Question

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

• A. $\frac{m{r}^2}{6\pi \eta }$
• B. $\sqrt{\left(\frac{6\pi mr\eta }{g^2}\right)}$
• C. $\frac{m}{6\pi \eta rv}$
• D. none of these

Answer 1

Answer: (D)

none of these

Since $\tau$ is time constant will have Dimension of Seconds($T$)

Option A has dimensions: $\frac{M{L}^2}{M{L}^{-1}{T}^{-2}}=L^3{T}^2$

Option B has dimensions: $\sqrt{\frac{\left(ML\right)\ast \left(M{L}^{-1}{T}^{-2}\right)}{L^2{T}^{-4}}}=M{L}^{-1}T$

Option C has dimensions: $\frac{M}{\left(M{L}^{-1}{T}^{-2}\right)\ast \left(L\right)\ast \left(L{T}^{-1}\right)}=\frac{T^3}{L}$

Hence, option D is correct.