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Chapter 6: Problem 11

A vessel contains oil (density \(=0.8 \mathrm{~g} / \mathrm{cm}^{3}\) ) over mercury (density \(\left.=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)\). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in \(\mathrm{g} / \mathrm{cm}^{3}\) is (1) \(3.3\) (2) \(6.4\) (3) \(7.2\) (4) \(12.8\)

Short Answer

Expert verified
The density of the sphere is 7.2 g/cm鲁 (option 3).

Step by step solution

01

Understand the setup

The problem involves a sphere floating at the interface of two liquids: oil and mercury. The sphere has half of its volume submerged in oil and the other half in mercury.
02

Apply Buoyancy Principle

For the sphere to be in equilibrium, the buoyant force must equal its weight. Buoyant force is the sum of the forces due to oil and mercury, which can be expressed as: \[ V_{ ext{in oil}} imes ho_{ ext{oil}} imes g + V_{ ext{in mercury}} imes ho_{ ext{mercury}} imes g = V_{ ext{total}} imes ho_{ ext{sphere}} imes g \] Since half the volume is in oil and the other half in mercury, this translates to: \[ \frac{V}{2} imes 0.8 + \frac{V}{2} imes 13.6 = V imes \rho_{ ext{sphere}} \]
03

Solve for Sphere Density

Simplify the equation from the previous step: \[ \frac{V}{2} \times 0.8 + \frac{V}{2} \times 13.6 = V \times \rho_{ ext{sphere}} \]\[ \frac{V}{2} (0.8 + 13.6) = V \times \rho_{ ext{sphere}} \]Cancel the common factor of \( V \) on both sides: \[ \frac{1}{2} (0.8 + 13.6) = \rho_{ ext{sphere}} \]Calculate the density: \[ \frac{1}{2} (14.4) = 7.2 \]
04

Conclusion

The calculated density of the sphere is \( \rho_{ ext{sphere}} = 7.2 \text{ g/cm}^3 \). This matches option (3) in the given choices.

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Key Concepts

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

Density
Density is a fundamental concept in understanding buoyancy and fluid mechanics. It is defined as the mass of a substance per unit volume and is usually expressed in grams per cubic centimeter (g/cm鲁) or kilograms per cubic meter (kg/m鲁).
Density can be calculated using the formula: \[ \rho = \frac{m}{V} \] where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume.
Density influences whether an object will float or sink when placed in a fluid. An object will float if its density is less than the density of the fluid it is placed in.
Floating objects
When discussing floating objects, there are key principles to remember. An object partially or fully submerged in a fluid floats due to the buoyant force, which acts upwards.
**Key Points about Floating Objects:**
  • Buoyant Force is dictated by Archimedes' Principle, which states that the upward force (buoyancy) on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
  • For an object to float, the buoyant force must be equal to or greater than the object's weight.
  • Partially submerged objects mean that the system is at equilibrium with the weight of the displaced fluid equaling the object鈥檚 weight.
In our exercise, the sphere floats because the buoyant force from both liquids (oil and mercury) matches the weight of the sphere.
Liquid density
Liquid density plays a crucial role in determining how objects behave when placed in a fluid. The density of each liquid is different and affects the distribution of buoyant forces. Each subsection of fluid contributes to the overall buoyancy:
  • Oil in this exercise has a density of 0.8 g/cm鲁 鈥� this means it exerts less buoyant force than more dense liquids.
  • Mercury is much denser at 13.6 g/cm鲁 and therefore provides a greater buoyant force compared to oil.
The combination of these two forces acting together maintains the sphere's balance and helps in calculating the sphere's density.
Equilibrium in fluids
Equilibrium occurs when all forces acting on an object are balanced, resulting in the object being at rest or moving with constant velocity. For an object floating in a fluid, this means the weight of the object is balanced exactly by the buoyant force.
For the sphere in the exercise, the equilibrium condition can be described by the equation: \[ F_{\text{buoyant }} = F_{\text{gravity }} \] The gravity force (\( F_{\text{gravity}} \)) acting downward is counterbalanced by the buoyant force acting upward, which consists of contributions from both oil and mercury.
This ensures the object remains stable and does not move vertically. Understanding these equilibrium conditions is crucial for predicting the behavior of objects in various liquids.

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Most popular questions from this chapter

An open vessel containing liquid is moving with constant acceleration \(a\) on a levelled horizontal surface. For this situation mark out the correct statement(s). (1) The maximum pressure is at the leftmost bottom corner. (2) Along a horizontal line within the liquid as we move from left to right the pressure decreases. (3) The pressure at all points on a line making an angle of \(\tan ^{-1}\left(\frac{a}{g}\right)\) with horizontal would be same. (4) Along a horizontal line within the liquid as we move from left to right, the pressure remains same.

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

Water from a tap emerges vertically downwards with an initial velocity \(v_{0}\). Assuming pressure is constant throughout the stream of water and the flow is steady, find the distance from the tap at which cross-sectional area of stream is half of the cross-sectional area of stream at the tap. (1) \(v_{0}^{2} / 2 g\) (2) \(3 v_{0}^{2} / 2 g\) (3) \(2 v_{0}^{2} / 2 g\) (4) \(5 v_{0}^{2} / 2 g\)

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}
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