Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

51.

What is the recommended modes of operation for encryption and decryption in the Secure Shell Transport Layer?(a) ECB(b) OFB(c) CBC(d) CFBI got this question by my college professor while I was bunking the class.The origin of the question is Secure Shell(SSH) topic in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct CHOICE is (C) CBC

The best explanation: The cipher BLOCK CHAINING mode is the recommended mode of OPERATION in the SSH Transport layer.

52.

The certificate message is required for any agreed-on key exchange method except _______________(a) Ephemeral Diffie-Hellman(b) Anonymous Diffie-Hellman(c) Fixed Diffie-Hellman(d) RSAI have been asked this question by my school teacher while I was bunking the class.This intriguing question originated from Secure Socket Layer in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» CORRECT answer is (b) Anonymous Diffie-Hellman

Explanation: The certificate MESSAGE is REQUIRED for any agreed-on KEY exchange method except Anonymous Diffie-Hellman.
53.

Broadcast address is given by –(a) 0.0.0.0(b) 127.x.x.x(c) 255.255.255.255(d) None of the mentionedThis question was addressed to me in an interview.My enquiry is from Designing Subnets in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct option is (c) 255.255.255.255

Easiest explanation: Broadcast address is given by 255.255.255.255. This address is the last address in the range of addresses, and is the address whose HOST portion is SET to all ONES.

54.

First Address of the third subblock –(a) 14.24.74.128/28(b) 14.24.74.127 /28(c) 14.24.74.192/27(d) 14.24.74.192/28I got this question by my school principal while I was bunking the class.Origin of the question is Designing Subnets topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right OPTION is (d) 14.24.74.192/28

To EXPLAIN: FIRST address = 14.24.74.192/28

First address = (any address) AND (NETWORK mask).

55.

First address of the second subblock –(a) 14.24.74.192/26(b) 14.24.74.128/26(c) 14.24.74.127/28(d) 14.24.74.67/27I had been asked this question at a job interview.This intriguing question originated from Designing Subnets topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right answer is (b) 14.24.74.128/26

The BEST EXPLANATION: First ADDRESS = 14.24.74.128/26

First address = (any address) AND (network MASK).

56.

Nsub3 =(a) 16(b) 14(c) 12(d) 10I got this question during a job interview.Query is from Designing Subnets topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct answer is (a) 16

Explanation: The number of addresses in the THIRD SUBBLOCK (10) is not a POWER of 2.

Thus, we allocate Nsub3 = 16 addresses.

57.

In an IP packet, the value of HLEN is 1000 in binary. How many bytes of options are being carried by this packet?(a) 16(b) 64(c) 32(d) 8This question was addressed to me in final exam.This is a very interesting question from IP Datagram Packet in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct OPTION is (c) 32

To EXPLAIN: HLEN = 1000 = 8.

8 x 4 = 32 bytes.

58.

In the “Opening a channel” phase what is the function of the “innitial window size” parameter?(a) It specifies the number of bytes in the IV(b) It specifies the maximum size of an individual packet(c) It specifies how many bytes of channel data can be sent(d) None of the mentionedThe question was asked during an interview.My question is from Secure Shell(SSH) topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right CHOICE is (c) It specifies how MANY bytes of CHANNEL data can be sent

Best EXPLANATION: The ‘innitial window size’ specifies how many bytes of channel data can be sent.

59.

Which is the lowest level in the SSH protocol stack?(a) SSH Transport Layer Protocol(b) TCP(c) IP(d) SSH User Authentication ProtocolI have been asked this question by my college director while I was bunking the class.My query is from Secure Shell(SSH) in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right answer is (c) IP

Easy EXPLANATION: The IP is the LOWEST LEVEL in SSH the protocol stack.

60.

An HTTP connection uses port _________ whereas HTTPS uses port ____________ and invokes SSL.(a) 40; 80(b) 60; 620(c) 80; 443(d) 620; 80This question was addressed to me by my college director while I was bunking the class.Query is from Transport Layer Security and HTTPS topic in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT answer is (C) 80; 443

Explanation: HTTP uses 80 PORTS, WHEREAS HTTPS uses 443 ports.
61.

