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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 rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity 'v' as shown in the figure . Explain the which loop do you expect the induced emf to the constant during the passage out of the field region . The magnetic field is normal to the loops . |
Answer» Solution :The INDUCED EMF is expected to be constant only in the CASE of the RECTANGULAR loop . In the case of circular loop , the rate of change of area of the loop during its passage out of ONE field region is not constant . Hence induced emf will vary accordingly. 
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| 2. |
Maxwell's equations describe the fundamental laws of : |
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Answer» ELECTRICITY only |
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| 3. |
A diode can be properly doped at the time of its manufacture, so that it have a shape break down voltage . The above diode is called |
| Answer» Answer :A | |
| 4. |
Find the effective resistance between A and B. |
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Answer» `2 OMEGA` The charge x passing between A and D will entirely PASS to DA2. So, we can detach the circuit at POINT D and shown in figure.   .
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| 5. |
Wheatstone bridge is most sensitive when the arms ratio is |
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Answer» EQUAL to ONE |
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| 6. |
The resistance R_(o) and R_(t) of a metallic wire at temperature 0^(@)C and t^(@)C are related as (alpha is the temperatureco-efficient of resistance). |
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Answer» `R_(t)=R_(o)(1+alpha t)` |
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| 7. |
Charge on alpha particle is: |
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Answer» `4.8 XX 10(-19)C` |
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| 8. |
In the circuit shown in the figure The steady state currentsi_(1)and i_(2) in the coils after the switch S is closed are |
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Answer» `i_(1)=(EL_(2))/(R(L_(1)+L_(2))` |
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| 9. |
Photoelectric emission from a given surface of metal can take place when the value of a 'physical quantity' is less than the energy of incident photon. The physical quantity is |
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Answer» THRESHOLD frequency. |
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| 10. |
(a) What is the magnitude of the orbital angular momentum in a state with l=3? (b) What is the magnitude of its largest projection on an imposed z axis? |
| Answer» SOLUTION :(a) `3.65xx10^(-34)J.s`, (B) `3.16xx10^(-34)J.s` | |
| 12. |
The r.m.s. velocity of gas at 127^@C is 200m/sec, its r.m.s. velocity at 527^@C will be nearly |
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Answer» 280 m/s |
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| 13. |
The anlge of dip at a place where horizontal and cvertical componentts are equalis |
| Answer» Answer :B | |
| 14. |
We have a fixed conductor having a cavity inside. There is a point charge q_(1) kept inside the cavity charge appearing at the inner surface of the cavity is q_(2). Charge appearing at outer surface of conductor is q_(3) and there lies another point charge q_(4) out side the conductor. Whole situation is shown in figure. q_(4). is held stationary by applying an external force vecF_(0) on it. Three points A, B and C are shown. Answer the following two questions. Q. If keeping the nature and position of q_(1) fixed, magnitude of q_(1) is changed, then choose the INCORRECT options(s) |
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Answer» External force REQUIRED to KEEP `q_(4)` stationary remains unchanged |
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| 15. |
We have a fixed conductor having a cavity inside. There is a point charge q_(1) kept inside the cavity charge appearing at the inner surface of the cavity is q_(2). Charge appearing at outer surface of conductor is q_(3) and there lies another point charge q_(4) out side the conductor. Whole situation is shown in figure. q_(4). is held stationary by applying an external force vecF_(0) on it. Three points A, B and C are shown. Answer the following two questions. Q. If keeping the magnitude and nature of q_(1) fixed position of q_(1) is changed inside the cavity. Then choose the correct option(s) |
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Answer» EXTERNAL force required to keep `q_(4)` STATIONARY REMAINS UNCHANGED |
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| 17. |
In the circuit, the current is to be measured. What is the value of the current if the ammeter shown : (b) is a galvanometer described in (i) but converted to an ammeter by a shunt resistance r_s = 0.02 Omega |
