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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. |
Three identical charged capacitors each of capacitance 5 muF are connected as shown in figure. Potential difference across capacitor (3), long time after the swithcesK_(1) and K_(2) are closed , is |
| Answer» Answer :D | |
| 2. |
Which of the following relation is correct for relative refractive index ? |
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Answer» `""^1n_2=""^1n_2xx""^3n_2` `=(n_1)/(n_3)XX(n_3)/(n_2)=(n_1)/(n_2)=""^1n_2` |
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| 3. |
Electromagnetic waves with wavelength(i)lambda_(1)is used in satellite communication.(ii)lambda_(2) is used to kill germs in water purifies.(iii)lambda_(3) is used to detect leakage of oil in underground pipelines.(iv)lambda_(4) is used to improe visibility in runways during fog and mist conditions.Identify and name the part of electromagnetic spectrum to which these radiations belong. |
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Answer» Solution :(i) In satellite communication microwaves are used so `lambda_(1)` is WAVELENGTH of microwave. (ii) Ultraviolet radiation are used to kill germs in water PURIFIER so `lambda_(2)` is wavelength of UV rays. (iii) X - rays are used to detect leakage of oil in underground PIPELINES so `lambda_(3)` is X - rays. (iv) Infrared rays are used to improve visiblity on runways during fog and mist condition so `lambda_(4)` is infrared rays. |
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| 4. |
Photo-electric emission is observed from a metalli surface for frequenceis f_(1) and f_(2) of the incident light. (f_(1)gtf_(2)) .If the maximum values of K.E. of photo-electrons emitted in two cases are in the ratio 1:n, then the threshold frequency of a metallic surface is ....... |
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Answer» `(f_(1)-f_(2))/(n-1)` `therefore (K_(2))/(K_(1))=(f_(2)-f_(0))/(f_(1)-f_(0))` `therefore (K_(2))/(K_(1))=(f_(2)-f_(0))/(f_(1)-f_(0))` `therefore (nK_(1))/(K_(1))=(f_(2)-f_(0))/(f_(1)-f_(0))` `therefore nf_(1)-nf_(0)=f_(2)-f_(0)` `therefore nf_(1)-f_(2)=(n-1)f_(0)` `therefore f_(0)=(nf_(1)-f_(2))/(n-1)` |
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| 5. |
Electromagnetic waves with wavelength(i)lambda_(1)is used in satellite communication.(ii)lambda_(2) is used to kill germs in water purifies.(iii)lambda_(3) is used to detect leakage of oil in underground pipelines.(iv)lambda_(4) is used to improe visibility in runways during fog and mist conditions.Write one more application of each. |
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Answer» Solution :(i) Microwaves is USED in RADAR. (ii) UV rays are used in LASIK eye SURGERY. (iii) X-rays is used to detect a fracture in bones. (iv)INFRARED is used in optical COMMUNICATION. |
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| 6. |
In thepreviousproblem, calculatethe averageenergydensityof magneticfield . |
| Answer» SOLUTION :`4.34 XX 10^(-12)Jm^(-3)` | |
| 7. |
(a) Students wants to use tow p-m jusnction diodes to convnet alternating currentinto direct current . Draw the labelled circuit she would use and explain how it works.(b) Give the truth tableand circuit symbolforNAND gate. |
Answer» Solution :(a) Toconvert ALTERNATING currentinto direct currentshe shoulduse fullwave rectifier. Thecircuit diagra and workingof full wave rectifier is as shown :  In this circuitthere are TWO diodes in ODER to rectifyboth the cycleof a . c `I//P`. For this centre taptransformer `(phi^("mer"))` is used . (i) Operationduringpositivehalfcycle : Duringpositive halfcycle `D_(1)`diode is forwardbiasd and hence, wil conduct `Id_(1)` flows through `R_(L)`. In this cyclediode. `D_(2)`is reverse baised and hence, does not conduct. Only `D_(1)`supplies the load current `I_(L) = Id_(I)` .  (ii) Operation during negative half-cycle : Inthe negavtive half cyclepolarity reverse and A becomes- veand `B^(-)` becomes - ve and `B^(-)`becomes`+ ve` hence, the diode `D_(1)`does notconduct but `D_(2)`conducts beingforwards biased currentin the ciruit is as shown DUE to `D_(2)` . 
