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

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Solution

let $0.96 \times 10^{- 10} = d$

$X c o m = \frac{m_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

$= \frac{- d s i n 52^{\circ} + d s i n 52^{\circ} + 16 \times 0}{1 + 1 + 16}$

= 0

$Y c o m = \frac{0 + 0 + 16 y_{3}}{18}$

$\frac{8}{9} \times 0.96 \times 10^{10} c o s 52^{\circ}$

hence coordinates are

$(0, \frac{8}{9} \times 0.9 \times 10^{- 10} c o s 52^{\circ})$

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

let 0.96×10−10=dXcom=m1+m2x2+m3x3m1+m2+m3=−d sin52∘+d sin52∘+16×01+1+16= 0Ycom=0+0+16y31889×0.96×1010cos52∘hence coordinates are(0,89×0.9×10−10cos52∘)