This logic circuit contains a NOT gate followed by an AND gate. The NOT gate inverts input \( A \), and the AND gate takes both inputs (inverted \( A \) and \( B \)).
The output \( Y \) will be LOW only when the combination of inputs results in a LOW output from the AND gate.
Key Observation:
- For an AND gate to output LOW, at least one of the inputs must be LOW.
- Therefore, when \( A \) is HIGH and \( B \) is LOW, the NOT gate will make \( A \) LOW, and the AND gate will output LOW.
Thus, the correct answer is (A): A is HIGH and B is LOW.
Match the LIST-I with LIST-II
LIST-I | LIST-II |
---|---|
A. Brillouin Zone | Provides the understanding of the origin of allowed and forbidden bands in solids. |
B. Extended Zone Scheme | The electrons in a crystal behave like free electrons for most of the \( k \) values except when it approaches \( n\pi/a \). |
C. Periodic Zone Scheme | The E-k curve for several values of \( n \) reduced into the first zone for a simple cubic lattice with vanishing potential. |
D. Reduced Zone Scheme | The E-K curve is not continuous and has discontinuities at \( k = \pm n\pi/a \), where \( n=1,2,3,\dots \). |
Choose the correct answer from the options given below:
At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)