/*! 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 49 Water rises in a capillary tube ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 49

Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by \(75 \times 10^{-4} \mathrm{~N}\), force due to the weight of the liquid. If the surface tension of water is \(6 \times 10^{-2} \mathrm{~N} / \mathrm{m}\), the inner circumference of the capillary must be (A) \(1.25 \times 10^{-2} \mathrm{~m}\) (B) \(0.50 \times 10^{-2} \mathrm{~m}\) (C) \(6.5 \times 10^{-2} \mathrm{~m}\) (D) \(12.5 \times 10^{-2} \mathrm{~m}\)

Short Answer

Expert verified
The inner circumference of the capillary tube is (A) \(1.25 \times 10^{-2} \mathrm{~m}\).

Step by step solution

01

Write down the given values

We are given: - Force due to weight of liquid = \(75 脳 10^{-4} \mathrm{~N}\) - Surface tension of water = \(6 脳 10^{-2} \mathrm{~N} / \mathrm{m}\)
02

Equate the forces and solve for the inner circumference

Write down the force due to weight of liquid equation and the force due to surface tension equation: \(F_w = V蟻g\) \(F_s = T 脳 C\) Since the forces are equal, we can set the two equations equal to each other and solve for the inner circumference \(C\): \(V蟻g = T 脳 C\) Now, we can isolate \(C\): \(C = \dfrac{V蟻g}{T}\) We know the values of \(F_w\) and \(T\), but we don't have the values for \(V\) and \(蟻\). However, we can rewrite the force due to the weight of liquid equation in terms of \(F_w\) and substitute: \(F_w = V蟻g\) \(V\rho g = 75 脳 10^{-4} \mathrm{~N}\) Now substituting this into the equation for \(C\): \(C = \dfrac{75 脳 10^{-4} \mathrm{~N}}{6 脳 10^{-2} \mathrm{~N} / \mathrm{m}}\)
03

Calculate the inner circumference

Divide the force by the surface tension to find the inner circumference: \(C = \dfrac{75 脳 10^{-4} \mathrm{~N}}{6 脳 10^{-2} \mathrm{~N} / \mathrm{m}} = 1.25 脳 10^{-2} \mathrm{~m}\) Thus, the correct answer is: (A) \(1.25 \times 10^{-2} \mathrm{~m}\)

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Key Concepts

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

Surface Tension
Surface tension is a property of liquids that arises from the cohesive forces between molecules on the surface. This force acts to minimize the surface area of a liquid, creating a 'skin-like' effect. In the context of capillary action, surface tension is responsible for the upward force that draws the liquid up the capillary tube. It works against gravity, which pulls the liquid down.

When a liquid comes into contact with a solid surface, such as the inner wall of a capillary, the adhesive forces between the liquid molecules and the solid can be stronger than the cohesive forces within the liquid. This causes the liquid to climb up the sides of the tube, a phenomenon we see with water in a capillary tube. The parameter for surface tension is denoted by the symbol 'T' and is measured in Newtons per meter (\( \text{N/m} \)).

The balance of forces is at the heart of capillary action physics problems. Ensuring a clear understanding of how surface tension functions can elucidate why certain liquids rise to particular heights in different capillary tubes.
Force Due to Weight of Liquid
The force due to the weight of a liquid is the gravitational force acting on the liquid's volume inside the capillary tube. It can be calculated using the formula \( F_w = V蟻g \), where \( F_w \) is the weight force, \( V \) is the volume of the liquid, \( 蟻 \) (rho) is the density of the liquid, and \( g \) is the acceleration due to gravity.

For liquids within a capillary tube, this force is countered by the surface tension force pulling the liquid upward. The precise measurement of the force due to weight of the liquid is crucial for solving physics problems involving equilibrium of forces, as seen in capillary action phenomena. In these problems, understanding the relationship between the liquid's volume, its density, and the force acting upon it due to gravity can greatly aid in comprehending the underlying physics.
Inner Circumference of Capillary
The inner circumference of a capillary tube is an essential factor in determining how far a liquid will rise due to capillary action. It is directly proportional to the force exerted by surface tension but inversely proportional to the force due to the weight of the liquid. To calculate the inner circumference (\( C \) of a capillary, the relationship \( F_s = T \times C \) is used, where \( F_s \) is the force due to surface tension, and \( T \) is the surface tension, as mentioned earlier.