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| 8. |
A nonconducting sphere with a cavity has volume charge density rho. O_1 and O_2 represent the two centres as shown. The electric fields inside the cavity is E_0. Now, an equal and opposite charge is given uniformly to the sphere on its outer surface. The magnitude of electric field inside the cavity becomes |
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Answer» zero |
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| 9. |
The final image in an astronomical telescope adjustment, a straingt black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnificationof thed telescope is |
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Answer» VIRTUAL and erect |
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| 10. |
Nuclear fission can be explained by |
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Answer» OPTICAL model of the nucleus |
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| 11. |
A plane mirror is placed with its plane at an angle of 30^@ with the y-axis. Plane of the mirror is perpendicular to the xy plane and the length of the mirror is 3 cm. An insect moves along X-axis starting from a distant point with a speed 2cm/s. Find the duration of the time for which the insect can see its own image in the mirror |
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Answer» |
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| 12. |
A 20kHz baseband signal is to be transmitted over a long distance. Comment on the size of the antenna required for this purpose. |
| Answer» SOLUTION :The SIZE of antenna should be comparable to wavelength of SIGNAL `(at LEAST lambda/4 )`. The WAVE length of 20kHz wave is 15km and antenna size should be at least 3.75km. | |
| 13. |
A variable frequency a.c. source is connected to a capacitor. Will the displacement current change if the frequency of the a.c. source is decreased ? |
| Answer» Solution :YES, the magnitude of displacement current changes [actually decreases] if the FREQUENCY of the a.c. SOURCE is DECREASED. | |
| 14. |
Suppose a 'n'-type wafer is created by dopin Si crystal having 5xx10^(28) atoms/m^(3) with 1 ppm concentration of As. On the surface 200 ppm Boron is added to create 'P' region in this wafer. Considering n_(i)=1.5xx10^(16)m^(-3)(i) Calculate the denisties of the charge carrier in the n & p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased. |
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Answer» Solution :Here 1 ppm = 1 part per million = 1 part in 10 lakh (i) For n-type semiconductor : (no. of Si atoms in 1 `m^(3)` volume) `to` (no. of As atoms in 1 `m^(3)` volume) `10^(6) to 1` `therefore 5xx10^(28) to `(?) `rArr` NUMBER density of PENTAVALENT As atoms (or) free electron number density). `n_(e )=(5xx10^(28))/(10^(6))=5xx10^(22)("free electron")/(m^(3))` Now `n_(e )n_(h)=n_(i)^(2)` `therefore n_(h)=(n_(i)^(2))/(n_(e ))` `=((1.5xx10^(16))^(2))/(5xx10^(22))` `therefore n_(h)=4.5xx10^(9)"hole"//m^(3)""...(1)` (ii) For p-type semiconductor: (no. of Si atoms in 1 `m^(3)` volume) `to` (no. of B atoms in 1 `m^(3)` volume) `10^(6) to 200` `therefore 5xx10^(28)to` (?) `rArr` Number density of trivalent B atoms (or hole number density). ` n_(h)=(200xx5xx10^(28))/(10^(6))` `therefore n_(h)=10^(25)("hole")/(m^(3))` Now `n_(e )n_(h)=n_(i)^(2)` `therefore n_(e )=(n_(i)^(2))/(n_(h))` `therefore n_(e )=((1.5xx10^(16))^(2))/(1xx10^(25))` `therefore n_(e )=2.25xx10^(7)("free electrons")/(m^(3))""...(2)` EQUATIONS (1) and (2) indicate that `n_(h) gt n_(e )` `rArr` In the reverse bias condition of given pn-junction, holes will contribute majority in producing reverse saturation current. |
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| 15. |
How much average power over complete cycle does an a.c. source supplied to a capacitor? |
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Answer» SOLUTION :AVERAGE POWER, `vec(P) = V_(rms) xx i_(rms) xx cos PHI` where `phi = 90^(@)` `:. vec(P) = V_(rms) xx i_(rms) xx cos 90^(@)` `= 0` |
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| 16. |
An object M is placed at a distance of 3 m from a mirror with its lower end at 2 m from ground as shown in Fig. There is a person at a distance of 4 m from object. Find minimum and maximum height of person to see the image of object |