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Answer» Solution :(b) RESISTANCE of the galvanometer as AMMETER is `(R_(0) r_s)/(R_(G) r_s) = (60Omega XX 0.02 OMEGA)/((60+0.02))= 0.02 Omega` Total resistance `R = 0.02Omega + 3OMEGA = 3.02Omega` Hence, `I = (3)/(302) = 0.99 A` |
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| 18. |
Describe how a semiconductor diode isused as a half wave rectifier. |
Answer» SOLUTION : (i) A half wave rectifier can be constructed with a single diode. The ac INPUT signal is fed to the primary coil of a transformer. The output signal is taken across the load resistance `R_(L)`. (ii) During POSITIVE half cycle, the diode is FORMED biased and current flows through the diode. (iii) During negative half cycle, diode biasedand no current flows through the load resistance. (IV)This means currentflows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positivehalf cycles. (v) Rectifier efficiency is definedas the ratio of output do power to the input ac power. `eta = ( P_(dc))/(P_(ac)) = (0.406R_(L))/(r_(f)+R_(L))` where `r_(f)=` Forward distance of a diode , `R_(L) = ` Load resistance The maximum efficiency of half wave rectifier is `40.6 %` |
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| 19. |
Nucleus ._(3)A^(7) has binding energy per nucleon of 10 MeV. It absorbs a proton and its mass increases by (90)/(100) times the mass of proton. Find the new binding energy of the nucleus so formed. [Take energy equivalent of proton = 930 MeV] |
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Answer» `=70+(1)/(100)(930)=79.MeV` |
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| 20. |
Light with an energy flux of 18 W//cm^(2) falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm^(2), the total momentum delivered (for complete absorption) during 30 minutes is: (a) 36xx10^(-5) kg m/s (b) 36xx10^(-4) kg m/s (c) 108xx10^(4) kg m/s (d) 1.08xx10^(7) kg m/s |
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Answer» <P>`36xx10^(-5)` kg m/s `therefore E=IAt` `=(20W)/(cm^(2))xx 30 cm^(2)xx30xx60 sec` `=108xx10^(4)` J Final momentum `P_(f)=(U)/(C )=(108xx10^(+4))/(3xx10^(8))` `therefore P_(f)=36xx10^(-4)NS` INITIAL momentum `P_(i)=0` `therefore` Momentum imparted to wall, `Delta P =P_(f)-P_(i)` `=36xx10^(-4)-0` `=36xx10^(-4)` kg m/s. |
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| 21. |
A parallel plate capacitor (condenser) has a certain capacitance (capacity). When 2/3 rd of the distance between the plates is filled with a dielectric, the (capacity) capacitance is found to be 2.25 times the initial capacitance. The dielectric constant of the dielectric |
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Answer» 1 |
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| 22. |
Sea water at frequency v=4xx10^8Hz has permittivity epsilon~~80epsilon_0. Permeability mu~~mu_0 and resistivity rho=0.25 Omega -m. Imaging a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t)=V_0 sin (2pivt). What fraction of the conduction current density is the displacement current density? |
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Answer» Solution :Let a be the distance between the plates. The applied voltageis given by `V(t)=V_0sin 2pivt`. The applied electric field, `E=(V(t))/d=(V_0)/d sin 2pivt` The conduction CURRENT DENSITY, `J_C=E/rho =(V_0)/(rhod) sin 2pivt=J_(0C) sin 2piv t` where `(V_0)/(rhod)=J_(0C)`=MAX. conduction current density. Displacement current density is given by `J_D=(I_D)/A=(in_0in_r d(phi_E)//dt)/A=(in_0 in_rd(EA))/(Adt)=in_0in_r (dE)/(dt) =in_0in_rd/(dt)((V_0)/d sin 2pivt)` `=in_0in_r (V_0)/d 2piv cos 2pivt =J_(0D) cos 2pivt` Where `in_0in_r (V_0)/dxx2piv=J_(0D)`=max. displacement current desity `:. (J_(0D))/(J_(0C))=(in_0 in_r V_0 2piv//d)/(V_0//(rhod))=2piv in_0in_r rho=4pi in_0 v in_r rho//2=1/(9xx10^9)xx(4xx10^8)xx80xx0.25//2=4/9` |
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| 23. |
In photoelectric effect the maximum kinetic energy of electrons emitted from the metal surface depends upon |
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Answer» INTENSITY of INCIDENT radiation |
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| 24. |
Which theory of light predicts the velocity of light in denser medium to be more than in a rarer medium? |
| Answer» SOLUTION :Newton.s CORPUSCULAR THEORY of LIGHT or corpuscular theory of light. | |
| 25. |
(A) : Long distance radio broadcasts use shortwave bands. (R) : Ionosphere reflects waves in these bands. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 26. |
A light rays is incident upon a prism minimum deviation position and suffers a deviation of 34^@. If the shaded half of the prism is knocked off, the ray will |
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Answer» SUFFER a deviation of `34^@` |
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| 27. |
The magnetic flux phi through a coil is varyingw.r.t. time 't' according to the relation phi=5t^(2)+4t+3 weber. Calculate the induced e.m.f. and current in the coil at t=3 sec. The resistance of the coil is 2Omega. |
