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Chapter 11: Problem 28

If air of weight \(w\) is filled in a empty balloon which weighs \(w_{1}\), the weight of balloon will become \(w_{2}\). Suppose the density of air inside and outside the balloon is same, then : (a) \(w_{2}=w_{1}+w\) (b) \(w_{2}=\sqrt{w_{1} w}\) (c) \(w_{2}=w_{1}\) (d) \(w_{2}=w_{1}-w\)

Short Answer

Expert verified
The correct answer is (a) \(w_{2}=w_{1}+w\).

Step by step solution

01

Understand the problem

We are given a balloon initially empty and it weighs \(w_{1}\). When air of weight \(w\) is added to it, the balloon's weight becomes \(w_{2}\). Our task is to find the correct relationship between \(w\), \(w_{1}\), and \(w_{2}\).
02

Analyze the given condition

The problem states that the density of air inside and outside the balloon is the same. This implies that the only change in weight comes from the addition of air. The weight of anything is simply the sum of its individual components, in this case, air weight and balloon weight.
03

Formulate the equation

Since air is being added to the balloon and the density does not change, we can deduce that the total weight of the balloon when filled with air is simply the initial weight of the balloon plus the weight of the air. Thus, the equation is \(w_{2} = w_{1} + w\).
04

Validate the equation with options

Review the options provided:(a) \(w_{2}=w_{1}+w\)(b) \(w_{2} = \sqrt{w_{1} w}\)(c) \(w_{2}=w_{1}\)(d) \(w_{2}=w_{1}-w\)The equation \(w_{2} = w_{1} + w\) from our previous step matches option (a), confirming that it is the correct choice.

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Key Concepts

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

Density of Air
Density is an essential concept in physics and understanding how objects interact with their surroundings. It measures how much mass is contained in a specific volume. For air, this means how much mass of air molecules is present within a cubic meter of space. The density of air can vary depending on conditions like temperature, pressure, and humidity.
In this exercise, we're considering the situation where the density inside the balloon and outside it is the same. This is an important factor because it implies that the air's mass per unit volume stays constant, whether inside or outside the balloon. Therefore, adding air only affects the total weight of the balloon due to the mass of the added air, and not through any variation in density.
Weight of Balloon
When considering the weight of an object, we account for the sum of the weights of all its components. Here, our balloon initially has its weight when empty, denoted as \(w_1\). When air with a certain weight \(w\) is added, the whole system's weight changes.
The crucial part is understanding that the weight of a filled balloon \(w_2\) will be the sum of the empty balloon's weight \(w_1\) and the weight of the air inside \(w\). This is a straightforward additive process because the introduced air has a specific weight, which combines with the balloon's original weight to give us the total.
Solving Equations
Equations are powerful tools that help us understand relationships between different quantities. In this problem, we were tasked with finding an appropriate equation that describes the balloon's weight after being filled with air. By analyzing the problem, we identified the sum of the air weight and the balloon weight as the overall weight of the balloon.
This leads us to a simple equation: \(w_2 = w_1 + w\). The provided options in the exercise represented different potential equations. By evaluating these options, we selected the equation that aligned with our logical deduction based on the concept of weight addition when density remains constant.
Understanding how to derive and validate equations like this is crucial in various physics problems, as it provides a methodical way to confirm our understanding of natural phenomena.

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Most popular questions from this chapter

Water flows along a horizontal pipe whose cross-section is not constant. The pressure is \(1 \mathrm{~cm}\) of \(\mathrm{Hg}\) where the velocity is \(35 \mathrm{~cm} / \mathrm{s}\). At a point where the velocity is \(65 \mathrm{~cm} / \mathrm{s}\), the pressure will be: (a) \(0.89 \mathrm{~cm}\) of \(\mathrm{Hg}\) (b) \(8.9 \mathrm{chi}, \sqrt{\mathrm{H}_{8}}\) (c) \(0.5 \mathrm{~cm}\) of \(\mathrm{Hg}\) (d) \(1 \mathrm{~cm}\) of \(\mathrm{Hg}\)

On a smooth inclined plane, making an angle \(\alpha\) with horizontal, a trolley containing a liquid of density \(\rho\) slides down. What is the angle of inclination \(\theta\) of free surface with horizontal? (a) \(\theta=-\alpha\) (b) \(\theta=\frac{\alpha}{3}\) (c) \(\theta=\frac{\alpha}{9}\) (d) \(\theta=\alpha\)

A tank accelerates upwards with acceleration \(a=1 \mathrm{~m} / \mathrm{s}^{2}\) contains water. A block of mass \(1 \mathrm{~kg}\) and -density \(0.8 \mathrm{~g} / \mathrm{cm}^{3}\) is held stationary inside the tank with the help of the string as shown in figure. The tension in the string is: (density of water \(=1000 \mathrm{~kg} / \mathrm{m}^{3}\) ) (a) \(T=2.2 \mathrm{~N}\) (b) \(T=2.75 \mathrm{~N}\) (c) \(T=3 \mathrm{~N}\) (d) \(T=2.4 \mathrm{~N}\)

Two identical cylindrical vessels with their bases at the same level each contains a liquid of density \(d\). The height of the liquid in one vessel is \(h_{1}\) and that in the other vessel is \(h_{2}\). The area of either base is \(A\). The work done by gravity in equalizing the levels, when the two vessels are connected is : (a) \(\left(h_{1}-h_{2}\right) g d\) (b) \(\left(h_{1}-h_{2}\right) g A d\) (c) \(\frac{1}{2}\left(h_{1}-h_{2}\right) g A d\) (d) \(\frac{1}{4}\left(h_{1}-h_{2}\right) g A d\)

A cylindrical vessel is filled with water to a height \(H . A\) vessel has two small holes in the side, from which water is rushing out horizontally and the two streams strike the ground at the same point. If the lower hole \(Q\) is \(h\) height above the ground, then the height of hole \(P\) above the ground will be: (a) \(2 h\) (b) \(\mathrm{H} / \mathrm{h}\) (c) \(H-h\) (d) \(\mathrm{H} / 2\)
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