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Between a plate of area \(100 \mathrm{~cm}^{2}\) and another plate of area \(100 \mathrm{~m}^{2}\) there is a \(1 \mathrm{~mm}\), thick layer of water, if the coefficient of viscosity of water is \(0.01\) poise, then the force required to move the smaller plate with a velocity \(10 \mathrm{~cm} \mathrm{~s}^{-1}\) with reference to large plate is (1) \(100 \mathrm{dyn}\) (2) \(10^{4} \mathrm{dyn}\) (3) \(10^{6} \mathrm{dyn}\) (4) \(10^{9}\) dyn

Short Answer

Expert verified
The force required is 100 dyn, which corresponds to option (1).

Step by step solution

01

Understand the Viscosity Formula

The force required to move a plate through a fluid is given by the viscosity formula: \( F = \eta \frac{A \cdot v}{d} \), where \( F \) is the force, \( \eta \) is the coefficient of viscosity, \( A \) is the area of the plate, \( v \) is the velocity, and \( d \) is the separation between the plates.
02

Convert Units Appropriately

The area of the smaller plate is given as \(100 \text{ cm}^2\), which can be used directly. The separation is \(1 \text{ mm} = 0.1 \text{ cm}\). The velocity \(v\) is already in \(10 \text{ cm s}^{-1}\). No conversion is required for these values.
03

Apply Values to Viscosity Formula

Plug the known values into the formula: \( \eta = 0.01 \text{ poise} \), \( A = 100 \text{ cm}^{2} \), \( v = 10 \text{ cm s}^{-1} \), \( d = 0.1 \text{ cm} \). Thus, the force is: \[ F = 0.01 \times \frac{100 \times 10}{0.1} \]
04

Calculate the Force

Simplify and calculate the expression: \[ F = 0.01 \times \frac{1000}{0.1} = 0.01 \times 10000 = 100 \text{ dyn} \].
05

Determine the Correct Answer

Compare the calculated result to the given options. The calculated force, \(100 \text{ dyn}\), corresponds to option (1).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plate Movement
Understanding how plates move through a fluid is essential in studying fluid mechanics. Here, we'll explore the scenario of a smaller plate moving relative to a larger stationary plate, with a thin layer of fluid, like water, in between them. This movement is akin to sliding one object over another, but with an intervening fluid that resists the motion due to viscous forces.

When we talk about plate movement, the velocity of the moving plate becomes crucial. In our exercise, the small plate moves at a speed of 10 cm/s. This velocity is necessary for calculating the force needed to overcome the fluid's viscosity. Imagine a tiny cushion of water particles reacting to the plate's movement and resisting it. This resistance applies a force back on the moving plate. The faster the plate moves, the greater the force required for the same thickness of fluid.
Coefficient of Viscosity
The coefficient of viscosity is a measure of a fluid's resistance to flow. In simpler terms, it quantifies how "thick" or "sticky" a fluid is. Water, with a low coefficient of viscosity, flows easily. In contrast, honey, with a much higher viscosity, flows slowly. The coefficient of viscosity gives us a way to express this behavior numerically.

In our exercise, the coefficient of viscosity for water is given as 0.01 poise. This number represents how much force is needed per unit area to move the fluid at a certain speed. A small coefficient means less resistance and, thus, less force required for movement. Parsing viscosity is crucial since it directly affects our calculations of force when combined with area and velocity. The lower the viscosity, like water, the easier it is to push a plate through it.
Force Calculation
Calculating the force required to move a plate through a fluid involves a formula that needs precise input values. The viscosity formula used is: \[ F = \eta \frac{A \cdot v}{d} \]
Here's what each term means in the context of our exercise:
  • \( F \): The force we need to find,
  • \( \eta \): The coefficient of viscosity, here it's 0.01 poise,
  • \( A \): The area of the moving plate, 100 cm²,
  • \( v \): The velocity of the plate, 10 cm/s,
  • \( d \): The distance or thickness of the fluid layer between plates, 0.1 cm.

Inserting these values: \[ F = 0.01 \times \frac{100 \times 10}{0.1} = 100 \text{ dyn} \]
By understanding each component, we see how conditions like increased area, velocity, or decreased separation impact the force magnitude needed to move the plate.
Fluid Mechanics
Fluid mechanics is the branch of physics studying fluids (liquids and gases) and the forces on them. It's a crucial part of understanding how fluids behave under various conditions and how they impact surroundings.

Several key concepts arise in fluid mechanics, like viscosity, laminar flow, turbulence, and pressure. In our specific exercise, we're dealing with a scenario that relies heavily on viscosity, illustrating how a fluid's internal friction affects objects moving through it.

Consider real-world applications where understanding viscosity and forces in fluids is essential, such as:
  • Designing efficient engines and pumps,
  • Creating accurate simulations in weather predictions,
  • Studying blood flow in medical science.
In essence, solving problems like the one in our exercise provides the foundation for analyzing and predicting the behavior of fluids in everyday and industrial applications. Understanding these basics helps in comprehending more complex phenomena that fluid dynamics encompasses.

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Most popular questions from this chapter

A cubical block (of side \(2 \mathrm{~m}\) ) of mass \(20 \mathrm{~kg}\) slides on inclined plane lubricated with the oil of viscosity \(\eta=10^{-1}\) poise with constant velocity of \(10 \mathrm{~m} / \mathrm{s}\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\). \(\mathrm{Th}_{\mathrm{h}}\) thickness of layer of liquid is \(\left(10^{-1}\right.\) poise \(\left.=10^{-2} \mathrm{~N} \mathrm{~s} \mathrm{~m}^{2}\right)\) (1) \(2.5 \mathrm{~mm}\) (2) \(4 \mathrm{~mm}\) (3) \(6 \mathrm{~mm}\) (4) \(5.5 \mathrm{~mm}\)

A water drop is divided into eight equal droplets. The pressure difference between inner and outer sides of the big drop (1) will be the same as for smaller droplet (2) will be half of that for smaller droplet (3) will be one-fourth of that for smaller droplet (4) will be twice of that for smaller droplet

A bowl whose bottom has round holes of diameter \(1 \mathrm{~mm}\) is filled with water. Assuming that surface tension acts only at holes, find the maximum height (in \(\mathrm{cm}\) ) up to which water can be filled in the vessel without leakage. (Given, surface tension of water \(=75 \times 10^{-3} \mathrm{Nm}^{-1}, g=10 \mathrm{~ms}^{-2}\) and density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\) ).

A soap film is made by dropping a circular frame of radjus \(b\) in soap solution. A bubble is formed by blowing air with speed \(v\) in the form of a cylinder. Air stops after striking surface of soap bubble. Density of air is \(\rho\). The radius of the bubble is \(R>b\). The radius \(R\) of the bubble when the soap bubble separates from the ring is \(\frac{\beta T}{\rho v^{2}}\), Find the value of \(\beta\). (where, surface tension of liquid is \(T\) ).

A circular ring has inner and outer radii equal to \(20 \mathrm{~mm}\) and \(50 \mathrm{~mm}\) respectively. Mass of the ring is \(m=0.8 \mathrm{~g}\) It gently pulled out vertically from a water surface by a sensitive spring. When the spring is stretched \(4 \mathrm{~cm}\) from its equilibrium position the ring is on verge of being pulled out from the water surface. If spring constant is \(k=0.9 \mathrm{Nm}^{-1}\) find the surface tension of water.

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