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[MCQ] Applied Physics – II

Module 1

1. The optical path of monochromatic light is the same if it travels 2 cm thickness of glass or 2.25 cm thickness of water. If the refractive index of water is 33, what is the refractive index of glass?
a) 2.5
b) 1.5
c) 3.5
d) 4.5
1

2. The absolute refractive indices of glass and water are 3/2 and 4/3. Determine the ratio of the speeds of light in glass and water.
a) 5:7
b) 9:8
c) 7:5
d) 8:9
2

3. The refractive index of glass is 1.5 and that of water is 1.3, the speed of light in water is 2.25 × 108 m/s. What is the speed of light in glass?
a) 7.95 × 108 m/s
b) 9.95 × 108 m/s
c) 1.95 × 108 m/s
d) 3.95 × 108 m/s
3

4. The speed of light in air is 3 × 108 m/s. If the refractive index of glass is 1.5, find the time taken by light to travel a distance of 10 cm in the glass.
a) 0.5 × 10-10 s
b) 5 × 10-10 s
c) 50 × 10-10 s
d) 500 × 10-10 s
4

5. The speed of yellow light in a certain liquid is 2.4 × 108 m/s. Find the refractive index of the liquid.
a) 1.25
b) 5.55
c) 6.25
d) 12.25
5

6. Which among the following isn’t a suitable phenomenon to establish that light is wave motion?
a) Interference
b) Diffraction
c) Reflection
d) Polarization
Answer: c
Explanation: Light undergoes interference, diffraction, and polarization. These phenomena establish that light is a wave motion. Therefore, out of the options, reflection isn’t a suitable phenomenon to establish that light is wave motion.

7. Identify the condition which is not necessary for two light waves to be coherent.
a) The two waves must be continuous
b) The two waves should be of the same frequency or wavelength
c) They should have a constant or zero phases difference
d) They two light sources should be narrow
Answer: d
Explanation: The essential conditions for two light waves to be coherent includes that the two waves must be continuous, the two waves should be of the same frequency or wavelength, and they should have a constant or zero phases difference. So, the unnecessary condition for two light waves to be coherent is the statement that the sources should be narrow.

8. Focal length of plane mirror is
a. At infinity
b. Zero
c. Negative
d. None of these
Answer   a

9. Image formed by plane mirror is
a. Real and erect
b. Real and inverted
c. Virtual and erect
d. Virtual and inverted
Answer   c

10. A concave mirror gives real, inverted and same size image if the object is placed
a. At F
b. At infinity
c. At C
d. Beyond C
Answer   c

11. Power of the lens is -40, its focal length is
a. 4m
b. -40m
c. -0.25m
d. -25m
Answer   c

12. A concave mirror gives virtual, refract and enlarged image of the object but image of smaller size than the size of the object is
a. At infinity
b. Between F and C
c. Between P and F
d. At E
Answer   c

13. In optics an object which has higher refractive index is called
a. Optically rarer
b. Optically denser
c. Optical density
d. Refractive index
Answer   b

14. The optical phenomena, twinkling of stars, is due to
a. Atmospheric reflection
b. Total reflection
c. Atmospheric refraction
d. Total refraction
Answer   c

15. Convex lens focus a real, point sized image at focus, the object is placed
a. At focus
b. Between F and 2F
c. At infinity
d. At 2F
Answer   c

16. The unit of power of lens is
a. Metre
b. Centimeter
c. Diopter
d. M-1
Answer   c

17. The radius of curvature of a mirror is 20cm the focal length is
a. 20cm
b. 10cm
c. 40cm
d. 5cm
Answer   b

18. How many lenses are used in Fresnel Diffraction?
a) Two Convex lenses
b) Two Concave lenses
c) One Convex lens
d) No lens used
Answer: d
Explanation: In Fresnel Diffraction, no lenses are used. The interference takes place between different light waves arriving at a point from the same wavefront. In this case, the phase of all the waves is not the same.

19. Which of the following is called the obliquity factor?
a) cosθ
b) sinθ
c) 1 + cosθ
d) 1 + sinθ
Answer: c
Explanation: The effect at a point fie to any particular zone is proportional to 1 + cosθ, called the obliquity factor. Naturally, the effect is maximum whenθ` = 0.677.

20. In Fresnel diffraction, the relative phase difference between the curved wavefront is ____________
a) Constant
b) Zero
c) Linearly increasing
d) Non=constant
Answer: d
Explanation: Since the radii of each half period zone are different, the distance traveled by each wavefront is different. Thus, the relative phase difference turns out to be non-constant.

21. What is the variation in the pattern observed for single slit Fresnel diffraction than single slit Fraunhofer Diffraction?
a) The pattern is not hyperbolic
b) The fringes are too thin
c) The region of minimum intensity is not completely dark
d) The fringes are colored
Answer: c
Explanation: The diffraction pattern due to the single slit Fresnel Diffraction is similar to single slit Fraunhofer Diffraction, except the regions of minimum intensity are not completely dark. This is so because there is never a complete destructive interference in Fresnel Diffraction.

22. The radius of the half period zone is proportional to ___________
a) The wavelength of light
b) The square root of the frequency of light
c) The square root of the wavelength light
d) The frequency of light
Answer: c
Explanation: We know that the formula for the radius of half period zone =nbλ−−−√, where n is a natural number. Thus, it is proportional to the square root of wavelength light and inversely proportional to the square root of the frequency of light.

23. Intensity at O due to the entire wavefront is x times that due to the first half period zone. What is x?
a) 4
b) 2
c) 1/2
d) 1/4
Answer: d
Explanation: The formula for intensity = m214
The intensity due to the first half period zone = m12
Thus, the intensity at O due to the entire wavefront is 1/4 times that due to the first half period zone.

24. Light of 5000 Å is incident on a circular hole of radius 1 cm. How many half period zones are contained in the circle if the screen is placed at a distance of 1 m?
a) 20
b) 200
c) 2000
d) 20000

Answer: d
Explanation: In this case, λ = 5000 Å = 5 X 10-5cm, b = 1m = 100cm
Therefore, Number of half period zones = 1λ
= 1/5 X 10-5
= 20000.

25. Light of 6000 Å is incident on a circular hole and is received on a screen 50 cm away. What is the radius of the hole, if the intensity of light on the screen is 4 times the intensity without the hole?
a) 0.025 cm
b) 0.047 cm
c) 0.054 cm
d) 0.089 cm
Answer: c
Explanation: The intensity will be 4 times than in its absence if the radius of the hole is equal to that of the first half period zone.
Therefore, radius, r = bλ−−√
Here, b = 50 cm and λ = 6000 Å = 6 X 10-5cm
r = 0.0548 cm.

26. The zone plate behaves like a ____________
a) Concave Lens with multiple foci
b) Convex Lens with multiple foci
c) Convex Lens with single foci
d) Concave Lens with single foci
Answer: b
Explanation: In a zone plate, a much brighter image of an object is obtained at the screen, which shows the converging action of a zone plate. Also, it’s equation resembles that of a lens. Thus, the zone plate behaves like a convex lens with multiple foci.

27. The radius of the first zone in a zone plate of focal length 25cm for a light of wavelength 5000 Å is ____
a) 0.01 cm
b) 0.02 cm
c) 0.03 cm
d) 0.04 cm
Answer: c
Explanation: Here, f1 = 25cm, λ=5000 Å = 5 X 10-5 cm.
Now using the formula, f=r2nλ
r1 = fλ−−√
= 0.03 cm.

28. What is the effective distance between the source of light and the screen in Fraunhofer Diffraction?
a) Focal length of the convex lens
b) Less than Focal Length of the convex lens
c) Greater than the focal length of the convex lens and less than infinite
d) Infinite
Answer: d
Explanation: In Fraunhofer Diffraction, the source of light and the screen are effectively placed at infinite distance. Two convex lenses are used for achieving such a condition. Thus, the incident waveform is plane and the secondary wavelets are in the same phase at every point in the plane of the aperture.

29. What happens with the Fraunhofer single slit diffraction pattern if the whole apparatus is immersed in water?
a) The Wavelength of light increases
b) Width of central maximum increases
c) Width of central maximum decreases
d) Frequency of light decreases
Answer: c
Explanation: As the whole apparatus is now immersed in water, the wavelength of the light will change.
λ‘=λμ
Therefore, as the refractive index of water is greater than the air, the wavelength of light will decrease.
Width of central maxima = 2λa
Therefore, as the wavelength decreases, the width of the central maxima decreases.

30. How shall the pattern change when white light is used in Fraunhofer Diffraction instead of monochromatic light?
a) The pattern will no longer be visible
b) The shape of the pattern will change from hyperbolic to circular
c) The colored pattern will be observed with a white bright fringe at the center
d) The bright and dark fringes will change position
Answer: c
Explanation: When white light is used instead of monochromatic light in Fraunhofer Diffraction, then the central maximum remains white as all seven wavelengths meet there in the same phase. The first minimum and second maximum will be formed by violet color due to its shortest wavelength while the last is due to the red color as it has the longest wavelength. Thus, a colored pattern is observed.
However, after the first few colored bands, the clarity of the band is lost, due to overlapping.

31. If the separation between the two slits in Double Slit Fraunhofer Diffraction is changed, what change will be observed in the diffraction pattern?
a) The fringe length will increase
b) The fringe length will decrease
c) Fringes will be colored
d) No change
Answer: d
Explanation: The separation between the two slits only affects the interference pattern in Double Slit Fraunhofer Diffraction. The diffraction pattern does not change.
For Diffraction, e sin θ = ±mλ
Where e = Width of slits
m = Any integer
For interference, (e + d) sin θ = ±nλ
Where e = Width of slits
d = Separation between the two slits
m = Any integer
Hence, there is no change in the Diffraction Pattern.

32. In Double Slit Fraunhofer Diffraction, some orders of interference pattern are missing. It is called _____
a) Missing Spectra
b) Absent Spectra
c) End Spectra
d) Emission Spectra
Answer: b
Explanation: In Double Slit Fraunhofer Diffraction, there are certain angles where the interference maxima and Diffraction minima overlap. These orders of interference pattern are missing in the pattern. It is known as Absent Spectra.

33. A screen is placed 2m away from the lens to obtain the diffraction pattern in the focal plane of the lens in a single slit diffraction experiment. What will be the slit width if the first minimum lies 5 mm on either side of the central maximum when plane light waves of wavelength 4000 Å are incident on the slit?
a) 0.16 mm
b) 0.26 mm
c) 0.36 mm
d) 0.46 mm
Answer: a
Explanation: Given: f = 2 m, x = 5 X 10-3m, λ = 4 X 10-7m, n=1
sin θ = nλa, we have
a = nλsinθ
= 1.6 X 10-4 m
= 0.16 mm.

34. Find the missing order for a double-slit Fraunhofer Diffraction pattern if the slit widths are 0.2 mm separated by 0.6 mm.
a) 1st, 5th, 9th, ….
b) 2nd, 6th, 10th, …
c) 3rd, 7th, 11th, ….
d) 4th, 8th, 12th, …
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35. If a is the width of the slits and b the distance between the slits, then a + b is called as _________
a) Opacities
b) Transparency
c) Grating Constant
d) Grating value
Answer: c
Explanation: The sum of the width of the slits and the distance between the slits is called the grating constant of the grating. It can also be described as the separation between the corresponding point of two adjacent transparencies.

36. The CDs available in the market act as ____________
a) Diffraction grating
b) Reflection grating
c) Transmission grating
d) Do not act as a grating
Answer: b
Explanation: The CDs have microscopic pits on its surface. It splits white light into its constituent colors and thus acts as a reflection grating. It is due to this that rainbow colors can be seen on CD surface.

37. Which of the following is the correct expression for the ratio of the intensity of principal maxima to the intensity of secondary maxima?
a) 1 + (N2-1) sin2β
b) 1/1 + (N2-1) sin2β
c) (N2-1) sin2β
d) 1/(N2-1) sin2β
Answer: a
Explanation: The expression 1 + (N2-1) sin2β is the relation for the ratio of the intensity of principal maxima to that of the secondary maxima. It shows that the intensity of secondary maximas are inversely proportional to N.

38. If there are N number of slits in a grating spectra, then there will be how many secondary maximas?
a) N
b) N – 1
c) N – 2
d) 2N
Answer: c
Explanation: Between any two consecutive principal maxima, there is a set of smaller secondary maxima and minima. Thus, there would be N – 1 secondary minimas and N – 2 secondary maximas.

39. If (a + b) < nλ, then nth order spectrum will be absent.
a) True
b) False
Answer: a
Explanation: We know that according to the grating rule,
(a + b) sinθ = nλ
Now, if (a + b) < nλ then sinθ > 1 which is not possible. Hence, in that case, the nth order spectrum will be absent.

40. The expression for the maximum number of orders in a grating spectrum is given by ____________
a) λ/ (a + b)
b) a/λ
c) b/λ
d) (a + b)/ λ
Answer: d
Explanation: We know that, (a + b) sinθ = nλ
Therefore, n = (a + b) sinθ/λ
Now, as the maximum value for sinθ is 1. Hence,
nmax = (a + b) /λ.

41. What is the relation between the dispersive power, D, of a grating and the order, n, of a spectrum?
a) D ∝ n
b) D ∝ N2
c) D ∝ 1/n
d) D ∝ 1/N2
Answer: a
Explanation: The dispersive power of a grating is given by:
∂θ/∂λ = n/ (a + b) cosθ
Thus, as we can see D ∝ n. Hence, higher the order, higher is the dispersive power.

42. A beam of monochromatic light is incident on a plane transmission grating having 5000 lines/cm and the second order spectral line is found to be diffracted at 30°. The wavelength of the light is _______
a) 4000 Å
b) 5000 Å
c) 6000 Å
d) 7000 Å
Answer: b
Explanation: Here, N = 5000, a + b = 1/5000 cm
Now using, (a + b) sinθ = nλ for second order spectrum we get,
λ = sin30°/ 2 X 5000 cm
λ = 5000 Å.

43. What is the highest order spectrum which may be seen with a monochromatic light of wavelength 5000 Å by means of a diffraction grating with 5000 lines/cm?
a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: Here, N = 5000. Therefore, a + b = 1/5000 cm
Now, as we know, Nmax = a + b/λ
= 1/5000 X 5000 X 10-8
= 4.

