Screw Gauge Viva Questions and Answers

Screw Gauge Viva Questions and Answers

(i) Name the instrument used for measuring the thickness of a piece of paper

Ans: Screw gauge.

(ii) What is meant by pitch of a screw gauge?

Ans: It is the distance advanced by the screw for one complete rotation of the head.

(iii) The pitch of a screw is 1 mm and the number of divisions on the head scale is 100. What is the least count?

Ans: Least count =  pitch/ No: of divisions on the head scale = 1/100 = 0.01 mm

(iv) What is zero error of a screw gauge?

Ans: If the zero of the head scale does not coincide with the zero of the pitch scale when the end of the movable screw is in contact with the stud, the screw gauge is said to have zero error.

(v) If the zero of the head scale is 5 divisions above the line of graduation of the pitch scale when the gap is closed, what is the zero error? What is the zero correction? Least count of the vernier = 0.01 mm

Ans: zero error = -5 x 0.01 = -0.05 mm and zero correction = +0.05 mm

(vi) If the zero of the head scale is 5 divisions below the line of graduation of the pitch scale when the gap is closed, what is the zero error and correction? Least count = 0.01 mm

Ans: zero error = +5 x 0.01 = +0.05 mm and zero correction = -0.05 mm

(vii) Why do we take a number of observations for the diameter of a wire?

Ans: Even if the wire appears uniform it may not be true. So to take the average value we take a set of readings

(viii) Why do you stop rotating the screw as soon as the ratchet begins to turn between the fingers?

Ans: When the ratchet begins to turn, the two faces are in contact or in contact with the body held between them. The device avoids undue pressure on the screw.

(ix) Aim of Screw Gauge Experiment

 

Ans: To find

(a) the diameter of a wire and

(b) thickness of an irregular glass plate using screw gauge and find their volumes.

 

(x) Theory of Screw Gauge Experiment

 

Ans: Dimension measured = P.S.R + fraction = P.S.R + (H.S.R x L.C);

where  P. S . R = Pitch Scale Reading,

H.S.R = Head Scale Reading

and L.C = Least count of the screw gauge

Volume of the wire =  πr²l; where r = radius of the wire and l = length of the wire

 

(xi) How to find the pitch of the screw?

 

Ans: Pitch = Distance moved/Number of rotations

 

(xii) How to find the least count of the screw gauge?

 

Ans: The least count of the screw is the distance moved by it when it is rotated through one division of the head scale.

Least count = Pitch/No: of divisions on the head scale.

 

(xiii) How to find the zero correction?

 

Ans: The screw is turned till the tip touches the stud. If the zero division does not coincide with zero of the pitch scale, then it is zero error. So a correction has to be applied. If it is not screwed, there is no zero correction. If it is over screwed, the zero correction is positive (+3 divisions). If it is under screwed, the zero correction is negative (-2 divisions).

 

(xiv) How to find the diameter and volume of the wire?

 

Ans: Diameter of the wire = P.S.R + fraction = P.S.R+ (corrected H.S.R. x L.C)

Volume (V) of the wire is calculated using the equation, V =  πr²l

 

(xv) How to find the thickness and volume of the glass plate?

 

Ans: The given glass plate is griped between the tip of the screw and the stud. The P.S.R and the H.S.R are noted as before. The thickness of the glass plate is

t = P.S.R + (corrected H.S.R) x LC

The glass plate is placed over a graph paper and its outline is traced on the graph paper. The area A of the glass plate is taken from the graph paper. The volume of the glass plate is calculated from the equation.

V = A x t

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Vernier Caliper Viva Questions and Answers

Vernier Caliper Viva Questions and Answers

(i) What is meant by least count of an instrument?

Ans: It is the smallest measurement that can be made with the given instrument

(ii) What is meant by least count of a vernier calipers?

Ans: It is the smallest length that can be measured with the instrument and it is equal to the difference between a main scale division and a vernier scale division.

(iii) The least count of a vernier is 0.001 cm. What is the order upto which it can measure length accurately?

Ans: It can measure accurately up to 10-3 cm.

(iv) What part of the vernier calipers is the vernier scale?

Ans: The sliding scale along the main scale is called vernier scale.

(v) Which is the instrument you will use to measure the internal and external diameter of a tube?

