMPARING with STANDARD EQUATION, we get `I_(0)=20`and `omega = 100 pi` `therefore I_(rms)=(I_(0))/(SQRT(2))=(20)/(sqrt(2))=10 sqrt(2)` Now, `omega = 100pi` or `2pi upsilon = 100 pi` or `upsilon = 50 Hz`. |
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| 24. |
The emf of a daniel cell is 1.08V. When the terminals of the cell are connected to resistance of 3Omega, the potential difference across the terminals is found to be 0.6 V. Then , the internal resistance of the cell is |
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Answer» `1.8Omega` |
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| 25. |
The polarising angle for a medium is found to be 60^(@).The critical angle of the medium is |
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Answer» a. `SIN^(-1)(1/2)` |
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| 26. |
Time period of cillations of a magnet of magnetic moment M and moment of inertia l in a vertical plane perpendicular to the magneitc meridian at a place where earth's horizontal and vertical component of magnetic field are B _(H) and B_(v) respectively is |
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Answer» `T = 2pi sqrt ((1)/(M (B_(v) ^(2) + B_(H) ^(2))))` |
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| 27. |
What is the value of shunt resistance required to convert a galvanometer of resistance 100 Omega into an ammeter of range 1 A ? Given: Full scale deflection of thegalvanometer is 5 mA. |
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Answer» `(5)/(9.95) OMEGA` |
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| 28. |
What is peak inverse voltage (PIV)? |
| Answer» Solution :The maximum REVERSE bias that can be applied before entering into the Zener region is called the PEAK INVERSE VOLTAGE. | |
| 29. |
In simplest terms, the energy of a wave is directly proportional to the square of Its |
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Answer» height |
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| 30. |
Which ofthe followingrespresentthe variation of conducatnes ofsolutionifweakbase NH_(4)OH is titrated with dilute HCl? |
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Answer»
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| 31. |
How do you compare the gravitational force to the electrostatic force of repulsion between two electrons. |
| Answer» Solution :LET ‘R’ be SEPARATION between two ELECTRONS. | |
| 32. |
If 10^9 electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body? |
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Answer» Solution :In one SECOND `10^9`electrons move out of the body. The charge given out in one second is `1.6 xx 10^(-19) xx 10^(9)C = 10^(-10) C`. The time required to accumulate a charge of 1 C can then be estimated to be `t = 1 C div (1.6 xx 10^(-10) C//s) = 6.25 xx 10^(9)s` `= 6.25 xx 10^9 div (365 xx 24 xx 3600) ` YEARS = 198 years. One coulomb is a very large UNIT for many PRACTICAL purposes. |
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| 33. |
10 drops of an oleic acid solution of concentration 1/400 cm^3 per cm^3 of alcohol, are dropped on a water surface. The circular film thus produced has radius 10 cm. Find the molecular size of oleic acid, if the radius of each drop is 1.4 mm. |
| Answer» SOLUTION :`t="NV"/"400A"` or `9.15xx10^(-9)` m | |
| 34. |
An object initially at rest explodes into three fragments A, B and C. The momentum of A is P and that of B is sqrt(3) P perpendicular to A. the momentumof C will be |
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Answer» <P>`(1 + sqrt(3))` P in a direction MAKING `120^@` with that of A |
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| 35. |
Unit and dimensional formula of linear charge density are....... |
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Answer» `CM, M^(0)L^(1)A^(1)T^(1)` `|lambda| =|q|/|l| = (A^(1)T^(1))` |
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| 36. |
In the network shown in figure. (i) calculate the current of the 6V battery and (ii) determine the potential difference between the points A and B. |
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Answer» |
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| 37. |
Four wires of same length and same material, whose diameters are in the ratio 4 : 3 : 2 :1, are clamped in such a way that each wire produces note of frequency double that of the preceding wire. If the tension in the first wire is 2 Kg-wt, then tension in the second wire will be - |
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Answer» 4.5 |
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| 38. |
