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(A) Let the terminal velocity $v_T$ be expressed as $v_T = k \cdot m^a \eta^b r^c g^d$,where $k$ is a dimensionless constant.The dimensional formulas

Vedclass · Accepted Answer

$v_T \propto \frac{mg}{\eta r}$

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(A) Let the terminal velocity $v_T$ be expressed as $v_T = k \cdot m^a \eta^b r^c g^d$,where $k$ is a dimensionless constant.The dimensional formulas are:$[v_T] = [L T^{-1}]$$[m] = [M]$$[\eta] = [M L^{-1} T^{-1}]$$[r] = [L]$$[g] = [L T^{-2}]$Substituting these into the equation:$[L T^{-1}] = [M]^a [M L^{-1} T^{-1}]^b [L]^c [L T^{-2}]^d$$[M^0 L^1 T^{-1}] = [M^{a+b} L^{-b+c+d} T^{-b-2d}]$Comparing the powers of $M, L,$ and $T$ on both sides:For $M$: $a + b = 0 \Rightarrow a = -b$For $T$: $-b - 2d = -1 \Rightarrow b + 2d = 1$For $L$: $-b + c + d = 1$Using Stokes' Law,the terminal velocity is $v_T = \frac{2r^2(\rho - \sigma)g}{9\eta}$. Since the mass $m$ of the ball is proportional to $r^3$ (i.e.,$m \propto r^3$),we substitute $r \propto m^{1/3}$ into the relation. Dimensional analysis shows that $v_T \propto \frac{mg}{\eta r}$ is the only relation consistent with the dimensions of velocity.

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