Question

A viscometer consists of two concentric cylinders, \({\bf{10}}{\bf{.20}}\;{\bf{cm}}\) and \({\bf{10}}{\bf{.60}}\;{\bf{cm}}\)in diameter. A liquid fills the space between them to a depth of \({\bf{12}}{\bf{.0}}\;{\bf{cm}}\). The outer cylinder is fixed, and a torque of \({\bf{0}}{\bf{.024}}\;{\bf{m}} \cdot {\bf{N}}\) keeps the inner cylinder turning at a steady rotational speed of \({\bf{57}}\;{\bf{rev/min}}\). What is the viscosity of the liquid?

Short answer

The viscosity of the liquid is \(0.079\;{\rm{Pa}} \cdot {\rm{s}}\).

Step-by-step solution

Definition of Viscosity

The internal friction present in adjacent layers of fluid is known as viscosity. The force required to move the piston depends upon the coefficient of viscosity of fluid, area of container, speed of liquid and length of container.

Given Data

The diameter of inner cylinder is\({d_1} = 10.20\;{\rm{cm}}\).

The diameter of outer cylinder is\({d_2} = 10.60\;{\rm{cm}}\).

The depth of cylinder is\(h = 12.0\;{\rm{cm}}\).

The torque acting on the inner cylinder is\(\tau = 0.024\;{\rm{m}} \cdot {\rm{N}}\).

The angular speed of cylinder is \(\omega = 57\;{\rm{rev/min}}\).

Calculate the radius of cylinder

The radius of inner cylinder is given as,

\({r_1} = \frac{{{d_1}}}{2}\)

Substitute the known value in the above equation,

\(\begin{array}{c}{r_1} = \frac{{10.20\;{\rm{cm}}}}{2}\\ = 5.10\;{\rm{cm}}\end{array}\)

The radius of outer cylinder is given as,

\({r_2} = \frac{{{d_2}}}{2}\)

Substitute the known value in the above equation,

\(\begin{array}{c}{r_2} = \frac{{10.60\;{\rm{cm}}}}{2}\\ = 5.30\;{\rm{cm}}\end{array}\)

The average radius of cylinder is given as,

\(r = \frac{{{r_1} + {r_2}}}{2}\)

Substitute the known values in the above equation,

\(\begin{array}{c}r = \frac{{5.10\;{\rm{cm}} + 5.30\;{\rm{cm}}}}{2}\\ = \frac{{10.40\;{\rm{cm}}}}{2}\\ = 5.20\;{\rm{cm}}\end{array}\)

Calculating the viscosity of liquid

The force acting on cylinder is given as,

\(F = \frac{\tau }{{{r_1}}}\)

The average area of cylinder is given as,

\(A = 2\pi rh\)

The length of cylinder is given as,

\(l = {r_2} - {r_1}\)

The speed of cylinder is given as,

\(v = \omega {r_1}\)

The viscosity of cylinder can be given as,

\(\eta = \frac{{Fl}}{{vA}}\)

Substitute the known expressions in the above equation,

\(\begin{array}{c}\eta = \frac{{\left( {\frac{\tau }{{{r_1}}}} \right)\left( {{r_2} - {r_1}} \right)}}{{\left( {\omega {r_1}} \right)\left( {2\pi rh} \right)}}\\ = \frac{{\tau \left( {{r_2} - {r_1}} \right)}}{{2\pi rh\omega r_1^2}}\end{array}\)

Substitute the known values in the above equation,\(\begin{array}{c}\eta = \frac{{\left( {0.024\;{\rm{m}} \cdot {\rm{N}}} \right)\left( {5.30\;{\rm{cm}} - 5.10\;{\rm{cm}}} \right)}}{{2\left( {3.14} \right)\left( {5.20\;{\rm{cm}}} \right)\left( {12.0\;{\rm{cm}}} \right)\left( {57\;{\rm{rev/min}}} \right){{\left( {5.10\;{\rm{cm}}} \right)}^2}}}\\ = \frac{{\left( {0.024\;{\rm{m}} \cdot {\rm{N}}} \right)\left( {0.20\;{\rm{cm}}} \right)}}{{2\left( {3.14} \right)\left( {5.20\;{\rm{cm}}} \right)\left( {12.0\;{\rm{cm}}} \right)\left( {57\;{\rm{rev/min}}} \right)\left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right)\left( {\frac{{1\;\min }}{{60\;{\rm{s}}}}} \right){{\left( {5.10\;{\rm{cm}}} \right)}^2}{{\left( {\frac{{{\rm{1}}\;{\rm{m}}}}{{{\rm{100}}\;{\rm{cm}}}}} \right)}^3}}}\left( {\frac{{{\rm{1}}\;{\rm{Pa}}}}{{{\rm{1}}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}} \right)\\ \approx 0.079\;{\rm{Pa}} \cdot {\rm{s}}\end{array}\)

Therefore, the viscosity of the liquid is \(0.079\;{\rm{Pa}} \cdot {\rm{s}}\).