The vector field \( \vec{v} \) is given in terms of its components along the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) directions. The path \( OABO \) forms a closed loop, and the line integral needs to be evaluated along this path.
By applying the line integral along the closed loop \( OABO \) and performing the necessary calculations, the result is:
\[ \frac{1}{4} (3\pi - 1) \]
Thus, the correct answer is (A).
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.)