A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a…

$$\begin{aligned} & \mathrm{T} \rightarrow \text { Time } \quad\left[\mathrm{T}^1\right] \\ & \mathrm{S} \rightarrow \text { Surface Tension } \mathrm{S}=\frac{\mathrm{F}}{\ell}=\frac{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^1\right]}=\left[\mathrm{M}^1 \mathrm{~T}^{-2}\right] \\ & \mathrm{A} \rightarrow \text { Area } \quad\left[\mathrm{L}^2\right] \\ & \rho \rightarrow \text { density }\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right] \\ & \mathrm{R} \rightarrow \text { Radius }\left[\mathrm{L}^1\right] \\ & \text { With Dimensional Analysis } \rightarrow \\ & \mathrm{T} \propto \mathrm{S}^\alpha \mathrm{A}^\beta \rho^\gamma \mathrm{R}^\delta \\ & {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{MT}^{-2}\right]^\alpha\left[\mathrm{L}^2\right]^\beta\left[\mathrm{ML}^{-3} \mathrm{~V}^\gamma[\mathrm{L}]^\delta\right.} \\ & {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{M}^{\alpha+\gamma}, \mathrm{L}^{2 \beta+\delta-3 \gamma}, \mathrm{~T}^{-2 \alpha}\right]} \\ & \Rightarrow-2 \alpha=1 \\ & \Rightarrow \alpha=-\frac{1}{2} \\ & \Rightarrow \alpha+\gamma=0 \\ & 2 \beta+\gamma-3 \gamma=0 \\ & 2 \beta+\delta-3\left(\frac{1}{2}\right)=0 \\ & 2 \beta+\delta=\frac{3}{2}\end{aligned}$$​T→ Time [T1]S→ Surface Tension S=ℓF​=[L1][M1 L1 T−2]​=[M1 T−2]A→ Area [L2]ρ→ density [M1 L−3]R→ Radius [L1] With Dimensional Analysis →T∝SαAβργRδ[M0 L0 T1]=[MT−2]α[L2]β[ML−3 Vγ[L]δ[M0 L0 T1]=[Mα+γ,L2β+δ−3γ, T−2α]⇒−2α=1⇒α=−21​⇒α+γ=02β+γ−3γ=02β+δ−3(21​)=02β+δ=23​​