The structure of a water molecule is shown in figure. Find the distance of the centre of mass of the molecule from the centre of the oxygen atom.

#### Solution

Let OX be the x-axis, OY be the Y-axis and O be the origin.

\[\text{ Mass of O atom, m}_1 = 16 \text{unit }\]

Let the position of oxygen atom be origin.

\[\Rightarrow x_1 = y_1 = 0\]

\[\text{ Mass of H atom ,m}_2 = 1 \text{unit}\]

\[ x_2 = - 0 . 96 \times {10}^{- 10} \sin 52^\circ\]

\[ y_2 = - 0 . 96 \times {10}^{- 10} \cos 52^\circ\]

\[\text{Now, m}_3 = 1 \text{unit}\]

\[ x_3 = 0 . 96 \times {10}^{- 10} \sin 52^\circ\]

\[ y_3 = - 0 . 96 \times {10}^{- 10} \cos 52^\circ \]

The X coordinate of the center of mass is given by:

\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}\]

\[ = \frac{16 \times 0 + 1 \times \left( - 0 . 96 \times {10}^{- 10} \sin 52^\circ\right) + 1 \times 0 . 96 \times {{10}^-}^{10} \sin 52^\circ] }{1 + 1 + 16} = 0\]

\[\text{ The Y coordinate of the center of mass is given by: }\]

\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}\]

\[ = \frac{16 \times 0 + 2 \times 0 . 96 \times {10}^{- 10} \cos 52^\circ]}{1 + 1 + 16}\]

\[ = \frac{2 \times 0 . 96 \times {10}^{- 10} \cos 52^\circ]}{18}\]

\[ = 6 . 4 \times {10}^{- 12} \text{m}\]

Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is (\[ = 6 . 4 \times {10}^{- 12} \text{m}\]).