The electron affinity values for each case are determined as follows:
(A) Be + e$^-$ $\to$ Be$^-$: Electron affinity is negative (E.A. = --ive).
(B) N + e$^-$ $\to$ N$^-$: Electron affinity is negative (E.A. = --ive).
(C) O + e$^-$ $\to$ O$^-$: Electron affinity is positive for O $\to$ O$^-$ (E.A. = +ive).
(D) Na + e$^-$ $\to$ Na$^-$: Electron affinity is positive (E.A. = +ive).
(E) Al + e$^-$ $\to$ Al$^-$: Electron affinity is positive (E.A. = +ive).
Thus, the correct answer is: A, B, and C only.
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: