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Chapter 8: Problem 73

An object is floating in a liquid, kept in a container. The container is placed in a lift. Choose the correct option(s). (A) Buoyant force increases as lift accelerates up. (B) Buoyant force decreases as lift accelerates up. (C) Buoyant force remains constant as lift accelerates. (D) The fraction of solid submerged into liquid does not change.

Short Answer

Expert verified
The correct option is (A) Buoyant force increases as lift accelerates up. This is because the effective gravity inside the lift increases when it accelerates upward, leading to an increase in buoyant force as given by the formula \(F_b = V_{submerged} \times \rho_{liquid} \times (g+a)\), where \(a\) is the upward acceleration of the lift.

Step by step solution

01

Option A: Buoyant force increases as lift accelerates up.

To determine if the buoyant force increases as lift accelerates up, let's recall the buoyant force formula: \(F_b = V_{submerged} \times \rho_{liquid} \times g\), where \(F_b\) is the buoyant force, \(V_{submerged}\) is the volume of the object submerged in the liquid, \(\rho_{liquid}\) is the density of the liquid, and \(g\) is the acceleration due to gravity. When the lift accelerates upward, the effective gravity inside the lift increases (i.e., \(g_{effective} = g + a\), where \(a\) is the upward acceleration of the lift). So, the buoyant force becomes \(F_b = V_{submerged} \times \rho_{liquid} \times (g+a)\). Since \(a>0\), the buoyant force increases when the lift accelerates up. Therefore, Option A is correct.
02

Option B: Buoyant force decreases as lift accelerates up.

We have already shown in Option A that the buoyant force increases when the lift accelerates upward. So, Option B is incorrect.
03

Option C: Buoyant force remains constant as lift accelerates.

Since we have concluded that the buoyant force increases as the lift accelerates upwards in Option A, the buoyant force cannot remain constant as the lift accelerates. Therefore, Option C is incorrect.
04

Option D: The fraction of solid submerged into liquid does not change.

When the lift accelerates upward, the force acting on the floating object increases, and the object experiences a greater upward force due to an increase in effective gravity. Since the object's weight remains constant, the buoyant force will also have to balance that weight. In this case, the floating object will undergo a change in the volume of liquid displaced, i.e., the fraction of the solid submerged into the liquid will change. Therefore, Option D is incorrect. In conclusion, the correct option is (A) Buoyant force increases as lift accelerates up.

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Key Concepts

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

Archimedes' Principle
Archimedes' Principle is a fundamental concept that explains why objects float or sink. It states that any object submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by that object. This principle is vital for understanding buoyancy and helps explain behaviors of objects in different fluids.
  • The buoyant force is given by the formula: \( F_b = V_{submerged} \times \rho_{liquid} \times g \), where:
    • \(V_{submerged}\) is the volume of the object submerged.
    • \(\rho_{liquid}\) is the density of the liquid.
    • \(g\) is the acceleration due to gravity.
  • Hydrodynamics or hydrostatics often use this principle to study floating or submerged bodies in fluids like water, oil, or air.
Understanding Archimedes' Principle enables us to predict whether an object will float or sink in a given liquid based on its density and volume. When the upward buoyant force balances the object鈥檚 weight, it floats.
Effective Gravity
When analyzing systems where there is acceleration, such as a lift moving up or down, understanding the concept of effective gravity becomes essential. Effective gravity refers to the apparent force of gravity on a system experiencing acceleration.
  • In an upward-accelerating lift, the effective gravity increases. This can be calculated as \( g_{effective} = g + a \), where
    • \(g\) is the standard acceleration due to gravity (approximately 9.81 m/s虏).
    • \(a\) is the acceleration of the lift.
  • Conversely, if the lift accelerates downwards, the effective gravity decreases.
This concept is crucial when calculating forces in accelerating systems, as it impacts both the weight experienced by objects and the buoyant force, affecting how objects behave in a fluid within the system.
Liquid Density
Liquid density is another fundamental factor in understanding buoyant forces and the overall interaction between fluids and submerged objects. Density is defined as mass per unit volume and is symbolized by \(\rho\).
  • Density formula: \( \rho = \frac{m}{V} \), where:
    • \(m\) is the mass of the liquid.
    • \(V\) is the volume of the liquid.
  • In the context of buoyancy, the density of the liquid affects the magnitude of the buoyant force experienced by objects submerged in it.
  • High-density liquids can exert larger buoyant forces on objects.
An object's buoyancy is determined not only by its material density but also by the density of the liquid in which it is submerged. Therefore, knowing the liquid鈥檚 density will help predict the object's behavior, such as how submerged it will be when floating.

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Most popular questions from this chapter

A large cylindrical tank has a hole of area \(A\) at its bottom. Water is poured in the tank by a tube of equal cross-sectional area \(A\) ejecting water at the speed \(v\). (A) The water level in the tank will keep on rising. (B) No water can be stored in the tank. (C) The water level will rise to a height \(v^{2} / 2 g\) and then stop. (D) The water level will oscillate.

One end of a uniform wire of length \(L\) and of weight \(W\) is attached rigidly to a point in the roof and a weight \(W_{1}\) is suspended from its lower end. If \(S\) is the area of cross-section of the wire, the stress in the wire at height ( \(3 L / 4\) ) from its lower end is(A) \(W_{1} / S\) (B) \(\left\\{W_{1}+(W / 4)\right\\} S\) (C) \(\left(W_{1}+\frac{3 W}{4}\right) / S\) (D) \(\left(W_{1}+W\right) / S\)

The weight of a body in air is \(100 \mathrm{~N}\). Its weight in water, if it displaces \(400 \mathrm{cc}\) of water is (A) \(90 \mathrm{~N}\) (B) \(94 \mathrm{~N}\) (C) \(98 \mathrm{~N}\) (D) None of these

If two soap bubbles of different radii are connected by a tube (A) Air flows from the bigger bubble to the smaller bubble till the sizes become equal. (B) Air flows from bigger bubble to the smaller bubble till the sizes are interchanged. (C) Air flows from the smaller bubble to the bigger bubble. (D) There is no flow of air.

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) \(\frac{\rho g h^{2} A}{4}\)
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