Question

The relationship between kinematic viscosity $\left(v\right)$ and dynamic viscosity $\left(\mu \right)$ is

• A. $v\left(\text{centistokes}\right)=\mu \left(\text{centiposie}\right)/\rho \left(g/c{m}^3\right)$
• B. $\mu \left(\text{centistokes}\right)=v\left(\text{centiposie}\right)/\rho \left(g/c{m}^3\right)$
• C. $\mu \left(\text{centipoise}\right)=\mu \left(\text{centistokes}\right)/\rho \left(g/c{m}^3\right)$
• D. $v\left(\text{stokes}\right)=\mu \left(\text{poise}\right)/\rho \left(g/c{m}^3\right)$

Answer 1

The correct option is A $v\left(\text{centistokes}\right)=\mu \left(\text{centiposie}\right)/\rho \left(g/c{m}^3\right)$

$\text{Kinematic viscosity(v)}=\frac{\text{Dynamic viscocity}\left(\mu \right)}{\text{Density}\left(\rho \right)}$

$\text{S.I unit of}v=m^2/s$

$\text{C.G.S unit of}v=c{m}^2/s\text{or stoke}$

$\text{S.I unit of}\mu =Ns/m^2$

$\text{C.G.S unit of}\mu =dynS/c{m}^2\text{or poise}$

$\text{S.I unit of}\rho =kg/m^3$

$\text{C.G.S unit of}\rho =g/c{m}^3$

$1\text{centistokes}={10}^{-2}\text{stokes}$

$1\text{centipoise}={10}^{-2}\text{poise}$

$\therefore \text{in option (a)}$

$v({10}^{-2}$ stokes$)$ $=\frac{\mu \left({10}^{-2}\text{poise}\right)}{\rho \left(g/c{m}^3\right)}$

$v\left(\text{stokes}\right)=\frac{\mu \left(\text{poise}\right)}{\rho \left(g/c{m}^3\right)}$

$\text{C.G.S}=\text{C.G.S}$

$\text{In both sides C.G.S system is followed.}$