Dimension of coefficient of viscosity

Viscosity is the internal resistance to flow that is possessed by a liquid. The liquids that flow slowly have high internal resistance. This happens because of the strong intermolecular forces. Thus, the dimension of the viscosity of liquid of these kinds are more viscous and hold high Viscosity. The dimension of viscosity in physics is conceptualized as quantifying the internal frictional force which arises between the adjacent layers of fluid, which are in relative motion. For example, when a fluid is forced through a tube, it flows quickly near the axis of the tube than near its walls. The liquids that flow rapidly have low internal resistance. This happens because of the weak intermolecular forces. Therefore, they have low viscosity or are less viscous. (Image will be uploaded soon) Working of Viscosity Let us know how viscosity works with an example. Consider a liquid flowing via a narrow tube. All parts of the liquids do not pass through the tube with similar velocity. Imagine the liquid is to be made of a large number of the thin cylindrical coaxial layers. The layers that are in contact with the walls of the tube are mostly stationary. As we travel from the wall towards the centre of the tube, the cylindrical layers’ velocity keeps on increasing until it is maximum at the centre. This is called laminar flow and is a type with a regular gradation of velocity in going from one to the layer besides to it. As we travel from the centre towards the walls, the layers’ vel...

Dimension Of Viscosity: Viscosity to resist the inner flow of liquid. Slower fluid has higher internal resistance. this can be because of the strong intermolecular forces. Therefore, the viscosity size of the liquid of those types is incredibly noticeable and holds a high Viscosity. The magnitude of the viscosity in physics is believed to live the inner frictional force that emerges between adjacent, moving layers of liquid. as an example, when fluid is forced through a tube, it flows faster along the axis of the tube than near its walls. this is often thanks to weak intermolecular forces. Therefore, they need low viscosity or low viscous. Working Of Viscosity: Viscosity to face up to the interior flow of liquid. Smooth fluid has high internal resistance. this is often thanks to the strong intermolecular potential. Therefore, the viscosity size of the liquid of those types is extremely noticeable and carries a high Viscosity. The magnitude of the viscosity in physics is believed to live the inner cohesive forces that occur between adjacent, moving layers of liquid. as an example, when fluid is forced through a tube, it flows faster along the axis of the tube than near its walls. this can be thanks to the weak intermolecular forces. Therefore, they need low viscosity or low viscous. f ∝ A (dv / dx) f = η A (dv / dx) When η is the same, the so-called coefficient of viscosity, and therefore the ‘dv / dx’ velocity gradient. If A = 1 cm sq, dx = 1, dv = 1 cm / sec, then f = η. ...

\( \newcommand\) • • • Dropping the Ball (Slowly) We’ve seen how viscosity acts as a frictional brake on the rate at which water flows through a pipe, let us now examine its frictional effect on an object falling through a viscous medium. To make it simple, we take a sphere. If we use a very viscous liquid, such as glycerin, and a small sphere, for example a ball bearing of radius a millimeter or so, it turns out experimentally that the liquid flows smoothly around the ball as it falls, with a flow pattern like: Figure \(\PageIndex\): The arrows show the fluid flow as seen by the ball. This smooth flow only takes place for fairly slow motion, as we shall see. If we knew mathematically precisely how the velocity in this flow pattern varied near the ball, we could find the total viscous force on the ball by finding the velocity gradient near each little area of the ball’s surface, and doing an integral. But actually this is quite difficult. It was done in the 1840’s by Sir George Gabriel Stokes. He found what has become known as Stokes’ Law: the drag force \(F\) on a sphere of radius a moving through a fluid of viscosity \(\eta\) at speed \(v\) is given by: \[ F= 6\pi a \eta v\] This drag force is directly proportional to the radius. That’s not obvious—one might have thought it would be proportional to the cross-section area, which would go as the square of the radius. The drag force is also directly proportional to the speed, not, for example to \(v^2\). Understanding Stoke...

Physical Quantities Symbol Mass M Time T Length L Electric Current A Temperature $\Theta$ Complete step-by-step solution: To find the dimension, it is necessary that we have knowledge of relation of given physical quantities with the other fundamental quantities. 1) For dimension first we describe what is viscosity and on which physical quantities it depends. It is the property of fluids (liquids and gases) that opposes the flow of fluids. Suppose liquid is in streamline flow and divides into different layer of molecules, with velocities $ < ...........$ Here we observe that, there is relative motion between any two layers, which are in contact. Hence the opposing (contact or friction) force acting between the layers, this opposing force (F) is directly proportional to the area (A) of layers, i.e. $$ (As the area increases the opposing force is also increased) Further we also observe that the layers of liquid which are in contact with solid surface has the relative velocity zero, as we go up the relative velocity of layers increases , thus between two successive layers at distance $(dx)$having relative velocity $(dv)$ Hence the relative velocity per unit distance i.e. velocity gradient\[ = \dfrac \right] \\ $ Note:- It is necessary to have the knowledge of dimensions with its limitations. Application of dimension is also useful for explanation of relation between physical quantities.

Dropping the Ball (Slowly) Michael Fowler, UVa Stokes’ Law We’ve seen how viscosity acts as a frictional brake on the rate at which water flows through a pipe, let us now examine its frictional effect on an object falling through a viscous medium. To make it simple, we take a sphere. If we use a very viscous liquid, such as glycerin, and a small sphere, for example a ball bearing of radius a millimeter or so, it turns out experimentally that the liquid flows smoothly around the ball as it falls, with a flow pattern like: (The arrows show the fluid flow as seen by the ball. This smooth flow only takes place for fairly slow motion, as we shall see.) If we knew mathematically precisely how the velocity in this flow pattern varied near the ball, we could find the total viscous force on the ball by finding the velocity gradient near each little area of the ball’s surface, and doing an integral. But actually this is quite difficult. It was done in the 1840’s by Sir George Gabriel Stokes. He found what has become known as Stokes’ Law: the drag force F on a sphere of radius a moving through a fluid of viscosity η at speed v is given by: F = 6 π a η v . Note that this drag force is directly proportional to the radius. That’s not obvious —one might have thought it would be proportional to the cross-section area, which would go as the square of the radius. The drag force is also directly proportional to the speed, not, for example to v 2 . Understanding Stokes’ Law with Dimensional A...