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Chapter 6: Problem 4

A spring balance reads \(W_{1}\) when a ball is suspended from it. A weighing machine reads \(W_{2}\) when a tank of liquid is kept on it. When the ball is immersed in the liquid, the spring balance reads \(W_{3}\) and the weighing machine reads \(W_{4}\). Then, which of the following are not correct? (1) \(W_{1}W_{3}\) (4) \(W_{2}>W_{4}\)

Short Answer

Expert verified
Options (1) and (4) are not correct.

Step by step solution

01

Understanding the Scenario

We have a spring balance initially reading the weight of a ball as \(W_1\). A weighing machine measures the weight of a tank with liquid as \(W_2\). When the ball is immersed in the liquid, the spring balance now reads \(W_3\), and the weighing machine reads \(W_4\).
02

Spring Balance Weight Analysis

The spring balance measures the apparent weight of the ball when it is in air and when it is immersed in liquid. The apparent weight \(W_3\) when the ball is immersed in the liquid will be less than the weight in air \(W_1\) due to the buoyant force exerted by the liquid. Thus, \(W_3 < W_1\).
03

Weighing Machine Weight Analysis

When the ball is immersed in the liquid, the liquid exerts an upward buoyant force on the ball, which increases the normal force on the weighing machine. As a result, the reading on the weighing machine \(W_4\) becomes greater than \(W_2\). Thus, \(W_4 > W_2\).
04

Evaluate Each Option

By analyzing the qualitative descriptions: - For option (1) \(W_1 < W_3\), it is incorrect because \(W_3\) is less than \(W_1\).- For option (2) \(W_1 > W_3\), it is correct because \(W_3\) is less than \(W_1\).- For option (3) \(W_2 < W_4\), it is correct because \(W_4\) is greater than \(W_2\).- For option (4) \(W_2 > W_4\), it is incorrect because \(W_4\) is greater than \(W_2\).

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Key Concepts

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

Spring Balance
A spring balance is a very useful tool for measuring the weight of an object. It works by detecting the force exerted by the object due to gravity. When a ball is hung from a spring balance, it records the force necessary to balance the gravitational pull on the ball, displaying it as weight \( W_1 \).

However, when the ball is immersed in a liquid, the situation changes dramatically. The liquid exerts an upward buoyant force on the ball, making the gravitational force appear weaker. Consequently, the spring balance now reads a lesser value, \( W_3 \), for the ball's weight. This happens because the buoyant force counters part of the gravitational force, making the ball seem lighter. As a result, the equation \( W_3 < W_1 \) holds true, which means the ball appears lighter when in a liquid compared to when it is in air.
Weighing Machine
A weighing machine, or a balance scale, is commonly used to measure the weight of objects. In this scenario, it's used to weigh a tank of liquid with and without a ball immersed.

Initially, the weighing machine reads \( W_2 \), the weight of the tank with liquid only. When the ball is immersed in the liquid, an interesting phenomenon occurs. The liquid exerts an upward buoyant force on the ball, and due to Newton's third law, there's an equal reaction force. This reaction force increases the normal force on the weighing machine, causing it to display a higher weight, \( W_4 \). Therefore, \( W_4 > W_2 \). This effect demonstrates how buoyancy can affect the readings on a weighing machine.
Apparent Weight
The concept of apparent weight is crucial in understanding how objects weigh under different circumstances. Apparent weight is essentially the weight an object seems to have when submerged in a fluid, factoring in the buoyant force exerted by the fluid.

When an object like a ball is submerged, the buoyant force partially offsets the gravitational force. This results in a lower apparent weight than the actual weight. In our scenario, the ball's actual weight \( W_1 \) is greater than its apparent weight \( W_3 \) when submerged. Conversely, in the case of the weighing machine, the presence of the buoyant force increases the apparent weight of the liquid-container system to \( W_4 \), which is more than \( W_2 \).
  • The concept of apparent weight is vital in fluid mechanics.
  • It helps explain phenomena like floating and sinking.
  • This concept is widely applied in designing ships and submarines.
Understanding apparent weight allows us to predict and adapt to changes in weight measurements due to buoyancy.

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Most popular questions from this chapter

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) \(7.5 \mathrm{~cm}\)

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 U-tube of base length ' \(l\) ' filled with the same volume of two liquids of densities \(\rho\) and \(2 \rho\) is moving with an acceleration ' \(a\) ' on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height \(h\) is given by (1) \(\frac{a}{2 g} l\) (2) \(\frac{3 a}{2 g} l\) (3) \(\frac{a}{g} l\) (4) \(\frac{2 a}{3 g} l\)

A uniform rod of length \(b=90 \mathrm{~cm}\) capable of turning about its end which is out of water, rests inclined to the vertical. If its specific gravity is \(5 / 9\), find the length immersed in water (in \(\mathrm{cm}\) ).

A uniform \(\operatorname{rod} A B, 12 \mathrm{~m}\) long weighing \(24 \mathrm{~kg}\), is supported at end \(B\) by a flexible light string and a lead weight (of very small size) of \(12 \mathrm{~kg}\) attached at end \(A\). The rod floats in water with one-half of its length submerged. Find the volume of the rod (in \(\times 10^{-3} \mathrm{~m}^{3}\) ) [Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), density of water \(\left.=1000 \mathrm{~kg} / \mathrm{m}^{3}\right]\)
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