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.
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Let OX be the x-axis, OY be the Y-axis and O be the origin. Mass of O atom, $m_1 = 16$ unit Let the position of oxygen atom be origin $\Rightarrow\text{x}_1=\text{y}_1=0$ Mass of H atom, $m_2 = 1$ unit $\text{x}_2=-0.96\times10^{-10}\text{sim}52^\circ$ $\text{y}_2=-0.96\times10^{-10}\cos52^\circ$ Now, $m_3 = 1$ unit $\text{x}_3=0.96\times10^{-10}\sin52^\circ$ $\text{y}_3=-0.96\times10^{-10}\cos52^\circ$ The X coordinate of the center or mass is given by: $\text{x}_\text{cm}=\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$ $=\frac{16\times0\times1\times(-0.96\times10^{-10}\text{sin}52^\circ)+1\times0.96\times10^{10}\sin52^\circ}{1+1+16}=0$ The Y coordinate of the center of mass is given by: $\text{y}_\text{cm}=\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2+\text{m}_3\text{y}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$ $=\frac{16\times0+2\times0.96\times10^{10}\cos52^\circ}{1+1+16}$ $=\frac{2\times0.96\times10^{-10}\cos52^\circ}{18}$ $=6.4\times10^{-12}\text{m}$ Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is $\text{s}(=6.4\times10^{1-12}\text{m})$
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