Which of the following is not a valid input to the PRF in SSLv3?(a) secret value(b) identifying label(c) initialization vector(d) secret valueThe question was asked during a job interview.The query is from Transport Layer Security and HTTPS topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct option is (C) initialization vector

The BEST I can explain: The PRF does not require an initialization vector.

62.

Which of the following are possible sizes of MACs?i) 12 Bytesii) 16 Bytesiii) 20 Bytesiv) 24 Bytes(a) i and iii(b) ii only(c) ii and iii(d) ii iii and ivI got this question in homework.The query is from Secure Socket Layer topic in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT answer is (c) ii and iii

For explanation I WOULD SAY: MACs can be 0, 16 or 20 Bytes.
63.

An IP packet has arrived with the first 8 bits as 01000010. What is the version?(a) 3(b) 4(c) 5(d) 6This question was addressed to me at a job interview.This intriguing question originated from IP Datagram Packet in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right answer is (b) 4

Easy explanation: The first 4-bits SHOW the VERSION i.e. 0100 -> version 4.

64.

Which one of the following is not a session state parameter?(a) Master Secret(b) Cipher Spec(c) Peer Certificate(d) Server Write KeyThis question was posed to me in an online interview.This intriguing question comes from Secure Socket Layer in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct option is (d) Server WRITE Key

For explanation: Session state is defined by the following PARAMETERS – Session IDENTIFIER, PEER CERTIFICATE, Compression method, Cipher spec, Master secret, Is resumable. Server Write Key falls under Connection State.

65.

Size of the sequence number is-(a) 32 bits(b) 48 bits(c) 56 bits(d) 64 bitsI have been asked this question in an interview for job.The query is from Secure Shell(SSH) in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct CHOICE is (a) 32 bits

The EXPLANATION: The SEQ# is 32 bits.

66.

What are the total number of addresses in Group 1?(a) 32768(b) 8192(c) 16384(d) None of the mentionedThis question was addressed to me in semester exam.The question is from Designing Subnets topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The CORRECT ANSWER is (c) 16384

For EXPLANATION: TOTAL addresses: 64×256 = 16384.

67.

In the SSL Protocol, each upper layer message if fragmented into a maximum of __________ bytes.(a) 2^16(b) 2^32(c) 2^14(d) 2^12The question was posed to me in my homework.I want to ask this question from Secure Socket Layer topic in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT choice is (c) 2^14

To elaborate: In the FRAGMENTATION process we OBTAIN blocks of 2^14 bytes which is compressedin the next step.
68.

What is the error (if any) in the following representation – 75.45.301.14?(a) There should be no leading zeros(b) We cannot have more than 4 bytes in an IPv4 address(c) Each byte should be less than or equal to 255(d) No errorI got this question during an interview.My doubt stems from IP Addressing in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» CORRECT answer is (c) Each BYTE should be LESS than or equal to 255

To ELABORATE: NONE.
69.

In the User Authentication Protocol, when the client requests for authentication, then the first byte ( in this format) has decimal value –(a) 80(b) 60(c) 50(d) 61The question was asked during an online exam.My enquiry is from Secure Shell(SSH) topic in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT OPTION is (C) 50

For EXPLANATION I would say: The first byte is ASSIGNED a value 50. This is the SSH user authentication message/byte.
70.

URL stands for –(a) Universal Remote Locator(b) Universal Resource Language(c) Uniform Resource Locator(d) Uniform Resource LanguageThe question was asked by my college professor while I was bunking the class.I need to ask this question from Transport Layer Security and HTTPS in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct CHOICE is (c) Uniform RESOURCE LOCATOR

The EXPLANATION: URL STANDS for Uniform Resource Locator.

71.

In TLS padding cann be upto a maximum of –(a) 79 bytes(b) 127 bytes(c) 255 bytes(d) none of the mentionedI had been asked this question during an interview for a job.The origin of the question is Transport Layer Security and HTTPS topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» CORRECT answer is (c) 255 bytes

Easiest explanation: PADDING can be UPTO a MAXIMUM of 255 bytes.
72.

What is the size of the RSA signature hash after the MD5 and SHA-1 processing?(a) 42 bytes(b) 32 bytes(c) 36 bytes(d) 48 bytesThe question was posed to me in exam.My question is based upon Secure Socket Layer in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The CORRECT choice is (c) 36 bytes

Easiest EXPLANATION: The size is 36 bytes after MD5 and SHA-1 PROCESSING.

73.

Which protocol is used to convey SSL related alerts to the peer entity?(a) Alert Protocol(b) Handshake Protocol(c) Upper-Layer Protocol(d) Change Cipher Spec ProtocolThis question was addressed to me by my college director while I was bunking the class.Asked question is from Secure Socket Layer in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT option is (a) ALERT Protocol

Easiest EXPLANATION: The Alert protocol is USED to convey SSL RELATED alerts to the peer entity.
74.

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 MD5I had been asked this question by my college director while I was bunking the class.The origin of the question is Secure Socket Layer in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right OPTION is (d) is the BYTE 0x36 repeated 48 times for MD5

For EXPLANATION: pad_1 = is the byte 0x36 repeated 48 times for MD5.

75.

In the SSL record protocol operation pad_2 is –(a) is the byte 0x36 repeated 40 times for MD5(b) is the byte 0x5C repeated 48 times for MD5(c) is the byte 0x5C repeated 48 times for SHA-1(d) is the byte 0x36 repeated 48 times for MD5I had been asked this question during an interview for a job.My question is from Secure Socket Layer topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right choice is (B) is the BYTE 0X5C repeated 48 times for MD5

Easy EXPLANATION: pad_2 = is the byte 0x5C repeated 48 times for MD5.

76.

The full form of SSL is(a) Serial Session Layer(b) Secure Socket Layer(c) Session Secure Layer(d) Series Socket LayerThe question was asked during an interview.This key question is from Secure Socket Layer topic in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT option is (b) Secure Socket LAYER

To explain I would say: SSL STANDS for Secure SOCKETS Layer.
77.

In the above figure, which of the above shaded block is transparent to end users and applications?(a) IP/IPSec(b) SSL(c) Kerberos(d) S/MIMEThe question was asked in homework.I'd like to ask this question from Secure Socket Layer in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT choice is (a) IP/IPSec

To explain I WOULD say: IP/IPSec is the Network LAYER which is TRANSPARENT to END users and applications.
78.

What is the first address of group 1?(a) 190.100.0.0/24(b) 190.100.0.0/16(c) 190.100.63.255/24(d) 190.100.0.255/16The question was posed to me by my school teacher while I was bunking the class.My question is based upon Designing Subnets in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct OPTION is (a) 190.100.0.0/24

Easiest EXPLANATION: Find the IP ADDRESS using basic NETWORKING concepts.

79.

The subnet mask nsub1=(a) 23(b) 25(c) 26(d) 27I had been asked this question in class test.This key question is from Designing Subnets topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» CORRECT ANSWER is (B) 25

The BEST explanation: nsub1= n+log2(N/Nsub1) = 25.
80.

Last Address –(a) 14.24.74.64/24(b) 14.24.74.127/24(c) 14.24.74.255/24(d) 14.24.74.256/24I got this question during an internship interview.My question is taken from Designing Subnets topic in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The CORRECT choice is (c) 14.24.74.255/24

The best I can explain: LAST Address = 14.24.74.255/24

Last address = (any address) OR [NOT (NETWORK MASK)].

81.

Address of the block is –(a) 2^14(b) 2^16(c) 2^8(d) 2^24The question was asked in my homework.My question is taken from Classful Addressing topic in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct ANSWER is (b) 2^16

The best I can EXPLAIN: Since 180 is between 128 and 191, it is CLASS B ADDRESS n=16.

N = 2^(32-n) = 2^16 = 65,536.

82.

