A spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in…

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 $(\tau)$. Dimensionally, $\tau$ can be represented by
  1. $\frac{m^{2}}{6 \pi n}$
  2. $\sqrt{\left(\frac{6 m n \eta}{g^{2}}\right)}$
  3. $\frac{m}{6 \pi n m}$
  4. None of these

Solution

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

Asked in: JEE Mains - Units and Dimensions - Test 2

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