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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 L-R combination is connected to a battery of emf 4 volt. If L = 0.1 H and R = 4.0 `Omega`, then the time taken to reach a current of 0.631 ampere is:-A. (1/40) secB. 0.4 secC. 1.6 secD. 0.63 sec |
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Answer» Correct Answer - A Final maximum current `=E/R=4/4=1` amp Instantaneous current , `I=E/R (1-e^(-Rt//L))` Here , I=0.6321 ampere 0.6321 = 1(1–e–Rt/L) 0.6321 = 1–e–Rt/L e–Rt/L = 1–0.6321 = 0.3679 = `1/e` or `"Rt"/L=1` or `t=L/R = 0.1/4=1/40` sec. |
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| 2. |
For oscillations to start in a circuit, the loop gain must be greater than 1 when the phase shift around the loop is a. 90 degrees b. 180 degrees c. 270 degrees d. 360 degrees |
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Answer» (d) 360 degrees |
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| 3. |
Two wires `A` and `B` of same material and same mass have radius `2r` and `r`. If resistance of wire `A` is `34 Omega`, then resistance of `B` will beA. 544 `Omega`B. 372 `Omega`C. 68 `Omega`D. 17 `Omega` |
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Answer» Correct Answer - A `R=rho l/A` and mass m = volume (V) x density (d) = (Al)d Since wires have same material so `rho` and d is same for both. Also they have same mass `rArr` Al = constant `rArr l prop 1/A` `rArr R_1/R_2 = l_1/l_2 xxA_2/A_1 =(A_2/A_1)^2=(r_2/r_1)^4` `rArr 34/R^2 = (r/"2r")^4 R_2= 544 Omega` |
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| 4. |
To vary the frequency of a Wien bridge, you can vary a. One resistor b. Two resistors c. Three resistors d. One capacitor |
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Answer» (b) Two resistors |
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| 5. |
The phase-shift oscillator usually has a. Two lead or lag circuits b. Three lead or fag circuits c. A lead-lag circuit d. A twin-T filter |
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Answer» (b) Three lead or fag circuits |
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| 6. |
A Wien bridge is sometimes called a a. Notch filter b. Twin-T oscillator c. Phase shifter d. Wheatstone bridge |
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Answer» (a) Notch filter |
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| 7. |
The typical input stage of an op amp has a a. Single-ended input and single-ended output b. Single-ended input and differential output c. Differential input and single-ended output d. Differential input and differential output |
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Answer» (c) Differential input and single-ended output |
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| 8. |
An oscillator always needs an amplifier with a. Positive feedback b. Negative feedback c. Both types of feedback d. An LC tank circuit |
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Answer» (a) Positive feedback |
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| 9. |
In a nonlinear op-amp circuit, thea. Op amp never saturatesb. Feedback loop is never openedc. Output shape is the same as the input shaped. Op amp may saturate |
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Answer» (d) Op amp may saturate |
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| 10. |
How many types of negative feedback are there? a. One b. Two c. Three d. Four |
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Answer» The correct answer is: (d) Four |
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| 11. |
The closed-loop input impedance in a VCVS amplifier is a. Usually larger than the open-loop input impedance b. Equal to the open-loop input impedance c. Sometimes less than the open-loop input impedance d. Ideally zero |
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Answer» (a) Usually larger than the open-loop input impedance |
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| 12. |
The loop gain AOLB a. Is usually much smaller than 1 b. Is usually much greater than 1 c. May not equal 1 d. Is between 0 and 1 |
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Answer» (b) Is usually much greater than 1 |
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| 13. |
With an ICVS amplifier, the circuit approximates an ideal a. Voltage amplifier b. Current-to-voltage converter c. Voltage-to-current converter d. Current amplifier |
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Answer» (b) Current-to-voltage converter |
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| 14. |
The open-loop voltage gain equals the a. Gain with negative feedback b. Differential voltage gain of the op amp c. Gain when B is 1 d. Gain at unity |
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Answer» (b) Differential voltage gain of the op amp |
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| 15. |
A VCVS amplifier approximates an ideal a. Voltage amplifier b. Current-to-voltage converter c. Voltage-to-current converter d. Current amplifier |
