Question

Turpentine oil is flowing through a capillary tube of length $l$ and radius $r$. The pressure difference between the two ends of the tube is $p$. The viscosity of oil is given by $\eta = \dfrac{{p\left( {r_{}^2 - x_{}^2} \right)}}{{4vl}}$ . Here $v$ is the velocity of oil at a distance $x$ from the axis of the tube. From this relation, the dimension formula of $\eta $ is
$
  {\text{A}}{\text{. }}\left[ {{\text{ML}}_{}^{{\text{ - 1}}}{\text{T}}_{}^{{\text{ - 1}}}} \right] \\
    \\
  {\text{B}}{\text{. }}\left[ {{\text{MLT}}_{}^{{\text{ - 1}}}} \right] \\
    \\
  {\text{C}}{\text{. }}\left[ {{\text{ML}}_{}^{\text{2}}{\text{T}}_{}^{{\text{ - 2}}}} \right] \\
    \\
  {\text{D}}{\text{. }}\left[ {{\text{M}}_{}^0{\text{L}}_{}^{\text{0}}{\text{T}}_{}^{\text{0}}} \right] \\
    \\
 $

Answer 1

Hint: Here, we can use the dimensions of different quantities used in the given formula of viscosity, $\eta $.

Complete step by step answer:
Given that the turpentine oil is flowing through a capillary tube of length $l$ and radius $r$.
Also we know that the pressure difference between the two ends of the tube is $p$. $v$ is the velocity of oil at a distance $x$ from the axis of the tube.
The dimension of length, $l = \left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{1}}{\text{T}}_{}^{\text{0}}} \right]$.
The dimension of radius, $r = \left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^1{\text{T}}_{}^{\text{0}}} \right]$ $ \Rightarrow r_{}^2 = \left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{2}}{\text{T}}_{}^{\text{0}}} \right]$.
The dimension of pressure, $p = \left[ {{\text{M}}_{}^{\text{1}}{\text{L}}_{}^{{\text{ - 1}}}{\text{T}}_{}^{{\text{ - 2}}}} \right]$.
The dimension of Velocity, $v = \left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{1}}{\text{T}}_{}^{{\text{ - 1}}}} \right]$.
The dimension of distance, $x = \left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^1{\text{T}}_{}^0} \right]$ $ \Rightarrow x_{}^2 = \left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{2}}{\text{T}}_{}^0} \right]$.
Substituting the values in the given dimension formula of viscosity, $\eta $
$\eta = \dfrac{{p\left( {r_{}^2 - x_{}^2} \right)}}{{4vl}}$ .
Also as 4 is a number it is a dimensionless quantity.
\[ \Rightarrow \eta = \dfrac{{\left[ {{\text{M}}_{}^{\text{1}}{\text{L}}_{}^{{\text{ - 1}}}{\text{T}}_{}^{{\text{ - 2}}}} \right]\left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{2}}{\text{T}}_{}^{\text{0}}} \right]}}{{\left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{1}}{\text{T}}_{}^{{\text{ - 1}}}} \right]\left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^{\text{1}}{\text{T}}_{}^{\text{0}}} \right]}}\]
\[ \Rightarrow \eta = \dfrac{{\left[ {{\text{M}}_{}^{{\text{1 + 0}}}{\text{L}}_{}^{{\text{ - 1 + 2}}}{\text{T}}_{}^{{\text{ - 2 + 0}}}} \right]}}{{\left[ {{\text{M}}_{}^{{\text{0 + 0}}}{\text{L}}_{}^{{\text{1 + 1}}}{\text{T}}_{}^{{\text{ - 1 + 0}}}} \right]}}\]
\[ \Rightarrow \eta = \dfrac{{\left[ {{\text{M}}_{}^{\text{1}}{\text{L}}_{}^{\text{1}}{\text{T}}_{}^{{\text{ - 2}}}} \right]}}{{\left[ {{\text{M}}_{}^{\text{0}}{\text{L}}_{}^2{\text{T}}_{}^{{\text{ - 1}}}} \right]}}\]
\[ \Rightarrow \eta = \left[ {{\text{M}}_{}^{{\text{1 - 0}}}{\text{L}}_{}^{{\text{1 - 2}}}{\text{T}}_{}^{{\text{ - 2 + 1}}}} \right]\]
\[ \Rightarrow \eta = \left[ {{\text{M}}_{}^{\text{1}}{\text{L}}_{}^{{\text{ - 1}}}{\text{T}}_{}^{{\text{ - 1}}}} \right]\]
Therefore, the dimension for viscosity $\eta = \dfrac{{p\left( {r_{}^2 - x_{}^2} \right)}}{{4vl}}$ \[ = \left[ {{\text{M}}_{}^{\text{1}}{\text{L}}_{}^{{\text{ - 1}}}{\text{T}}_{}^{{\text{ - 1}}}} \right]\]
Hence (A) \[\left[ {{\text{M}}_{}^{\text{1}}{\text{L}}_{}^{{\text{ - 1}}}{\text{T}}_{}^{{\text{ - 1}}}} \right]\] is the required answer.

Additional information:
$\eta $ is called viscosity of the liquid and it depends upon the nature of the liquid.
The coefficient of viscosity is defined as the viscous force per unit area per unit velocity gradient.
SI unit of $\eta $ is $\dfrac{{{\text{Ns}}}}{{{\text{m}}_{}^{\text{2}}}}$
CGS unit of $\eta $ is $\dfrac{{{\text{dyne - s}}}}{{{\text{cm}}_{}^{\text{2}}}}$or poise.

If a force of one dyne per square centimeter exists between two layers of fluid with unit velocity gradient, then its coefficient of viscosity is one poise. The viscosity of liquid decreases with temperature, while that of gases increases with temperature.

Note: It is also important to make sure that units of all quantities are considered in the same system of units. The resistance of the fluid by gradual deformation by shear or stress is known as viscosity.