Answer»  `(h_(min))/(10=2/3)` `:. H_(min)=(20)/(3)m` `(h_(max))/(10)=(4/3)` `:. h_(max)=(40/3 m)`. |
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| 17. |
An electron movingaroungthe nucleus withan angularmomentum l has a magnetic moment |
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Answer» `(E)/(m)L` |
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| 18. |
What is the Be Broglie wavelength of an electron with kinetic energy of 120 eV. |
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Answer» Solution :GIVEN, `KE = 120 eV, m = 9.1xx10^(-3)kg, e = 1.6xx10^(-19)C` `lambda = (12.27)/(sqrt(V)) Å = (12.27)/(sqrt(120)) Å = 0.112xx10^(-9) m THEREFORE lambda = 0.112 NM`. |
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| 19. |
What is the nature of magnetic field in a moving coil galvannometer ? |
| Answer» SOLUTION :BASED on Wheatstone.s BRIDGE | |
| 20. |
(a) Why are infra-red waves often called heat waves ? Explain. (b) What do you understand by the statement, ..Electromagnetic waves transport momentum.. ? |
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Answer» Solution :(a) Infrared waves are PRODUCED by hot bodies and are characterised by their heating property.Therefore, infrared waves are often CALLED heat waves. (b) Like all travelling waves, the electromagnetic waves also transport momentum. CONSIDER a plane perpendicular to the direction of propagation of an electromagnetic wave. If electric charges are there on this plane they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic waves. The charges THUS acquire energy and momentum from the waves and thus electromagnetic waves transport momentum. |
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| 21. |
Give an expression for the magnetic field on the axis of a circular current loop. What is the value at its centre? |
| Answer» Solution :On the AXIS `B= (mu_0Ia^2)/(2(a^2 +x^2)^(3//2))` and at the CENTRE, `B= (mu_0I)/(2a)` | |
| 22. |
Double-convex lenses are to be manufactured from a glass of refractive index 1.53, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ? |
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Answer» SOLUTION :`(1)/(F) = (N - 1) ((1)/(R_(1)) - (1)/(R_(2)))` f= 20 cm , n = 1.55, `R_(1) = R, R_(2) = -R"therefore (1)/(20)= (1.55 - 1) ((1)/(R) + (1)/(R)) = 0.55 XX (2)/(R)` `R = 0.55 xx 2 x 20 = 22 ` cm |
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| 23. |
The enegry is transferred form a source of constant voltage V to a consumer by means of long straight coaxial cable with negligible active resistance. The consumed current is I. Final the enegry flux across the cross-section of the cable. The conductive sheath is supposed to be thin. |
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Answer» Solution :The electric FIELD `(………rarr)` and the magnetic field `(H rarr)` are as shown. The electirc field by Gauss's therorem is LIKE `E_(r) = (A)/(r )` Intergrating `varphi = A` in `(r_(2))/(r )` so `A = (V)/(In(r_(2))/(r_(1)) (r_(2) gt r_(1))` Then `E = (V)/(rIn(r_(2))/(r_(1)))` Magnetic field is `H_(theta) = (I)/(2pir)` The Poynting vector `S` is along the `Z` axis and non zero between then `(r_(1) lt r ltr_(2))`. the total power FLUX is `= int_(r_(1))^(r_(2)) (IV)/(2pir^(2)In(r_(2))/(r_(1))).2PI r dr = Iv` 
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| 24. |
A sphere of brass is suspended to a vertical spring and oscillates with frequency f'. The ball is now immersed is non-viscous liquid whose density is (1)/(10) th of density of brass. If the sphere remains vibrating in the liquid the frequency f now will be : |
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Answer» `f' = f` When BALL is immersed in the liqiud then m and k remain the same so frequencyy also remains the same. Thus correct choice is (a). |
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| 25. |
The output of AND gate is 1, so …… |
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Answer» both input are zero. From TRUTH table of AND gate. |
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| 26. |
What is the de Broglie wavelength associated with an electron, accelerated through a potential differnece of 100 volts? |