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Answer» (a) Voltage amplifier |
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| 16. |
In a VCVS amplifier, any decrease in open-loop voltage gain produces an increase in a. Output voltage b. Error voltage c. Feedback voltage d. Input voltage |
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Answer» (b) Error voltage |
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| 17. |
The closed-loop input impedance with an ICVS amplifier is a. Usually larger than the open-loop input impedance b. Equal to the open-loop input impedance c. Sometimes less than the open-loop impedance d. Ideally zero |
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Answer» (d) Ideally zero |
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| 18. |
An ICVS amplifier has no output voltage. A possible trouble is a. No positive supply voltage b. Open feedback resistor c. No feedback voltage d. Shorted load resistor |
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Answer» (d) Shorted load resistor |
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| 19. |
An ICVS amplifier is saturated. A possible trouble is a. No supply voltages b. Open feedback resistor c. No input voltage d. Open load resistor |
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Answer» (b) Open feedback resistor |
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| 20. |
An ICVS amplifier has no output voltage. A possible trouble is a. No negative supply voltage b. Shorted feedback resistor c. No feedback voltage d. Open load resistor |
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Answer» (b) Shorted feedback resistor |
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| 21. |
An ICIS amplifier is saturated. A possible trouble is a. Shorted load resistor b. R2 is open c. No input voltage d. Open load resistor |
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Answer» (b) R2 is open |
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| 22. |
If the collector resistor is shorted, the ac output voltage will a. Decrease b. Increasec. Remain the same d. Equal zero |
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Answer» (d) Equal zero |
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| 23. |
As the temperature increases, the current gain a. Decreases b. Remains the same c. Increases d. Can be any of the above |
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Answer» (d) Can be any of the above |
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| 24. |
If the base resistor is shorted, the transistor will probably be a. Saturated b. In cutoff c. Destroved d. None of the above |
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Answer» (c) Destroved |
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| 25. |
When the base resistor decreases, the collector voltage will probably a. Decrease b. Stay the same c. Increase d. Do all of the above |
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Answer» (a) Decrease |
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| 26. |
Ignoring the bulk resistance of the collector diode, the collector-emitter saturation voltage is a. 0 b. A few tenths of a volt c. 1 V d. Supply voltage |
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Answer» The correct answer is: a. 0 |
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| 27. |
If the base resistor is very small, the transistor will operate in the a. Cutoff region b. Active region c. Saturation region d. Breakdown region |
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Answer» (c) Saturation region |
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| 28. |
Explain why it is not safe to go out with an umbrella when there is heavy raine lightning or thunder. |
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Answer» Large amount of charges get built up in the clouds in rainy and stormy weather. 1. Sometimes these charges are transferred to the earth in the form of a bolt of lightning. 2. Wire of umbrella is made up of metal and cloth is made up of synthetic material. 3. If lightning strikes, then umbrella catches it and there are chances of fire and it would be dangerous. 4. So it is not safe to go out with an umbrella when there is heavy rain. |
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| 29. |
Which of the statements given in Exercise 14.1 is true for p-type semiconductors. |
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Answer» The correct statement is (d). |
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| 30. |
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? |
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Answer» The correct statement is (c). Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge |
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| 31. |
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. |
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Answer» Input frequency = 50 Hz Output frequency = 50 Hz |
