/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A thin metal disc of radius \(r\... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin metal disc of radius \(r\) float on water surface and bends the surface downwards along the perimeter making an angle \(\theta\) with vertical edge of the disc. If the disc displaces a weight of water \(w\) and surface tension of water is \(T\), then the weight of metal dise is (a) \(2 \pi \tau+w\) (b) \(2 \pi r T \cos \theta-w\) (c) \(2 \pi T \cos \theta+w\) (d) \(w-2 \pi T \cos \theta\)

Short Answer

Expert verified
(b) \(2 \pi r T \cos \theta - w\)

Step by step solution

01

Understanding the Concept of Surface Tension Force

The surface tension force around the perimeter of the disc acts vertically upwards. Its magnitude can be calculated using the formula for surface tension forces: \(F = 2 \pi r T \cos \theta\). This force is due to the tension at every point along the circular edge of the disc.
02

Establishing the Forces Acting on the Disc

The disc is in equilibrium because it is floating. Thus, the upward forces (due to surface tension) must balance the downward forces (due to the weight of the disc and the displaced water). This gives the equation: \(w + W_{disc} = 2 \pi r T \cos \theta\), where \(W_{disc}\) is the weight of the disc.
03

Solving for the Weight of the Metal Disc

Using the equilibrium relationship from Step 2, rearrange to solve for the weight of the disc. Thus, \(W_{disc} = 2 \pi r T \cos \theta - w\). This calculation shows the weight of the disc when considering surface tension and the displaced water.

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Key Concepts

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

Floating Objects
When an object floats on the surface of a fluid, it is supported by buoyant forces that arise due to the fluid beneath it. A key aspect to remember about floating objects is that they must displace a volume of fluid whose weight is equal to the weight of the object. For the thin metal disc in our problem, it floats because the downward force due to its weight is perfectly balanced by the upward buoyant force exerted by the displaced water.
However, it's important to acknowledge the role of surface tension here. Surface tension acts along the perimeter of the disc, providing an additional upward stabilizing force. This phenomenon is why even small and dense objects, like the thin metal disc, can float if surface tension is sufficiently large. As a result, surface tension complements buoyancy to keep the disc afloat.
  • Objects float by displacing a fluid volume equal to their own weight.
  • Surface tension provides an extra upward force, complementing buoyancy.
  • The disc is in equilibrium, with both downward and upward forces balancing out.
Equilibrium in Fluids
Equilibrium in fluids involves understanding how forces balance each other when an object is submerged or floating. Let's consider our metal disc example. For an object to stay in equilibrium, the total upward forces must equal the total downward forces. In this case, two major forces contribute to equilibrium: the weight of the water displaced by the disc and the vertical component of the surface tension around the disc's edge.
When calculating equilibrium, it's necessary to account for the forces acting on the entire system. The weight of the disc and the displaced water must be equaled by the surface tension force acting along the disc's perimeter. Mathematically, we set up an equation: \( w + W_{disc} = 2 \pi r T \cos \theta \), where \(w\) is the weight of the displaced water and \(2 \pi r T \cos \theta\) represents the surface tension force.
  • Total upward forces equal total downward forces in equilibrium.
  • Both surface tension and displaced water contribute to upward forces.
  • Understanding equilibrium involves summing all forces acting on an object.
Forces on Submerged Objects
When objects are submerged or partially immersed in fluids, several forces come into play. For the metal disc in our scenario, submerged forces include the weight of the disc, the surface tension acting at its perimeter, and the buoyant force from the displaced water. Each of these forces affects how the disc remains afloat.
The surface tension force acts upwards, counterbalancing the downward weight. This tension is derived from the liquid's cohesive forces creating a film-like effect at the surface. Meanwhile, the buoyant force results from the pressure difference between the displaced water at different depths. The specific angles and geometry of the disc further influence these resultant forces, as depicted in the given exercise.
In summary, the equilibrium of the disc occurs when the upward forces perfectly counteract the downward weight, showcasing the fascinating interplay between surface tension, buoyancy, and gravitational forces.
  • Multiple forces determine the stability of submerged objects.
  • Surface tension contributes to upward stabilization.
  • Buoyant forces arise from fluid displacement resulting in pressure differences.

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Most popular questions from this chapter

A drop of water breaks into two droplets of equal size. In this process, which of the following statements is correct? (a) The sum of the temperatures of the two droplets together is equal to temperature of the original drop (b) The sum of the masses of the two droplets is equal to mass of drop (c) The sum of the radii of the two droplets is equal to the radius of the drop (d) The sum of the surface areas of the two droplets is equal to the surface area of the original drop

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