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Chapter 8: Problem 45

A thin square plate of side \(5 \mathrm{~cm}\) is suspended vertically from a balance so that lower side just dips into water with side to surface. When the plate is clean \(\left(\theta=0^{\circ}\right)\), it appears to weigh \(0.044 \mathrm{~N}\). But when the plate is greasy \(\left(\theta=180^{\circ}\right)\), it appears to weigh \(0.03 \mathrm{~N}\). The surface tension of water is (1) \(3.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (2) \(7.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (3) \(14.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (4) \(1.08 \mathrm{~N} / \mathrm{m}\)

Short Answer

Expert verified
The surface tension of water is \( 0.07 \text{ N/m} \).

Step by step solution

01

Understanding the Problem

We need to find the surface tension \( T \) of water using the given data about a square plate dipping in water. The plate's apparent weight changes depending on whether it is clean or greasy due to different surface tension forces. When clean, \( \theta = 0^\circ \) and when greasy, \( \theta = 180^\circ \).
02

Analyze the Clean Plate Scenario

For the clean plate, the water wets the plate fully. The apparent weight, \( W_1 \), is the actual weight minus the upward force due to surface tension, such that \( W_1 = W - 4 \, T \, L \). Given that \( W_1 = 0.044 \, \text{N} \), and \( L = 0.05 \, \text{m} \).
03

Analyze the Greasy Plate Scenario

For the greasy plate, the water doesn't wet the plate, \( \theta = 180^\circ \), and the force due to surface tension is zero in effect. The apparent weight \( W_2 \) is the same as the actual weight, so \( W_2 = 0.03 \text{ N} = W \).
04

Calculate the Actual Weight

As the greasy plate's apparent weight equals its actual weight, we have \( W = 0.03 \text{ N} \).
05

Solve for Surface Tension T

For the clean plate, using the equation from Step 2, we have \( 0.044 = 0.03 + 4 \, T \, 0.05 \). Simplifying, we find \( 0.014 = 0.2 \times T \). So, \( T = \frac{0.014}{0.2} = 0.07 \text{ N/m} \).
06

Conclusion

The calculated surface tension \( T = 0.07 \text{ N/m} \) matches option (2) in the given list of options.

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Key Concepts

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

apparent weight
Apparent weight is the perceived weight of an object when it is submerged in or supported by a fluid. It can differ from the actual gravitational weight due to additional forces such as buoyancy or surface tension. In the context of the square plate dipping into water:
  • When the plate is clean, it experiences an upward force due to the surface tension of water because the wetting angle, \( \theta \), is \( 0^\circ \). This means water completely wets the plate, reducing its apparent weight.
  • When the plate is greasy, the wetting angle is \( 180^\circ \) and water does not adhere to it. The apparent weight is closer to the actual weight of the plate because the upward force due to surface tension is negligible.
Understanding apparent weight helps to explore how surface tension affects objects, making it crucial for calculations based on mechanics imposed in exams like JEE Advanced.
wetting angle
The wetting angle, also known as the contact angle \( \theta \), measures how a liquid interacts with a solid surface. It's the angle formed between the liquid's interface and the solid surface. This angle indicates how well a liquid wets or adheres to a surface:
  • \( \theta = 0^\circ \): The liquid fully wets the surface, spreading completely over it. This was the case when the plate was clean.
  • \( \theta = 180^\circ \): The liquid does not wet the surface at all, forming a bead. This was the scenario with the greasy plate in the exercise.
The degree of wetting is key in determining the forces due to surface tension, affecting the apparent weight of submerged objects. Knowledge of wetting angles enables students to predict how different surfaces will interact with liquids, a core concept in physics problems involving fluids.
square plate
In problems involving fluid mechanics and surface tension, the shape and size of an object play a crucial role. The exercise involved a square plate, which is particularly useful in experiments because:
  • Its symmetric shape simplifies calculations of forces as surface tension acts uniformly along its edges.
  • The square plate's side length can be directly used in formulas, such as calculating the force exerted by surface tension: \( F = T \cdot 4L \), where \( L \) is the side length.
Understanding why certain shapes are chosen in physics problems helps clarify concepts and refine problem-solving skills. When dealing with JEE Advanced Mechanics or similar exams, recognizing patterns and geometries is often pivotal in simplifying complex problems.
JEE Advanced Mechanics
The JEE Advanced exam is known for testing a deep understanding of physics concepts and their applications. Mechanics, a major component of the exam, often includes problems involving:
  • Surface tension and its effects on solid objects.
  • Wetting angles and how they influence surface interactions.
  • Diverse shapes such as squares, which affect fluid mechanics outcomes.
To excel in JEE Advanced Mechanics, students must grasp these abstract concepts and apply them confidently in problem-solving. This involves mastering equations, understanding physical principles, and recognizing the underlying mechanics in given situations.

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Most popular questions from this chapter

The lower end of a capillary tube is at a depth of \(12 \mathrm{~cm}\) and water rises \(3 \mathrm{~cm}\) in it. The mouth pressure required to blow an air bubble at the lower end will be \(x \mathrm{~cm}\) of water column, where \(x\) is (1) 12 (2) 15 (3) 3 (4) 9

A rain drop starts falling from a height of \(2 \mathrm{~km}\). It falls with a continuously decreasing acceleration and attains its terminal velocity at a height of \(1 \mathrm{~km}\). The ratio of the work done by the gravitational force in the first half and the second half of the drops journey is (1) \(1: 1\) and the times of fall of the drop in the two halves is \(a: 1\) (where \(a>1\) ) (2) \(1: 1\) and the times of fall of the drop in the two halves is \(a: 1\) (where \(a<1\) ) (3) \(a: 1\) (where \(a>1\) ) and the times of fall of the drop in the two halves is \(1: 1\) (4) \(a: 1\) (where \(a<1\) ). and the times of fall of the drop in the two halves is \(1: 1\)

A plate of area \(2 \mathrm{~m}^{2}\) is made to move horizontally with a speed of \(2 \mathrm{~ms}^{-1}\) by applying a horizontal tangential force over the free surface of a liquid. The depth of the liquid is \(1 \mathrm{~m}\) and the liquid in contact with the bed is stationary. Coefficient of viscosity of liquid \(=0.01\) poise. Find the tangential force needed to move the plate (in \(\mathrm{N}\) ).

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

A long thin walled capillary tube of mass \(M=10 \mathrm{~g}\) and radius \(r=7 \mathrm{~mm}\) is partially immersed in a liquid of surface tension \(T=0.75 \mathrm{Nm}^{-1} .\) The angle of contact for the liquid and the tube wall is \(60^{\circ} .\) How much force is needed to hold the tube vertically? Neglect buoyancy force on the tube.
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