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## Interview MCQ Set 1

1. When two fluids of the heat exchanger are separated by a plane wall, the thermal resistance comprises
(i) Convection resistance due to the fluid film at the inside surface
(ii) Conduction resistance
(iii) Convection resistance due to the fluid film at the outside surface
Identify the correct option
a) 1 and 2
b) 1, 2 and 3
c) 2 and 3
d) 1 and 3

Answer: b [Reason:] It consists of all the above i.e. convection resistance and conduction resistance.

2. Figure represents the block diagram of a heat exchanger. There were some aspects in the design and performance analysis of a heat exchanger. Identify the correct one a) The coolant picks up heat = m h c c (t c1 – t c2)
b) The hot fluid gives up heat = m h c h (t h1 – t c2)
c) The coolant picks up heat = m c c h (t h1 – t c2)
d) The hot fluid gives up heat = m h c h (t h1 – t h2)

Answer: d [Reason:] The structure of the heat exchanger transfers heat from the hot fluid to the coolant.

3. The heat loss from unpainted aluminum side of a house has been calculated on the presumption that overall coefficient of heat transfer is 5 W/m 2 K. Later, it was discovered that the air pollution levels are such that fouling factor on this side is of the order of 0.0005 m2 K/W. Find overall heat transfer coefficient of dirty side
a) 2.9875 W/m2 K
b) 3.9875 W/m2 K
c) 4.9875 W/m2 K
d) 5.9875 W/m2 K

Answer: c [Reason:] R = 1/U DIRTY – 1/U CLEAN.

4. What is the value of fouling factor for engine exhaust?
a) 0.002 m2 K/W
b) 0.003 m2 K/W
c) 0.004 m2 K/W
d) 0.005 m2 K/W

Answer: a [Reason:] Its unit is m2 hr K/kcal and it represents the reciprocal of the scale coefficient i.e. heat transfer.

5. What is the value of fouling factor for industrial liquids?
a) 0.0004 m2 K/W
b) 0.0003 m2 K/W
c) 0.0002 m2 K/W
d) 0.0001 m2 K/W

Answer: c [Reason:] Its unit is m2 hr K/kcal and it represents the reciprocal of the scale coefficient i.e. heat transfer.

6. After being in service for a period of six months, a heat exchanger transforms 10% less heat than it does what new. Determine the effective fouling factor in terms of its clean overall heat transfer coefficient. It may be presumed that the heat exchanger operates between the same temperature differentials and that there is no change in the effective surface area due to scale build up
a) 0.13/ U CLEAN
b) 0.11/ U CLEAN
c) 0.09/ U CLEAN
d) 0.07/ U CLEAN

Answer: b [Reason:] Q CLEAN/Q DIRTY = U CLEAN/U DIRTY.

7. In a counter flow heat exchanger, water flowing through a tube of 10 cm inner diameter is heated by steam condensing on the outside of the tube. The convective film coefficient on the water and steam side are estimated to be 12000 and 20000 k J/m2 hr degree. Neglecting tube thickness and its resistance to heat flow, workout the overall heat transfer coefficient for the heat exchanger
a) 4500 k J/m2 hr degree
b) 5500 k J/m2 hr degree
c) 6500 k J/m2 hr degree
d) 7500 k J/m2 hr degree

Answer: d [Reason:] U = 1/h i + h o.

8. A heat exchanger to preheat oil for a furnace was designed without considering the possibility of scale formation, and the overall heat transfer coefficient based on the fuel oil side was 3200 k J/m2 hr degree. What would be the corrected coefficient of heat transfer if a fouling factor of 0.00025 m2 hr degree/k J for the fuel oil is taken into account?
a) 1777.78 k J/m2 hr degree
b) 1666.78 k J/m2 hr degree
c) 1555.78 k J/m2 hr degree
d) 1444.78 k J/m2 hr degree

Answer: a [Reason:] U = 1/ (R + 1/h S).

