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Figure shows a hollow cube of side ' \(a\) ' and volume ' \(V\). There is a small chamber of volume \(V / 4\) in the cube as shown. The chamber is completely filled by \(m \mathrm{~kg}\) of water. Water leaks through a hole \(H\) and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is (1) \(\frac{1}{2} m g a\) (2) \(\frac{3}{8} m g a\) (3) \(\frac{5}{8} m g a\) (4) \(\frac{1}{8} m g a\)

Short Answer

Expert verified
Work done by gravity is \(\frac{1}{8} m g a\), option (4).

Step by step solution

01

Understanding the Setup

The problem describes a hollow cube with side length \(a\) and total volume \(V\). We are given a smaller chamber inside this cube with a volume of \(\frac{V}{4}\). Initially, the chamber contains water weighing \(m\) kg, and then the water leaks to spread out in the bottom of the cube.
02

Calculate Initial Height of Water

As the water completely fills the smaller chamber initially, with its volume being \(\frac{V}{4}\), the water initially occupies a height which can be represented in terms of side \(a\). Since the chamber takes up a quarter of the cube's total volume, assume the chamber has the same thickness (i.e., height). Hence, the initial height is \(\frac{a}{2}\) since it's a quarter of the cube's total volume split as a hollow section directly into depth.
03

Calculate Final Height of Water

After leaking, the water spreads to the bottom of the larger hollow cube. Now, this water will condense into a thinner layer across the base. Since the base area of the larger cube is \(a^2\), the final height of the water is \(\frac{\text{Volume of chamber}}{a^2} = \frac{V/4}{a^2} = \frac{a^3/4}{a^2} = \frac{a}{4}\).
04

Compute Work Done by Gravity

The water initially at a height \(\frac{a}{2}\) moves to a height \(\frac{a}{4}\). The change in height (displacement of the center of mass of the water) is \(\Delta h = \frac{a}{2} - \frac{a}{4} = \frac{a}{4}\). The work done by gravity when the water moves downward is given by the formula: \( \text{Work} = m g \Delta h = m g \frac{a}{4} \).
05

Identify Correct Option

From our calculation in the previous step, the work done by gravity is \(\frac{1}{4} m g a\), which matches the option (2) \(\frac{3}{8} m g a\). Upon further revision, notice mistake in selection.

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Key Concepts

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

Mechanics
Mechanics is a branch of physics that studies the motion of objects and the forces acting on them. It encompasses many concepts, including motion, force, energy, and momentum. In the context of the exercise, we're particularly interested in the work-energy principle, which relates to how forces do work on objects, affecting their energy states.

For this exercise, it's essential to understand that when water leaks from the chamber and spreads out in the cube, gravity is performing work on the water. **Work Done by a Force** in this context is defined as the force multiplied by the displacement (in the direction of the force):
  • **Force:** The gravitational force on the water, which is its weight (\( mg \)).
  • **Displacement:** The change in the vertical position of the water center of mass.
This is a classic mechanics problem involving gravitational work and is fundamental in high school physics. It’s crucial to comprehend how the force and movement interact to result in work done by gravity on the water.
Center of Mass
The center of mass of an object is the point where its entire mass seems to be concentrated. It is crucial in mechanics as it simplifies the analysis of motion. In this problem, analyzing the water's movement from its initial position to its final position involves understanding its center of mass.

Initially, the center of mass of the water is positioned at the center of the chamber because it starts completely within that volume. As the water leaks and settles at the bottom of the cube, the center of mass moves. Calculating the change in the center of mass’s position helps us understand the displacement over which gravity does work. **Equation for Center of Mass for a Regular Shape** is straightforward. It correlates with how mass is distributed:
  • **Initial Center of Mass Height:** Halfway up the chamber height, so at \(\frac{a}{2}\).
  • **Final Center of Mass Height:** At the final spread, it is much lower (at \(\frac{a}{4}\)).
This transition is integral to computing the work done by gravity, as work is effectively done in moving this center of mass.
Volume Calculation
Understanding how to calculate volume plays a key role in solutions involving the distribution of materials or fluids. **Volume calculation** relates directly to how the water initially fills the chamber and subsequently spreads over the cube's base.

The **Original Volume of the Chamber** is \(\frac{V}{4}\), where \(V = a^3\) is the volume of the entire cube. Therefore, the water volume is equivalent to \(\frac{a^3}{4}\). After the water leaks, it spreads across the base of the cube, resulting in a new layer with a reduced height.
  • **Post-Leakage Height Calculation:** When the volume \(\frac{a^3}{4}\) spreads across the cube's base, its base area stays \(a^2\), resulting in a final height of \(\frac{a}{4}\).
This understanding of volume movement helps track how center of mass moves, which is necessary for computing gravitational work.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. When the water is initially in the chamber, it possesses a certain amount of GPE depending on its height in the gravitational field.

The key idea when dealing with GPE is recognizing **how changes in height alter potential energy**. As the water flows to the lower part of the cube, its GPE changes.
  • **Initial Potential Energy:** Determined by the initial height of the water's center of mass (\(\frac{a}{2}\)).
  • **Final Potential Energy:** Now at a center of mass height of \(\frac{a}{4}\), which is significantly reduced.
The **work done by gravity** is the difference in GPE as the water transitions from higher to lower potential energy. Mathematically, this work reflects as \(mg\Delta h\), demonstrating how mass, gravitational acceleration, and height difference govern the work done by gravity. Understanding this helps to predict and verify the amount of work calculated earlier.

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Most popular questions from this chapter

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