Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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?

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

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

$v_T \propto \frac{\eta m}{gr}$

$v_T \propto \frac{mgr}{\eta}$

Step-by-Step Solution

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

Terminal Velocity ( $v_T$ ): Being a velocity, its dimensions are $[L T^{-1}]$ .
Mass ( $m$ ): Dimension is $[M]$ .
Acceleration due to gravity ( $g$ ): Dimensions are $[L T^{-2}]$ .
Radius ( $r$ ): Dimension is $[L]$ .
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.

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