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| 32. |
For transistor action, which of the following statements are correct: (a) Base, emitter and collector regions should have similar size and doping concentrations.(b) The base region must be very thin and lightly doped.(c) The emitter junction is forward biased and collector junction is reverse biased.(d) Both the emitter junction as well as the collector junction are forward biased. |
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Answer» The correct statement is (b),(c). For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reverse-biased. |
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| 33. |
In an unbiased p-n junction, holes diffuse from the p-region to n-region because(a) free electrons in the n-region attract them.(b) they move across the junction by the potential difference.(c) hole concentration in p-region is more as compared to n-region.(d) All the above. |
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Answer» The correct statement is (c). |
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| 34. |
When a forward bias is applied to a p-n junction, it (a) raises the potential barrier.(b) reduces the majority carrier current to zero.(c) lowers the potential barrier.(d) None of the above. |
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Answer» The correct statement is (c). |
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| 35. |
If `x=a(theta+sintheta)` and `y=a(1-costheta)`, find `dy//dx`.A. `(sintheta)/(1+costheta)`B. `(costheta)/(1+sintheta)`C. `(1+costheta)/(sintheta)`D. `(sintheta)/(1-costheta)` |
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Answer» Correct Answer - 1 `(dx)/(d theta)=a(1+costheta),(dy)/(d theta)=a(0+sintheta)` `(dy)/(dx)=(dy)/(d theta)//(d theta)/(dx)=(asintheta)/(a(1+costheta))` |
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| 36. |
A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: A. a positive charge B. a negative charge C. no appreciable charge D. a charge whose sign depends on what part of the inner surface it touched E. a charge whose sign depends on where the small hole is located in the conductor |
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Answer» C. no appreciable charge |
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| 37. |
Charge on an originally uncharged conductor is separated by holding a positively charged rod very closely nearby, as shown in figure. Assume that the induced negative charge on the conductor is equal to the positive charge q on the rod. Then the flux through surface `S_1` is A. zeroB. `q//epsilon_(0)`C. `-e//epsilon_(0)`D. none of these |
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Answer» Correct Answer - B `phi = (q_("in"))/(epsilon_(0))` |
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| 38. |
The position of a particle moving along x-axis varies eith time t as `x=4t-t^(2)+1`. Find the time interval(s) during which the particle is moving along positive x-direction. |
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Answer» If the particle moves along positive x-direction, its x-coordinate must increase with time t. x-coordinate will increase with time t if `(dx)/(dt) gt 0`. `(dx)/(dt)=4 -2t` `(dx)/(dt) gt 0rArr 4-2t gt 0rArr t lt 2` Hence, the particle moves in positive x-direction during time-interval `0 lt t lt 2`. |
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| 39. |
If in a certain code language : (1) 'Tom Kun Sud' means 'Dogs are barking' (2) 'Kun Jo Mop' means 'Dogs and horses' (3) 'Mut Tom Ko' means 'Donkeys are mad'. Which word in that language means 'barking' ? (A) Ko (B) Kun (C) Jo (D) Sud |
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Answer» Correct option (D) Sud Explanation: As Dog = Kun, are = Tom, So, Sud = barking. |
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| 40. |
In a certain code language, '134' means „good and tasty‟; '478' means „see good pictures‟ and '729' means 'pictures are faint'. Which of the following digits sands for 'see' ? (A) 9 (B) 2 (C) 1 (D) 8 |
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Answer» Correct option (D) 8 Explanation: By summing up the given data we can conclude that good is coded as 4 and picture is coded as 7. So coding of see is option D. |
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| 41. |
Fill up the blanks using suitable prepositions on your own.1. The soldier climbed ______ a horse and rode away. 2. They have been here ______ a long time. 3. Kumaravel has lived in this city _____ 2012. 4. The paper was published _____ an International journal. 5. When will you return _____ home? 6. One ______ the four students wrote the answers correctly. 7. This fruit is _____ the Mexican capital. 8. The head offce is ____ Nungambakkam. It is College Road. As you go the station, the office is the right side. 9. The sailors were taken _____ the forest and made to walk 10 miles. 10. The girl standing _____ me was sneezing. |
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Answer» 1. on 2. for 3. since 4. in 5. back 6. of 7. from 8. at; on; through; at 9. through for 10. before |