| Answer» SOLUTION :`34 V, 17 A` | |
| 28. |
The interference patterns is obtained with two coherent light sources of intensity ratio n. I the interference pattern, the ratio (I_("max")-I_("min"))(I_("max")+I_("min")) will be...... |
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Answer» `(SQRT(n))/((n+1)^(2))` `:. (I_(1))/(I_(2))=n` `(A_(1)^(2))/(A_(2)^(2))=n "" [ :. I PROP A^(2)]` `:.(A_(1))/(A_(2))= sqrt(n)` Taking COMPONENDO and dividendo `(A_(1)+A_(2))/(A_(1)-A_(2))=(sqrt(n)+1)/(sqrt(n)-1)` `:. ((A_(1)+A_(2))^(2))/((A_(1)-A_(2))^(2))=(n+2sqrt(n)+1)/(n-2sqrt(n)+1)` `:. (I_("max"))/(I_("min"))=(n+2sqrt(n)+1)/(n-2sqrt(n)+1)` Taking componendo and dividendo again `(I_("max")-I_("min"))/(I_("max")-I_("min))=(n+2sqrt(n)+1-n+2sqrt(n)+1)/(n+2sqrt(n)+1+n-2sqrt(n)+1)` `=(4sqrt(n))/(2(n+1))=(2sqrt(n))/(n+1)` |
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| 29. |
On a particular day, the maximum frequency reflected from the ionosphere is 10 MHz. On another day, it was found to increase to 11MHz. Calculate the ratio of the maximum electron densities of the ionosphere on the two days. Point out a plausible explanation for this. |
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Answer» 1.21 `N_("max")=(v_(c)^(2))/(81)=v_(c)^(2)rArr((N_(2))_("max"))/((N_(1))_("max"))=((11)/(10))^(2)=(121)/(100)=1.21` |
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| 30. |
Find the capacitane value of C if the equivalent capacity between A and B in the given figure is 1muF. |
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Answer» |
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| 31. |
भारत का स्थलीय सीमा रेखा की लंबाई कितना किलोमीटर है? |
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Answer» 6,100 KM. |
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| 32. |
A particle of mass m, oscillates with SHM between points x_(1) and x_(2), the equilibrium position being O. Its P.E. is plotted. It will be as given below in the graph. |
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Answer»
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| 33. |
Each atom in the periodic table has a unique set of spectral lines. Which one of the following statements is the best explanation for this observation? |
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Answer» |
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| 34. |
When notes on Photodiode |
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Answer» Solution :(i) A p-n JUNCTION diode which converts an optical signal into ELECTRIC current is known photodiode. Thus, the operation of photodiode is exactly opposite to that of an LED. (ii) The direction of arrows indicates that the light is incident on the photo diode. Construction: (i) Construction : (ii) The device consists of a p-n junction semiconductor made of photosensitive material kept safely inside a plastic case in Figure. (ii) It has a small transparent window that allows light to be incident on the p-n junction. (iii) Photodiodes can generate current when the p-n junction is exposed to light and hence are CALLED as light SENSORS.  Working: (i) When a photon of sufficient energy (hv) strikes the depletion region of the diode some ofthe valence band electronics are.elevated into conduction band , in turn holes are developed in the valence band. (ii) This CREATES electron -hole pairs. (iii) The amount of electron -hole pairs generated depends on the intensity of light incident on the p-n junction. (iv) These electrons and holes are swept across the p-n junction by the electric field created by reverse voltage before recombination takes place. (v) Thus, holes move towards the n-side and electrons toward the p-side. (vi) When the external circuit is made, the electrons flow through the external circuit and constitute the photocurrent. (vii)When the external circuit is made, the electrons flow through the external circuit and constitute the photocurrent. (viii) When the incident light is zero, there exists a reverse current which is negligible. (ix) This reverse current in the absence of any incident light is called dark current and is due to the thermallygenerated minority carriers. |
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| 35. |
Two rings X and Y are placed in such a way that their axes are along the X and the Y axes respectively and their centres are at the origin. Both the rings X and Y have the same radii of 3.14 cm. If the current through X and Y rings are 0.6 A and 0.8 A respectively, find the value of the resultant magnetic field at the origin. (mu_(0)=4pixx10^(-7)SI) |
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Answer» Solution :1. Magnetic field PRODUCED in the X-ring due to CURRENT `I_(1)=0.6A` is, `B_(1)=(mu_(0)I_(1))/(2r)""...(1)` 2. Magnetic field produced in the Y-ring due to current `I_(2)=0.8A` is `B_(2)=(mu_(0)I_(2))/(2r)""...(2)` 3. Resultant magnetic field produced NEAR origin is, `vecB=vecB_(1)+vecB_(2)` `thereforeB=sqrt(B_(1)^(2)+B_(2)^(2))` From equation (1) and (2), `thereforeB=sqrt(((mu_(0)I_(1))/(2r))^(2)+((mu_(0)I_(2))/(2r))^(2))` = `mu_(0)/(2r)sqrt(I_(1)^(2)+I_(2)^(2))` = `(4pixx10^(-7))/(2xx3.14xx10^(-2))sqrt((0.6)^(2)+(0.8)^(2))` = `2xx10^(-5)sqrt(0.36+0.64)` = `2xx10^(-5)T` `thereforeB=2xx10^(-5)T` |
