The graph between terminal velocity (v_t) (al | vedclass

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(A) The terminal velocity $v_t$ of a spherical body of radius $r$ and density $\rho$ falling through a fluid of density $\sigma$ and viscosity $\eta$

Vedclass · Accepted Answer

Straight line with positive slope

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(A) The terminal velocity $v_t$ of a spherical body of radius $r$ and density $\rho$ falling through a fluid of density $\sigma$ and viscosity $\eta$ is given by the formula:$v_t = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$Here,$g$ is the acceleration due to gravity,$\rho$ is the density of the sphere,$\sigma$ is the density of the fluid,and $\eta$ is the coefficient of viscosity.Since $g, \rho, \sigma,$ and $\eta$ are constants for a given system,we can write:$v_t = K r^2$,where $K = \frac{2g(\rho - \sigma)}{9\eta}$ is a constant.This equation is of the form $y = mx$,which represents a straight line passing through the origin with a positive slope $K$ when $v_t$ is plotted against $r^2$.

The graph between terminal velocity $(v_t)$ (along $y$-axis) and the square of the radius $(r^2)$ (along $x$-axis) of a spherical body of density $\rho$ falling through a fluid of density $\sigma$ is:

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