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| 42. |
` H^(+), He^(+) and O^(++)` all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity . The masses of ` H^(+), He^(+) and O^(2+)` are `1 amu, 4 amu and 16 amu` respectively . ThenA. `H^(+)` will be deflected the mostB. `O^(2+)` will be deflected the mostC. `He^(+_` and O^(2+)` will be deflected equally.D. All will be deflected equally |
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Answer» Correct Answer - A::C `K=1/2 mv^(2) implies v=sqrt(2K//m), r=(mv)/(qB)=(sqrt(2mK))/(qB)` `r_(H^+)=sqrt(2mK((eB), r_(He^+))=(sqrt(2(4m)K))/(eB))=2r_(H^+)` `r_(O^(2+)) = (sqrt(2(16m)K))/(2eB)=2r_(H^+)` Lasser the radius, more will be the deflection. |
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| 43. |
A plane rectangular loop is placed in a magnetic field. The emf induced in the loop due to this field is `epsilon_(1)` whose maximum value is `epsilon_(im)`. The loop was pulled out of the magnetic field at a variable velocity. Assume that `vec(B)` is uniform and constant `epsilon_(1)` is plotted against t as shown in the graph. Which of the following are/is correct statement(s): A. `epsilon_(im)` is independent of rate of removal of coil from the field.B. The total chage that passes through any point of the loop in the process of complete removal of the loop does not depend on velocity of removal.C. The total are under the curve `(epsilon_(i) vs t)` is independent of rate of removal of coilfrom the field.D. The area under the curve is dependent on the rate of remval of the coil. |
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Answer» Correct Answer - B::C `|epsilon_(i)|=|-(d phi)/(dt)|=|epsilon_(i)|alpha |-(d phi)/(dt)|` and also area under the curve gives change in flux `=phi_(f)-pho_(i)=int d phi= int epsilon_(i)dt` Total charge that will flow=`Delta phi//R`. |
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| 44. |
A solenoid of resistance R and inductance L has a piece of soft iron inside it. A battery of emf E and of negligible internal resistance is connected across the solenoid as shown in Fig. At any instant, the piece of soft iron is pulled out suddenly so that inductance of the solenoid decrease to `etaL(eta lt 1)` with battery remaining connected. The work done to pull out the soft iron piece isA. `(eta LE^(2))/(2R^(2))`B. `((1-eta)LE^(2))/(2R^(2))`C. `((1-eta)LE^(2))/(etaR^(2))`D. `((1-eta)LE^(2))/(2etaR^(2))` |
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Answer» Correct Answer - D `L_(1)=L, L_(2)=etaL, phi=L_(1)I_(1)=L_(2)I_(2)I_(1)=E//R` `implies I_(2)=(L_1)/(L_2)I_(1)=I_(1)//eta` `W=U_(f)-U_(i)=1/2 L_(2)I_(2)^(2)-1/2L_(1)I_(1)^(2)` `1/2etaL((I_1)/(eta))^(2)-1/2 LI_(1)^(2)=(LE^2)/(2etaR^(2))[1-eta]`. |
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| 45. |
In the given arrangement, the space between a pair of co-axial cylindrical conductors is evacuated. The outer cylinder, called anode, may be given a positive potential V relative to the inner cylinder. A static homogeneous magnetic field `vec(B)` parallel to the cylinder axis, directed out of plane of figure is aslo present. induced charges in the conductors are neglected. We study the dynamics of electrons with rest mass m and charge e. The electrons are released at the surface of inner cylinder. Consider the following two cases Case 1: Firstly the potential V is turned on, but `vec(B) = 0`. An electron with negligible velocity is ejected at the surface of inner cylinder. It is found to hit the anode. Case 2: Now V = 0 but `vec(B)` is present. An electron starts out with an initial velocity `vec(v_(0))` in radial direction. For magnetic field larger than critical value `B_(c)` the electron will not reach the anode. Considering case 2, the trajectory of electron will beA. straight lineB. CircularC. ParabolicD. Helical |
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Answer» Correct Answer - B When `V=0`, the electron moves in homogenous static field B. The magnetic field acts orthogonal to the velocity and hence electron moves in a circle. Hence choice (b) is correct. |
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| 46. |
In the given arrangement, the space between a pair of co-axial cylindrical conductors is evacuated. The outer cylinder, called anode, may be given a positive potential V relative to the inner cylinder. A static homogeneous magnetic field `vec(B)` parallel to the cylinder axis, directed out of plane of figure is aslo present. induced charges in the conductors are neglected. We study the dynamics of electrons with rest mass m and charge e. The electrons are released at the surface of inner cylinder. Consider the following two cases Case 1: Firstly the potential V is turned on, but `vec(B) = 0`. An electron with negligible velocity is ejected at the surface of inner cylinder. It is found to hit the anode. Case 2: Now V = 0 but `vec(B)` is present. An electron starts out with an initial velocity `vec(v_(0))` in radial direction. For magnetic field larger than critical value `B_(c)` the electron will not reach the anode. Considering the case 1, the trajectory of electron will beA. straight lineB. CircularC. ParabolicD. Helical |
