The wavenumber ($\tilde{\nu}$) is given by:
\[ \tilde{\nu} = \frac{1}{\lambda}, \]
where $\lambda$ is the wavelength in cm.
Step 1: Convert wavelength to cm
\[ \lambda = 5800 \, \text{\AA} = 5800 \times 10^{-8} \, \text{cm}. \]
Step 2: Calculate wavenumber
\[ \tilde{\nu} = \frac{1}{5800 \times 10^{-8}} = \frac{1}{5.8 \times 10^{-5}} = 17241 \, \text{cm}^{-1}. \]
Step 3: Express as $x \times 10 \, \text{cm}^{-1}$
\[ 17241 \, \text{cm}^{-1} = 1724 \times 10 \, \text{cm}^{-1}. \]
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: