The subnet mask 255.255.255.192
(A) extends the network portion to 16 bits
(B) extends the network portion to 26 bits
(C) extends the network portion to 36 bits
(D) has no effect on the network portion of an IP address
Default sub-net mask for Class C is 255.255.255.192
(192)10 = (11000000)2
Since, 192 is written as 11000000, it has 2 sub-nets and remaining all hosts.
So, for first three octets, 24 bits are fixed and for last octet 2 bits are fixed, i.e. 24 + 2 = 26 bits
Option (B) is correct.
Quiz of this Question
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