Question

Q1. A cylinder of 0.12m radius rotates concentrically inside a fixed hollow cylinder of 0.13 m radius.
Both the cylinders are 0.3 m long. Determine the viscosity of the liquid which fills the space between
the cylinders if a torque of 0.88Nm is required to maintain an angular velocity of 2π

Answer 1

The viscosity of the liquid is 4.4 Ns/m²

Explanation:

Given:

A cylinder of radius 0.12 m rotates concentrically inside a fixed hollow cylinder of radius 0.13 m

Length of both the cylinders = 0.13 m

Torque between the cylinders = 0.88 Nm

Angular velocity of rotation = 2π

To find out:

The viscosity of the liquid that fills between the cylinders

Solution:

$\tau=0.88$ Nm

$\omega=2\pi$ rad/s

$l=0.3$ m

The force of this torque will be acting tangentially

Therefore,

$\tau=F\times r$

$\implies 0.88=F\times 0.13$

$\implies F=\frac{0.88}{0.13}$ N

From Newton's law of viscosity

$F=-\eta A\frac{dv}{dr}$

$\implies \eta=\frac{F}{A(dv/dr)}$

The velocity from the surface of inner cylinder to outer cylinder will vary linearly

Thus,

$\frac{dv}{dr}=\frac{v}{r}=\frac{r\omega}{r}=\omega=2\pi$

$A=2\pi rl$

$\implies A=2\pi \times 0.13\times 0.3$

Thus,

$\eta=\frac{0.88/0.13}{2\pi\times 0.13\times 0.3\times 2\pi}$

$\implies \eta=4.4$ Ns/m²

Hope this answer is helpful.

Know More:

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Answer 2

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