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NEET · Physics · STD 11 - 1. units,dimensions and measurement

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

  1. A \(a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-1, \gamma=+1, \delta=\frac{3}{2}\)
  2. B \(a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=-\frac{1}{2}, \delta=\frac{5}{2}\)
  3. C \(a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}\)
  4. D \(a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-\frac{1}{2}, \gamma=\frac{1}{2}, \delta=\frac{7}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}\)

Step-by-step Solution

Detailed explanation

\(T \rightarrow\) Time \(\left[ T ^1\right]\)
\(S \rightarrow\) Surface Tension \(S =\frac{ F }{\ell}=\frac{\left[ M ^1 L^1 T^{-2}\right]}{\left[ L ^1\right]}=\left[ M ^1 T^{-2}\right]\)
\(A \rightarrow\) Area \(\left[ L ^2\right]\)
\(\rho \rightarrow\) density \(\left[ M ^1 L^{-3}\right]\)
\(R \rightarrow\) Radius \(\left[ L ^1\right]\)
With Dimensional Analysis \(\rightarrow\)
\(T \propto S ^\alpha A ^\beta \rho^\gamma R ^\delta\)
\(\left[ M ^0 L^0 T^1\right]=\left[ MT ^{-2}\right]^\alpha\left[ L ^2\right]^\beta\left[ ML ^{-3} V^\gamma[ L ]^\delta\right.\)
\(\left[ M ^0 L^0 T^1\right]=\left[ M ^{\alpha+\gamma}, L ^{2 \beta+\delta-3 \gamma}, T^{-2 \alpha}\right]\)
\(\Rightarrow-2 \alpha=1\)
\(\Rightarrow \alpha=-\frac{1}{2}\)
\(\Rightarrow \alpha+\gamma=0\)
\(2 \beta+\gamma-3 \gamma=0\)
\(2 \beta+\delta-3\left(\frac{1}{2}\right)=0\)
\(2 \beta+\delta=\frac{3}{2}\)
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