Question

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.
Figure

Answer 1

The distance of the center of mass of the molecule from the center of the oxygen atom is   [0 , 0.520 × 10⁻¹⁰ ]

Explanation:

Step 1:

Given data in the question  

m₁ = 1 gm                            

m₂ = 1 gm      

x₁ = - (0.96 x 10⁻¹⁰) sin 52°    

x₂ = - (0.96 x 10⁻¹⁰) sin 52°    

x₃ = 0

y₁ = 0                                      

y₂ = 0                           

y₃ = c(0.96 x 10⁻¹⁰) cos 52°

Step 2:

So , The place of the mass center is

=   $\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} , \frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

=    $\frac{[(- (0.96 \times 10^{-10}) \times sin 52^\circ+ (0.96 \times 10^{-10}) sin 52^\circ+16 \times 0 )}{(1+1+16)} , \frac{(0+0+16)(0.96 \times 10^{-10}) cos 52^\circ)}{(1+1+16)}$

=    $\frac{[(- (0.96 \times 10^{-10}) \times 0.788 + (0.96 \times 10^{-10}) 0.788+0 )}{(18)} , \frac{(16)(0.96 \times 10^{-10}) cos 52^\circ)}{(18)}$

=   $[0 ,\frac{8}{9} (0.96 \times 10^{-10}) cos 52^\circ ]$  

=     $[0 ,0.88 (0.96 \times 10^{-10}) 0.615]$

The place of center of the mass =   [0  ,0.520 × 10⁻¹⁰]  

In a water molecule( $H_2O$ )  the distance of the center of mass of the molecule from the center of the oxygen ($O_2$) atom is   [0 , 0.520 × 10⁻¹⁰ ]  

Answer 2

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$0.0520 \times 10^{10}$