The given equation is Laplace's equation in Cartesian coordinates:
$ \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0 $
We need to transform this equation into polar coordinates ($r, \theta$), where $x = r \cos \theta$ and $y = r \sin \theta$.
The Laplacian operator ($\nabla^2$) in polar coordinates is given by the formula:
$ \nabla^2 u = \frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}} $
Setting the Laplacian in polar coordinates equal to zero gives the transformed Laplace equation:
$ \frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}} = 0 $
This matches Option A.
| List-I | List-II |
| Electronic Configuration | First Ionisation energy (kJ mol$^{-1}$) |
| (A). ns$^2$ | (I). 2100 |
| (B). ns$^2$np$^1$ | (II). 1400 |
| (C). ns$^2$np$^3$ | (III). 800 |
| (D). ns$^2$np$^6$ | (IV). 900 |
| List-I | List-II |
| Spectroscopy | Property |
| (A). Raman | (I). Polarizability |
| (B). FTIR | (II). Dipole Moment |
| (C). UV-Visible | (III). Absorbance |
| (D). NMR | (IV). Spin |