44. A grating has 16000 per inch over a length of 5 inches. What will be the smallest wavelength difference for a light of wavelength 6000 Å?
a) 0.01 Å
b) 0.02 Å
c) 0.03 Å
d) 0.04 Å
Answer: c
Explanation: N = 16000 X 5 = 80000, n = 2
Wavelength = 6 X 10-5 cm
Resolving power = nN = 2 X 80000 = 160000
Smallest wavelength difference = λ/nN
= 0.0375 Å.

45. Which of the following is the correct expression for the resolving power of a grating?
a) (nN + 1)/λ
b) nN/λ
c) nN/λ + 1
d) nN
Answer: d
Explanation: The expression for the resolving power of a grating is given by: λλ = nN
Hence, it is proportional to the number of slots on the grating and the order of the spectrum. It is independent of the grating constant.

46. What is the SI unit of the Resolving power of a plane transmission grating?
a) m-1
b) cm-1
c) s-1
d) No SI unit
Answer: d
Explanation: Mathematically, the resolving power of a plane transmission grating can be defined as the ratio of the mean wavelength of a pair of spectral lines and the wavelength difference between them. As both the quantities have the same unit, resolving power has no unit.

47. The resolving power of a grating is inversely proportional to the wavelength of the incident light.
a) True
b) False
Answer: a
Explanation: As we know, the resolving power of a grating = nN
Now, n = (e + d) sinΘ / λ, where e + d is the width of the ruled surface of the grating.
Thus, Resolving power = N (e + d) sinΘ / λ.
Hence, the resolving power is inversely proportional to the wavelength of incident light.

48. Our eyes see two objects as separate, only if the angle subtended by them at the eye is greater than _________
a) 30 seconds
b) 1 minute
c) 2 minute
d) 10 seconds
Answer: b
Explanation: Human eyes can see two objects as separate only when the angle subtended by them on the eye is greater than 1 minute. This is the minimum angle of resolution of a normal human eye. This process of observing two objects as separate is called resolution.

49. The resolving power of a grating is directly proportional to grating constant.
a) True
b) False
Answer: b
Explanation: For a grating, the resolving power = nN, where n is the order of the spectrum and N is the total number of slits on the grating. It is independent of the grating constant.

50. If a light is incident on a grating with 5000 lines/cm, then the angular separation of the two lines (5000 Å and 5006 Å) in first order spectrum is ______
a) 0.01°
b) 0.02°
c) 0.03°
d) 0.04°
Answer: b
Explanation: Here, N = 5000. Therefore, a + b = 1/5000 cm
Now we know, (a + b) sinΘ = nλ
For λ = 5000 Å, sinΘ1 = 5 X 10-5 X 5000
Θ1 = 14.47°
For λ = 5006 Å, sinΘ2 = 5.006 X 10-5 X 5000
Θ2 = 14.49°
Angular separation = Θ2 – Θ1 = 0.02°.

51. Light is incident normally on a grating of width 5 X 10-3 m with 2500 lines. What is the resolving power of the grating in the second order spectrum?
a) 2500
b) 5000
c) 1250
d) 500
Answer: b
Explanation: The resolving power of a grating = nN, where n is the order of the spectrum and N is the number of slits on the grating.
Thus, Resolving power = 2 X 2500
= 5000.

52. What is the minimum number of lines per cm in a 2.5 cm wide grating spectrum which will just resolve two sodium lines (5890 Å and 5896 Å) in the first order spectrum?
a) 98
b) 196
c) 392
d) 694
52

53. A grating has 16000 per inch over a length of 5 inches. What will be the smallest wavelength difference for a light of wavelength 6000 Å?
a) 0.01 Å
b) 0.02 Å
c) 0.03 Å
d) 0.04 Å
Answer: c
Explanation: N = 16000 X 5 = 80000, n = 2
Wavelength = 6 X 10-5cm
Resolving power = nN = 2 X 80000 = 160000
Smallest wavelength difference = λ/nN
= 0.0375 Å.

54. The following pattern was observed by a plane diffraction grating. How are the two images resolved?
54
a) Not resolved
b) Just Resolved
c) Partially resolved
d) Well resolved
Answer: b
Explanation: As stated in Rayleigh’s criteria, when the central maxima of one image coincide with the first minimum of the other image, the two images are said to be just resolved. In the fig, this condition is satisfied. Hence, the two lines are just resolved.

Module 2

1. Which of the following is a unique property of laser?
a) Directional
b) Speed
c) Coherence
d) Wavelength
Answer: c
Explanation: Coherence is an important characteristic of laser beam because in laser beams, the wave trains of the same frequency are in phase/ Due to high coherence it results in extremely high power.

2. Which of the following is an example of optical pumping?
a) Ruby laser
b) Helium-Neon laser
c) Semiconductor laser
d) Dye laser
Answer: a
Explanation: The atoms of Ruby are excited with the help of photons emitted by an external optical source. The atoms absorb energy from photos and raises to excited state. Therefore Ruby laser is an example of optical pumping.

3. When laser light is focussed on a particular area for a long time, then that particular area alone will be heated.
a) True
b) False
Answer: a
Explanation: Laser beam has very high intensity, directional properties and coherence. When it is focussed on a particular area for a long time, then the area alone will be heated and the other area will remain as such. This is called thermal effect.

4. Calculate the wavelength of radiation emitted by an LED made up of a semiconducting material with band gap energy 2.8eV.
a) 2.8 Å
b) 4.3308 Å
c) 5548.4 Å
d) 4430.8 Å
Answer: d
Explanation: E = hc/ʎ
Therefore, ʎ = hc/E
ʎ = 4430.8 Å.

5. Which of the following can be used for the generation of laser pulse?
a) Ruby laser
b) Carbon dioxide laser
c) Helium neon laser
d) Nd- YAG laser
Answer: d
Explanation: Since Nd YAG laser has a higher thermal conductivity than other solid state lasers, it lends itself for the generation of laser pulses at a higher pulse repetition rate or a quasi continuous wave operation.

6. What is the need to achieve population inversion?
a) To excite most of the atoms
b) To bring most of the atoms to ground state
c) To achieve stable condition
d) To reduce the time of production of laser
Answer: a
Explanation: When population inversion is achieved, the majority of atoms are in the excited state. This causes amplification of the incident beam by stimulated emission. Thus the laser beam is produced.

7. Laser is called as a non-material knife.
a) False
b) True
Answer: b
Explanation: In laser surgery, without knife, bloodless operation, cutting tissues etc can be made, hence laser is called non-material knife.

8. DVD uses the laser.
a) True
b) False
Answer: a
Explanation: A DVD player contains a laser. By moving the lens longitudinally, different depths can be reached in the disc. In order to make room for a lot of information on every disc, the beam has to be focused on as small an area as possible. This cannot be done with any other light source.

9. Which of the following is used in atomic clocks?
a) Laser
b) Quartz
c) Maser
d) Helium
Answer: c
Explanation: Before laser maser was used. It stood for microwave amplification by stimulated emission of radiation. This was based on Albert Einstein’s principle of stimulated emission. It was used in the atomic clock.

10. Which of the following can be used in the vibrational analysis of structure?
a) Maser
b) Quarts
c) Electrical waves
d) Laser
Answer: d
Explanation: Laser can be used in the vibrational analysis of the structure. This is because when a structure under test begins to vibrate a distinctive pattern begins to emerge.

11. Calculate the number of photons, from green light of mercury (ʎ = 4961 Å), required to do one joule of work.
a) 4524.2×1018/m3
b) 2.4961×1018/m3
c) 2.4961/m3
d) 2.4961/m
Answer: b
Explanation: E = hc/ʎ
E = 4.006×10-19 Joules
Number of photons required = (1 Joule)/(4.006×10-19)
N = 2.4961×1018/m3.

12. What is the principle of fibre optical communication?
a) Frequency modulation
b) Population inversion
c) Total internal reflection
d) Doppler Effect
Answer: c
Explanation: In optical fibres, the light entering the fibre does not encounter any new surfaces, but repeatedly they hit the same surface. The reason for confining the light beam inside the fibres is the total internal reflection.

13. What is the other name for a maximum external incident angle?
a) Optical angle
b) Total internal reflection angle
c) Refraction angle
d) Wave guide acceptance angle
Answer: d
Explanation: Only this rays which pass within the acceptance angle will be totally reflected. Therefore, light incident on the core within the maximum external incident angle can be coupled into the fibre to propagate. This angle is called a wave guide acceptance angle.

14. A single mode fibre has low intermodal dispersion than multimode.
a) True
b) False
Answer: a
Explanation: In both single and multimode fibres the refractive indices will be in step by step. Since a single mode has less dispersion than multimode, the single mode step index fibre also has low intermodal dispersion compared to multimode step index fibre.

15. How does the refractive index vary in Graded Index fibre?
a) Tangentially
b) Radially
c) Longitudinally
d) Transversely
Answer: b
Explanation: The refractive index of the core is maximum along the fibre axis and it gradually decreases. Here the refractive index varies radially from the axis of the fibre. Hence it is called graded index fibre.

16. Which of the following has more distortion?
a) Single step-index fibre
b) Graded index fibre
c) Multimode step-index fibre
d) Glass fibre
Answer: c
Explanation: When rays travel through longer distances there will be some difference in reflected angles. Hence high angle rays arrive later than low angle rays. Therefore the signal pulses are broadened thereby results in a distorted output.

17. In which of the following there is no distortion?
a) Graded index fibre
b) Multimode step-index fibre
c) Single step-index fibre
d) Glass fibre
Answer: a
Explanation: The light travels with different speeds in different paths because of the variation in their refractive indices. At the outer edge it travels faster than near the centre But almost all the rays reach the exit end at the same time due to the helical path. Thus, there is no dispersion in the pulses and hence the output is not a distorted output.

18. Which of the following loss occurs inside the fibre?
a) Radiative loss
b) Scattering
c) Absorption
d) Attenuation
Answer: b
Explanation: Scattering is a wavelength dependent loss. Since the glass used in the fabrication of fibres, the disordered structure of glass will make some vibrations in the refractive index inside the fibre. This causes Rayleigh scattering.

19. What causes microscopic bend?
a) Uniform pressure
b) Non-uniform volume
c) Uniform volume
d) Non-uniform pressure
Answer: d
Explanation: Micro-bends losses are caused due to non-uniformities inside the fibre. This micro-bends in fibre appears due to non-uniform pressures created during the cabling of fibre.

20. When more than one mode is propagating, how is it dispersed?
a) Dispersion
b) Inter-modal dispersion
c) Material dispersion
d) Waveguide dispersion
Answer: b
Explanation: When more than one mode is propagating through a fibre, then inter modal dispersion will occur. Since many modes are propagating, they will have different wavelengths and will take different time to propagate through the fibre.

21. A fibre optic telephone transmission can handle more than thousands of voice channels.
a) True
b) False
Answer: a
Explanation: Optical fibre has larger bandwidth hence it can handle a large number of channels for communication.

22. Which of the following is known as fibre optic back bone?
a) Telecommunication
b) Cable television
c) Delay lines
d) Bus topology
Answer: d
Explanation: Each computer on the network is connected to the rest of the computers by the optical wiring scheme called bus topology, which is an application known as fibre optic back bone.

23. During Population inversion, which of the following processed is dominant?
a) Stimulated Absorption
b) Stimulated Emission
c) Spontaneous Emission
d) Spontaneous Absorption
Answer: b
Explanation: LASER is a short form of Light Amplification by Stimulated Emission of Radiations. Stimulated Emission is the process by which amplification of radiations takes place. Hence, Laser operation requires stimulated emission to be dominant.

24. The relationship between N1 and N2 for stimulated emission to be dominant is ___________
a) N1 = N2
b) N1 > N2
c) N2 > N1
d) No such relationship
Answer: c
Explanation: In this case, the population of atoms in the exited state is more than that in the ground state. This condition is known as population inversion and in this case, stimulated emission is dominant.

25. During pumping, the atoms are exited to ___________
a) Higher Exited States
b) Lower Energy states
c) Meta Stable states
d) Not Excited
Answer: c
Explanation: Pumping is the process in which the atoms from the lower energy states are excited to meta-stable states by some external source so as to create population inversion.

26. Which of the following is a four-level laser?
a) ND: YAG
b) Ruby
c) He-Ne
d) Semiconductor laser
Answer: c
Explanation: He-Ne laser is a four-level laser. Whenever an electric discharge is passed through the gas, the helium atoms gets to higher energy states, as the concentration of helium atoms is higher to achieve population inversion.

27. The ratio of N2 and N1 is given by ___________
23
23-1

28. What will be the relative population of atoms in a ruby layer that produces a light beam of wavelength 6943 Å at 300 K.
a) 5 X 10-31
b) 6 X 10-31
c) 7 X 10-31
d) 8 X 10-31
28

29. Between which two layers, population inversion takes place?
29

a) E1 and E2
b) E1 and E3
c) E2 and E2
d) Neither energy levels
Answer: a
Explanation: Between E2 and E1, there are more electrons in E2 level and it also has higher energy. Hence, population inversion exists between E2 and E1.

30. Optical pumping happens via ___________
a) Spontaneous emission
b) Spontaneous Absorption
c) Stimulated emission
d) Stimulated Absorption
Answer: d
Explanation: In optical pumping, a highly intense radiation from an optical flash tube is incident on the active medium, raising the atoms to a higher energy state through stimulated absorption.

31. Pumping is done in order to achieve __________
a) Steady state
b) Population inversion
c) Equilibrium
d) Photon emission
Answer: b
Explanation: Pumping is the process of raising the atoms or molecules of the active medium to a higher energy state so that the stage of population inversion can be achieved.

32. Which pumping method is used in He-Ne laser?
a) Optical Pumping
b) Electrical Excitation
c) Chemical Pumping
d) Direct Conversion
Answer: b
Explanation: Generally, in He-Ne laser, an electric discharge is used to excite the atoms of the active medium. This process is known as Electrical Excitation and is normally used in gas lasers.

33. Direct conversion of electrical energy into radiation takes place in ___________
a) Ruby laser
b) ND: YAG
c) LED
d) He-Ne laser
Answer: c
Explanation: In Light Emitting Diodes (LED) and semiconductor lasers, direct conversion of electrical energy into radiation takes place. This process of pumping is called as Direct conversion.

34. At least 2 levels are required in a crystal to achieve population inversion.
a) True
b) False
Answer: b
Explanation: Population inversion takes place in crystals with a minimum three-level system. Two level system is not suitable for achieving population inversion because in that case the population of both the levels would remain same as B12 = B21.

35. Which of the following is a four-level laser?
a) ND: YAG
b) Ruby
c) He-Ne
d) Semiconductor laser
Answer: c
Explanation: He-Ne laser is a four-level laser. Whenever an electric discharge is passed through the gas, the helium atoms gets to higher energy states, as the concentration of helium atoms is higher to achieve population inversion.

36. The meta-stable state have a larger lifetime then lower energy state.
a) True
b) False
Answer: a
Explanation: The meta-stable states have an unusually long lifetime. It is due to this that the number of atoms if higher in meta-stable state then in lower energy state, which causes population inversion.

37. After pumping, Which level will the atom jump to?
3801
a) Exited State
b) Meta-stable state
c) Higher than exited state
d) Remain in ground state
Answer: a
Explanation: After pumping, the atom will get excited to the exited state. After that, as the lifetime of the exited state is lower, it will drop to the meta-stable state.

38. The output of a laser has pulse duration of 30 ms and average output power of 1 W per pulse. How much energy is released per pulse if wavelength is 6600 Å?
a) 0.001 J
b) 0.002 J
c) 0.003 J
d) 0.004 J
Answer: c
Explanation: As we know, Energy = Power X Time
= 1 W X 30 X 10-3s
= 0.003 J.

39. What is the region enclosed by the optical cavity called?
a) Optical Region
b) Optical System
c) Optical box
d) Optical Resonator
Answer: d
Explanation: The optical cavity resembles an oscillator as it provides feedback of the photons by reflection, at the mirrors. Therefore, the area enclosed inside the optical cavity is called optical resonator.

40. What is the relationship between B21 and B12?
a) B12 > B21
b) B12 < B21
c) B12 = B21
d) No specific relation
Answer: c
Explanation: B21 is the coefficient for the stimulated emission while B12 is the coefficient for stimulated absorption. Both the processes are mutually reverse processes and their probabilities are equal. Therefore, B12 = B21.

41. Which of the following Einstein’s coefficient represents spontaneous emission?
a) A12
b) A21
c) B12
d) B21
Answer: b
Explanation: A21 represents the spontaneous emission of photons. A12 signifies spontaneous absorption.B12 is for stimulated absorption while B21 is for stimulated emission.

42. The correct expression for the rate of stimulated emission is _______________
a) Rse = A21N2
b) Rse = A21uN2
c) Rse = B21N2
d) Rse = B21uN2
Answer: d
Explanation: The stimulates emission is directly proportional to the energy density u, of the external radiation field. Also, stimulated emission rate increases with the increase in number N2 of exited atoms.

43. Which Einstein’s coefficient should be used in this case?
43

a) A12
b) A21
c) B12
d) B21

Answer: d
Explanation: The given figure shows stimulated emission. Hence, the Einstein coefficient for stimulated emission is B21. If it had been spontaneous emission, then A21 would have been used.
44. Which law is used for achieving the relation between the Einstein’s coefficients?
a) Heisenberg’s Uncertainty Principle
b) Planck’s radiation law
c) Einstein’s equation
d) Quantum law

48

45. Which of the following is a four-level laser?
a) ND: YAG
b) Ruby
c) He-Ne
d) Argon laser
Answer: c
Explanation: He-Ne laser is a four-level laser. Whenever an electric discharge is passed through the gas, the helium atoms gets to higher energy states, as the concentration of helium atoms is higher. It was one of the first successfully operated laser.

46. The difference between He-Ne Laser is __________
a) It gives pulsed output
b) It gives a non-continuous laser beam
c) It gives a continuous laser beam
d) No difference
Answer: c
Explanation: As we know, the ruby laser beam is discontinuous and pulsed. However, the He-Ne laser beam is a continuous laser beam. Also, it is a four-level laser while ruby laser is a three-level laser.

47. He-Ne laser is a type of ____________
a) Solid laser
b) Liquid laser
c) Gas laser
d) Diode laser
Answer: c
Explanation: He-Ne laser is the most widely used laser. It was the first continuous laser generated. Gas lasers are less prone to damage by overheating, as compared to solid laser.

48. Which pumping method is used in He-Ne laser?
a) Optical Pumping
b) Electrical Excitation
c) Chemical Pumping
d) Direct Conversion
Answer: b
Explanation: Generally, in He-Ne laser, an electric discharge is used to excite the atoms of the active medium. This process is known as Electrical Excitation and is normally used in gas lasers.

49. The He-Ne laser operates at a wavelength of ____________
a) 540 nm
b) 632 nm
c) 690 nm
d) 717 nm
Answer: b
Explanation: The helium-neon laser operates at a wavelength of 632.8 nanometers (nm), in the red portion of the visible spectrum. It is the most widely used gas laser.

50. The output of a He-Ne laser has pulse duration of 15 ms and average output power of 1 W per pulse. How much energy is released per pulse?
a) 5 mJ
b) 10 mJ
c) 15 mJ
d) 20 mJ
Answer: c
Explanation: As we know, Energy = Power X Time
= 1 W X 15 X 10-3 s
= 15 mJ.

51. When the transition takes place from En6 -> En5, what is the wavelength of produced beam?
a) 6328 Å
b) 33913 Å
c) 11523 Å
d) 7550 Å
Answer: b
Explanation: When the lasing transition is from En6 -> En5, it produces a beam of 33913 Å. The opper level is same in this case and in the 6328 Å transition.

52. He-Ne laser is used in Holography.
a) True
b) False
Answer: a
Explanation: He-Ne laser is highly coherent and monochromatic. Due to this, it is used in holography, spectrometers, prints, scanners, etc. It is widely used in Laboratories.

53. The number of photons emitted for a 2.5 mW He-Ne laser is __________
a) 4.9 X 1015
b) 5.9 X 1015
c) 6.9 X 1015
d) 7.9 X 1015
Answer: b
Explanation: We know, for a He-Ne laser, λ = 6328 Å = 6.328 X 10-7 m
Power = 2.5 mW = 2.5 X 10-3 E
Energy = Power X Time = 0.0025 X 1
= 0.0025
Number of photons in each pulse = Energy X λ / hc
= 0.0025 X 6.6 X 10-7/6.6 X 10-34 X 3 X 108
= 7.9 X 1015.

54. What is the wavelength of the emitted laser by a carbon dioxide?
a) 9.4 μm
b) 10.6 μm
c) 11.4 μm
d) 12.5 μm
Answer: b
Explanation: The radiations emitted in a CO2 laser has a wavelength of 10.6 μm. It is a four-level laser. The transition takes place between the different vibrational states of the molecule.

55. Which of the following is a characteristic of semiconductor lasers?
a) Output in Visible region
b) High Efficiency
c) Output in UV region
d) Pulsed output
Answer: b
Explanation: Semiconductor lasers such as GaAs, InP, InSb etc. are used extensively because of their high efficiency. Also, they can be employed in optical communications with ease.

56. In the CO2 molecular gas laser, transition takes place between the ______________
a) Molecular states
b) Atomic states
c) Vibrational states
d) Energy states
Answer: c
Explanation: In the CO2 molecular gas laser, transition takes place between the vibrational states of Carbon dioxide molecules. It is a very efficient laser.

57. Which of the following is a four-level laser?
a) ND: YAG
b) Ruby
c) Argon laser
d) CO2 laser
Answer: c
Explanation: The CO2 laser is a very efficient laser. It is a four-level laser and it operates at 10.6 μm in the far IR region. The active medium is a gas mixture of CO2, N2 and He.

58. Which of the following gas is not a part of the active medium in a CO2 laser?
a) CO2
b) N2
c) He
d) O2
Answer: d
Explanation: The active medium of a CO2 laser consists of a mixture of CO2, N2 and He. It is the vibrational transition in the CO2 that results in the lasing action.

59. The highest-powered CO2 laser had a power of _________
a) 1 W
b) 10 W
c) 1000 W
d) 10000 W
Answer: c
Explanation: Light from a CO2 laser is powerful enough to cut many materials, including cloth, wood and paper; the most powerful CO2 lasers are used for machining steel and other metals. The highest-powered CO2 lasers run over 1,000 W.

60. The active medium of a semiconductor diode is the junction of the forward biased P-N diode.
a) True
b) False
Answer: a
Explanation: When the p-side of a semiconductor diode is connected to the positive terminal of the battery and the n-side to the negative terminal, the diode is said to be forward biased. A semiconductor laser is essentially a semiconductor diode, where the active medium is the forward biased p-n junction.

61. Semiconductors lasers do not need mirrors to form optical cavity.
a) True
b) False
Answer: a
Explanation: Semiconductor lasers do not require two external mirrors to form an optical cavity. Reflection from the cleaved ends of the semiconductor is enough to produce lasing.

62. The following figure shows the laser spectrum for ________
9
a) CO2 Laser
b) Fabry-Pot Semiconductor Laser
c) DFB semiconductor laser
d) FBG semiconductor laser
Answer: b
Explanation: Fabry-Pot semiconductor laser is a type of semiconductor laser which is characterized by the use of the cleavage plane of a laser crystal for reflection of the light emitted in the active layer. It’s lasing spectrum observed is as shown in the figure.

63. Where is ND: YAG most commonly used?
a) Cosmetic Surgery
b) Welding
c) Photography
d) Optical Communications
Answer: a
Explanation: ND: YAG is most commonly used for cosmetic energy because it has the property of maximum energy absorption by the target (hair or lesion) with minimum absorption by the surrounding skin structures.

64. In which region of the electromagnetic spectrum, does the semiconductor laser lies?
a) Visible Region
b) UV Region
c) Microwave Region
d) Infrared Region
Answer: d
Explanation: In the gallium arsenide laser, the wavelength of the emitted laser is around 845 nm to 905 nm, which lies in the infrared region of the electromagnetic spectrum.

65. The following graph shows the physical gain for which kind of laser?
12
a) CO2 Laser
b) Semiconductor Laser
c) Helium Laser
d) Ruby Laser
Answer: b
Explanation: The given figure shows the physical gain for a semiconductor laser. The upper band is the conduction band while the lower one is the valence band. The lasing action is taking place between the two bands.

66. Lasers are used for welding of wires because they can be focused onto a fine spot.
a) True
b) False
Answer: a
Explanation: Laser beams are highly directional with almost no convergence. Thus, they can be focused onto a fine spot with ease. Due to this, they are used in welding of fine wires, contacts in miniature assemblies, drilling holes etc.

67. Where is ND: YAG most commonly used?
a) Cosmetic Surgery
b) Welding
c) Photography
d) Optical Communications
Answer: a
Explanation: ND: YAG is most commonly used for cosmetic energy because it has the property of maximum energy absorption by the target (hair or lesion) with minimum absorption by the surrounding skin structures.

68. The information carrying capacity of laser is enormous due its large _________
a) Coherence
b) Bandwidth
c) Directionality
d) Intensity
Answer: b
Explanation: Laser has a large bandwidth. The rate at which the information can be transmitted is proportional to bandwidth and the bandwidth is proportional to carrier frequency. Because of these properties, Laser is widely used as optical carrier signal.

69. Which characteristic of LASER allows it to be used in holography?
a) Coherency
b) Directionality
c) Intensity
d) Monochromaticity
Answer: a
Explanation: The production of an image in a hologram takes place via a process called reconstruction. In this process, the image is “reconstructed” in the form of a hologram. This reconstruction if possible, via LASER as they are highly coherent.

70. What is the region enclosed by the optical cavity called?
a) Optical Region
b) Optical System
c) Optical box
d) Optical Resonator
Answer: d
Explanation: The optical cavity resembles an oscillator as it provides feedback of the photons by reflection, at the mirrors. Therefore, the area enclosed inside the optical cavity is called optical resonator.

71. What is the full form of LASER?
a) Light Absorbent and Stimulated Emission of Radiations
b) Light Absorbing Solar Energy Resource
c) Light Amplification by Stimulated Emission of Radiations
d) Light Amplification of Singular Emission of Radiations
Answer: c
Explanation: LASER is a short form of Light Amplification by Stimulated Emission of Radiations. Stimulated Emission is the process by which amplification of radiations takes place. Hence the meaning of LASER that the light is amplified by stimulating the emission of radiations.

72. In Stimulated Absorption, what is the lifetime of atoms ground state?
a) 1 second
b) 1 minute
c) 1 hour
d) Infinity
Answer: d
Explanation: At the ground state, the atoms are perfectly stable. They are under no excessive force that might lead to become unstable. All the forces are balanced. Thus, as the atom is stable in ground state, its lifetime is infinity.

73. Phonons are __________
a) Quanta of energy
b) Quanta of light waves
c) Quanta of sound waves
d) Quanta of heat
Answer: c
Explanation: Phonons are the quanta of sound waves. When energy is provided, the lattice absorbs energy and gets excited to a higher state. When it de-excites to ground state, it releases radiation in sound-wave region, known as phonons.

74. Which of the following is not a characteristic of LASERS?
a) Monochromatic
b) Coherent
c) Divergent
d) Intense
Answer: c
Explanation: The lasers are highly directional having almost no divergence. The output beam of laser has a well-defined wave front due to which it can be focused on a point.
Lasers are highly intense compared to ordinary light. They are monochromatic and coherent.

Module 3

1. Coulomb is the unit of which quantity?
a) Field strength
b) Charge
c) Permittivity
d) Force
Answer: b
Explanation: The standard unit of charge is Coulomb. One coulomb is defined as the 1 Newton of force applied on 1 unit of electric field.

2. Coulomb law is employed in
a) Electrostatics
b) Magnetostatics
c) Electromagnetics
d) Maxwell theory
Answer: a
Explanation: Coulomb law is applied to static charges. It states that force between any two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. Thus it is employed in electrostatics.

