Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

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. - Physics

Short Note

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 ($= 6 . 4 \times {10}^{- 12} \text{m}$).

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HC Verma Class 11, 12 Concepts of Physics 1
Chapter 9 Centre of Mass, Linear Momentum, Collision
Q 2 | Page 159