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Answer» Correct Answer - A In the absence of magnetic field, the electron moves in a straight hit to ht the anode due to the presence of electrostatic force. Hence choice (a) is correct. |
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| 47. |
The path of a charged particle in a uniform magnetic field depends on the angle `theta` between velocity vector and magnetic field, When `theta is 0^(@) or 180^(@), F_(m) = 0` hence path of a charged particle will be linear. When `theta = 90^(@)`, the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius `r = (mv)/(qB)`. The time period for circular path will be `T = (2pim)/(qB)` When `theta` is other than `0^(@), 180^(@) and 90^(@)`, velocity can be resolved into two components, one along `vec(B)` and perpendicular to B. `v_(|/|)=cos theta` `v_(^)= v sin theta` The `v_(_|_)` component gives circular path and `v_(|/|)` givestraingt line path. The resultant path is a helical path. The radius of helical path `r=(mv sin theta)/(qB)` ich of helix is defined as `P=v_(|/|)T` `P=(2 i mv cos theta)` `p=(2 pi mv cos theta)/(qB)` A charged particle moves in a uniform magnetic field. The velocity of particle at some instant makes acute angle with magnetic field. The path of the particle will beA. A stralight lineB. A circleC. A helix with uniform pitchD. A helix with non-uniform pitch |
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Answer» Correct Answer - C In the given question, the angel between `vec(V) and vec(B)` are acute, hence path will be helical. In the direction of magnetic field, particle moves with constant speed and hence the pitch is uniform. So choice (c) is correct. |
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| 48. |
The path of a charged particle in a uniform magnetic field depends on the angle `theta` between velocity vector and magnetic field, When `theta is 0^(@) or 180^(@), F_(m) = 0` hence path of a charged particle will be linear. When `theta = 90^(@)`, the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius `r = (mv)/(qB)`. The time period for circular path will be `T = (2pim)/(qB)` When `theta` is other than `0^(@), 180^(@) and 90^(@)`, velocity can be resolved into two components, one along `vec(B)` and perpendicular to B. `v_(|/|)=cos theta` `v_(^)= v sin theta` The `v_(_|_)` component gives circular path and `v_(|/|)` givestraingt line path. The resultant path is a helical path. The radius of helical path `r=(mv sin theta)/(qB)` ich of helix is defined as `P=v_(|/|)T` `P=(2 i mv cos theta)` `p=(2 pi mv cos theta)/(qB)` Two ions having masses in the ratio 1:1 and charges 1:2 are projected from same point into a uniform magnetic field with speed in the ratio 2:3 perpendicular to field. The ratio of radii of circle along which the two particles move is :A. `4:3`B. `2:3`C. `3:2`D. `3:4` |
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Answer» Correct Answer - A `R prop (mv)/(q)` So, `R_(1):R_(2)=(m_1)/(m_2) xx (V_1)/(V_2) xx (q_2)/(q_1) = 1/2 xx 2/3 xx 2/1 = 4/3` `(R_1)/(R_2)=4/3` So, choice (a) is correct. |
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| 49. |
The path of a charged particle in a uniform magnetic field depends on the angle `theta` between velocity vector and magnetic field, When `theta is 0^(@) or 180^(@), F_(m) = 0` hence path of a charged particle will be linear. When `theta = 90^(@)`, the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius `r = (mv)/(qB)`. The time period for circular path will be `T = (2pim)/(qB)` When `theta` is other than `0^(@), 180^(@) and 90^(@)`, velocity can be resolved into two components, one along `vec(B)` and perpendicular to B. `v_(|/|)=cos theta` `v_(^)= v sin theta` The `v_(_|_)` component gives circular path and `v_(|/|)` givestraingt line path. The resultant path is a helical path. The radius of helical path `r=(mv sin theta)/(qB)` ich of helix is defined as `P=v_(|/|)T` `P=(2 i mv cos theta)` `p=(2 pi mv cos theta)/(qB)` Which particle will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field.A. `Li^(+)`B. electronC. ProtonD. He^(+)` |
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Answer» Correct Answer - A Frequency of revolution is given by `v=(qB)/(2 pi m) or v prop 1/m` So choice (a) is correct. |
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| 50. |
किस प्रतिशत वार्षिक चक्रवृद्धि ब्याज की दर से रुपए 1000 की धनराशि दो वर्षा मे 1 ,102 . 50 हो जाएगी|A. 0.05B. 0.055C. 0.06D. 0.065 |
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Answer» Correct Answer - a (a) In these type of question go through options to save your valuable time. Option (a) `rArr` Rate of interest = 5% `rArr` Amount = Principal `(1 + ("rate")/(100))` `rArr 1102.5 = 1000(1 + (5)/(100))^(2)` `rArr (1102.5)/(1000) = (441)/(400)` `rArr 1.1025 =1.1025` `rArr ` L.H.S = R.HS Options (a) is correct |
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