The image presents three problems. Problem 11 asks for the binary equivalent of the hexadecimal number $F52_{16}$. Problem 12 provides the ASCII code for the character 'C' and asks for the ASCII code for the word 'BEG'. Problem 13 presents a logic circuit composed of a NOR gate followed by a NOT gate, and asks for the equivalent logic gate.

1. Problem Description

The image presents three problems.

Problem 11 asks for the binary equivalent of the hexadecimal number $F52_{16}$.

Problem 12 provides the ASCII code for the character 'C' and asks for the ASCII code for the word 'BEG'.

Problem 13 presents a logic circuit composed of a NOR gate followed by a NOT gate, and asks for the equivalent logic gate.

2. Solution Steps

Problem 11:

To convert a hexadecimal number to binary, we convert each hexadecimal digit to its 4-bit binary equivalent.

$F_{16} = 15_{10} = 1111_2$

$5_{16} = 5_{10} = 0101_2$

$2_{16} = 2_{10} = 0010_2$

Thus, $F52_{16} = 111101010010_2$.

Problem 12:

We are given that the ASCII code for 'C' is $1000011_2$. We can determine the ASCII codes for 'B', 'E', and 'G' by using the fact that ASCII codes are sequential for letters.

'A' = 'C' - 2

'B' = 'C' - 1 = $1000011_2 - 1 = 1000010_2$

'C' = $1000011_2$

'D' = 'C' + 1 = $1000100_2$

'E' = 'C' + 2 = $1000101_2$

'F' = 'C' + 3 = $1000110_2$

'G' = 'C' + 4 = $1000111_2$

Therefore, the ASCII code for 'BEG' is $1000010_2$, $1000101_2$, $1000111_2$.

Problem 13:

The circuit consists of a NOR gate followed by a NOT gate. A NOR gate outputs 1 only if both inputs are 0, otherwise it outputs

0. A NOT gate inverts the input.

Let A and B be the inputs to the NOR gate. Then the output of the NOR gate is $\overline{A + B}$.

The NOT gate inverts this output, so the final output C is $\overline{\overline{A + B}} = A + B$.

Thus, the equivalent gate is an OR gate.

3. Final Answer

Problem 11: The binary equivalent of $F52_{16}$ is $111101010010_2$. The answer is (3).

Problem 12: The ASCII code for 'BEG' is $1000010$ $1000101$ $1000111$. The answer is (1).

Problem 13: The equivalent logic gate is an OR gate. The answer is (3).