Question

Turpentine oil is flowing through a tube of length ${}^{\prime }{l}^{\prime }$ and radius ${}^{\prime }{r}^{\prime }$ . The pressure difference between the two ends of the tube is $P$. The viscosity of oil is given by $\eta =\frac{P(r^2-x^2)}{4vl}$, where $v$ is the velocity of oil at a radial distance '$x$' from the axis of the tube. The dimensions of $\eta$ are

• A. $\left[M{L}^{\circ }{T}^{\circ }\right]$
• B. $\left[ML{T}^{-1}\right]$
• C. $\left[M{L}^2{T}^{-2}\right]$
• D. $\left[M{L}^{-1}{T}^{-1}\right]$

Answer 1

The correct option is D $\left[M{L}^{-1}{T}^{-1}\right]$

Pressure, $P=\frac{F}{A}$
$\Rightarrow \left[P\right]=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

Also, $[r^2-x^2]=\left[L^2\right](\because$ '$r$' and '$x$' are both distances)

$\Rightarrow \left[\eta \right]=\left[\frac{P(r^2-x^2)}{4vl}\right]=\frac{\left[M{L}^{-1}{T}^{-2}\right]\left[L^2\right]}{\left[L{T}^{-1}\right]\left[L\right]}$
$\therefore \left[\eta \right]=\left[M{L}^{-1}{T}^{-1}\right]$