Q109 / 180 · NEET 2025 · Physics
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 gas of density $\rho$ and takes a spherical shape of radius $R$. When the gas is allowed to flow freely out of it, its radius changes from $R$ to 0 (zero) in time $T$. If the speed $\eta(r)$ of gas coming out of the balloon depends on $r$ as $r^\alpha$ and $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$ then
- A$a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-1, \gamma=+1, \delta=\frac{3}{2}$
- B$a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=-\frac{1}{2}, \delta=\frac{5}{2}$
- Correct answerC$a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}$
- D$a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-\frac{1}{2}, \gamma=\frac{1}{2}, \delta=\frac{7}{2}$
Solution
$\begin{aligned} & \mathrm{T} \rightarrow \text { Time } \quad\left[\mathrm{T}^1\right] \\ & \mathrm{S} \rightarrow \text { Surface Tension } \mathrm{S}=\frac{\mathrm{F}}{\ell}=\frac{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^1\right]}=\left[\mathrm{M}^1 \mathrm{~T}^{-2}\right] \\ & \mathrm{A} \rightarrow \text { Area } \quad\left[\mathrm{L}^2\right] \\ & \rho \rightarrow \text { density }\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right] \\ & \mathrm{R} \rightarrow \text { Radius }\left[\mathrm{L}^1\right] \\ & \text { With Dimensional Analysis } \rightarrow \\ & \mathrm{T} \propto \mathrm{S}^\alpha \mathrm{A}^\beta \rho^\gamma \mathrm{R}^\delta \\ & {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{MT}^{-2}\right]^\alpha\left[\mathrm{L}^2\right]^\beta\left[\mathrm{ML}^{-3} \mathrm{~V}^\gamma[\mathrm{L}]^\delta\right.} \\ & {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{M}^{\alpha+\gamma}, \mathrm{L}^{2 \beta+\delta-3 \gamma}, \mathrm{~T}^{-2 \alpha}\right]} \\ & \Rightarrow-2 \alpha=1 \\ & \Rightarrow \alpha=-\frac{1}{2} \\ & \Rightarrow \alpha+\gamma=0 \\ & 2 \beta+\gamma-3 \gamma=0 \\ & 2 \beta+\delta-3\left(\frac{1}{2}\right)=0 \\ & 2 \beta+\delta=\frac{3}{2}\end{aligned}$
Hence, the correct answer is option: 'C'.