9. A copper pipe (k = 350 W/m K) having inner diameter of 1.75 cm and 2.0 cm outside diameter conveys water and the oil flows through the annular passage between this pipe and a steel pipe. On the water side, the film coefficient is 4600 W/m2 K and the fouling factor is 0.00034 m2 K/W. The corresponding values for the oil side are 1200 W/m2 K and 0.00086 m2 K/W. Estimate the overall heat transfer coefficient between the oil and water
a) 235.16 W/m2 K
b) 335.16 W/m2 K
c) 435.16 W/m2 K
d) 535.16 W/m2 K
d) 535.16 W/m2 K

Answer: c [Reason:] A fouling factor represents the reciprocal of the scale coefficient i.e. heat transfer.

10. Unit of fouling factor is
a) m2 K/kcal
b) m2 hr K/kcal
c) m2 hr/kcal
d) m2 hr K

Answer: b [Reason:] The fouling factors are used in the design of heat exchangers. It is the overall heat transfer coefficient.

## Interview MCQ Set 2

1. In CCMP a 128-bit AES Key is used for both integrity and confidentiality. The scheme uses a ___________ bit packet number to construct a nonce to prevent __________ attacks.
a) 48, replay
b) 64, replay
c) 48, transient
d) 64, transient

Answer: a [Reason:] The scheme uses a 48-bit packet number to construct a nonce to prevent replay attacks.

2. HMAC SHA-1 produces a hash of length-
a) 180 bits
b) 240 bits
c) 160 bits
d) 40 bits

Answer: c [Reason:] The hash length and key length in HASH SHA-1 is 160 bits.

3. In the IEEE 802.11i PRF, a nonce is generated by the following expression –
nonce = PRF (Random Number, “InitCounter”, MAC || Time, Len)
What is the value of the ‘Len’ (desirable number of pseudorandom bits)?
a) 64
b) 128
c) 256
d) 512

Answer: c [Reason:] Nonce = PRF (Random Number, “InitCounter”, MAC || Time, Len).

4. Nonce = PRF (Random Number, “InitCounter”, MAC || Time, Len)
What is the Key ‘K’ in the above expression?
a) InitCounter
b) Random Number
c) PRF
d) Time

Answer: c [Reason:] PRF is the secret key.

5. What is ‘Len’ (desirable number of pseudorandom bits) field in the case of CCMP PTK computation?
a) 256
b) 384
c) 568
d) 298

Answer: b [Reason:] The ‘Len’ value in the case of CCMP is 128+128+128 = 384 bits.

6. The message input to the IEEE 802.11 pseudorandom Function consists of 4 items concatenated together. Which of the following option is not a part of this message input –
a) the parameter B
b) a byte with value FF (11111111)
c) the parameter A
d) counter value i

Answer: b [Reason:] The byte concatenated has a value 00h.

7. A __________ is a secret key shared by the AP and a STA and installed in some fashion outside the scope of IEEE 802.11i.
a) pre-shared key
b) pairwise transient key
c) master session key
d) key conformation key

Answer: a [Reason:] A pre-shared key(PSK) is a secret key shared by the AP and a STA and installed in some fashion outside the scope of IEEE 802.11i.

8. The Pairwise master key (PMK) can be derived from either PSK or MSK.
a) True
b) False

Answer: a [Reason:] If a PSK is used then the PSK becomes the PMK; if an MSK is used, then the PMK is derived from the MSK using truncation. PSK – Pre-shared Key MSK – Master shared key.

9. The pairwise transient key (PTK) is derived from –
a) Key Confirmation Key (KCK)
b) Temporal Key (TK)
c) Pairwise master Key (PMK)
d) Pre-Shared Key (PSK)

Answer: c [Reason:] The pairwise transient key (PTK) is derived from the Pairwise master Key (PMK).

10. The pairwise transient key (PTK) consists of ______________ keys.
a) 3
b) 4
c) 5
d) 2

Answer: a [Reason:] The pairwise transient key (PTK) consists of three keys, which are KCK KEK and TK.

