Calculate the energy released when 1000 small water drops each of same radius \({10}^{{-{7}}}{m}\) coalesce to form one large drop. The surface tension of water is \({7.0}\times{10}^{{-{2}}}{N}/{m}\).

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Calculate the energy released when 1000 small water drops each of same radius \({10}^{{-{7}}}{m}\) coalesce to form one large drop. The surface tension of water is \({7.0}\times{10}^{{-{2}}}{N}/{m}\).

Answer: A、B

Let r be the radius drops and \({R}\) of bigger one. Equanting the initial and final volume, we have

\(\frac{{{4}}}{{{3}}}\pi{R}^{{{3}}}={\left({1000}\right)}{\left(\frac{{{4}}}{{{3}}}\pi{r}^{{{3}}}\right)}\)

or, \({R}={10}{r}={\left({10}\right)}{\left({10}^{{-{7}}}\right)}{m}\)

or, \({R}={10}^{{-{6}}}{m}\)

Further, the water drops have only one free surface. Therefore,

\(\Delta{A}={4}\pi{R}^{{{2}}}-{\left({1000}\right)}{\left({4}\pi{r}^{{{2}}}\right)}\)

\(={4}\pi{\left[{\left({10}^{{-{6}}}\right)}^{{{2}}}-{\left({10}^{{{3}}}\right)}{\left({10}^{{-{7}}}\right)}^{{{2}}}\right]}\)

\(=-{36}\pi{\left({10}^{{-{12}}}\right)}{m}^{{{2}}}\)

Here, negative sign implies that surface area is decreasing. Hence, energy released in the process. \({U}={T}{\left|\Delta{A}\right|}={\left({7}\times{10}^{{-{2}}}\right)}{\left({36}\pi\times{10}^{{-{12}}}\right)}{J}\)

\(={7.9}\times{10}^{{-{12}}}{J}\).

Let r be the radius drops and \({R}\) of bigger one. Equanting the initial and final volume, we have

\(\frac{{{4}}}{{{3}}}\pi{R}^{{{3}}}={\left({1000}\right)}{\left(\frac{{{4}}}{{{3}}}\pi{r}^{{{3}}}\right)}\)

or, \({R}={10}{r}={\left({10}\right)}{\left({10}^{{-{7}}}\right)}{m}\)

or, \({R}={10}^{{-{6}}}{m}\)

Further, the water drops have only one free surface. Therefore,

\(\Delta{A}={4}\pi{R}^{{{2}}}-{\left({1000}\right)}{\left({4}\pi{r}^{{{2}}}\right)}\)

\(={4}\pi{\left[{\left({10}^{{-{6}}}\right)}^{{{2}}}-{\left({10}^{{{3}}}\right)}{\left({10}^{{-{7}}}\right)}^{{{2}}}\right]}\)

\(=-{36}\pi{\left({10}^{{-{12}}}\right)}{m}^{{{2}}}\)

Here, negative sign implies that surface area is decreasing. Hence, energy released in the process. \({U}={T}{\left|\Delta{A}\right|}={\left({7}\times{10}^{{-{2}}}\right)}{\left({36}\pi\times{10}^{{-{12}}}\right)}{J}\)

\(={7.9}\times{10}^{{-{12}}}{J}\).

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