Does the angle of deviation of a small angled prism depend on angle of incidence ? |
| Answer» SOLUTION :No, for a SMALL angled PRISM angle of deviation DELTA = ( MU - 1 | |
| 39. |
The figure shows a circuit consisting of four capacitors. Find the effective capacitance between X and Y. |
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Answer» |
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| 40. |
which one of the following is (A-B)uu(B-A)=? |
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Answer» `(AuuB)uu(A-B)` |
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| 41. |
Two small identical circular loops, marked (1) and (2), carrying equal currents are placed with the geometrical axes perpendicular to each other as shown in the fig. Find the fig. Find the magnitude and direction of the net magnetic field at the net magnetic field at the point O. |
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Answer» Solution :We know that the magnetic field at a point on the axial line of a SMALL current LOOP of radius R is given by `B = (mu_0 I R^2)/(2 x^3)` `:.` Magnetic field at point O due to current loop number 1 `B_1 = (mu_0 I R^2)/(2x^3) ` along + ve X - axis. and magnetic field at point O due to current loop number 2 `B_2 = (mu_0 I R^2)/(2 x^3) ` along + ve Y-axis As `B_1 and B_2` are in mutually perpendicular directions (fig.) , the resulatant magnetic field subtends an angle B from horizontal, where `tan beta= (B_2)/(B_1) = 1 implies beta = 45^@` |
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| 42. |
Assuming that the silicon diode having resistance of 20Omega , the current through the diode is (knee voltage 0.7 V) : |
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Answer» 0 mA `impliesi=6.5xx10^(-3)=6.5` mA |
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| 43. |
A body of density D_(1) and mass M is moving downward in glycerine of density D_(2) with constant velocity what is the viscous force acting on it. |
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Answer» `MG(1-(D_(2))/(D_(1)))` `T=2pisqrt((l)/(g_("eff"))) "Here",g_("eff")=(g^(2)+(W^(2))/(m^(2)C^(2)))^(1/2)` |
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| 44. |
Addition of Oxygen is |
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Answer» Oxidation |
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| 45. |
While tuning in a certain broad cast station with a receiver, we are actually |
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Answer» varying the LOCAL oscillator |
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| 46. |
If f_0 and f_e, are respective focal lengths of objective and eye-piece of compound microscope, then ...... |
| Answer» Solution :`f_0 lt f_e` | |
| 47. |
Give the sign of the potentiaol energy difference of a small negative charge between the points Q and P,A and B. |
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Answer» <P> Solution :A small negative charge will be attracted towards POSITIVE charge. The negative charge moves from higher petential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive.similarly `(P.E.)_(A) GT (P.E.)_(B)` and hence sign of potential energy differences is positive. |
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| 48. |
A particle moves on a circle of radius r with centripetal acceleration as function of time as a_e = k^2 rt^2 where k is a positive constant. Find the resultant acceleration. |
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Answer» `KR^2` |
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| 49. |
A solenoid of 2m long & 3cm diameter has 5 layers of winding of 500 turns per metre length in each layer & carries a current of 5A. Find intensity of magnetic field at the centre of the solenoid. |
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Answer» Solution :For LONG SOLENOID at the CENTRE `B=mu_(0)NI` `H=B/(mu_(0))=ni=(500)5xx5=1.25xx10^(4)A//m` |
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| 50. |
The adjacent graph shows the extra extension (Deltax) of a wire of length 1m suspended from the top of a roof at one end with an extera load Deltaw connected to the other end If the cross sectional area of the wire is 10^(-5)m^(2) calculate the Young's modulus of the meterial of the wire (A) 2 xx 10^(11) N//m^(2) (B) 2 xx 10^(-11)N//m^(2) (c) 3 xx 10^913) N//m^(3) (D) 2 xx 10^(16)N//m^(2) . |
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Answer» SOLUTION :`DELTAL=((l_0)/(AY))Deltawslope=(l_(0))/(AY)=(1xx10^(-4))/(20)rArr(1)/((10^(-6))Y)=(1xx10^(-4))/(20)` `Y = 20 xx 10^(10) = 2 xx 10^(11) N//m^(2)` . |
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