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| 36. |
The angle between two vectors vecA=vecB is theta. The resultant of vecA and vecB making an angle (theta)/(2) with vecA. Then: |
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Answer» `A=B` |
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| 37. |
A series LCR circuit with R = 20 Omega, L = 1.5 H and C = 35 muF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? |
| Answer» SOLUTION :2,000 W | |
| 38. |
The total energy of a particle, executing S.H.M. is : |
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Answer» `prop X` So the CORRECT choiceis ( c ). |
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| 39. |
The height vertically above the earth's surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the earth) |
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Answer» 8R |
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| 40. |
Excess pressure inside a drop of water of radius 2mm is 70 N//m^2. The same in a drop of radius 4mm will be (T_(water) = 70 xx 10^-3 N//m) |
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Answer» `55 N//m^2` |
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| 41. |
A wave disturbance in a medium is desiribed byY = 0.02 cos (50pit + pi/2). Where x and y are in m and t in sec. The values in Column-II are in SI units. |
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Answer» |
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| 42. |
Answer the following questions regarding earth's magnetism The angle of dip at alocation in southern India is about18^(@). Would you expect a greater or smaller dip angle in Britain ? |
| Answer» SOLUTION :Britain is situated for NORTH compared to India . So the angle of dip will be GREATER than`18^(@)`. | |
| 43. |
A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance |
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Answer» INCREASES as R increases for `r ltR` |
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| 44. |
The ratio of force between two small charged sphere(i)in air(ii)in a medium ofdielectric constant'K' is |
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Answer» `K^2:1` |
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| 45. |
The angle between velocity and acceleration of a particle describing uniform circular motion is |
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Answer» `45^(@)` Althought the speed does not VARY the particle is accelerating because the velocity its direction at every point on circular track.  The acceleration is centripetal, which is perpendicular to motion at every point and acts along THERADIUS and directed towards the centre of the curved circular path. |
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| 46. |
A solenoid of 2.5 m length and 4.0 cm diameter possesses 100 turns per cm .A current of 5 A is flowing through it . The magnetic induction at axis inside the solenoid is : |
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Answer» `(2pixx10^(-2)) (WB)//m^2` |
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| 47. |
For a sphericla shell |
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Answer» If potential inside itis zero then it necessarily electrically neutral |
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| 48. |
Choose the correct ray diagram of a thin equi- convex lens which is cut from upper half as shown in the figure. |
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Answer» |
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| 49. |
An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency ? |
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Answer» 0.2 `therefore ` % INCREASE = `(v. -v)/(v )xx 100 = ((6)/(5) - 1) xx 100 = 20%` correct choice is (a) . |
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| 50. |
An oil drop with 10 excess electrons is held stationary under a constant electric field of 3 xx 10^(4) NC^(-1) in Millikan oil drop experiment. The density of the oil is 1.26 xx 10^(3) kg m^(-3). Calculate the radius of the drop. |
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Answer» Solution :Data SUPPLIED, r=radius of the droplet, d=density of oil drop `=1.26 xx 10^(3) kg m^(-3)` `g=9.8 m//s^(2), ` n=no. of excess electrons =10 `e=1.6 xx 10^(-19)C""E=3 xx 10^(4) NC^(-1)` As the oil drop is STATIONARY, weight of the droplet=force due to ELECTRIC field `i.e 4/3 pir^(3) dg=ne E, r^(3) =(3ne E)/(4PI dg) =(3 xx 10 xx 1.6 xx 10^(-19) xx 3 xx 10^(4))/(4 xx 3.14 xx 1.26 xx 10^(3) xx 9.8)= 0.9285 xx 10^(-18)` `r=(0.9285 xx 10^(-18))^(1/3)=0.9756 xx 10^(-6)=9.76 xx 10^(-7) m` |
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