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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2024 · 31 January · Evening Shift · Physics · Semiconductors

The output of the given circuit diagram is

The output of the given circuit diagram is

The output of the given circuit diagram is

Options

  1. ATwo strings (A, B) having linear densities μ_ A =2 × 10^ -4 ~kg / m and, μ_ B =4 × 10^ -4 ~kg / m and lengths L_ A =2.5
  2. BTwo strings (A, B) having linear densities μ_ A =2 × 10^ -4 ~kg / m and, μ_ B =4 × 10^ -4 ~kg / m and lengths L_ A =2.5
  3. CTwo strings (A, B) having linear densities μ_ A =2 × 10^ -4 ~kg / m and, μ_ B =4 × 10^ -4 ~kg / m and lengths L_ A =2.5
  4. DTwo strings (A, B) having linear densities μ_ A =2 × 10^ -4 ~kg / m and, μ_ B =4 × 10^ -4 ~kg / m and lengths L_ A =2.5

Correct answer

C. Two strings (A, B) having linear densities μ_ A =2 × 10^ -4 ~kg / m and, μ_ B =4 × 10^ -4 ~kg / m and lengths L_ A =2.5

Step-by-step solution

The output of the given circuit diagram is

If A=0;A¯=1A = 0; \overset{¯}{A} = 1

A=1;A¯=0A = 1; \overset{¯}{A} = 0

B=0;B¯=1B = 0; \overset{¯}{B} = 1

B=1;B¯=0B = 1; \overset{¯}{B} = 0

Using De Morgan's theorem, we can write 

Y=(A+B¯)+(A¯+B)¯=(A+B¯)¯⋅(A¯+B)¯=(A¯⋅B)⋅(A⋅B¯)=0Y = \overset{¯}{\left(A + \overset{¯}{B}\right) + \left(\overset{¯}{A} + B\right)} = \overset{¯}{\left(A + \overset{¯}{B}\right)} \cdot \overset{¯}{\left(\overset{¯}{A} + B\right)} = \left(\overset{¯}{A} \cdot B\right) \cdot \left(A \cdot \overset{¯}{B}\right) = 0

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