As the dragging viscous force of air on balloon is neglected so there is net Buoyant force $=V \rho g$
$=$ Vol. of air displaced $\times$ net density upward $\times g$
$=V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g($ upward $)$
Let $a$ be the upward acceleration of ballon then where, $m=$ Mass of balloon
$V=$ volume of balloon
$\rho_{\mathrm{He}}=$ Density of helium
$\rho_{\text {air }}=$ Density of air
So, $V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g=m a=$ upthrust
$V\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g=\left(m \frac{d v}{d t}\right)$
Integrating both side with respect to $t$.
$$
\begin{aligned}
&m d v=V\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) d t \\
&m v=V\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g t \\
&v=\frac{V}{m}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g t \\
&\text { K.E. of balloon }=\frac{1}{2} m v^2 \\
&\frac{1}{2} m v^2=\frac{1}{2} m \frac{V^2}{m^2}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right)^2 g^2 t^2 \\
&\quad=\frac{1}{2 m} V^2\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right)^2 g^2 t^2
\end{aligned}
$$
If the balloon rises to a height $h$ from
$$
\begin{aligned}
&s=u t+\frac{1}{2} a t^2 ;(u=0) \\
&{\left[\therefore a=\frac{V}{m}\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g\right]}
\end{aligned}
$$
So, $h=\frac{1}{2} a t^2=\frac{1}{2} \frac{V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right)}{m} g t^2$
From Eqs. (iii) and (ii),
$$
\begin{aligned}
& \frac{1}{2} m v^2=\left[V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) \mathrm{g}\right]\left[\frac{1}{2 m} V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g t^2\right] \\
=& V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g h \\
\frac{1}{2} m v^2=V \rho_{\text {air }} g h-V \rho_{\mathrm{He}} g h
\end{aligned}
$$
Rearranging the terms,
$$
\begin{aligned}
&\frac{1}{2} m v^2+V \rho_{\mathrm{He}} g h=V \rho_{\text {air }} h g \\
&\mathrm{KE}_{\text {balloon }}+\mathrm{PE}_{\text {balloon }}=\text { Change in PE of air. }
\end{aligned}
$$
Hence, as the balloon goes up, an equal volume of air comes down, increase in PE and KE of the balloon is at the cost of PE of air.