The circumference can be computed if the forces are known and balanced, which leads to an equilibrium state where the liquid ceases to rise. This effect is encapsulated in the given problem, where the correct inner circumference of the capillary tube is determined by dividing the upward force by the surface tension. The importance of the inner circumference sheds light on the capillary action's sensitivity to the tube's dimension, which is a key concept in fluid dynamics and various applications such as inkjet printing and microfluidics.

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Most popular questions from this chapter

The amount of work done in increasing the size of a soap film \(10 \mathrm{~cm} \times 6 \mathrm{~cm}\) to \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\) is \(\left(\mathrm{S} . \mathrm{T}_{.}=\right.\) \(\left.30 \times 10^{-3} \mathrm{~N} / \mathrm{m}\right)\) (A) \(2.4 \times 10^{-2} \mathrm{~J}\) (B) \(1.2 \times 10^{-2} \mathrm{~J}\) (C) \(2.4 \times 10^{-4} \mathrm{~J}\) (D) \(1.2 \times 10^{-4} \mathrm{~J}\)

A metal wire of length \(L\), area of cross-section \(\mathrm{A}\), and Young's modulus \(Y\) is stretched by a variable force \(F\) such that \(F\) is always slightly greater than the elastic forces of resistance in the wire. When the elongation of the wire is \(\ell\) (A) The work done by \(F\) is \(\frac{Y A \ell^{2}}{2 L}\) (B) The work done by \(F\) is \(\frac{Y A \ell^{2}}{L}\) (C) The elastic potential energy stored in the wire is \(\frac{Y A \ell^{2}}{2 L}\) (D) No heat is produced during the elongation

A rectangular block of density \(\rho\), base area \(A\), and height \(h\) is kept on a spring. The lower end of spring is fixed on the bottom of an empty vessel of base area \(2 A\). The block compresses the spring by \(h / 4\) at equilibrium. The vessel is then slowly filled by a liquid of density\(2 \rho\) till the spring becomes relaxed. The block is then slowly pushed inside the liquid till it is immersed completely. Work done to push the block completely inside is \(W_{1}\), work done by gravity on the block is \(W_{2}\), and work done by upthrust is \(W_{3}\). Match the following based on the above statements. Column-I \(\quad\) Column-II (A) \(\left|W_{1}\right|\) (1) \(\frac{5 \rho g h^{2} A}{8}\) (B) \(\left|W_{2}\right|\) (2) \(\frac{\rho g h^{2} A}{8}\) (C) \(\left|W_{3}\right|\) (3) \(\frac{3 \rho g h^{2} A}{4}\) (4) \(\frac{\rho g h^{2} A}{4}\)

If the terminal speed of a sphere of gold (density \(=\) \(19.5 \mathrm{kgm}^{-3}\) ) is \(0.2 \mathrm{~ms}^{-1}\) in a viscous liquid (density \(=\) \(1.5 \mathrm{kgm}^{-3}\) ), find the terminal speed of a sphere of silver (density \(=10.5 \mathrm{~kg} / \mathrm{m}^{-3}\) ) of the same size in the same liquid. [2006] (A) \(0.4 \mathrm{~ms}^{-1}\) (B) \(0.133 \mathrm{~ms}^{-1}\) (C) \(0.1 \mathrm{~ms}^{-1}\) (D) \(0.2 \mathrm{~ms}^{-1}\)

Two rods of identical dimensions, with Young's modulus \(Y_{1}\) and \(Y_{2}\) are joined end to end. The equivalent Young's modulus for the composite rod is (A) \(\frac{2 Y_{1} Y_{2}}{Y_{1}+Y_{2}}\) (B) \(\frac{Y_{1} Y_{2}}{Y_{1}+Y_{2}}\) (C) \(\frac{1}{2\left(Y_{1}+Y_{2}\right)}\) (D) \(Y_{1}+Y_{2}\)
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