11. Which Hashing algorithm is used to derive the PTK for PMK?
a) SHA-1
b) SHA-2
c) SHA-3
d) MD-5

Answer: a [Reason:] SHA-1 is used to derive PMK from the PTK.

## Interview MCQ Set 3

1. On which port will the server generally listen to for SMTP?
a) port 35
b) port 63
c) port 25
d) port 65

Answer: c [Reason:] The server would have to listen on port 25 for the simple mail transfer protocol.

2. In the alert protocol the first byte takes the value 1 or 2 which corresponds to ________ and _________ respectively.
a) Select, Alarm
c) Warning, Alarm
d) Warning, Fatal

Answer: d [Reason:] The first byte takes the value warning(1) or fatal(2) to convey the severity of the message.

3. In terms of Web Security Threats, “Impersonation of another user” is a Passive Attack.
a) True
b) False

Answer: b [Reason:] Passive attacks include eavesdropping on network traffic between browser and server and gaining access to information on a website that is supposed to be restricted. Active attacks include impersonating another user, altering messages in transit between client and server, altering information on a website.

4. In the SSL record protocol operation pad_1 is :
a) is the byte 0x36 repeated 40 times for MD5
b) is the byte 0x5C repeated 40 times for MD5
c) is the byte 0x5C repeated 48 times for SHA-1
d) is the byte 0x36 repeated 48 times for MD5

Answer: d [Reason:] pad_1 = is the byte 0x36 repeated 48 times for MD5.

5. In the Phase 2 of the Handshake Protocol Action, the step server_key_exchange is not needed for which of the following cipher systems?
a) Fortezza
b) Anonymous Diffie-Hellman
c) Fixed Diffie-Hellman
d) RSA

Answer: c [Reason:] The Fixed Diffie-Helmann does not require the server_key_exchange step in the handshake protocol.

6. Which key exchange technique is not supported by SSLv3?
a) Anonymous Diffie-Hellman
b) Fixed Diffie-Hellman
c) RSA
d) Fortezza

Answer: d [Reason:] Fortezza is not supported in SSLv3.

7. An HTTP connection uses port __________ whereas HTTPS uses port __________ and invokes SSL.
a) 40; 80
b) 60; 620
c) 80; 443
d) 620; 80

Answer: c [Reason:] HTTP uses 80 ports, whereas HTTPS uses 443 ports.

8. SSID stands for
a) Secure Service Identifier
b) Secure Set Independent Device
c) Secure Set Identifier
d) Service Set Independent Device

Answer: c [Reason:] SSID stands for Secure Set Identifier.

9. “When an attacker is able to eavesdrop on network traffic and identify the MAC address of a computer with network previleges.” Which type of Wireless network threat would you classify this under?
a) Identity Theft
b) Man in the middle attack
c) Network Injection
d) Accidental Association

Answer: a [Reason:] This threat falls under Identity Theft.

10. Loopback address is given by:
a) 0.0.0.0
b) 127.x.x.x
c) 255.255.255.255
d) None of the mentioned

Answer: b [Reason:] Loopback address is given by 127.x.x.x.

## Interview MCQ Set 4

1. RC phase shift oscillators contains a minimum of _________ Phase shift network.
a) 1
b) 2
c) 3
d) 0

Answer: c [Reason:] RC phase shift oscillator contains a minimum of three phase shift network. There is can be also four and it increases stability of oscillation.

2. One phase shift network of an RC phase shift oscillator contain __________ capacitor
a) 1
b) 2
c) 3
d) 0

Answer: a [Reason:] One phase shift network of an RC phase shift oscillator contains one capacitor and one resistor.

3. One phase shift network of an RC phase contain _________ inductor
a) 1
b) 2
c) 3
d) 0

Answer: d [Reason:] One phase shift network of an RC phase shift oscillator contains one capacitor and one resistor. There is no need of an inductor.