3. Find the force between 2C and -1C separated by a distance 1m in air(in newton).
a) 18 X 106
b) -18 X 106
c) 18 X 10-6
d) -18 X 10-6
Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109.

4. Two charges 1C and -4C exists in air. What is the direction of force?
a) Away from 1C
b) Away from -4C
c) From 1C to -4C
d) From -4C to 1C
Answer: c
Explanation: Since the charges are unlike, the force will be attractive. Thus the force directs from 1C to -4C.

5. Find the force of interaction between 60 stat coulomb and 37.5 stat coulomb spaced 7.5cm apart in transformer oil(εr=2.2) in 10-4 N,
a) 8.15
b) 5.18
c) 1.518
d) 1.815
Answer: d
Explanation: 1 stat coulomb = 1/(3 X 109) C
F = (1.998 X 1.2488 X 10-16)/(4∏ X 8.854 X 10-12 X 2.2 X (7.5 X 10-2)2) = 1.815 X 10-4 N.

6. Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN.
a) 1.44
b) 2.44
c) 1.404
d) 2.404
Answer: c
Explanation: Before the charges are brought into contact, F = 11.234 μN.
After charges are brought into contact and then separated, charge on each sphere is, (q1 + q2)/2 = 0.5nC
On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN.

7. The Coulomb law is an implication of which law?
a) Ampere law
b) Gauss law
c) Biot Savart law
d) Lenz law
Answer: b
Explanation: The Coulomb law can be formulated from the Gauss law, using the divergence theorem. Thus it is an implication of Gauss law.

8. Two small diameter 10gm dielectric balls can slide freely on a vertical channel. Each carry a negative charge of 1μC. Find the separation between the balls if the lower ball is restrained from moving.
a) 0.5
b) 0.4
c) 0.3
d) 0.2
Answer: c
Explanation: F = mg = 10 X 10-3 X 9.81 = 9.81 X 10-2 N.
On calculating r by substituting charges, we get r = 0.3m.

9. A charge of 2 X 10-7 C is acted upon by a force of 0.1N. Determine the distance to the other charge of 4.5 X 10-7 C, both the charges are in vacuum.
a) 0.03
b) 0.05
c) 0.07
d) 0.09
Answer: d
Explanation: F = q1q2/(4∏εor2) , substituting q1, q2 and F, r2 = q1q2/(4∏εoF) =
We get r = 0.09m.

10. For a charge Q1, the effect of charge Q2 on Q1 will be,
a) F1 = F2
b) F1 = -F2
c) F1 = F2 = 0
d) F1 and F2 are not equal
Answer: b
Explanation: The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.

11. Divergence theorem is based on
a) Gauss law
b) Stoke’s law
c) Ampere law
d) Lenz law
Answer: a
Explanation: The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.

12. The Gaussian surface for a line charge will be
a) Sphere
b) Cylinder
c) Cube
d) Cuboid
Answer: b
Explanation: A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.

13. The Gaussian surface for a point charge will be
a) Cube
b) Cylinder
c) Sphere
d) Cuboid
Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.

14. A circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?
a) 3
b) 2
c) 1
d) 0
Answer: d
Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.

15. The total charge of a surface with densities 1,2,…,10 is
a) 11
b) 33
c) 55
d) 77
Answer: c
Explanation: Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of 1,2,…,10,i.e, 1+2+…+10 = 10(11)/2 = 55.

16. The work done by a charge of 10μC with a potential 4.386 is (in μJ)
a) 32.86
b) 43.86
c) 54.68
d) 65.68
Answer: b
Explanation: By Gauss law principles, W = Q X V = 10 X 10-6 X 4.386 = 43.86 X 10-6 joule.

17. The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)
a) 12.74
b) 13.47
c) 12.47
d) 13.74
Answer: c
Explanation: The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b = 2m and a = 1m. We get V = 12.47 X 109 volts.

18. Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.
a) 18π
b) 24π
c) 36π
d) 72π
Answer: d
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.

19. Gauss law cannot be used to find which of the following quantity?
a) Electric field intensity
b) Electric flux density
c) Charge
d) Permittivity
Answer: d
Explanation: Permittivity is constant for a particular material(say permittivity of water is 1). It cannot be determined from Gauss law, whereas the remaining options can be computed from Gauss law.

20. Gauss law for magnetic fields is given by
a) Div(E) = 0
b) Div(B) = 0
c) Div(H) = 0
d) Div(D) = 0
Answer: b
Explanation: The divergence of magnetic flux density is always zero. This is called Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in any magnetic field.

21. Choose the best definition of a dipole.
a) A pair of equal and like charges located at the origin
b) A pair of unequal and like charges located at the origin
c) A pair of equal and unlike charges separated by a small distance
d) A pair of unequal and unlike charges separated by a small distance
Answer: c
Explanation: An electric dipole generally refers to two equal and unlike (opposite signs) charges separated by a small distance. It can be anywhere, not necessarily at origin.

22. The potential due to a dipole at a point P from it is the
a) Sum of potentials at the charges
b) Difference of potentials at the charges
c) Multiplication of potentials at the charges
d) Ratio of potentials at the charges
Answer: b
Explanation: The total potential at the point P due to the dipole is given by the difference of the potentials of the individual charges.
V = V1 + (-V2), since both the charges are unlike. Thus V = V1 – V2.

23. Calculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.
a) 0.02
b) 0.04
c) 0.06
d) 0.08
Answer: b
Explanation: The dipole moment of charge 2C and distance 2cm will be,
M = Q x d. Thus, M = 2 x 0.02 = 0.04 C-m.

24. Find the angle at which the potential due a dipole is measured, when the distance from one charge is 12cm and that due to other is 11cm, separated to each other by a distance of 2cm.
a) 15
b) 30
c) 45
d) 60
Answer: d
Explanation: Here, the two charges are separated by d = 2cm.
The distance from one charge (say Q1) will be R1 = 11cm. The distance from another charge (say Q2) will be R2 = 12cm. If R1 and R2 is assumed to be parallel, then R2 – R1 = d cos θ. We get 1 = 2cos θ and cos θ = 0.5. Then θ =
cos-1(0.5) = 60.

25. Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.
a) 0
b) Unity
c) ∞
d) -∞
Answer: a
Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr2). When the angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will become zero.

26. For two charges 3C and -3C separated by 1cm and are located at distances 5cm and 7cm respectively from the point P, then the distance between their midpoint and the point P will be
a) 5.91
b) 12.6
c) 2
d) 9
Answer: a
Explanation: For a distant point P, the R1 and R2 will approximately be equal.
R1 = R2 = r, where r is the distance between P and the midpoint of the two charges. Thus they are in geometric progression, R1R2=r2
Now, r2 = 5 x 7 = 35. We get r = 5.91cm.

27. Calculate the distance between two charges of 4C forming a dipole, with a dipole moment of 6 units.
a) 1
b) 1.5
c) 2
d) 2.5
Answer: b
Explanation: The dipole moment is given by, M = Q x d. To get d, we rearrange the formula d = M/Q = 6/4 = 1.5units.

28. The potential due to the dipole on the midpoint of the two charges will be
a) 0
b) Unity
c) ∞
d) -∞
Answer: c
Explanation: The potential due a dipole at a point P will be V = m cos θ/(4πεr2).
Now it is given that potential on the midpoint, which means P is on midpoint, then the distance from midpoint and P will be zero. When r = 0 is put in the above equation, we get V = ∞. This shows that the potential of a dipole at its midpoint will be maximum/infinity.

29. Dipoles in any electric field undergo
a) Magnetism
b) Electromagnetism
c) Magnetisation
d) Polarisation
Answer: d
Explanation: Dipoles in any pure electric field will undergo polarisation. It is the process of alignment of dipole moments in accordance with the electric field applied.

30. Dipole moments are used to calculate the
a) Electric field intensity
b) Polarisation patterns
c) Strength of the dipole in the field
d) Susceptibility
Answer: b
Explanation: Dipole moment implicates the strength of the dipole in the electric field. They are then used to compute the polarisation patterns based on the applied field. Once the polarisation is determined we can find its susceptibility. Though all options seem to be correct, the apt answer is to calculate polarisation, provided applied field is known.

31. The first Maxwell law is based on which law?
a) Ampere law
b) Faraday law
c) Lenz law
d) Faraday and Lenz law
Answer: d
Explanation: The first Maxwell equation states that Curl(E) = -dB/dt. It is based on the emf concept. Thus it is derived from the Faraday and Lenz law.

32. The benefit of Maxwell equation is that
a) Any parameter can be calculated
b) Antenna can be designed
c) Polarisation of the wave can be calculated
d) Transmission line constants can be found
Answer: a
Explanation: The Maxwell equation relates the parameters E, D, H, B. When one parameter is known the other parameters can be easily calculated. In other words, it is used to relate an electric field parameter with its equivalent magnetic field.

33. The correct sequence to find H, when D is given is
a) D-E-B-H
b) D-B-E-H
c) It cannot be computed from the data given
d) D-H
Answer: a
Explanation: There is no direct relation between D and H, so the option D-H is not possible. Using the formula D = εE, the parameter E can be computed from D. By Maxwell equation, Curl(E) = -dB/dt, the parameter B can be calculated. Using the formula B = μH, the parameter H can be calculated. Thus the sequence is D-E-B-H.

34. The curl of the electric field intensity is
a) Conservative
b) Rotational
c) Divergent
d) Static
Answer: b
Explanation: The curl of electric field intensity is Curl(E). From Maxwell law, the curl of E is a non-zero value. Thus E will be rotational.

35. Which of the following identities is always zero for static fields?
a) Grad(Curl V)
b) Curl(Div V)
c) Div(Grad V)
d) Curl(Grad V)
Answer: d
Explanation: The curl of gradient of a vector is always zero. This is because the gradient of V is E and the curl of E is zero for static fields.

36. Find the Maxwell first law value for the electric field intensity is given by A sin wt az
a) 0
b) 1
c) -1
d) A
Answer: a
Explanation: The value of Maxwell first equation is Curl(E). The curl of E is zero. Thus for the given field, the value of Maxwell equation is zero. Thus the field is irrotational.

37. Find the electric field applied on a system with electrons having a velocity 5m/s subjected to a magnetic flux of 3.6 units.
a) 15
b) 18
c) 1.38
d) 0.72
Answer: b
Explanation: The electric field intensity is the product of the velocity and the magnetic flux density. Thus E = v x B, on substituting v = 5 and B = 3.6, we get E = 5 x 3.6 = 18 units.

38. Which of the following relations holds good?
a) Bq = ILE
b) E = ILBq
c) Eq = ILB
d) B = ILEq
Answer: c
Explanation: The force of a electrostatic field in given by F = Eq. The force on a conductor is given by F = BIL. In the case when a charge exists on a conductor, both the forces can be equated. Thus Eq = BIL is true.

39. When the Maxwell equation is expressed in frequency domain, then which substitution is possible?
a) d/dt = w/j
b) d/dt = j/w
c) d/dt = jw
d) Expression in frequency domain is not possible
Answer: c
Explanation: The conversion of time to frequency domain in Maxwell equation is given by the Fourier Transform. Differentiation in time gives jw in frequency domain. Thus d/dt = jw in frequency domain.

40. Calculate the emf of a material having a flux linkage of 2t2 at time t = 1second.
a) 2
b) 4
c) 8
d) 16
Answer: b
Explanation: The emf of a material is given by Vemf = -dλ/dt. On substituting λ = 2t2, the emf is 4t. At t = 1 sec, the emf will be 4 units.

41. Calculate the emf of a material having flux density 5sin t in an area of 0.5 units.
a) 2.5 sin t
b) -2.5 cos t
c) -5 sin t
d) 5 cos t
Answer: d
Explanation: The emf can be written as Vemf = -d(∫B.ds)/dt. It can be written as Vemf = -B= -5sin t, since the integration and differentiation gets cancelled.

42. To find D from B, sequence followed will be
a) B-E-D
b) B-H-D
c) E-H-D
d) E-B-D
Answer: a
Explanation: Using Maxwell equation, from B we can calculate E by Curl(E) = -dB /dt. From E, D can be calculated by D = εE. Thus the sequence is B->E->D.

43. Maxwell second equation is based on which law?
a) Ampere law
b) Faraday law
c) Lenz law
d) Coulomb law
Answer: a
Explanation: The second Maxwell equation is based on Ampere law. It states that the field intensity of a system is same as the current enclosed by it, i.e, Curl(H) = J.

44. The Maxwell second equation that is valid in any conductor is
a) Curl(H) = Jc
b) Curl(E) = Jc
c) Curl(E) = Jd
d) Curl(H) = Jd
Answer: a
Explanation: For conductors, the conductivity parameter σ is significant and only the conduction current density exists. Thus the component J = Jc and Curl(H) = Jc.

45. In dielectric medium, the Maxwell second equation becomes
a) Curl(H) = Jd
b) Curl(H) = Jc
c) Curl(E) = Jd
d) Curl(E) = Jd
Answer: a
Explanation: In dielectric medium conductivity σ will be zero. So the current density has only the displacement current density. Thus the Maxwell equation will be Curl(H) = Jd.

46. Find the displacement current density of a material with flux density of 5sin t
a) 2.5cos t
b) 2.5sin t
c) 5cos t
d) 5sin t
Answer: c
Explanation: The displacement current density is the derivative of the flux density. Thus Jd = dD/dt. Put D = 5sin t in the equation, we get Jd = 5cos t units.

47. Find the conduction current density of a material with conductivity 200units and electric field 1.5 units.
a) 150
b) 30
c) 400/3
d) 300
Answer: d
Explanation: The conduction current density is given by Jc = σE, where σ = 200 and E = 1.5. Thus we get, Jc = 200 x 1.5 = 300 units.

48. Calculate the conduction density of a material with resistivity of 0.02 units and electric intensity of 12 units.
a) 300
b) 600
c) 50
d) 120
Answer: b
Explanation: The conduction density is given by Jc = σE, where σ is the inverse of resistivity and it is 1/0.02 = 50. Thus we get, Jc = 50 x 12 = 600 units.