Ans: Vernier calipers

(vi) Apparatus of Vernier Calipers

Ans: The vernier calipers, the given cylinder and the metallic rectangular block. The vernier calipers consists of a main scale and a small subscale called vernier scale. The main scale is usually graduated in mm. The vernier scale is a sub-scale which can slide along the main scale and can be fixed at any position with the help of a screw. Usually there will be 10 divisions on the vernier scale which are equivalent to 9 mm on the main scale. The main scale and the vernier scale are provided with jaws at their ends. When the jaws are in contact, the zero of the vernier coincides with the zero of the main scale.

(vii) Theory of Vernier Calipers

Ans: When a body is gripped between the jaws, the main scale reading (M.S.R) is the reading on the main scale just before the zero mark of the vernier scale and the vernier scale reading (V.S.R) is the number of the vernier division which coincides with some division of the main scale.

If L. C is the least count of the vernier calipers,

Dimension measured = M.S.R + fraction = M.S.R + (V.S.R x L.C)

Volume of a cylinder = πr²l;

where r is the radius and l is the length of the cylinder.

Volume of the rectangular block = lbh;

where l, b and h are length, breadth and height of the block.

If r is the internal radius and h is the depth of the calorimeter,

Internal volume of the calorimeter = πr²h

(viii) Procedure of Vernier Calipers

To find the least count (LC) of the vernier calipers

The least count is the difference between a main scale division and a vernier scale division. if (n — 1) main scale divisions are divided into n vernier scale divisions then,

Least count = 1/n x 1 m.s.d

(a) To measure the dimensions of the cylinder

To find the length of the cylinder, it is gripped lengthwise between the jaws. The main scale reading (M.S.R) immediately before the zero of the vernier, and the division of the vernier (V.S.R) coinciding with any of the main scale division are noted.

The length of the cylinder, l = M.S.R. + a fraction = M.S.R. + (V.S.R x L.C)

The experiment is repeated by keeping the vernier calipers at different positions of the cylinder and the average length is calculated.

Similarly the mean diameter of the cylinder is determined. The radius (r) of the cylinder is calculated from its diameter.

Volume of the cylinder = πr²l

(b) To find the volume of the given rectangular block of known mass by measuring its dimensions with the vernier calipers and hence to find its density.

The length (I), breadth (b) and the height (h) of the rectangular block is determined as in the case of the measurement of the length of the cylinder.

Volume of the rectangular block, V = l x b x h

If m is the mass of the block, its density, d = m/V

(c) To find the internal radius (r), depth (h) and volume (V) of the calorimeter.

The upper ends of the jaws are put inside the calorimeter and open them till each of them touches the inner wall of the calorimeter. The main scale reading (M.S.R) and the vernier scale reading (V.S.R) are noted. The inner diameter of the calorimeter = M.S.R + (V.S.R x L.C). The experiment is repeated keeping the projections of the jaws tightly inside the calorimeter at different positions and the average diameter is calculated. The inner radius r of the calorimeter is calculated from its diameter.

The end of the main scale strip is kept on the upper edge of the calorimeter. The vernier scale is pushed over the main scale till the tip of the pointer attached to the back of the vernier touches the bottom of the calorimeter. The M.S.R and V.S.R are noted. The depth h of the calorimeter = M.S.R + (V.S.R x L.C). The experiment is repeated and the average depth is calculated. 

The volume V of the calorimeter is calculated from the equation,

V = πr²h

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Microwave Engineering Viva Questions and Answers

1. Give the range of frequency Band of Microwave?

Microwaves are signals in the form of electromagnetic waves with wavelengths ranging from 1m to 1mm. The frequency range of microwaves is from 3 GHz to 300 GHz. The different frequency range bands are given below.

2. Why the Micro wave got the name as microwaves?

The meaning of microwaves is the small waves used for communication. Microwaves can travel in straight line. So in satellite communication, microwaves are most commonly used.

3. Discuss in detail about Electromagnetic Frequency Spectrum?

The different ranges of frequency in spectrum are given below:

4. List the major advantages of Microwaves?

The major advantages of microwaves are:

a. Microwave Supports larger bandwidth, so microwaves are widely used for point-to-point communications.

b. By using microwaves, it is possible to attain antenna gain more.

c. Since the bandwidth is more, high data transmission rates are possible.

d. As the frequency increases, the size of antenna gets reduced

e. Low power consumption.

f. Effect of fading is less.

g. Gives effective reflection area in the radar systems.

h. Used in Satellite and terrestrial communications

i. Effective spectrum usage.

