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Chapter 6: Problem 63

A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is \(r\) and the speed of revolution is \(n\) rotations/second, find the difference in height of the liquid at the centre of vessel and its sides. (1) \(h=\frac{2 \pi^{2} n^{2} r^{2}}{g}\) (2) \(h=\frac{4 \pi^{2} n^{2} r^{2}}{g}\) (3) \(h=\frac{\pi^{2} r^{2} n^{2}}{g}\) (4) \(h=\frac{\pi^{2} r^{2} n^{2}}{2 g}\)

Short Answer

Expert verified
Option (1): \( h = \frac{2 \pi^2 n^2 r^2}{g} \).

Step by step solution

01

Understand the Problem

A cylindrical vessel contains liquid and is rotated at the rate of \(n\) revolutions per second around its axis. Due to rotation, the liquid level rises at the sides of the vessel and is lower at the center. We need to find the formula for the difference in the height of the liquid at the center and the side.
02

Define the Physical Condition

When the vessel is rotating, the liquid surface forms a parabolic curve due to centrifugal force. With radius \(r\), angular velocity \(\omega = 2\pi n\), the difference in height (\(h\)) between the center and the sides of the liquid is determined by the balance of forces.
03

Apply the Mathematical Formula

The difference in height, \(h\), due to rotation is given by the formula:\[ h = \frac{\omega^2 r^2}{2g} \]Substituting \(\omega = 2\pi n\) into the equation gives:\[ h = \frac{(2\pi n)^2 r^2}{2g} = \frac{4\pi^2 n^2 r^2}{2g} \]Simplifying, we have:\[ h = \frac{2\pi^2 n^2 r^2}{g} \]
04

Choose the Correct Option

From the simplification in Step 3, we find that the expression for the height difference is \( h = \frac{2 \pi^2 n^2 r^2}{g} \). This corresponds to option (1).

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Key Concepts

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

Cylindrical Vessel
Imagine a simple cylindrical cup or container. That's essentially what a cylindrical vessel is. It is a 3D shape with a uniform, straight, circular cross-section, like a can or a glass. The cylindrical vessel is often used in physics problems involving liquids. This is because its shape helps in simplifying calculations, especially when dealing with rotational dynamics.
In our exercise, the cylindrical vessel holds a liquid that gets affected by rotation. Because of its symmetry along the axis and the constant radius, the problem can be reduced to understanding how the motion and forces affect the liquid inside the vessel.
Rotational Motion
When an object spins or rotates around a central axis, it experiences rotational motion. In this exercise, the cylindrical vessel is rotated around its axis. The speed of this rotation is described by the number of rotations per second \(n\).
Rotational motion brings about several effects, particularly on fluids. It generates centrifugal forces, which push the liquid outward toward the sides of the vessel. This happens because every particle of the liquid is moving in a circular path, trying to maintain its motion in a straight line due to inertia. Thus, the liquid rises at the vessel's sides.
Parabolic Liquid Surface
When the cylindrical vessel rotates, the liquid inside does not simply rotate uniformly. It actually adjusts its surface to form a parabolic shape. This is due to the centrifugal forces that act more on the liquid as you move away from the axis, causing it to rise more dramatically at the sides.
The parabolic liquid surface is a result of a balance between gravitational pull (pulling down) and the centrifugal force (pushing outwards). This phenomenon is essential to comprehend since the difference in height we'd be calculating in the exercise is directly connected to this parabolic arrangement.
Angular Velocity
Angular velocity represents how fast something rotates. It is denoted here by \(\omega\), and for the cylindrical vessel, it can be calculated as \((2\pi n)\). This is because each full rotation is \(2\pi\) radians, and the vessel rotates at \(n\) revolutions per second.
Angular velocity is key to solving the exercise since it helps determine how significantly the centrifugal force affects the liquid. The higher the angular velocity, the steeper the parabolic curve of the liquid's surface becomes. This information is critical in determining the height difference of the liquid at the center versus its edges.

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Most popular questions from this chapter

For Problems 26-28 A small spherical ball of radius \(r\) is released from its completely submerged position (as shown in the figure) in a liquid whose density varies with height \(h\) (measured from the bottom) as \(\rho_{L}=\rho_{0}\left[4-\left(3 h / h_{0}\right)\right]\). The density of the ball is \((5 / 2) \rho_{0} .\) The height of the vessel is \(h_{0}=12 / \pi^{2}\) Consider \(r \ll h_{0}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\). Where will the ball come in the equilibrium? (1) At a depth \(h_{0} / 2\) from top (2) At the bottom of the vessel (3) At a depth \(3 h_{0} / 4\) from top (4) The ball will never be at equilibrium

A lawn sprinkler has 20 holes, each of cross-sectional area \(2.0 \times 10^{-2} \mathrm{~cm}^{2}\), and is connected to a hose pipe of crosssectional area \(2.4 \mathrm{~cm}^{2}\). If the speed of water in the hose pipe is \(1.5 \mathrm{~ms}^{-1}\), the speed of water as it emerges from the holes is (1) \(2.25 \mathrm{~ms}^{-1}\) (2) \(4.5 \mathrm{~ms}^{-1}\) (3) \(9 \mathrm{~m} \mathrm{~s}^{-1}\) (4) \(18 \mathrm{~m} \mathrm{~s}^{-1}\)

For Problems 31-32 A U-tube of uniform cross section contains a liquid of density d. Diameter of the tube is \(D\). Height of liquid in the left arm and also in the right arm is \(h\) and length of horizontal portion of the tube and hence the horizontal column of liquid is \(L\). Consider the following situations. (i) The tube is given a uniform acceleration \(a\) towards left. (ii) The tube is mounted on a horizontal table that is made to rotate with a uniform angular speed \(\omega\) with one of the arms (left or right) on the axis of rotation. Take \(M\) as the total mass of the liquid and \(g\) as the acceleration due to gravity. Difference in height between the liquid columns in the vertical arms (left and right) in situation (i) will be (1) \(\sqrt{\frac{L a}{g}} \times h\) (2) \(\frac{L^{2} h^{2} d a}{M g}\) (3) \(\frac{L a}{g}\) (4) \(\frac{L a D}{g h}\)

A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. The ratio of radius of cavity to that of outer radius of the sphere must be (1) \(\frac{7^{1 / 3}}{2}\) (2) \(\frac{5^{1 / 3}}{2}\) (3) \(\frac{9^{1 / 3}}{2}\) (4) \(\frac{3^{1 / 3}}{2}\)

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weighs \(0.25 \mathrm{~kg}\) in air and is observed to displace \(0.015\) litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, \(19.3\) and \(8.9\). The approximate percentage of copper in the ornament is (1) \(5 \%\) (2) \(10 \%\) (3) \(15 \%\) (4) \(25 \%\)
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