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Answer» Solution :ACCELERATING POTENTIAL `V = 100 V`. The DE Broglie wavelength `lambda` is `lambda=h//p=(1.227)/(sqrt(V))=nm` `lambda=(1.227)/(sqrt(100))nm=0.123nm` The de Broglie wavelength associated with an electron in this case is of the order of X-ray WAVELENGTHS. |
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| 27. |
A circular loop of radius r is made of a wire of circular cross section of diameter ‘a’. When current I flows through the loop the magnetic flux linked with the loop due to self induced magnetic field is given by phi = mu_(0)r [ln((16r)/(a)) - (7)/(4)]l. The resistivity of the material of teh wire is rho and r gt gt a.Switch S is closed at time t = 0 so as to connect the loop to a cell of emf V. Find the current in the loop at time t. |
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Answer» Where `tau = (mu_(0)a^(2))/(8rho) [In ((16r)/(a)) - (7)/(4)]` |
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| 28. |
In intrinsic semiconductorthe drift velocity of holes and electron are respectiveltyv_(h) and v_(e ) then, …….. |
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Answer» `v_(E ) lt v_(h)` Hole is steadyin covalent BOND. Whereas electrons are in motion. |
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| 29. |
Heavy water is used in a nuclear reactor |
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Answer» to SLOW down the neutron |
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| 30. |
Electric field lines about a negative point charge are |
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Answer» CIRCULAR, ANTICLOCKWISE |
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| 31. |
Discuss the process of nuclear fusion and how energy is generated in stars. |
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Answer» Solution :Nuclear Fusion : When two or more light nuclei `(A lt 20)` combine to form a heavier nucleus,then it is called nuclear fusion. In the nuclear fusion, the mass of the resultant nucleus is less than the sum of the masses or original light nuclei..The mass difference appears as energy .The nuclear fusion neveroccurs at room temperatureunlike nuclear fission. It is because when two ligt nuclei come closer to combine, it is strongly repelled by the coulomb repulsive force. To overcome this repulsion, the two light nuclei must have enough kinetic energy to movecloser to each other such that the nuclear force becomes effective.This can be achieved if the temperature is very much greater than the vaue `10^(7)` K.When the surrounding temperature REACHES around `10^(7)`K, lighter nuclei start fusing to form heavier nuclei and this resulting reaction is called thermonuclear fusion reaction. Energy generation in stars: The natural place where nuclear fusion occurs in the core of the stars, since its temperature is of the order of `10^(7)K`. In fact , the energy generation in every star is only through thermonuclear fusion.Most of the stars including our Sun fuse hydrogen into helium and some stars even fuse helium into heavier ELEMENTS. The early stage of a stars is in the form of cloud and dust.Due to their own gravitational pull, thses clouds fall inward.As a result, its gravitational potential energy is converted to kinetic energy and finally into heat. When the temperature is high enough to initiate the thermonuclear fusion, they start to release enormous energy which tends to stabilize the starand prevents it from further collapse. The sun.s interior temperature is around `1.5 xx 10^(7)K` .The sun is converting `6 xx 10^(11)Kg` hydrogen into helium every second and it has enough hydrogen such that these fusion lasts for another 5 billion years. When the hydrogen is burnt out, the sun will enter INOT new phases called red giant where helium will fuse to BECOME carbon. During this stage, sun will expand greatly in sizeand all its planets will be engulfed in it. According to Hans Bethe, the sun is powered by proton- proton cycle of fusion reaction.This cycle consists of three steps and the first two steps are as follows: `""_(1)^(1)"H" + ""_(1)^(1)"H" to ""_(1)^(2)"H" + e^(+) + v` ....(1) ` ""_(1)^(1)"H" + ""_(1)^(2)"H" to ""_(2)^(3)"H" + gamma`....(2) A NUMBER of reactions are possible in the third step. But the dominant one is `""_(2)^(3)"H" + ""_(2)^(3)"H" to ""_(2)^(4)"H" + ""_(1)^(1)"H" + ""_(1)^(1)"H"` The overall energy production in the above reactions is about 27 MeV. The radiation energy we received from the sun is due to these fusion reactions. |
