Question:

An air bubble of radius $1 \mathrm{~cm}$ in water has an upward acceleration $9.8 \mathrm{~cm} \mathrm{~s}^{-2}$. The density of water is $1 \mathrm{gm}$ $\mathrm{cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $\left(\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^{2}\right)$

Solution:

(3) Given:

Radius of air bubble $=1 \mathrm{~cm}$,

Upward acceleration of bubble, $a=9.8 \mathrm{~cm} / \mathrm{s}^{2}$,

$\rho_{\text {water }}=1 \mathrm{~g} \mathrm{~cm}^{-3}$

$F_{\text {buoyant }}-m g=m a \Rightarrow m=\frac{F_{\text {buoyant }}}{g+a}$

$\therefore m=\frac{\left(V \rho_{\omega} g\right)}{g+a}=\frac{V \rho_{\omega}}{1+\frac{a}{g}}=\frac{(4.19) \times 1}{1+\frac{9.8}{980}}=\frac{4.19}{1.01}=4.15 \mathrm{~g}$