4. One phase shift network of an RC phase contain __________ resistor
a) 1
b) 2
c) 3
d) 0

Answer: a [Reason:] One phase shift network of an RC phase shift oscillator contains one capacitor and one resistor.

5. Phase shift provided by one phase shift network in RC phase shift oscillator in 3 stage is
a) 180 degrees
b) 60 degrees
c) 120 degrees
d) 90 degrees

Answer: b [Reason:] Phase shift provided by one phase shift oscillator in RC phase shift oscillator in 3 stage is 60 degrees. It is 180/number of stages.

6. Total phase shift provided by all phase shift networks in RC phase shift oscillator
a) 180 degrees
b) 60 degrees
c) 120 degrees
d) 360 degrees

Answer: a [Reason:] To satisfy Barkhausen’s criteria total phase shift should be 360 degrees. 180 degrees is produced by amplifier and rest by phase shift network even it is three stage or four stage.

7. The phase shift network will produce a phase shift of 180 degrees at
a) Three different frequencies
b) One frequency
c) Two different frequencies
d) Infinitely many frequencies

Answer: b [Reason:] The phase shift oscillator will produce a phase shift of 180 degrees only at a particular frequency by which it is meant to oscillate.

8. Which of the following is not a reason for beginning oscillations in RC phase shift oscillator?
a) Phase shift network
b) Noise inherent in transistor
c) Minor variations in the voltage DC source
d) None of the mentioned

Answer: a [Reason:] Phase shift network doesn’t initialize the oscillation it just phase shift a given frequency. Oscillation will start by noise inherent of transistor or by minor voltage variations in DC source.

9. Amplifier gain for RC phase shift oscillation, to obey Barkhausen’s criteria should be minimum of
a) 43
b) 4
c) 10
d) 29

Answer: d [Reason:] Amplifier gain should be minimum of 29 or else Aβ will be less than one.

10. Phase shift provided by one phase shift network in RC phase shift network in 4-stage will be
a) 180 degrees
b) 45 degrees
c) 60 degrees
d) 90 degrees

Answer: b [Reason:] Phase shift provided by one phase shift oscillator in RC phase shift oscillator in 4 stage is 45degrees. It is 180/number of stages.

11. Frequency of oscillation for three section RC phase shift network is given by
a) 1/(ᴨ√6 RC)
b) 2/(ᴨ√6 RC)
c) 1/(2ᴨ√6 RC)
d) 1/(2√6 RC)

Answer: c [Reason:] The frequency of oscillation is given by the equation 1/(2ᴨ√6 RC). It can be reduced as 0.65/RC.

12. Distortion level in the output of RC phase shift network will be less than
a) 1%
b) 2%
c) 5%
d) 10%

Answer: c [Reason:] Distortion in output of phase shift oscillator is less 5%.

13. Which of the following is not true for an RC phase shift oscillator?
a) Not Bulky
b) Less costly
c) Effective for oscillation less than 10KHz
d) Pure sine wave output is possible

Answer: d [Reason:] Since only one frequency can fulfill barkhausen’s criteria requirement positive feedback occurs only for one frequency. Hence pure sine wave is not possible.

14. The feedback factor for RC phase shift oscillator is
a) 1/18
b) 1/29
c) 1/11
d) 1/33

Answer: b [Reason:] The feedback factor for RC phase shift oscillator is 1/29 and to satisfy Barkhausen’s criteria amplifier gain should be greater than 29.

15. What will be oscillator frequency, if phase shift network of RC phase shift oscillator contains a capacitor of 7nF and a resistance of 10K?
a) 928 Hz
b) 1KHz
c) 1.2KHz
d) 895Hz

Answer: a [Reason:] Oscillator frequency can be given by the equation .

