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(D) The equation of motion for a spherical body falling in a viscous medium is given by $m \frac{dv}{dt} = mg - 6\pi\eta rv$. Rearranging this,we get

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(D) The equation of motion for a spherical body falling in a viscous medium is given by $m \frac{dv}{dt} = mg - 6\pi\eta rv$. Rearranging this,we get $\frac{dv}{dt} = g - \frac{6\pi\eta r}{m} v$. This is a linear differential equation of the form $\frac{dv}{dt} + \frac{6\pi\eta r}{m} v = g$. The time constant $	au$ for such a system is the reciprocal of the coefficient of $v$,which is $	au = \frac{m}{6\pi\eta r}$. Now,let us check the dimensions of the given options: $1$. Dimension of $\frac{mr^2}{6\pi\eta} = \frac{[M][L^2]}{[ML^{-1}T^{-1}]} = [L^3T]$. $2$. Dimension of $\sqrt{\frac{6\pi mr\eta}{g^2}} = \sqrt{\frac{[M][L][ML^{-1}T^{-1}]}{[LT^{-2}]^2}} = \sqrt{\frac{M^2T^{-1}}{L^2T^{-4}}} = \sqrt{M^2L^{-2}T^3} = [ML^{-1}T^{1.5}]$. $3$. Dimension of $\frac{m}{6\pi\eta rv} = \frac{[M]}{[ML^{-1}T^{-1}][L][LT^{-1}]} = \frac{[M]}{[MT^{-1}L]} = [L^{-1}T]$. Since none of these expressions result in the dimension of time $[T]$,the correct option is $D$.

$A$ spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

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