49. In the conversion of line integral of H into surface integral, which theorem is used?
a) Green theorem
b) Gauss theorem
c) Stokes theorem
d) It cannot be converted
Answer: c
Explanation: To convert line integral to surface integral, i.e, in this case from line integral of H to surface integral of J, we use the Stokes theorem. Thus the Maxwell second equation can be written as ∫H.dl = ∫∫J.ds.

50. An implication of the continuity equation of conductors is given by
a) J = σ E
b) J = E/σ
c) J = σ/E
d) J = jwEσ
Answer: a
Explanation: The continuity equation indicates the current density in conductors. This is the product of the conductivity of the conductor and the electric field subjected to it. Thus J = σE is the implication of the continuity equation for conductors.

51. Find the equation of displacement current density in frequency domain.
a) Jd = jwεE
b) Jd = jwεH
c) Jd = wεE/j
d) Jd = jεE/w
Answer: a
Explanation: The displacement current density is Jd = dD/dt. Since D = εE and in frequency domain d/dt = jw, thus we get Jd = jwεE.

52. The total current density is given as 0.5i + j – 1.5k units. Find the curl of the magnetic field intensity.
a) 0.5i – 0.5j + 0.5k
b) 0.5i + j -1.5k
c) i – j + k
d) i + j – k
Answer: b
Explanation: By Maxwell second equation, the curl of H is same as the sum of conduction current density and displacement current density. Thus Curl(H) = J = 0.5i + j – 1.5k units.

53. At dc field, the displacement current density will be
a) 0
b) 1
c) Jc
d) ∞
Answer: a
Explanation: The DC field refers to zero frequency. The conduction current is independent of the frequency, whereas the displacement current density is dependent on the frequency, i.e, Jd = jwεE. Thus at DC field, the displacement current density will be zero.

54. Both the conduction and displacement current densities coexist in which medium?
a) Only conductors in air
b) Only dielectrics in air
c) Conductors placed in any dielectric medium
d) Both the densities can never coexist
Answer: c
Explanation: Conduction density exists only for good conductors and displacement density is for dielectrics in any medium at high frequency. Thus both coexist when a conductor is placed in a dielectric medium.

55. Find the curl of E when B is given as 15t.
a) 15
b) -15
c) 7.5
d) -7.5
Answer: b
Explanation: From Maxwell first law, we get Curl of E as the negative derivative of B with respect to time. Thus Curl(E) = -dB/dt. On substituting B= 15t and differentiating, Curl(E) = -15 units.

56. The charge build up in a capacitor is due to
a) Conduction current density
b) Displacement current density
c) Polarisation
d) Magnetization
Answer: b
Explanation: The capacitor consists of a dielectric placed between two conducting plates, subjected to a field. The current due to a dielectric is always due to the displacement current density.

57. The surface integral of which parameter is zero?
a) E
b) D
c) B
d) H
Answer: c
Explanation: The divergence of the magnetic flux density is always zero. By Stokes theorem, the surface integral of B is same as the volume integral of the divergence of B. Thus the surface integral of B is also zero.

58. Harmonic electromagnetic fields refer to fields varying sinusoidally with respect to time. State True/False.
a) True
b) False
Answer: a
Explanation: Fields that varying sinusoidally with respect to time are called as harmonic fields. An example for harmonic fields is A sin wt.

59. When electric potential is null, then the electric field intensity will be
a) 0
b) 1
c) dA/dt
d) –dA/dt
Answer: d
Explanation: The electric field intensity is given by E = -Grad(V)- dA/dt, where V is the electric potential and A is the magnetic vector potential. When V is zero, then E = -dA/dt.

60. The gradient of the magnetic vector potential can be expressed as
a) –με dV/dt
b) +με dE/dt
c) –με dA/dt
d) +με dB/dt
Answer: a
Explanation: The gradient of A is the ratio of the negative gradient of electric potential to the speed of light c. We can write c = 1/√(με). Thus grad(A) = -με dV/dt is the required expression.

61. Find the time constant of a capacitor with capacitance of 2 microfarad having an internal resistance of 4 megaohm.
a) 2
b) 0.5
c) 8
d) 0.25
Answer: c
Explanation: The time constant of capacitor is given by T = RC, where R = 4×106 and C = 2×10-6. Thus T = 4×106 x2x10-6 = 8 seconds.

62. Which components exist in an electromagnetic wave?
a) Only E
b) Only H
c) Both E and H
d) Neither E or H
Answer: c
Explanation: In an electromagnetic wave, the electric and magnetic components coexist. They propagate perpendicular to each other and to the direction of propagation in space.

63. The propagation of the electromagnetic waves can be illustrated by
a) Faraday law
b) Ampere law
c) Flemming rule
d) Coulomb law
Answer: c
Explanation: By Flemming’s rule, when the thumb and the middle finger represent the inputs (say current and field respectively), then the fore finger represents the output (force, in this case). The EM propagation can be illustrated by this rule.

64. Which one of the following laws will not contribute to the Maxwell’s equations?
a) Gauss law
b) Faraday law
c) Ampere law
d) Curie Weiss law
Answer: d
Explanation: The Gauss law, Faraday law and the Ampere law are directly used to find the parameters E, H, D, B. Thus it contributes to the Maxwell equations. The Curie Weiss law pertains to the property of any magnetic material. Thus it is not related to the Maxwell equation.

65. For time varying currents, the field or waves will be
a) Electrostatic
b) Magneto static
c) Electromagnetic
d) Electrical
Answer: c
Explanation: For stationary charges, the field is electrostatic. For steady currents, the field is magneto static. But for time varying currents, the field or waves will be electromagnetic.

66. According to Faraday’s law, EMF stands for
a) Electromagnetic field
b) Electromagnetic force
c) Electromagnetic friction
d) Electromotive force
Answer: d
Explanation: The force in any closed circuit due to the change in the flux linkage of the circuit is called as electromotive force EMF. This phenomenon is called as Faraday’s law.

67. Calculate the emf when the flux is given by 3sin t + 5cos t
a) 3cos t – 5sin t
b) -3cos t + 5sin t
c) -3sin t – 5cos t
d) 3cos t + 5sin t
Answer: b
Explanation: The electromotive force is given by Vemf = -dλ/dt. Thus Vemf = -dλ/dt = -(3cos t – 5sin t) = -3cos t + 5sin t.

68. The induced voltage will oppose the flux producing it. State True/False.
a) True
b) False
Answer: a
Explanation: According to Lenz law, the induced voltage acts in such a way that it opposes the flux producing it. This is indicated by a negative sign.

69. Calculate the emf when a coil of 100 turns is subjected to a flux rate of 0.3 tesla/sec.
a) 3
b) 30
c) -30
d) -300
Answer: c
Explanation: The induced emf is given by Vemf = -dλ/dt = -Ndψ/dt. Thus emf will be -100 x 0.3 = -30 units.

70. Find the force due to a current element of length 2cm and flux density of 12 tesla. The current through the element will be 5A.
a) 1 N
b) 1.2 N
c) 1.4 N
d) 1.6 N
Answer: b
Explanation: The force due to a current element is given by F = BI x L. Thus F = 12 x 5 x 0.02 = 1.2 units.

71. Which of the following statements is true?
a) E is the cross product of v and B
b) B is the cross product of v and E
c) E is the dot product of v and B
d) B is the dot product of v and E
Answer: a
Explanation: The electric field is the cross product of the velocity and the magnetic field intensity. This is given by Lorentz equation.

72. The time varying electric field E is conservative. State True/False.
a) True
b) False
Answer: b
Explanation: The time varying electric field E(t) is not a closed path. Thus the curl will be non-zero. This implies E(t) is not conservative and the statement is false.

73. When the conduction current density and displacement current density are same, the dissipation factor will be
a) Zero
b) Minimum
c) Maximum
d) Unity
Answer: d
Explanation: Dissipation factor refers to the tangent of loss angle. It is the ratio of conduction current density to displacement current density. When both are same, the loss tangent or the dissipation factor will be unity.

74. The point form of Ampere law is given by
a) Curl(B) = I
b) Curl(D) = J
c) Curl(V) = I
d) Curl(H) = J
Answer: d
Explanation: Ampere law states that the line integral of H about any closed path is exactly equal to the direct current enclosed by that path. ∫ H.dl = I The point form will be Curl (H) = J.

75. The Ampere law is based on which theorem?
a) Green’s theorem
b) Gauss divergence theorem
c) Stoke’s theorem
d) Maxwell theorem
Answer: c
Explanation: The proof of the Ampere’s circuital law is obtained from Stoke’s theorem for H and J only.

76. Electric field will be maximum outside the conductor and magnetic field will be maximum inside the conductor. State True/False.
a) True
b) False
Answer: a
Explanation: At the conductor-free space boundary, electric field will be maximum and magnetic field will be minimum. This implies electric field is zero inside the conductor and increases as the radius increases and the magnetic field is zero outside the conductor and decreases as it approaches the conductor.

77. Find the magnetic flux density of a finite length conductor of radius 12cm and current 3A in air( in 10-6 order)
a) 4
b) 5
c) 6
d) 7
Answer: b
Explanation: The magnetic field intensity is given by H = I/2πr, where I = 3A and r = 0.12. The magnetic flux density in air B = μ H, where μ = 4π x 10-7.Thus B = 4π x 10-7 x 3/2π x 0.12 = 5x 10-6 units.

78. Calculate the magnetic field intensity due to a toroid of turns 50, current 2A and radius 159mm.
a) 50
b) 75
c) 100
d) 200
Answer: c
Explanation: The magnetic field intensity is given by H = NI/2πrm, where N = 50, I = 2A and rm = 1/2π. Thus H = 50 x 2/2π x 0.159 = 100 units.

79. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is above the sheet.
a) -6
b) 12k
c) 60
d) 6
Answer: d
Explanation: The magnetic field intensity when the normal component is above the sheet is Hx = 0.5 K, where K = 12. Thus we get H = 0.5 x 12 = 6 units.

80. Find the magnetic field intensity due to an infinite sheet of current 5A and charge density of 12j units in the positive y direction and the z component is below the sheet.
a) 6
b) 0
c) -6
d) 60k
Answer: c
Explanation: The magnetic intensity when the normal component is below the sheet is Hy = -0.5 K, where K = 12.Thus we get H = -0.5 x 12 = -6 units.

81. Find the current density on the conductor surface when a magnetic field H = 3cos x i + zcos x j A/m, for z>0 and zero, otherwise is applied to a perfectly conducting surface in xy plane.
a) cos x i
b) –cos x i
c) cos x j
d) –cos x j
Answer: b
Explanation: By Ampere law, Curl (H) = J. The curl of H will be i(-cos x) – j(0) + k(-z sin x) = -cos x i – zsin x k. In the xy plane, z = 0. Thus Curl(H) = J = -cos x i.

82. When the rotational path of the magnetic field intensity is zero, then the current in the path will be
a) 1
b) 0
c) ∞
d) 0.5
Answer: b
Explanation: By Ampere law, Curl(H) = J. The rotational path of H is zero, implies the curl of H is zero. This shows the current density J is also zero. The current is the product of the current density and area, which is also zero.

83. Find the magnetic field intensity when the current density is 0.5 units for an area up to 20 units.
a) 10
b) 5
c) 20
d) 40
Answer: a
Explanation: We know that ∫ H.dl = I. By Stoke’s law, we can write Curl(H) = J. In integral form, H = ∫ J.ds, where J = 0.5 and ds is defined by 20 units. Thus H = 0.5 x 20 = 10 units.

Module 4

1. A rod of length 5 m is moving at a speed of 0.6 c. To an observer sitting perpendicular to the direction of motion, the length appears to be _______________
a) 5 m
b) 4 m
c) 3 m
d) 2 m
Answer: a
Explanation: In Lorentz Length transformation, there is no change in the dimensions of the objects in the direction perpendicular to the direction of motion. Thus, to the observer, the length remains the same.

2. If an object reaches the speed of light, it’s length changes to ___________
a) Infinite
b) Double of the value
c) Half of the value
d) Zero
1

3. A 20-year-old person goes at a high speed in a rocket on his birthday. when he comes back to earth after 1 earth year, he would be ___________
a) 1 year older
b) 2 years older
c) A few months older
d) Same age
Answer: c
Explanation: When a person is going around at high speed, time dilation takes place. For that person, the time starts running slowly. Thus, as 1 earth year has passed away but for that person, it must have been only a few months.

4. The length of a rod seems shorter to an observer when it moves in a specific direction. What change would he observe when the direction of rod changes by 180o?
a) The rod becomes even smaller
b) The length of the rod increases
c) The length of the rod remains the same
d) The rod has the length equal to its proper length
3

5. An object of length 1 m is moving at speed 0.5c. To an observer at rest relative to the object, the length of the object seems to be ___________
a) 0.86 m
b) 0.5 m
c) 1 m
d) 0.14 m
Answer: c
Explanation: The observer at rest relative to the object does not notice any kind of contraction of length of the object. It is so because the scale with which he measures will also get contracted by the same amount.

6. How fast does a rocket have to move relative to an observer for its length to be contracted to 95% of its original length?
a) 0.5 c
b) 0.4 c
c) 0.3 c
d) 0.2 c
5

7. A cube with sides of proper length lo is viewed from a reference frame moving with velocity v, parallel to an edge of the cube. The expression for the volume of the cube for the observer is __________
7
7-1

8. A rod of length 1m moves with a speed of 0.5 c. How much length contraction takes place?
a) 50 %
b) 14 %
c) 10 %
d) 35 %
8

9. A particle with a lifetime of 2 X 10-6 s moves through the laboratory with a speed of 0.9 c. It’s lifetime, as measured by an observer in the laboratory, is ___________
a) 2 X 10-6 s
b) 3.2 X 10-6 s
c) 4.6 X 10-6 s
d) 5.4 X 10-6 s
9

10. All the given particles have a lifetime of 1 microsecond. Which of them will survive the longest?
10

a) A
b) B
c) C
d) D
10-1

11. The Cartesian system is also called as
a) Circular coordinate system
b) Rectangular coordinate system
c) Spherical coordinate system
d) Space coordinate system
Answer: b
Explanation: The other name for Cartesian is rectangular system, which is given by (x,y,z).