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| 32. |
Who looked out at the young trees? |
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Answer» The poet |
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| 33. |
In problem No. 133, the phase difference between two points separated by time interval 0.2098 is |
| Answer» Solution :`DELTA = (2pi X)/(lamda) = (2pi Delta t)/(T) = 2pi` | |
| 34. |
A charge is to be divided into two small parts. What should be the value of the charges on the parts so that the force between the parts will be maximum? |
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Answer» Solution :Let Q and (Q-q) be the charges on those bodies FORCE between the charges `F= 1/(4 pi in_0) ((Q - q)q)/(r^2)`, For F to be maximum `(dF)/(dq) = 0` `d/(dq)(Q - q)q = 0 implies (Q - q) q = 0 implies (Q- q) + (-q) = 0 implies Q - 2 q = 0, q = Q//2`. Thus the CHARGE should be divided equally on the objects. |
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| 35. |
A wire having Young's modulus 1.2 xx 10^11 N/m^2 is subjected to the stress of 2.4 xx 10^7 N /m^2 . If the length of the wire is 10m, the extension produced in it |
| Answer» Answer :A | |
| 36. |
Explain why sound produced by vibration of tuning fork becomes louder when its stem is placed in contact with the tube? |
| Answer» Solution : The entire air COLUMN with the tube vibrates with the FREQUENCY of the FORK. Now due to resonance it PRODUCES a louder sound. | |
| 37. |
A steel cylinder was cooled in liquid nitrogen (72 K) and fitted without play into a nickel-chrome steel shell at room temperature (20^@C). The internal radius of the shell is 25 mm and the external radius is 35 mm. Neglecting deformation of the cylinder find the stress in the shell and the nature of its deformation. |
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Answer» <P> `p = sigma_1 = E_1 epsilon= E_1 (Delta r_1)/(r) = E_1 alpha_1 Deltat` Here `E_1` is the Young modulus for steel, `alpha_1`is the THERMAL expansion coefficient for steel. Hence`E_1 alpha_1 Delta tralpha h = sigma_2 had, ` and the stress in the shell is `sigma = E_1 alpha_1 r Deltat//d` 
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| 38. |
At what angle does a diver see the setting sun ? |
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Answer» At about `41^(@)` to the horizon  `MU = (4)/(3) = (1)/(sinC)` `sin C = (3)/(4)` `THEREFORE C = 90^(@) - 41^(@) = 49^(@)` |
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| 39. |
The magnetic potential at a point on the line inclined at 30^@ with axis of a short magnet is 1.5xx 10^(-5) J/(Am). What will be the distance of point centre of the magnet if the magnetic moment is 1.732 Am^2. |
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Answer» 0.01 cm |
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| 40. |
Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abeda. The temperature of gas at c and d are 300 and 500 K. How much will be the heat absorbed by the gas during the process? |
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Answer» `400 R" In " 2` For process ` a to b, `(P - CONSTANT ) `W_(a to b) = P.dV = nRdT = - 400 R` For process `b to c `(T - constant ) `W_(b to c) = - 2R (300 ) ` In 2 For process` c to d` (T - constant) `W_(c to d) = + 400 R` For process `d to a ` (T - constant) `W_(d to a) = + 2R (500 )` In 2 Net work `(Delta W) = W_(a to b) + W_(b to c) + W_(c to d) + W_(d to a)` ` Delta W = 400 R " In " 2` `:. Delta Q = Delta U + Delta W` ` :. Delta Q = 400 R" In " 2` |
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| 41. |
Define the term 'wavefront of light'. A plane wave front AB propagating from denser medium (1) into a rarer medium (2) is incident on the surface P_(1)P_(2) separating the two media as shown in figure. Using Huygen's principle, draw the secondary wavelets and obtain the refracted wavefront in the diagram. |
Answer» Solution :Awavefront is defined as a surface of constant PHASE. Alternately, a wavefront is the CONTINUOUS locus of all those POINTS which are in same phase at any instant of time. SECONDARY WAVELETS and the refracted wavefront is shown in the following diagram.
