An Air Bubble Of A Radius 1cm In Water Has An Upward Acceleration ...

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According to Newtons 2nd law

B-mg=ma

m=\frac{B}{g+a}

Given

\begin{array}{l} \text { Volume } =\mathrm{V}=\frac{4 \pi}{3} \mathrm{r}^{3}=\frac{4 \pi}{3} \times(1)^{3}=4.19 \mathrm{~cm}^{3} \\ \mathrm{a}=9.8 \mathrm{~cm} / \mathrm{s}^{2} \end{array}

And

B=\mathrm{V} \rho_{w} \mathrm{g}

So

\begin{array}{l} \mathrm{m}=\frac{\left(\mathrm{V} \rho_{w} \mathrm{g}\right)}{\mathrm{g}+\mathrm{a}}=\frac{\mathrm{V} \rho_{\mathrm{w}}}{1+\frac{\mathrm{a}}{\mathrm{g}}} =\frac{(4.19) \times 1}{1+\frac{9.8}{980}}=\frac{4.19}{1.01}=4.15 \mathrm{gm} \end{array}

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avinash.dongre

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