5.  List the major applications of Microwaves?

Microwaves can be used in wide variety of applications like Military, Food industry, radio astronomy, medical applications, radar, semiconductor processing techniques,  spectroscopy etc.

6. Give the frequency band of Infrared Rays (IR)?

The range of IR of the visible spectrum at is from 700 nanometers (nm) to 1 millimeter (mm). This leads to a frequency range of approximately 430 THz down to 300 GHz.

7. Discuss the band of visible Light?

The frequency band of light is from about 4 × 10¹⁴ to 8 × 10¹⁴ cycles per second, or hertz (Hz) and wavelengths ranges from 740 nanometers (nm) or 2.9 × 10⁻⁵ inches, to 380 nm (1.5 × 10⁻⁵ inches).

8. Mention the band of  X Rays?

The frequency band of X rays ranges from10 picometres to 10 nanometres, and the frequencies ranges from 30 petahertz to 30 exahertz (3×10¹⁶ Hz to 3×10¹⁹ Hz).

9. What is the band of gamma Rays?

The frequency range of gamma rays are from 10²⁰ - 10²⁴ and the wavelength is in the range < 10 - 12 m

10. Define Phase velocity and Group velocity?

The velocity having a wave packet that travels is referred as group velocity. Whereas, the velocity having the phase of a wave that travels is referred as phase velocity.

11. The standard frequency used in Microwave Oven for heating is?

The most modern microwave ovens use a frequency of 2450 MHz (2.45 GHz) for heating.

12. Give the speed of Electromagnetic waves in free space?

In free space, the electromagnetic waves has a speed same as that of light, c = 3 x 10⁸ m/s.

13. Explain about the propagation of EM waves?

When an electric charge (q) vibrates, the electromagnetic waves are generated. This generated wave has both an electric and a magnetic component. Light waves are the example of EM waves.

14. What is Faraday's law?

Faraday's law is defined as a current that would be induced in a conductor which is exposed to a varying magnetic field

15. Define Gauss law?

Gauss's law is defined as the net flux of an electric field that is closed in an enclosed surface is directly proportional to the enclosed electric charge.

16. Define Ampere's law?

Ampere’s law states that “the magnetic field caused by an electric current would be proportional to the size of electric current having a constant of proportionality corresponding to the permeability of free space”.

17. Define TEM wave?

Transverse electromagnetic (TEM) waves are those waves with a mode of propagation of waves, while the electric and magnetic field lines are transverse (normal wave) to the direction of propagation.

18. What you understand by Transverse Electric (TE) wave?

TE wave is an electromagnetic wave propagating in a medium, including free space, or any other medium in such a manner that the electric field vector is directed entirely transverse (perpendicular), to the general forward direction of propagation.

19. What is mean by Transverse magnetic (TM) waves?

TM waves, also known as E waves are characterised by the fact that the magnetic vector (H vector) is always perpendicular to the direction of propagation.

20. What you understand about EM wave?

Electromagnetic waves (EM) waves are the type of waves which are created by a result of vibrations between an electric field and a magnetic field. That is we can say that the EM waves are composed of oscillating magnetic and electric fields.

21. Discuss about the need of Transmission line?

The main use of transmission line is to reduce power loss, dissipated as heat because of the resistance of the conductors. For long distance power transmissions, high voltage transmission lines are mostly used. 

22. Discuss about the types of transmission lines?

The different types of transmission lines are:

a. Wave Guide

b. Fiber Optic

c. Balanced two wire

d. Co-axial cable

e. Microstrip

23. What is single stub matching?

In order to match any complex load to a transmission line, stub matching is used. Stub matching is group of shorted or opened segments of the line, which are connected in parallel or in series with the line at an appropriate distances from the load.

24. List the main advantages of microwave integrated circuits (MIC)?

The main advantages of MIC are:

a. Many (thousands) of devices can easily be fabricated in a single MMIC at the same time.

b. Overall cost of MMIC is less.

c. Mismatches in the MMIC between components are less.

d. Since the distances between the components are less, the signal delay is less in a MMIC.

e. The wire bond reliability issues are less.

25. Discuss about reflection coefficient?

Reflection coefficient can be defined as a parameter that describes how much of a wave is reflected by an impedance discontinuity in the transmission medium. In other words, reflection coefficient gives the amount of reflected waves in a transmission medium due to the impedance mismatching. 