## Interview MCQ Set 5

1. Which of the following graphs will be appropriate to describe output V of the circuit given below?
The voltage VB is 1V and input to the circuit Vin is 5sint. The resistance R is 1K.
(Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: d [Reason:] In the circuit VB is in the forward direction for diode. At positive cycle of Vin diode will be forward biased up to Vin = 0.3V. Up to 0.3V output is 0.3V and after this diode is reverse biased and output follows input. In negative cycle diode is always forward biased and output will be equal to 1-0.7 = 0.3V.

2. Which of the following graphs will be appropriate to describe output V of the circuit given below?
The voltage VB is 1V and input to the circuit Vin is 5sint. The resistance R is 1K.
(Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: c [Reason:] In the circuit VB is in the forward direction for diode. At positive cycle of Vin diode will be forward biased up to Vin = 0.3V. Up to 0.3V output is 0.7V and after this diode is reverse biased and output follows input. In negative cycle diode is always forward biased and output will be equal to 0.7V in reverse direction. In the following diagram red represents input and green represents output. 3. Which of the following graphs will be appropriate to describe output V of the circuit given below?
The voltage VB is 1V and input to the circuit Vin is 5sint. The resistance R is 1K.
(Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: b [Reason:] In the circuit VB is in the reverse direction for diode. At positive cycle of Vin diode will be forward biased from Vin = 1.7V. Up to 1.7V output is 5sint-1V and after this diode is forward biased and output will be 0.7V. In negative cycle diode is always reverse biased and output will be equal to 5sint-1.

4. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1V and input to the circuit V is 5sint. Assume both diodes are identical. (Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: d [Reason:] In the circuit V1 and V2 are in the forward direction for diodes. Since these are parallel we can consider this as a single diode and source. At positive cycle of Vin diode will be forward biased up to Vin = 0.3V. Up to 0.3V output is 0.3V and after this diode is reverse biased and output follows input. In negative cycle diode is always forward biased and output will be equal to 1-0.7 = 0.3V.

5. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1V and input to the circuit V is 5sint. Assume both diodes are identical. (Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: b [Reason:] In this circuit diodes are placed opposite to each other. So output will be determined by diode which is in forward bias. At negative cycle first diode will be forward biased hence output will be 0.3V. At positive cycle diode will be reverse bias up to 1.7V so input follows output and after this it is constant and equals to 1.7V.

6. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1V and input to the circuit V is 5sint. Assume both diodes are identical. (Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: c [Reason:] Since both diodes are in reverse mode with respect to V1 and V2, at negative V output follows input. At positive cycle up to 1.7V diodes are in reverse bias mode and after this they becomes forward biased and output becomes a constant and equals to 1.7V.

7. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1.5V and input to the circuit V is 5sint. Assume both diodes are identical. (Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: b [Reason:] In this circuit diodes are placed opposite to each other. So output will be determined by diode which is in forward bias. At negative cycle first diode will be forward biased hence output will be 0.3V. At positive cycle diode will be reverse bias up to 1.7V so input follows output and after this it is constant and equals to 1.7V.

8. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1V and input to the circuit V is 5sint.
(Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: a [Reason:] At positive cycle of V, V1 opposes hence output will be 5sint-1. For negative half cycle of V, output will be 0.3V because diode is already in forward bias.

9. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1V and input to the circuit V is 5sint.
(Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: a [Reason:] For negative half cycle diode is forward biased and hence output is 0.3V. For positive half cycle output will be (5sint+1)/2(Since voltage divided in between two 1K, third one has no effect). Hence maximum voltage will be 6/2 = 3V.

10. Which of the following graphs will be appropriate to describe output Vout of the circuit given below?
The voltage V1 is 1V, V2 is 1V and input to the circuit V is 5sint.
(Use constant voltage drop model for diode and take cut-in voltage as 0.7V) a) b) c) d) Answer: d [Reason:] For negative half cycle of V and up to 3.4V in positive half cycle since diode is in reverse bias output will be (5sint-1)/2 Since voltage divided in between two 1K, third one has no effect). Hence minimum voltage will be -4/2 = -2V. After 3.4V output becomes 1.7V. 