12. The volume of a parallelepiped in Cartesian is
a) dV = dx dy dz
b) dV = dx dy
c) dV = dy dz
d) dV = dx dz
Answer: a
Explanation: The volume of a parallelepiped is given by product of differential length, breadth and height.

13. A charge is placed in a square container. The position of the charge with respect to the origin can be found by
a) Spherical system
b) Circular system
c) Cartesian system
d) Space coordinate system
Answer: c
Explanation: Since the container possesses dimensions of a square (length, breadth and height), it can be found by Cartesian system.

14. The scalar factor of Cartesian system is unity. State True/False.
a) True
b) False
Answer: a
Explanation: The range of Cartesian system is one to infinity. Thus the minimum scalar value of the system is unity.

15. The angular separation between the vectors A = 4i + 3j + 5k and B = i – 2j + 2k is (in degrees)
a) 65.8
b) 66.8
c) 67.8
d) 68.8
Answer: c
Explanation: The dot product the vector is 8. Angle of separation is cos θ = 8/ (7.07 X 3) = 0.377 and θ = cos-1(0.377) = 67.8.

16. The Cartesian coordinates can be related to cylindrical coordinates and spherical coordinates. State True/False.
a) True
b) False
Answer: a
Explanation: All the coordinate systems are inter-convertible and all the vector operations are applicable to it.

17. Transform the vector A = 3i – 2j – 4k at P(2,3,3) to cylindrical coordinates
a) -3.6j – 4k
b) -3.6j + 4k
c) 3.6j – 4k
d) 3.6j + 4k
Answer: a
Explanation: Convert the Cartesian form to cylindrical form by formula and substitute the points to get -3.6j – 4k.

18. The spherical equivalent of the vector B = yi + (x + z)j located at (-2,6,3) is given by
a) (7,64.62,71.57)
b) (7,-64.62,-71.57)
c) (7,-64.62,71.57)
d) (7,64.62,-71.57)
Answer: d
Explanation: Substitute the points in the vector and convert the Cartesian to cylindrical form to get radius as 7, plane angle1 as 64.62 and plane angle2 as -71.57.

19. Which of the following criteria is used to choose a coordinate system?
a) Distance
b) Intensity
c) Magnitude
d) Geometry
Answer: d
Explanation: The coordinate system is chosen based on the geometry of the given problem. From a point charge +Q, the electric field spreads in all 360 degrees. The calculation of electric field in this case will be spherical system.

20. Vector transformation followed by coordinate point substitution and vice-versa, both given the same result. Choose the best answer.
a) Possible, when the vector is constant
b) Possible, when the vector is variable
c) Possible in all cases
d) Not possible
Answer: a
Explanation: The order of vector transformation and point substitution will not affect the result, only when the vector is a constant.

Module 5

1. Which of the following is an example of top-down approach for the preparation of nanomaterials?
a) Gas phase agglomeration
b) Molecular self-assembly
c) Mechanical grinding
d) Molecular beam epitaxy
Answer: c
Explanation: Mechanical grinding is an example of top-down approach for the preparation of nanomaterials. All the other options are the example of bottom-up approach.

2. Which of the following is an example of bottom-up approach for the preparation of nanomaterials?
a) Etching
b) Dip pen nano-lithography
c) Lithography
d) Erosion
Answer: b
Explanation: Dip pen nanolithography is an example of bottom-down approach for the preparation of nanomaterials. All the other options are the example of top-down approach.

3. The properties like melting point, solubility, color, etc changes on varying the ___________
a) Size
b) Composition
c) Surface properties
d) None of the mentioned
Answer: a
Explanation: The properties like melting point, solubility, color, etc changes on varying the size of the particles. These all are the physical properties and depends upon the physical characteristics of the particle.

4. The properties like dispersibility, conductivity, etc changes on varying the ___________
a) Size
b) Composition
c) Surface properties
d) None of the mentioned
Answer: c
Explanation: The properties like dispersibility, conductivity, etc changes on varying the surface properties of the nanoparticle. These all are the chemical properties and depend upon the surface characteristics of the particle.

5. Quantum confinement results in ___________
a) Energy gap in semiconductor is proportional to the inverse of the square root of the size
b) Energy gap in semiconductor is proportional to the inverse of the size
c) Energy gap in semiconductor is proportional to the square of size
d) Energy gap in semiconductor is proportional to the inverse of the square of size
Answer: d
Explanation: The energy gap in a semiconductor is proportional to the inverse of the square of the size. This effect is a result of quantum confinement.

6. Which of the following is the principal factor which causes the properties of nanomaterials to differ significantly from other materials?
a) Size distribution
b) Specific surface feature
c) Quantum size effects
d) All of the mentioned
Answer: d
Explanation: Size distribution, specific surface feature and quantum size effects are the principal factor which causes the properties of nanomaterials to differ significantly from other materials.

7. Select the incorrect statement from the following options.
a) Self-assembly is a top-down manufacturing technique
b) In self-assembly, weak interactions play very important role
c) Self-assembling molecules adopt an organised structure which is thermodynamically more stable than the single, unassembled components
d) Compared to the isolated components, the self-assembled structure has a higher order
Answer: a
Explanation: Self-assembly is a bottom-down manufacturing technique. All the other options are correct. In self-assembly, weak interactions play very important role, self-assembling molecules adopt an organised structure which is thermodynamically more stable than the single, unassembled components.

8. Which of the following is the application of nanotechnology to food science and technology?
a) Agriculture
b) Food safety and biosecurity
c) Product development
d) All of the mentioned
Answer: d
Explanation: The application of nanotechnology to food science and technology are agriculture, food safety and biosecurity, product development, food processing and ingredient technologies.

9. What are the advantages of nano-composite packages?
a) Lighter and biodegradable
b) Enhanced thermal stability, conductivity and mechanical strength
c) Gas barrier properties
d) All of the mentioned
Answer: d
Explanation: The advantages of nano-composite packages are-Lighter and biodegradable, enhanced thermal stability, conductivity, mechanical strength and gas barrier properties.

10. The efficiency of today’s best solar cell is about ___________
a) 15-20%
b) 40%
c) 50%
d) 75%
Answer: b
Explanation: The efficiency of today’s best solar cells is about 40%. A solar cell or photovoltaic cell is an electrical device that converts the energy of light directly into electricity by the photovoltaic effect.

11. The four types of Artificial nanomaterials are __________
a) Carbon-based, non-metallic, composites and ceramics
b) Carbon-based, metallic, composites and ceramics
c) Carbon-based, non-metallic, composites and dendrimers
d) Carbon-based, metallic, composites and dendrimers
Answer: d
Explanation: Artificial Nanomaterials can be divided into four categories – Carbon-based, metal-based, dendrimers and composites. The other types of nanomaterials are called natural nanomaterials.

12. Solution of pure buckminsterfullerene has a colour of ___________
a) Green
b) Purple
c) Pink
d) Yellow
Answer: b
Explanation: Solutions of pure C60 have a deep purple colour. They are a class of allotropes of carbon which conceptually are graphene sheets rolled into spheres.

13. HyFn stands for __________
a) Hydrated Fluorine
b) Hydrolysed Fluorine
c) Hydrolysed Fullerene
d) Hydrated Fullerene
Answer: d
Explanation: HyFn stands for Hydrated fullerene. It is a stable, highly hydrophilic, super-molecular complex consisting of C60 molecules.

14. Nano sized polymers built from branched units are called __________
a) Dendrimers
b) Composites
c) Carbon-based materials
d) Metal-based materials
Answer: a
Explanation: The Nano-sized polymers built from branched units are called dendrimers. The dendrimers can be used to perform many chemical functions, such as catalysis, as it has numerous chain ends.

15. Which property of nanoparticles provides a driving force for diffusion?
a) Optical Properties
b) High surface area to volume ratio
c) Sintering
d) There is no such property
Answer: b
Explanation: The Nanoparticles provide a high driving force for diffusion, as it has a high surface area to volume ratio. This driving force is even higher at elevated temperatures.

16. The colour of the nano gold particles is __________
a) Yellow
b) Orange
c) Red
d) Variable
Answer: d
Explanation: The colour of the nano gold particle varies with the size of the particles. It shows different colours like orange, red, purple, or greenish.

17. On both ends of the CNTs, which carbon nanostructure is placed?
a) Graphite
b) Diamond
c) C60
d) Benzene
Answer: c
Explanation: Carbon nanotubes, CNTs, are nanostructures with large application potential. Its structure consists of a single sheet of graphite rolled into a tube. The ends of the CNTs are capped with C60 hemispheres.

18. When semiconductors are reduced to nanometres they become pure conductors.
a) True
b) False
Answer: b
Explanation: When semiconductors are reduced to the nano form their chemical properties changes significantly and they become insulators, as there is no more space for free electrons to move.

19. Quantum dots can be used in _________
a) Crystallography
b) Optoelectronics
c) Mechanics
d) Quantum physics
Answer: b
Explanation: Quantum dots are basically semiconductor nanoparticles that show a particular colour on illumination by a light. They have unique electrical and optical properties. Due to this, they are widely used in optoelectronics.

20. Vesicle is a type of __________
a) Nanostructure
b) Nanoparticle
c) Nanocrystal
d) Supramolecular system
Answer: d
Explanation: Vesicle is a bubble of liquid within another liquid, a supramolecular assembly made up of man different molecules. They can be formed naturally as well as artificially.

21. TEM and SEM are the same microscopy techniques.
a) True
b) False
Answer: b
Explanation: Both Transmission Electron Microscope (TEM) and Scanning Electron Microscope (SEM) use electrons to generate images but they differ by the mode of image generation. TEM uses electrons that pass through the sample whereas SEM uses electrons that are reflected or knocked off from the sample.

22. The resolving power of TEM is derived from _______________
a) electrons
b) specimens
c) power
d) ocular system
Answer: a
Explanation: The resolving power of a transmission electron microscope is derived from the wave-like property of electrons that pass through the specimen. In SEM, the electrons reflect back from the specimen.

23. The cathode of transmission electron microscope consists of a ____________________
a) tungsten wire
b) bulb
c) iron filament
d) gold wire
Answer: a
Explanation: The cathode of a transmission electron microscope (TEM) is located on top of the column, it contains a tungsten wire filament that is heated to provide the source of electrons.

24. The resolution attainable with standard TEM is less than the theoretical value.
a) True
b) False
Answer: a
Explanation: The resolution that can be attained with a standard transmission electron microscope is about two orders of magnitude less than the theoretical value. This is due to spherical aberration of electron-focusing lenses.

25. During TEM, a vacuum is created inside the _________________________
a) room of operation
b) specimen
c) column
d) ocular system
Answer: c
Explanation: To prevent the premature scattering of electrons by collision with the gas molecules, a vacuum is generated through which the electrons travel, in the column prior to operation.

26. Which of the following component of TEM focuses the beam of electrons on the sample?
a) ocular lens
b) condenser lens
c) stage
d) column
Answer: b
Explanation: The condenser lens focuses the electron beam on to the specimen, in case of transmission electron microscope. The specimen is supported on the grid holder and placed inside the column.

27. Image formation in electron microscope is based on ___________________________
a) column length
b) electron number
c) differential scattering
d) specimen size
Answer: c
Explanation: In case of the electron microscope, the image formation is based on the differential scattering of the electrons by parts of the specimen. The scattering of electrons is proportional to the size nuclei of the atoms that make up the sample.

28. The biological materials have little intrinsic capability to ____________________
a) scatter electrons
b) stain
c) remain viable
d) be captured
Answer: a
Explanation: The insoluble materials of cells contain atoms of low atomic number such as carbon, hydrogen, oxygen and nitrogen. The biological materials therefore have very little intrinsic capability of scattering the electrons.

29. Glutaraldehyde is a ________________
a) metal
b) fixative
c) non-metal
d) atomic species
Answer: b
Explanation: Glutaraldehyde and osmium tetroxide are common fixatives used in the transmission electron microscopy for the fixation of biological specimens. They stain as well as keep the sectioned specimens in a state of similarity with the living counterpart.

30. Osmium is a ___________________
a) non metal
b) heavy metal
c) alloy
d) light metal
Answer: b
Explanation: Osmium is a heavy metal that reacts with fatty acids leading to the preservation of membranes. Osmium tetroxide is used as a fixative in transmission electron microscopy.

31. In TEM, the tissue is stained by floating on drops of ______________________
a) hydrocarbons
b) slow-molecular weight stains
c) heavy metal soutions
d) oil immersion
Answer: c
Explanation: The tissue is stained by floating on drops of uranyl acetate and lead citrate (heavy metal solutions). These solutions when bound to the specimen, provide the density required to scatter the electron beam.

32. Shadow casting is a technique of visualizing ___________________
a) isolated particles
b) mounts
c) shoot tips
d) root tips
Answer: a
Explanation: Shadow casting is a technique of viewing isolated particles. The particles are made to cast shadows after their placement in sealed chambers. The chamber contains a filament of carbon and a heavy metal.

33. What is the average particle size of ultra-fine grinders?
a) 1 to 20 µm
b) 4 to 10 µm
c) 5 to 200 µm
d) 50 to 100 µm
Answer: a
Explanation: The particle averaging 1 to 2 µm in size with substantially all particles through a standard 325 mesh screen.

34. Which of the following uses impact and attrition?
a) Ball mill
b) Tumbling mill
c) Ultra-fine grinders
d) Hammer mill
Answer: c
Explanation: Grinding takes place due to impact and attrition, it consists set of swing hammers, rotor, classifier wheels and fan wheels.

35. What is the feed size for a Ultra-Fine grinder?
a) > 3⁄4 inches
b) < 3⁄4 inches c) > 1⁄2 inches
d) < 1⁄2 inches
Answer: b
Explanation: The feed size of ultra-fine grinder is less than 3⁄4 inches and the product sizes between 1 to 20 micro meter passing through 100 mesh.

36. The fluid energy mill involves __________
a) Pressure
b) Temperature
c) Fluid
d) Density
Answer: a
Explanation: The fluid energy mill usually uses compressed air and super-heated steam admitted at a pressure of 7 atm through nozzles.

