A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time, the velocity of the ball attains a constant value known as terminal velocity $v_T$. The terminal velocity depends on the mass of the ball $m$, $\eta$, $r$, and acceleration due to gravity $g$. Which of the following relations is dimensionally correct?

To determine which relation is dimensionally correct, we check the dimensions of each physical quantity involved:

1. Terminal Velocity ($v_T$): Being a velocity, its dimensions are $[L T^{-1}]$ .
2. Mass ($m$): Dimension is $[M]$ .
3. Acceleration due to gravity ($g$): Dimensions are $[L T^{-2}]$ .
4. Radius ($r$): Dimension is $[L]$ .
5. Coefficient of Viscosity ($\eta$): From Stokes' Law ($F = 6\pi\eta rv$), we have $\eta = \frac{F}{6\pi rv}$. The dimensions of Force ($F$) are $[M L T^{-2}]$ . Thus, the dimensions of $\eta$ are $\frac{[M L T^{-2}]}{[L][L T^{-1}]} = [M L^{-1} T^{-1}]$ .

Now, test the dimensions of the expression in Option A: $\frac{mg}{\eta r}$

• Numerator ($mg$): $[M][L T^{-2}] = [M L T^{-2}]$.
• Denominator ($\eta r$): $[M L^{-1} T^{-1}][L] = [M T^{-1}]$.
• Ratio: $\frac{[M L T^{-2}]}{[M T^{-1}]} = [L T^{-1}]$.

Since the dimensions of $\frac{mg}{\eta r}$ match the dimensions of velocity ($v_T$), the relation $v_T \propto \frac{mg}{\eta r}$ is dimensionally correct.