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Chapter 6: Problem 46

A solid uniform ball of volume \(V\) floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is \(\rho_{1}\) and that of lower one is \(\rho_{2}\) and the specific gravity of ball is \(\rho\left(\rho_{1}<\rho<\rho_{2}\right)\). The fraction of the volume of the ball in the upper liquid is (1) \(\frac{\rho_{2}}{\rho_{1}}\) (2) \(\frac{\rho_{2}-\rho}{\rho_{2}-\rho_{1}}\) (3) \(\frac{\rho-\rho_{1}}{\rho_{2}-\rho_{1}}\) (4) \(\frac{\rho_{1}}{\rho_{2}}\)

Short Answer

Expert verified
The fraction of the volume of the ball in the upper liquid is \(\frac{\rho - \rho_1}{\rho_2 - \rho_1}\) (option 3).

Step by step solution

01

Understand Archimedes' principle

For the ball to float at the interface of two liquids, the weight of the ball equals the buoyant force exerted by the liquids. This buoyant force is the sum of the buoyant forces from both, the upper and lower liquids.
02

Express the buoyant force from each liquid

Let's denote the volume of the ball submerged in the upper liquid as \(V_1\) and in the lower liquid as \(V_2\), with \(V_1 + V_2 = V\). The buoyant force from the upper liquid is \(F_1 = \rho_1 g V_1\) and from the lower liquid is \(F_2 = \rho_2 g V_2\). Here, \(g\) is the acceleration due to gravity.
03

Write the equation of equilibrium

The weight of the ball is \(\rho g V\). According to Archimedes' principle, the total buoyant force \(F_1 + F_2 = \rho g V\). So, \(\rho_1 g V_1 + \rho_2 g V_2 = \rho g V\).
04

Express in terms of volume fractions

Express the terms in the equilibrium equation relative to the full volume \(V\): \(\rho_1 V_1 + \rho_2 V_2 = \rho V\). Introduce the volume fraction \(x\) of the ball submerged in the upper liquid: \(V_1 = xV\), hence \(V_2 = (1-x)V\).
05

Solve for the fraction of ball in upper liquid

Substitute the expressions for \(V_1\) and \(V_2\) into the equilibrium equation: \(\rho_1 xV + \rho_2 (1-x)V = \rho V\). Cancel \(V\) from all terms: \(\rho_1 x + \rho_2 (1-x) = \rho\). Simplify and solve for \(x\): \(x = \frac{\rho - \rho_2}{\rho_1 - \rho_2}\).
06

Verify solution choice

The calculated fraction, \(x = \frac{\rho - \rho_1}{\rho_2 - \rho_1}\), corresponds to option (3), after adjusting the reference by considering \(x\) as the fraction in the upper liquid.

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Key Concepts

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

Buoyant Force
Buoyant force is one of the fascinating ideas in physics that explains why objects float or sink when placed in a fluid. This force is the upward force exerted by the fluid that opposes the weight of an object immersed in it. Essentially, when you submerge an object in a fluid, the fluid pushes from below with a force equal to the weight of the fluid displaced by the object.

According to Archimedes' Principle, for an object to float, the buoyant force must equal the gravitational force pulling the object down. If you have ever observed a boat floating on water or a helium balloon rising in the air, you have witnessed buoyant force in action.

In the context of the exercise, the buoyant force is at play twice. The ball is floating at the interface of two liquids, hence two buoyant forces are experienced: one from the upper liquid with specific gravity \( \rho_1 \) and another from the lower liquid with \( \rho_2 \). This requires that the total buoyant force (sum of both forces) equals the gravitational force acting on the ball. This balance is crucial for the ball to float calmly.
Specific Gravity
Specific gravity is a dimensionless quantity and is greatly helpful in comparing the density of a substance to the density of a reference substance (usually water). It is calculated by dividing the density of the material by the density of the reference material.

The specific gravity offers insight into whether an object will float or sink. If the specific gravity is less than 1, the object will float (assuming water as the reference). If it is greater than 1, the object will sink.

In the exercise, the specific gravity values of the liquids and the ball are critical for solving the problem. The ball needs to have a specific gravity \( \rho \) that is between the two liquids: \( \rho_1 < \rho < \rho_2 \). This ensures that the ball partially floats in the upper liquid and partially in the lower liquid, influencing how we calculate the level of submersion.
Equilibrium of Floating Objects
Equilibrium is the state where the sum of the forces, and thus accelerations, on an object, is zero. For floating objects, like our ball, equilibrium implies that the weight of the object is perfectly balanced by the buoyant force acting on it, allowing it to float without sinking or rising.

This principle was applied in our exercise to determine the fraction of the ball submerged in each liquid. By setting the total buoyant force (from the two fluids) equal to the weight of the ball, we were able to construct an equation. Solving this equilibrium equation allows us to find the volume fraction of the ball immersed in each liquid.

Achieving equilibrium, in this context, is not just about floating but floating steady in two layers of liquids. This meticulous balance provides stability and dictates the floating behavior of the ball at the interface, demonstrating how principles of physics work to create equilibrium in nature.

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Most popular questions from this chapter

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) \(h=\frac{\pi^{2} r^{2} n^{2}}{2 g}\)

A tank is filled upto a height \(h\) with a liquid and is placed on a platform of height \(h\) from the ground. To get maximum range \(x_{m}\) a small hole is punched at a distance \(y\) from the free surface of the liquid. Then (1) \(x_{m}=2 h\) (2) \(x_{m}=1.5 h\) (3) \(y=h\) (4) \(y=0.75 h\)

For Problem 29-30 In a U-tube, if different liquids are filled then we can say that pressure at same level of same liquid is same. In a U-tube \(20 \mathrm{~cm}\) of a liquid of density \(\rho\) is on left hand side and \(10 \mathrm{~cm}\) of another liquid of density \(1.5 \rho\) is on right hand side. In between them there is a third liquid of density \(2 \rho\). What is the value of \(h ?\) (1) \(5 \mathrm{~cm}\) (2) \(2.5 \mathrm{~cm}\) (3) \(2 \mathrm{~cm}\) (4) \(7.5 \mathrm{~cm}\)

An object is weighted at the North Pole by a beam balance and a spring balance, giving readings of \(W_{B}\) and \(W_{S}\) respectively. It is again weighed in the same manner at the equator, giving reading of \(W_{B}^{\prime}\) and \(W_{s}^{\prime}\) respectively. Assume that the acceleration due to gravity is the same everywhere and that the balances are ante sensitive: (1) \(W_{B}=W_{S}\) (2) \(W_{B}^{\prime}=W_{S}^{\prime}\) (3) \(W_{B}=W_{B}^{\prime}\) (4) \(W_{S}^{\prime}

A container filled with liquid up to height \(h\) is placed on a smooth horizontal surface. The container is having a small hole at the bottom. As the liquid comes out from the hole, the container moves in a backward direction with acceleration \(a\) and finally, when all the liquid is drained out, it acquires a velocity \(v\). Neglect mass of the container. In this case (1) both \(a\) and \(v\) depend on \(h\) (2) only \(a\) depends on \(h\) (3) only \(v\) depends on \(h\) (4) neither \(a\) nor \(v\) depends on \(h\)
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