37. At what speed stirred mill operates?
a) 50 to 500 rpm
b) 5 to 10 rpm
c) 10 to 500 rpm
d) 100 to 1500 rpm
Answer: d
Explanation: In a stirred mill, a central paddle wheel or impellor armature stirs the media at speed from 100 to 1500 rpm.

38. What does colloid mill produces?
a) Emulsions and Solid dispersions
b) Wet and dry products
c) Coarse and fine particles
d) Broad particles
Answer: a
Explanation: Colloidal suspensions, emulsions and solid dispersions are produced by means of colloid mills or dispersions mills.

39. The clearance between rotor and stator is ______
a) 5 to 10 mm
b) 1 to 2.5 mm
c) 0 to 1.25 mm
d) 5 to 15 mm
Answer: c
Explanation: The mill consists of a flat rotor and stator, the clearance between the rotor and stator is generally 1 to 1.25 mm.

40. The power consumption of colloid mill is _______
a) Very low
b) Optimum
c) Critical
d) Very high
Answer: d
Explanation: The power consumption is very high and the material should therefore be ground as fine as possible before it is fed to mill.

41. The speed of a rotary knife cutter is ___
a) 100 to 500 revmin
b) 200 to 900 revmin
c) 400 to 600 revmin
d) 50 to 60 revmin
Answer: c
Explanation: A rotary knife cutter contains a horizontal rotor turning at 200 to 900 revolutions per minute, which is cylindrical chamber.

42. ______ indicates the impact of resistance.
a) Toughness
b) Rigidity
c) Brittleness
d) Fatigue
Answer: a
Explanation: Toughness indicates the impact of resistance. It is the inverse of friability and brittleness, it is also a deciding factor for commutation.

43. For high sensitivity or selectivity environmental sensors to sense the gaseous chemical like _________
a) CO2
b) NO3
c) O2
d) NO
Answer: d
Explanation: For high sensitivity or selectivity environmental sensors to sense the gaseous chemical like NO. The other gases are that are to be sensed is CO, NO, NO2 and O3 in high traffic environments are fabricated.

44. The nano materials are used in the light emitted electro luminescence devices.
a) True
b) False
Answer: a
Explanation: The nano particles are used in the light electro luminescence devices. The find application in flat panel display technologies like television, computer monitor, etc.

45. The synthesized magnetic nano particles from _________ have been found to self-arrange automatically.
a) Zinc
b) Copper
c) Iron
d) Zirconium
Answer: c
Explanation: The synthesized magnetic nano particles from iron and palladium have been found to self arrange automatically. These materials are extensively used in the manufacture of magnetic devices.

46. The nano particles from iron and palladium are used to produce ________
a) Magnets
b) Magnetic lens
c) Magneto meters
d) Magnetic storage devices
Answer: d
Explanation: The nano particles from iron and palladium were synthesized and they are used to produce the magnetic storage devices. They produce only tetra byte storage capabilities.

47. Nano particles target the rare _______ causing cells and remove them from blood.
a) Tumour
b) Fever
c) Infection
d) Cold
Answer: a
Explanation: Nano particles target the rare tumour causing cells and remove them completely from the blood stream. They are also used in the many other drugs.

48. ___________ is the field in which the nano particles are used with silica coated iron oxide iron oxide.
a) Magnetic applications
b) Electronics
c) Medical diagnosis
d) Structural and mechanical materials
Answer: c
Explanation: Medical diagnosis is the field in which the nano particles are used with silica coated iron oxide iron oxide. They are embedded with magnetic colloidal particles sent into the blood stream.

49. DNA detection through the ___________ by using the oligonucleotide functionalised gold nano crystals is developed.
a) Colorimetric
b) Diathermy
c) Electro therapy
d) Treatment tables
Answer: a
Explanation: DNA detection through the colorimetric technique by using the oligonucleotide functionalised gold nano crystals is developed. The nano particles are where anti bodies react and binds the hormone and move rapidly.

50. Coating the nano crystals with the ceramics is carried that leads to ________
a) Corrosion
b) Corrosion resistant
c) Wear and tear
d) Soft
Answer: b
Explanation: Coating the nano crystals with the ceramics is carried that leads to the corrosion resistant and hard and wear resistant and ambient ductility.

51. The __________ to the ceramics are superior coatings.
a) Nano particles
b) Nano powder
c) Nano crystals coating
d) Nano gel
Answer: c
Explanation: The nano particles coatings to the ceramics are superior coatings. They make the ceramics corrosion resistant.

52. _________ of ceramic components are easier through nano structuring.
a) Lubrication
b) Coating
c) Fabrication
d) Wear
Answer: c
Explanation: The fabrication of the ceramics is easier through the nano structuring. Fabrication is the process of producing the things.

53. By nano scale distribution of the _______ in matrix improves the life and performance.
a) Carbide
b) Tungsten
c) Hydrides
d) Nitrites
Answer: b
Explanation: By the nano scale distribution of the tungsten in matrix. The matrix contains tungsten carbide that improves the life and performance of cutting tool materials.

54. Industrial catalysts should have _________ surface area.
a) High
b) Low
c) Moderate
d) No
Answer: a
Explanation: Industrial catalysts should have the high surface area. They should also have the capacity to attach any material to their surface.

55. The extensively used nano particles as catalyst is_________
a) Silver
b) Copper
c) Gold
d) Cerium
Answer: c
Explanation: The extensively used nano particles as catalyst are gold. Some of them are molybdenum, cerium oxide and nickel.

56. Due to _________ tensile strength some of the nano materials are used in air crafts.
a) High
b) Low
c) Moderate
d) No
Answer: a
Explanation: Dye to the high tensile strength some of the nano materials are used in air crafts. One of them is carbon nano tubes. They are used in the air crafts.

57. Fabrics are extensively made out of nano materials like ___________
a) Carbon nano tubes
b) Fullerenes
c) Mega tubes
d) Polymers
Answer: b
Explanation: Fabrics are extensively made out of nano materials like fullerenes. The sports goods and cleaning products are also made out of them.

58. What’s the procedure in Top-down fabrication method?
a) Nano-particles -> Powder -> Bulk
b) Powder -> Bulk – > Nano-particles
c) Bulk -> Powder – > Nano-particles
d) Nano-particle – > Bulk -> Powder
Answer: c
Explanation: Top-down approach is the one in which a material of regular size is converted into a nano-particle. In the bottom-up approach, the atoms are joined to form nano-particles.

59. Which of the following is an example of Bottom Up approach?
a) Attrition
b) Colloidal dispersion
c) Milling
d) Etching
Answer: b
Explanation: Colloidal dispersion is an example of bottom up approach in the synthesis of Nano particles. Attrition, milling and etching are typical top down methods.

60. For milling operations, what kind of environment is preferred?
a) Acidic
b) Basic
c) Active
d) Inert
Answer: d
Explanation: Milling is the process of particle size reduction with the objective of mixing or blending and change of particle size. An inert environment is preferred for this process.

61. What kind of metals are used for milling operations?
a) Soft and brittle
b) Soft and elastic
c) Hard and brittle
d) Hard and elastic
Answer: c
Explanation: For the milling operation, normally hard brittle materials with fracture, deform and cold weld are used. The reason for choosing dense materials is the fact that the kinetic energy of balls depends upon their mass and velocity.

62. The following flow chart is for which method?
62
a) Milling
b) Attrition
c) Pattering
d) Microfabrication
Answer: d
Explanation: The given process is the process of microfabrication. The water is prepared and then photoresist is applied. The product is exposed to the UV light which after a series of processes, results in the fabrication of nanomaterials.

63. CVD stands for ____________
a) Carbon vapour density
b) Chemical vapour density
c) Chemical vapour deposition
d) Carbon vapour deposition
Answer: c
Explanation: Chemical vapour deposition, or CVD, is a chemical process used to produce high-purity, high-performance solid materials. The process is often used to produce carbon nanotubes.

64. Photolithography is a type of patterning technique.
a) True
b) False
Answer: a
Explanation: Photolithography has been the predominant patterning technique for a long time. It will require the use liquid immersion and a host of resolution enhancement technologies.

65. Chemical solution deposition is also known as ____________
a) Sol-gel
b) CVD
c) Plasma spraying
d) Laser pyrolysis

Answer: a
Explanation: Sol-gel, or chemical solution deposition, is used primarily for the fabrication of material starting from a chemical solution that acts as the precursor for an integrated network.

66. Typical precursor used in sol-gel are ____________
a) Metal oxides
b) Metal dioxides
c) Metal alkoxides
d) Metal fluorides
Answer: c
Explanation: Metal alkoxides and metal chlorides are basically used as precursors in sol-gel. Furthermore, a colloidal suspension is formed when they undergo hydrolysis or poly-condensation.

67. The following is a bottom-up process.
67
a) True
b) False
Answer: a
Explanation: Sol-gel is a bottom-up approach in which a precursor is used to for wither a network or a colloidal suspension, so as to form desired nanoparticles.

68. Choose the best definition of a dipole.
a) A pair of equal and like charges located at the origin
b) A pair of unequal and like charges located at the origin
c) A pair of equal and unlike charges separated by a small distance
d) A pair of unequal and unlike charges separated by a small distance
Answer: c
Explanation: An electric dipole generally refers to two equal and unlike (opposite signs) charges separated by a small distance. It can be anywhere, not necessarily at origin.

69. The potential due to a dipole at a point P from it is the
a) Sum of potentials at the charges
b) Difference of potentials at the charges
c) Multiplication of potentials at the charges
d) Ratio of potentials at the charges
Answer: b
Explanation: The total potential at the point P due to the dipole is given by the difference of the potentials of the individual charges.
V = V1 + (-V2), since both the charges are unlike. Thus V = V1 – V2.

70. Calculate the dipole moment of a dipole with equal charges 2C and -2C separated by a distance of 2cm.
a) 0.02
b) 0.04
c) 0.06
d) 0.08
Answer: b
Explanation: The dipole moment of charge 2C and distance 2cm will be,
M = Q x d. Thus, M = 2 x 0.02 = 0.04 C-m.

71. Find the angle at which the potential due a dipole is measured, when the distance from one charge is 12cm and that due to other is 11cm, separated to each other by a distance of 2cm.
a) 15
b) 30
c) 45
d) 60
Answer: d
Explanation: Here, the two charges are separated by d = 2cm.
The distance from one charge (say Q1) will be R1 = 11cm. The distance from another charge (say Q2) will be R2 = 12cm. If R1 and R2 is assumed to be parallel, then R2 – R1 = d cos θ. We get 1 = 2cos θ and cos θ = 0.5. Then θ =
cos-1(0.5) = 60.

72. Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.
a) 0
b) Unity
c) ∞
d) -∞
Answer: a
Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr2). When the angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will become zero.

73. For two charges 3C and -3C separated by 1cm and are located at distances 5cm and 7cm respectively from the point P, then the distance between their midpoint and the point P will be
a) 5.91
b) 12.6
c) 2
d) 9
Answer: a
Explanation: For a distant point P, the R1 and R2 will approximately be equal.
R1 = R2 = r, where r is the distance between P and the midpoint of the two charges. Thus they are in geometric progression, R1R2=r2
Now, r2 = 5 x 7 = 35. We get r = 5.91cm.

74. Calculate the distance between two charges of 4C forming a dipole, with a dipole moment of 6 units.
a) 1
b) 1.5
c) 2
d) 2.5
Answer: b
Explanation: The dipole moment is given by, M = Q x d. To get d, we rearrange the formula d = M/Q = 6/4 = 1.5units.

75. The potential due to the dipole on the midpoint of the two charges will be
a) 0
b) Unity
c) ∞
d) -∞
Answer: c
Explanation: The potential due a dipole at a point P will be V = m cos θ/(4πεr2).
Now it is given that potential on the midpoint, which means P is on midpoint, then the distance from midpoint and P will be zero. When r = 0 is put in the above equation, we get V = ∞. This shows that the potential of a dipole at its midpoint will be maximum/infinity.

76. Dipoles in any electric field undergo
a) Magnetism
b) Electromagnetism
c) Magnetisation
d) Polarisation
Answer: d
Explanation: Dipoles in any pure electric field will undergo polarisation. It is the process of alignment of dipole moments in accordance with the electric field applied.

77. Dipole moments are used to calculate the
a) Electric field intensity
b) Polarisation patterns
c) Strength of the dipole in the field
d) Susceptibility
Answer: b
Explanation: Dipole moment implicates the strength of the dipole in the electric field. They are then used to compute the polarisation patterns based on the applied field. Once the polarisation is determined we can find its susceptibility. Though all options seem to be correct, the apt answer is to calculate polarisation, provided applied field is known.

Module 6

1. Which of the following is correct for tactile sensors?
a) Touch sensitive
b) Pressure sensitive
c) Input voltage sensitive
d) Humidity sensitive
Answer: a
Explanation: Tactile sensors are those which sensitive to touch.

2. Change in output of sensor with change in input is ____________
a) Threashold
b) Slew rate
c) Sensitivity
d) None of the mentioned
Answer: c
Explanation: Sensitivity of a sensor is the change in output for a change in input.

3. Which of the following can be cause for non-zero output when zero input?
a) Bias
b) Slew
c) Offset
d) Offset or bias
Answer: d
Explanation: For ideal condition, zero input produces zero output.

4. Sensitivity of a sensor can be depicted by _______________
a) Niquist plot
b) Pole- zero plot
c) Bode plot
d) None of the mentioned
Answer: c
Explanation: Bode plot can be used for describing the sensitivity of a sensor.

5. Which of the following error is caused by a reversal of measured property?
a) Hysterisis
b) Noise
c) Digitization error
d) Quantization error
Answer: a
Explanation: Digitization error is caused by a reversal of measured value.

6. Smallest change which a sensor can detect is ____________
a) Resolution
b) Accuracy
c) Precision
d) Scale
Answer: a
Explanation: Resolution is the smallest change a sensor can detect.

7. Thermocouple generate output voltage according to ____________
a) Circuit parameters
b) Humidity
c) Temperature
d) Voltage
Answer: c
Explanation: Thermocouple is a device which is capable of producing output voltage according to input temperature.

8. Sensor is a type of transducer.
a) True
b) False
Answer: a
Explanation: Sensor is a device which enables measurement of input value.

