A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy ? You can neglect viscous drag of air and assume that density of air is constant.

$m=$ Mass of balloon

$\mathrm{V}=$ Volume of balloon

$\rho_{\mathrm{He}}=$ Density of helium

$\rho_{\text {air }}=$ Density of air

Volume $\mathrm{V}$ of balloon displaces volume $\mathrm{V}$ of air.

So, $\mathrm{V}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) \mathrm{g}=m a=m \frac{d v}{d t}$

Integrating with respect to $t$,

$\Rightarrow \mathrm{V}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g t=m v$

$\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{\mathrm{V}^{2}}{m^{2}}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right)^{2} g^{2} t^{2} \ldots \text { (ii) }$

$=\frac{1}{2 m} \mathrm{~V}^{2}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right)^{2} g^{2} t^{2}$

If the balloon rises to a height $\mathrm{h}$,

from $\mathrm{s}=u t+\frac{1}{2} a t^{2}$

$\therefore h=\frac{1}{2} a t^{2}$

$=\frac{1}{2} \frac{\mathrm{V}\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right)}{m} g t^{2} \quad(\because u=0)$

From $(iii)$ and $(ii)$

$\begin{aligned} \frac{1}{2} m v^{2} =\left[\mathrm{V}\left(\rho_{\mathrm{a}}-\rho_{\mathrm{He}}\right) g t^{2}\left[\frac{1}{2 \mathrm{~m}} \mathrm{~V}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g t^{2}\right]\right.\\ =\mathrm{V}\left(\rho_{\mathrm{a}}-\rho_{\mathrm{He}}\right) g h \end{aligned}$