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| 42. |
When the temperature of a copper coin is raised by 100^@C , its diameter increase by 0.20% To two significant figures, give the percent increase in ( a) the area of a face ,( b) the thickness, (c) the volume , and ( d) the mass of the coin. ( e) Calculate the coefficient of linear expansion of the coin. |
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Answer» |
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| 43. |
An n - type semiconductor has resistivity 0.1Omegam . The number of donor atoms which must be added to achieve this is ( mu=0.05 m^2V^(-1)s^(-1) ) : |
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Answer» `1.25xx10^(17)` |
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| 44. |
Length of a very long bar magnet is kept along the axis of a circular loop of radius r in such a manner that North Pole of magnet lies at the centre of circular loop. Magnitude of magnetic field due to magnet at any point on the circumference of loop is B. Current i flows through the loop. Magnetic force acting on circular loop is |
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Answer» `2pi i rB`along the plane of loop ` F= int idl B sin 90^@` ` rArr F = IB intdl` ` rArr F= iB int dl` ` rArr F = iB(2pi r)` `rArr F = 2pi r B , ` along the axis of loop . |
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| 45. |
Assertion :- When a conducting rod moves in uniform transversemagnetic field with uniform speed which is perpendicular to its length , then potential difference may developed across its ends . Reason :- In any conductor , free electrons and free positive ions are available . |
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Answer» If the ASSERTION & Reason are True& the Reason is a correct EXPLANATION of the Assertion . |
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| 46. |
Following can be used as a rectifier. |
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Answer» PURE SEMICONDUCTOR |
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| 47. |
Describe qualitatively the path of a charged particle moving in a uniform magnetic field with initial velocity a. Parallel to the field b. Perpendicular to the field and C. At an arbitrary angle with the field direction. |
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Answer» Solution :a. STRAIGHT line B. Circular PATH C. Helix |
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| 48. |
An alternating voltage V = 400 sin (500omegat)V is connected to a 0.2 kOmegaresistor, what will be the rms value of current ? |
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Answer» 14.14 A `V_m`=400 V `therefore V_"RMS"=V_m/sqrt2` `I_"rms"=V_"rms"/R=V_m /(sqrt2R)=400/(sqrt2xx200)` `=2/sqrt2=sqrt2`=1.414 A |
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| 49. |
A metal wire of circular cross-section has a resistance R. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes R_2 then R_2 : R_1 is |
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Answer» `1:1` here` RHO ` and Vare CONSTANT`thereforeR propl^2therefore(R_1)/( R_2) =(l_(1)^(2))/( l_(2)^(2))` given`l_2 = 2l_1`. Hence` (R_1)/( R_2)=(l_1^(2) )/( 4l_(1)^(2)) = 1/4implies R_2: R_1= 4 :1` |
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