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(A) The equation of motion for a sphere falling in a viscous medium is given by $m \frac{dv}{dt} = mg - 6\pi \eta rv$. At terminal velocity $v_t$,the

Vedclass · Accepted Answer

$\frac{m}{6\pi \eta r}$

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(A) The equation of motion for a sphere falling in a viscous medium is given by $m \frac{dv}{dt} = mg - 6\pi \eta rv$. At terminal velocity $v_t$,the acceleration is zero,so $mg = 6\pi \eta rv_t$. Substituting this,we get $m \frac{dv}{dt} = 6\pi \eta r (v_t - v)$. Rearranging,$\frac{dv}{v_t - v} = \frac{6\pi \eta r}{m} dt$. Integrating from $v=0$ to $v=0.63v_t$,we find the time constant $	au = \frac{m}{6\pi \eta r}$. Checking dimensions: $[	au] = [T]$. Dimension of $\frac{m}{6\pi \eta r} = \frac{[M]}{[ML^{-1}T^{-1}][L]} = \frac{[M]}{[MT^{-1}]} = [T]$. Thus,option $(a)$ is dimensionally correct.

$A$ 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 $(v_t)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

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