What is the error (if any) in the following representation – 111.56.045.78?(a) There should be no leading zeros(b) We cannot have more than 4 bytes in an IPv4 address(c) Each byte should be less than or equal to 255(d) No errorI got this question in a national level competition.I would like to ask this question from IP Addressing in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The CORRECT choice is (a) There should be no LEADING ZEROS

For EXPLANATION I WOULD say: There should be no leading zeros.

83.

The certificate_request massage includes two parameters, one of which is-(a) certificate_extension(b) certificate_creation(c) certificate_exchange(d) certificate_typeThe question was asked during a job interview.Enquiry is from Secure Socket Layer topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct choice is (d) certificate_type

To explain I would say: The certificate_request massage INCLUDES TWO parameters : certificate_type and certificate_authorities.

84.

In the alert protocol the first byte takes the value 1 or 2 which corresponds to _________ and _________ respectively.(a) Select, Alarm(b) Alert, Alarm(c) Warning, Alarm(d) Warning, FatalI got this question in semester exam.Question is taken from Secure Socket Layer in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» CORRECT answer is (d) Warning, Fatal

For EXPLANATION I would SAY: The first byte takes the VALUE warning(1) or fatal(2) to CONVEY the severity of the message.
85.

In the above question, what is the subnetwork mask for each network?(a) 27(b) 29(c) 28(d) 26I got this question in quiz.Asked question is from Designing Subnets topic in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT option is (C) 28

Explanation: nsub1 = nsub2 = nsub3 = nsub4 = n + LOG2(N/Nsub) =26+ log2(64/16)= 28.
86.

Number of addresses :(a) 64(b) 128(c) 256(d) 512I got this question in an interview for job.This is a very interesting question from Designing Subnets topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT ANSWER is (C) 256

The BEST explanation: N = 2^(32-24) = 256.
87.

Each block in class A contains _____________ addresses.(a) 2^16(b) 2^24(c) 2^8(d) 2^14The question was asked in a job interview.My question is from Classful Addressing topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct option is (B) 2^24

To explain: Each block in CLASS A CONTAINS 2^24 ADDRESSES.

88.

Each block in class B contains __________ addresses.(a) 2^8(b) 2^24(c) 2^14(d) 2^16This question was posed to me by my college director while I was bunking the class.My query is from Classful Addressing in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right ANSWER is (d) 2^16

Explanation: Each BLOCK in CLASS B CONTAINS 2^16 ADDRESSES.

89.

Calculation of the certificate_verify in TLS involves the use of a finished_label. The finished_label is the string-(a) client finished for the client(b) client finished for the client, server finished for the server(c) server finished for the server(d) client finished for the server, server finished for the clientThis question was addressed to me in an interview for internship.I'd like to ask this question from Transport Layer Security and HTTPS in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right answer is (B) CLIENT FINISHED for the client, server finished for the server

Easiest explanation: The finished_label is the string client finished for the client, server finished for the server.

90.

Which is the key exchange algorithm used in CipherSuite parameter?(a) RSA(b) Fixed Diffie-Hellman(c) Ephemeral Diffie-Hellman(d) Any of the mentionedThis question was addressed to me by my college professor while I was bunking the class.Origin of the question is Secure Socket Layer in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct answer is (d) Any of the mentioned

For explanation I would say: We can use either of the following for the CipherSuite KEY exchange-

i) RSA

ii) Fixed Diffie-Hellman

iii) Ephemeral Diffie-Hellman

iv) ANONYMOUS Diffie-Hellman

v) FORTEZZA.

91.

Which protocol is used for the purpose of copying the pending state into the current state?(a) Alert Protocol(b) Handshake Protocol(c) Upper-Layer Protocol(d) Change Cipher Spec ProtocolThis question was posed to me in an interview for internship.Asked question is from Secure Socket Layer in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The CORRECT CHOICE is (d) CHANGE Cipher Spec PROTOCOL

Explanation: The Change Cipher Spec Protocol is used for this action.

92.

What is the prefix length for group 2?(a) 14(b) 7(c) 25(d) None of the mentionedI had been asked this question in unit test.The doubt is from Designing Subnets topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The CORRECT choice is (c) 25

The explanation: 128 customers, each needs 128 addresses

Suffix LENGTH = 7 (2^7=128)

Prefix length = 32-7= 25.

93.

Last Address is –(a) 180.8.255.255.(b) 180.8.255.0.(c) 180.12.0.255.(d) 180.9.255.255.I had been asked this question at a job interview.I would like to ask this question from Classful Addressing topic in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct CHOICE is (a) 180.8.255.255.

Explanation: KEEP the leftmost 16 BITS and set the rightmost 16 bits all to 1s. Last ADDRESS = 180.8.255.255.

94.

First Address is –(a) 180.8.0.0(b) 180.7.64.0(c) 180.8.1.256(d) 180.12.0.0I got this question in an interview for internship.I'd like to ask this question from Classful Addressing topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct choice is (a) 180.8.0.0

For explanation I WOULD say: KEEP the leftmost 16 bits and SET the rightmost 16 bits all to 0S. First address = 180.8.0.0, in which 16 is the VALUE of n.

95.

Convert the following dotted-decimal notation to binary notation – 111.56.45.78(a) 01101111 00111000 00101101 01001110(b) 11101111 00111000 00101101 10001110(c) 10000000 00001011 00000011 00011111(d) 10000001 00001011 00001011 11101111I got this question at a job interview.The query is from IP Addressing topic in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right ANSWER is (a) 01101111 00111000 00101101 01001110

Easy explanation: Replacing each decimal NUMBER with its BINARY equivalent we GET 01101111 00111000 00101101 01001110.

96.

SSH_MSG_USERAUTH_FALIURE is represented by byte-(a) 50(b) 51(c) 52(d) None of the mentionedI had been asked this question in unit test.The origin of the question is Secure Shell(SSH) topic in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct choice is (b) 51

For explanation I would say: SSH_MSG_USERAUTH_FALIURE is REPRESENTED by 51. This happens in cases when the SERVER (1) either rejects the AUTHENTICATION request or (2) accepts the request but REQUIRES one or more additional authentication methods.

97.

In the Key generation procedure, the Encryption key from the server to client is represented as:(a) HASH(K||H||”C”||session_id)(b) HASH(K||H||”B”||session_id)(c) HASH(K||H||”F”||session_id)(d) HASH(K||H||”D”||session_id)I had been asked this question in final exam.This is a very interesting question from Secure Shell(SSH) topic in portion Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Right OPTION is (d) HASH(K||H||”D”||session_id)

To EXPLAIN I would say: The INITIALIZATION Vector from the server to client is given by HASH(K||H||”D”||session_id).

98.

In the Key generation procedure, the Integrity Key from the client to server is represented as:(a) HASH(K||H||”E”||session_id)(b) HASH(K||H||”C”||session_id)(c) HASH(K||H||”B”||session_id)(d) HASH(K||H||”D”||session_id)I have been asked this question at a job interview.Origin of the question is Secure Shell(SSH) in division Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

Correct CHOICE is (a) HASH(K||H||”E”||session_id)

The explanation is: The INITIALIZATION VECTOR from the CLIENT to server is GIVEN by HASH(K||H||”E”||session_id).

99.

The correct order of operations in the SSH Transport Layer Protocol Packet Formation is –(a) Compression; Encryption; Padding and Addition of seq bits(b) Compression; Padding and Addition of seq bits; Encryption(c) Encryption; Padding and Addition of seq bits; Compression(d) Padding and Addition of seq bits; Encryption; CompressionI have been asked this question at a job interview.Enquiry is from Secure Shell(SSH) in chapter Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer»

The correct answer is (b) COMPRESSION; PADDING and ADDITION of seq bits; Encryption

For EXPLANATION: The correct order of operation is – Compression; Padding and Addition of seq bits; Encryption.

100.

The client_key_exchange message uses a pre master key of size –(a) 48 bytes(b) 56 bytes(c) 64 bytes(d) 32 bytesI have been asked this question by my college director while I was bunking the class.This intriguing question originated from Secure Socket Layer in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer» RIGHT ANSWER is (a) 48 bytes

Explanation: The client_key_exchange message uses a PRE master key of SIZE 48 bytes.