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NEET Physics QUESTION #1050
Question 1
A balloon of surface tension \(S\), outlet area \(A\), gas density \(\rho\), radius \(R\). Gas flows out: radius changes from \(R\) to 0 in time \(T\). Speed \(v(r)\propto r^a\) and \(T\propto S^\alpha A^\beta \rho^\gamma R^\delta\). Then:
  • \(a=-\frac{1}{2},\alpha=-\frac{1}{2},\beta=-1,\gamma=\frac{1}{2},\delta=\frac{7}{2}\)✔️
  • \(a=\frac{1}{2},\alpha=\frac{1}{2},\beta=-\frac{1}{2},\gamma=\frac{1}{2},\delta=\frac{7}{2}\)
  • \(a=\frac{1}{2},\alpha=\frac{1}{2},\beta=-1,\gamma=+1,\delta=\frac{3}{2}\)
  • \(a=-\frac{1}{2},\alpha=-\frac{1}{2},\beta=-1,\gamma=-\frac{1}{2},\delta=\frac{5}{2}\)
Correct Answer Explanation
Pressure inside balloon \(=\frac{4S}{r}\). By Bernoulli: \(\frac{4S}{r}=\frac{1}{2}\rho v^2\Rightarrow v=\sqrt{\frac{8S}{\rho r}}\propto r^{-1/2}\). So \(a=-\frac{1}{2}\). For time: \(T\propto \frac{R}{v(R)}\cdot f(geometry)\). Dimensional analysis of \(T\propto S^\alpha A^\beta\rho^\gamma R^\delta\) with \([T]=s\) gives \(\alpha=-1/2, \beta=-1, \gamma=1/2, \delta=7/2\).