Question

A 15 cm diameter vertiCal cylinder rotates concentri
inside another Cylinder of diameter 15.10 cm
Both Cylinders are 25 cm hligh. the space between
the Cylinders is filled with a liquid whose
Viscosity is unknown if a torque of 12.0 Nm is
requided to rotate the inner Cylinder at 100 rpm.
Determine the viscosity of the liquid.​

Answer 1

Answer:

The viscosity of the liquid is approximately 0.000127 Ns/m².

Explanation:

We can use the following formula to calculate the liquid's viscosity:

τ = 2πηN(R₁² - R₂²)

where τ denotes the torque needed to rotate the inner cylinder,η denotes the liquid's viscosity, N denotes the rotational speed in revolutions per minute (RPM), R1 denotes the inner cylinder's radius, and R2 denotes the outer cylinder's radius.

The dimensions given can first be converted to metres:

• Inner cylinder diameter: 15 cm (0.15 m).
• The outer cylinder's diameter is 15.10 cm (0.151 m).
• Each cylinder's height is 25 cm, or 0.25 metres.

After that, we can determine the cylinders' radii:

• Inner cylinder radius, R1 = 0.15/2 = 0.075 m
• outer cylinder radius, R2 = 0.151/2 = 0.0755 m

We are given that the torque required to rotate the inner cylinder at 100 RPM is 12.0 Nm. Thus, we can substitute the given values into the formula and solve for η:

12.0 Nm = 2πη(100 RPM)(0.075² m² - 0.0755² m²)

η = 0.000127 Ns/m²

Therefore, the viscosity of the liquid is approximately 0.000127 Ns/m².

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