A spherical body of mass m and radius r is al | vedclass

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Question

$\left[\frac{m r^{2}}{6 \pi \eta}\right]=\left[\frac{M L^{2}}{M L^{-1} T^{-1}}\right]=\left[L^{3} T\right]$

As we have $[\eta]=\left[M L^{-1} T^{-1

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

$\left[\frac{m r^{2}}{6 \pi \eta}\right]=\left[\frac{M L^{2}}{M L^{-1} T^{-1}}\right]=\left[L^{3} T\right]$

As we have $[\eta]=\left[M L^{-1} T^{-1}\right]$ $\left[\left(\frac{6 \pi m r \eta}{g^{2}}\right)^{1 / 2}\right]=\left[\left(\frac{M L M L^{-1} T^{-1}}{L^{2} T^{-4}}\right)^{1 / 2}\right]$

$\left[\frac{m}{6 \pi \eta r v}\right]=\left[\frac{M}{M L^{-1} T^{-1} L L T^{-1}}\right]=\left[L^{-1} T^{2}\right]$

Thus, none of the given expressions have the dimensions of time.

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 time constant $\left( \tau \right)$. Dimensionally $\tau $ can be represented by

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