26. Discuss about the different types of losses in transmission lines?

The losses occurred in transmission lines are mainly classified into three types. They are:

a. Copper Losses

b. Dielectric Losses

c. Radiation (Induction) Losses

EDC Lab Viva Questions with Answers

Electronic Devices and Circuits Lab Viva Questions with Answers

                                                          

1. Differentiate Electronics and Electrical Engg.?

We can say that electronics is the branch of science which deals with vacuum tubes, semi conductor materials like diodes, transistor etc. The maximum allowed voltage range of electronics engg is about 0 to 30 V. Electrical is the branch of science which deals with the currents in conductors, in other words, we can say that electrical engg deals with 0 to 230 V, 50 Hz operating devices. Operation of devices like motors, generators etc are included in electrical engg.  

2. Classify materials in terms of conductivity?

By the property of conduction of electricity, materials can be mainly classified as conductors, semiconductors and insulators. In conductors, the valance band (VB) and conduction band (CB) overlaps each other (i.e., energy gap is zero), so the electrons can easily move from valance band to the conduction band. Examples are Gold, Copper, Silver etc. In insulators, the energy gap is very wide (about 3 to 5 eV) and hence the electrons cannot flow from VB to CB. In semiconductors, the energy gap is narrower so the electrons from VB can move to CB by absorbing energy. The energy gap of Si is 1.1 eV and for Ge it is 0.7 eV.

3. Differentiate Semiconductor and conductors?

Semiconductor has four electrons in its VB. The energy gap of semiconductors is about 1 eV. So at normal condition the electrons cannot move from VB to CB. By giving some external energy to the electrons, electrons can move from VB to CB. The best semiconductors are Si, Ge, etc. The conductors have more than four electrons in its VB. The energy gap of conductors is 0 eV. So the electrons can easily move from VB to CB. Best conductors are Ag, Al, Cu.

4. Compare intrinsic and extrinsic Semiconductors?

Intrinsic semiconductors are the pure form of semiconductors. At room temperature, the conductivity of intrinsic semiconductor is zero. At room temperature, its VB is totally filled and the CB is empty. When some heat energy is supplied (as temperature increases), electrons can jump to CB and can move randomly. As temperature increases, the conductivity increases and the resistivity decreases. Extrinsic semiconductors are formed by intentionally adding impurities on it. At room temperature, the extrinsic semiconductor is very little conductive. The impurities added can be trivalent (resulting P type materials) or pentavalent (resulting N type materials).

5. Explain different types of extrinsic semiconductors and discuss their formation?

The extrinsic semiconductors are made of by adding impurities on fourth group elements of periodic table (Si, Ge etc). There are mainly two types of extrinsic semiconductors- P Type and N Type. P types are formed by adding trivalent impurities (third group elements in Periodic table such as arsenic, antimony, phosphorus) to the fourth group elements. N type semiconductors are formed by adding pentavalent impurities (fifth group elements like aluminum, boron etc) to the fourth group elements.

6. Discuss the process of doping?

Simply we can say that doping is the process of adding impurities. Extrinsic semiconductors are made by doping process. The two types of doped semiconductor materials are P type and N type materials.

7. Discuss the process of formation of PN Junction?

A P-N junction (technically termed as diode) is a piece of semiconductor material in which it’s one half is doped by P type impurity and the other half is doped by N type impurity. P-N junction was developed in the year 1936. The main advantage of P-N junction is it passes current in only one direction. So we can say that PN junction act as a switch in electronic circuits. The main application of PN junction is rectifier circuits.

8. Explain about cut in voltage (Barrier Potential)?

The cut in voltage is the minimum voltage in which a PN junction starts its conduction. For Silicon diode, it is 0.7 Volts and for Ge diode is 0.3 V. That is when the external applied voltage crosses the barrier potential, the diode conducts electrons.

9. Discuss about Forward bias and Reverse bias conditions?

If the externally applied voltage of the anode (P) is greater than the cathode (N) voltage in a PN junction diode, the diode is said to be in forward biased condition. If the anode voltage is less than the cathode voltage, then the diode is in reverse biased condition.

10. Compare between Diffusion and Drift currents?

The current formed by the diffusion of holes or electrons is termed as the diffusion current. In other words we can say that due to the non uniform concentration of charged particles, the transportation of charge occurs and this current is termed as diffusion current. Diffusion current doesn’t need an external applied electric field. Drift current is the current occurred due to the effect of an externally applied electric field. That is the motion of charged particles due to the effect of an externally applied electric field. If there is no electric field, the drift current becomes zero.