9. Which of the following is not an analog sensor?
a) Potentiometer
b) Force-sensing resistors
c) Accelerometers
d) None of the mentioned
Answer: d
Explanation: All of the mentioned devices are analog sensors.

10. Measured property have no relation with error.
a) True
b) False
Answer: a
Explanation: Error of a system is independent of the measured value.

11. Pressure is the _________
a) force per unit area
b) mass per unit area
c) force per unit volume
d) mass per unit volume
Answer: a
Explanation: Pressure at a point is defined as the force acting per unit area. It is measured at some given point over a surface.

12. Pressure measurement devices make use of ________
a) non-elastic member
b) elastic member
c) bendable member
d) non-bendable member
Answer: b
Explanation: Pressure sensors employ elastic member at the input stage to detect or sense the pressure variations. Elastic members are usually of various forms and convert the pressure into mechanical displacement.

13. Output of electrical transducer is ________
a) inversely proportional to displacement
b) proportional to square of displacement
c) proportional to displacement
d) constant
Answer: c
Explanation: The output of an electrical transducer is proportional to displacement. Displacement is measured using electrical transducers.

14. In general how many pressure sensitive devices are there?
a) 6
b) 20
c) 10
d) 4
Answer: d
Explanation: Usually there are four pressure sensitive devices. They are as follows:
• Diaphragms
• Capsule
• Bourdon tube
• Bellows.

15. Diaphragms in a pressure sensor are of ________
a) 2 types
b) 5 types
c) 10 types
d) 20 types
Answer: a
Explanation: Generally in a pressure sensor we have four pressure sensitive devices. Diaphragm is a type of a pressure sensitive device. They are of two types:
• Flat type
• Corrugated type.

16. Temperature is the only consideration while selecting a diaphragm.
a) True
b) False
Answer: b
Explanation: While selecting a suitable diaphragm for sensing the pressure the following factors are considered important.
• Temperature range
• Shock and vibration
• Frequency response requirements.

17. Capsule type of pressure sensor consists of ________
a) 6 dissimilar diaphragms
b) 4 identical diaphragms
c) 2 identical diaphragms
d) 8 dissimilar diaphragms
Answer: c
Explanation: Capsule is a type of a pressure sensor. It comprises of two identical annular corrugated metal diaphragms that are sealed together to form a shell like enclosure.

18. Bourdon tubes are ________
a) very highly sensitive to shock
b) not sensitive to shock
c) less sensitive to shock
d) more sensitive to shock
Answer: d
Explanation: Bourdon tube is one type of pressure sensor. It is more sensitive to shock and vibrations as compared to diaphragms. Bourdon tube can be used for precision measurements of pressure up to 3 MN/m2.

19. Bellows have 5 to 20 convolutions.
a) True
b) False
Answer: a
Explanation: Bellow is a type of pressure sensor. They have about 5 to 20 convolutions. The number of convolutions depends on the pressure range, displacement, and operating temperature.

20. Sensitivity in a capsule is increased ________
a) through parallel connection of capsules
b) through series connection of capsules
c) through series and parallel connection of capsules
d) by not connecting them at all
Answer: b
Explanation: Capsule is a type of pressure sensor. In a capsule, we can increase the sensitivity by connecting two or more capsules in series. The resultant displacement is equal to the number of capsules.

21. Piezoelectric transducer is used for measuring __________
a) non-electrical quantities
b) electrical quantities
c) chemical quantities
d) any quantity
Answer: a
Explanation: A piezoelectric transducer is used for measuring non-electrical quantities such as vibration, acceleration, pressure and the intensity of sound.

22. Piezoelectric crystals __________
a) float on water
b) dissolve in water
c) are not soluble in water
d) absorb water
Answer: b
Explanation: A piezoelectric crystal dissolves in water. It is fully soluble in water. When the temperature is humid, the piezoelectric crystals gets dissolved in water.

23. Piezoelectric crystals produce _________
a) no voltage
b) low voltage
c) high voltage
d) very high voltage
Answer: c
Explanation: Piezoelectric crystals can be used spark ignition engines. They are also used in electrostatic dust filters and produce high voltage at low current.

24. Piezoelectric transducer consists of _________
a) copper rod
b) aluminum wire
c) gold crystal
d) quartz crystal
Answer: d
Explanation: A piezoelectric transducer consists of a quartz crystal. It comprises of silicon and oxygen arranged in a crystal structure of SiO2.

25. When a compressive force is applied to a quartz crystal then ____________
a) positive charges are induced
b) negative charges are induced
c) no charge is induced
d) both positive and negative charges are induced
Answer: a
Explanation: When a quartz crystal is subjected to compressive stress, positive charges are induced in one side of the crystal while negative charges are induced on the other side of the crystal.

26. In kitchen applications a piezoelectric crystal is used for _________
a) skimming milk
b) lighting a gas stove
c) grinding
d) mixing
Answer: b
Explanation: A piezoelectric lighter is used for lighting a gas based stove in kitchen applications. The pressure induced on the piezoelectric sensor creates an electric signal leading to a spark.

27. A piezoelectric transducer has a _________
a) very high sensitivity
b) low sensitivity
c) high sensitivity
d) zero sensitivity
Answer: c
Explanation: The sensitivity is high in a piezoelectric transducer. A piezoelectric transducer can be used as a sensor. It can also be used in an accelerometer due to its good frequency response.

28. A piezoelectric transducer is used as an ignition source for a cigarette.
a) True
b) False
Answer: a
Explanation: Cigarettes use piezoelectric transducers as a source of ignition. They are also used in the measurement of sonar, microphone, pressure, displacement and force.

29. Microphone converts light into heat.
a) True
b) False
Answer: b
Explanation: A microphone is used to convert the pressure induced in the form of sound waves into electric signal. The electrical signal is then amplified to produce louder sound.

30. A quartz crystal is _________
a) a chemical transducer
b) a photoelectric transducer
c) not a self-generating transducer
d) a self-generating transducer
Answer: d
Explanation: The quartz crystal is a self-generating transducer. It does not need any electric voltage for operation. The quartz crystal becomes short in length due to an applied electric field in the opposite direction.

31. Optical fiber sensors are electrically ____________
a) active
b) passive
c) active as well as passive
d) cannot be determined
Answer: b
Explanation: Optical fiber sensors are electrically passive and consequently immune to electromagnetic disturbances. They are geometrically flexible and corrosion resistant. They can be miniaturized and are most suitable for telemetry applications.

32. Optical fibers are not immune to ________
a) electronic disturbances
b) magnetic disturbances
c) ambient light interference
d) electromagnetic disturbances
Answer: c
Explanation: Optical fibre sensors are non-electrical and hence are free from electrical interference usually associated with electronically based sensors. Ambient light can interfere. Consequently, the sensor has to be applied in a dark environment or must be optically isolated.

33. Optical fiber sensors are not immune to electromagnetic disturbances.
a) True
b) False
Answer: b
Explanation: Optical fiber sensors are electrically passive and consequently immune to electromagnetic disturbances. They are geometrically flexible and corrosion resistant. They can be miniaturized and are most suitable for telemetry applications.

34. In which of the following optic fiber sensor the fiber is simply used to carry light to and from an external optical device where the sensing takes place?
a) extrinsic fiber optic sensor
b) energized fiber optic sensor
c) all fibers are used to simply carry light to and from the external optical devices
d) intrinsic fiber optic sensor
Answer: a
Explanation: In an extrinsic fiber optic sensor fiber is simply used to carry light to and from an external optical device where the sensing takes place. In an intrinsic fiber optic sensor, one or more of the physical properties of the fiber undergo a change.

35. On the bases of application of optic fiber sensor, which of the following is not considered to be the classification of fiber optic sensor?
a) biomedical/photometric sensors
b) physical sensors
c) thermal sensors
d) chemical sensors
Answer: c
Explanation: The variations in the returning light are sensed using a photodetector. Such sensors monitor variations either in the amplitude or frequency of the reflected light. Two of the most important physical parameters that can be advantageously measured using fibre optics are temperature and pressure.

36. The type of sensor that detects the analyte species directly through their characteristic spectral properties is called _____________
a) chemical sensor
b) thermal sensor
c) light sensor
d) spectroscopic Sensors
Answer: d
Explanation: Spectroscopic Sensors is the one that detects the analyte species directly through their characteristic spectral properties. In these sensors, the optical fibre functions only as a light guide, conveying light from the source to the sampling area and from the sample to the detector. Here, the light interacts with the species being sensed.

37. How many coils are required to make LVDT?
a) 4
b) 6
c) 3
d) 2
Answer: c
Explanation: Total 3 coils are required in LVDT. One centered coil which is the energizing or primary coil connected to the sine wave oscillator. The other two coils are the secondary coils so connected that their outputs are equal in magnitude but opposite in phase.

38. A chemical transduction system is interfaced to the optical fibre at its end. This type of sensor is called?
a) chemical sensor
b) thermal sensor
c) photoelectric sensor
d) light sensor
Answer: a
Explanation: In the chemical sensors, a chemical transduction system is interfaced to the optical fibre at its end. In operation, interaction with analyte leads to a change in optical properties of the reagent phase, which is probed and detected through the fibre optic. The optical property measured can be absorbance, reflectance or luminescence.

39. Which devices measures gases or liquid?
a) Proximity sensor
b) Pressure sensor
c) Temperature sensor
d) Touch sensor
Answer: b
Explanation: The pressure sensor is a device for pressure measurement of gases or liquid. Pressure is an expression of the force required to stop a fluid from expanding.

40. Which sensor measures the pressure relative to perfect vacuum?
a) Absolute pressure sensor
b) Gauge pressure sensor
c) Vacuum pressure sensor
d) Differential pressure sensor
Answer: a
Explanation: Pressure sensor can be classified in terms of pressure ranges they measure, temperature ranges of operation, and most importantly the type of pressure they measure. Absolute pressure sensor is a sensor that measures the pressure relative to perfect vacuum.

41. Which sensor measure the pressure relative to atmospheric pressure?
a) Absolute pressure sensor
b) Gauge pressure sensor
c) Vacuum pressure sensor
d) Differential pressure sensor
Answer: b
Explanation: This sensor measures the pressure relative to atmospheric pressure. A tire pressure gauge is an example of gauge pressure measurement; when it indicates zero, then the pressure it is measuring is the same as the ambient pressure.

42. Barometer is which type of sensor________
a) Pressure sensor
b) Touch sensor
c) Temperature sensor
d) Humidity sensor
Answer: a
Explanation: Barometers and pressure gauges are the most popular sensors used for IOT ecosystem. Barometers are an absolute help in weather forecasting as in accurately measures the ambient air.

43. Touch screen devices use which sensor?
a) Touch sensor
b) Temperature sensor
c) Humidity sensor
d) Pressure sensor
Answer: d
Explanation: Pressure sensors are ultimate solution for IOT as they can be used in various areas such as touch screen devices.

44. What is the stability of pressure sensor?
a) /-0.75% FS
b) /-0.5% FS
c) /- 0.35% FS
d) /-0.125% FS
Answer: c
Explanation: One of the main feature of pressure sensor is its stability which is /-0.35% FS(Full Scale).

45. What is the operating voltage of pressure sensor?
a) 3.5v
b) 1.5v
c) 5v
d) 3v
Answer: d
Explanation: The pressure sensor has and absolute pressure of 1.2 bar to 10 bar. The typical operating system is 3v. With a supply voltage of 5v they offer sensitivities of between 15 mV/bar and 80 mV/bar, depending on type.

46. The feature of c39 is its low insertion height.
a) True
b) False
Answer: a
Explanation: The c39 is especially suitable for IOT and consumer applications. One feature of c39 is its low insertion height of just 0.24mm, which makes the low-profile MEMS pressure sensor die ideal for applications in smartphones and wearables.

47. What is the absolute pressure of c39?
a) 2.0 bar
b) 1.5 bar
c) 1.2 bar
d) 1.35 bar
Answer: c
Explanation: The c39 is designed for an absolute pressure of 1.2 bar and, like the c33 series, offers long-term stability of /-0.35% FS.

48. Which pressure standard is used for gauge?
a) Dead-weight tester
b) Manometer
c) Pressure switches
d) Stain gauge pressure sensor
Answer: b
Explanation: A mercury manometer is a simple pressure standard and may be used for gauge, differential, and absolute measurements with a suitable reference.

49. Pressure measured relative to perfect vacuum is termed as ______
a) Absolute Pressure Measurement
b) Differential Pressure Measurement
c) Gauge Pressure Measurement
d) Both Absolute and Differential
Answer: a
Explanation: Pressure measured relative to perfect vacuum is termed as Absolute Pressure Measurement. Perfect vacuum is a condition where there is no matter present in the atmosphere and hence nil air pressure exits in that region.

50. In which pressure measurement, pressure of two distinct positions are compared?
a) Absolute Pressure Measurement
b) Differential Pressure Measurement
c) Gauge Pressure Measurement
d) Both Absolute and Differential
Answer: b
Explanation: In differential pressure measurement, pressure of two distinct positions are compared. For example, pressure difference calculated by measuring it at different floors of a tall building will give us differential pressure.

51. ______ can be defined as a subtype of differential pressure measurement.
a) Absolute Pressure Measurement
b) Differential Pressure Measurement
c) Gauge Pressure Measurement
d) Both Absolute and Differential
Answer: c
Explanation: Gauge Pressure Measurement is defined as subtype of differential pressure measurement where we compare pressure pressure at any point to the current atmospheric pressure. Gauge Pressure measurement is used in applications like tire pressure of blood pressure measurement.

52. The computer devices and smart phones that have _______ with pressure sensor.
a) Automotive Industry
b) Industrial
c) Aviation
d) Touch screen
Answer: d
Explanation: The computer devices and smart phones that have touch screen display with pressure sensor. Whenever slight pressure is applied on touch screen through a finger, the sensor determines where it has been applied.

53. In which place the pressure sensor is needed for monitoring gases and their partial pressures.
a) Touch screen
b) Automotive Industry
c) Industrial
d) Aviation
Answer: c
Explanation: Pressure sensors are used to monitor gases and their partial pressures in industrial units so that the large chemical reactions take place in precisely controlled environmental conditions.

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