11. Describe the meaning of 1N4007?

IN4007 is actually represents a diode. The expansion of the diode is

IN means it is a single junction diode

400x indicates the voltage, current and power

4007 represents the reverse voltage from 50V to 1000V, the maximum forward current is 1A

12. Justify the Diode current Equation?                     

The diode equation can be derived as

I=I_o(e^(V/ηVT)-1),

Where, I_o is the Reverse saturation current, and the value of η=1 for germanium, and 2 for silicon.

13. What you mean by Peak inverse voltage (PIV)?

PIV is the maximum affordable reverse voltage of a diode. When the applied reverse voltage exceeds the PIV, the PN junction gets damaged (diode goes to breakdown condition). It is also known as Peak Reverse Voltage (PRV). The PIV is also termed as reverse breakdown voltage.

14 Discuss about Reverse saturation current?

The reverse saturation current (Ico) is produced due to the diffusive flow of minority charge carriers, when a diode is at reverse biased condition. At R-B condition, the holes will get attracted to the negative terminal of the applied cell and the electrons will get attracted to the positive terminal of the cell. Hence the depletion width increases and the flow of charges through the junction becomes zero. But there exists thermally produced some electron-hole pairs. These thermally generated electron-hole pairs are less in number and the electrons produced are repelled by the –ve terminal and the holes are repelled by the +ve terminal. This will cause a small current passes through the PN junction. This current is very small and is almost constant.      

15. Discuss about the testing of diode by a multimeter?

The working condition of diode can be easily tested by a multimeter. For testing a diode, the anode (P side) of the diode is connected to the positive terminal of multimeter and the cathode (N side) of the diode is connected to the negative terminal of the multimeter. If the readed voltage is in the range of 0.3 to 0.7 Volts, then the diode is in good condition.

16. Mention the capacitive effects formed in a P-N junction.

There are mainly two types of capacitance formed in a PN junction.

a. Diffusion Capacitance: The diffusion capacitance occurs in forward biased condition.

b. Transition capacitance: Transition capacitance occurs in the reverse biased condition.

17.Discuss about the Break down condition, compare avalanche Break down and Zener Break down?              

The breakdown occurs due to the impact of ionization of electron-hole pair in a PN junction diode is termed as the Avalanche breakdown.

At the reverse biased voltage condition, the electron hole pair breaks in a zener diode and this breakdown is termed as zener breakdown.

18. Discuss the effect of temperature on reverse saturation current and barrier voltage?

As the temperature increases, the reverse saturation current also increases. As the temperature raise by 10 degrees, the reverse saturation current doubles.

As the temperature increases, the barrier voltage decreases. For every temperature increase of 1 degree, the barrier voltage decreases by 2.5 mv

19. Differentiate between static and dynamic resistances?

The ratio of voltage and current (V-I ratio) at the Q-point(operating point) is the static resistance.

The inverse to the slop of the forward bias characteristic curve at operating (Q point) is termed as the dynamic resistance.

  

20. Mention the applications of  PN Diode and Zener diode?

The main applications of a PN junction diode are: Rectifier, clipper, clamper, switch etc.

Voltage regulator circuit is the main application of zener diode.

21.What is the major difference of PN junction and zener diode?

The main difference of PN junction diode and zener diode is the PN junction is lightly dopped and the zener diode is heavily dopped.

22. Zener diode is not used in forward biased condition. justify your answer?

The response of a zener diode in forward bias condition is same as that of a PN junction diode. So generally the zener diode is used in reverse bias state with zener breakdown.

23.Discuss the effect of temperature on Zener diode?

The effect of temperature on zener diode depends on the breakdown voltage. If the breakdown voltage (V_z) is less than 6V (breakdown voltage), V_z is inversely proportional to temperature.

The breakdown voltage V_z is directly proportional to temperature, if the the breakdown voltage is greater than 6V

24. Give the advantage of silicon over germanium?

The breakdown voltage of silicon is more than germanium. The reverse saturation current of silicon is less than that of germanium. Since the raw material is sand, silicon diode is cheaper than germanium diode. The cut-off voltage of silicon diode is 0.7 V and that of germanium is 0.3V.