Apparent weight of body immersed in liquid

AI Thread Summary
The discussion focuses on the apparent weight of a body immersed in different liquids, analyzing how it changes based on the liquid's density relative to the body's density. In scenarios (a), (b), and (c), the apparent weight transitions from positive to zero to negative as the body becomes fully immersed, indicating whether it sinks or floats. The upthrust is only accurately calculated when the body is fully immersed, and the relationship between the body’s immersion depth and upthrust is emphasized. The participants clarify that the density order of the liquids is correct, and the behavior of the body in each liquid can be inferred from the apparent weight graph. Understanding these principles helps determine the buoyancy behavior of the object in various liquids.
brotherbobby
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Homework Statement
A rectangular block is pushed face-down into three liquids, in turn. The apparent weight ##W_{\text{app}}## of the block versus the depth ##h## in the three liquids is plotted in the graph shown below.

Rank the liquids according to their greatest weight per unit volume, greatest first.
Relevant Equations
Apparent weight of a body partially (or wholly) immersed in liquid ##w'_B = w_B - U## where ##w_B## is the (original) weight of the body and ##U## is the upthrust which is equal to the weight of the liquid diplaced : ##U = \Delta W_L = \rho_L V_B g##. Here ##\rho_L \; \text{and} \; V_B## are the density of the liquid and the volume of the body, respectively.
1580206450466.png


I have to assume that ##h## is the height of the body. The graph above shows how the apparent weight of the body changes as it is immersed into the liquid.

In (a), after immersing the whole height of the body, the apparent weight ##W_{\text{app}} > 0##. Hence the upthrust ##U < w_B \Rightarrow (\rho_L)_a < \rho _B##. If released, the body would sink into the liquid (a).

In (b), the apparent weight of the body is 0 by the time it is wholly immersed : ##W_{\text{app}} = 0##. Hence the upthrust ##U = w_B \Rightarrow (\rho_L)_b = \rho _B##. If released, the body would just sink into the liquid (b), and float entirely immersed.

In (c), the apparent weight of the body is negative by the time it is wholly immersed : ##W_{\text{app}} < 0##. Hence the upthrust ##U > w_B \Rightarrow (\rho_L)_c > \rho _B##. If released, the body would float in the liquid, displacing an amount of liquid equal to its weight. The (height) depth of the body above the liquid is not ##h## but the height corresponding to where the line in (c) cuts the ##x## axis, where it's apparent weight is zero.

From above, we find that ##\boxed{\color{red}{(\rho_L)_c > (\rho_L)_b > (\rho_L)_a}}##.

Is my answer correct?
 
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Apart from assuming that h is the height of the body, when you are told it is the depth of immersion, yes. You are not told the height of the body, so you don't know at what point it will be totally immersed, but the order of the densities is correct.

Also U = ρLVBg is only correct when the body is fully immersed. If it is partially immersed, the volume you should use is the volume of that part of the body which is immersed.
Note that this means that at the point where the body becomes totally immersed, the slope of the lines in your graph changes to become horizontal. This point is not reached in the experiment described. But that doesn't affect your right answer to the question.
 
mjc123 said:
Apart from assuming that h is the height of the body, when you are told it is the depth of immersion, yes. You are not told the height of the body, so you don't know at what point it will be totally immersed, but the order of the densities is correct.

Also U = ρLVBg is only correct when the body is fully immersed. If it is partially immersed, the volume you should use is the volume of that part of the body which is immersed.
Note that this means that at the point where the body becomes totally immersed, the slope of the lines in your graph changes to become horizontal. This point is not reached in the experiment described. But that doesn't affect your right answer to the question.

Yes thank you. Calling the upthrust ##U = \rho_L V_B g## is only valid if the body is wholly immersed, or else the body's entire volume does not 'engage' with the liquid to displace it. Hence, upthrust is a function of the depth of a body immersed, so we can write I suppose ##U= U(d)##. It's a small point that books miss out on.

Yes, the upthrust of a liquid on a body is independent of position if the body was wholly immersed into it. Hence, as you said, the graphs in my question would be horizontal beyond ##h##.

Thank you, those were important points.
 
How does one know that in liquid a, the object sinks, in liquid b, it floats, and in liquid c it floats and displaces liquid equal to its own weight? I couldn't tell by looking at the graph. Thank you in advance!
 
Welcome to PF.

ducsinhsn said:
How does one know that in liquid a, the object sinks, in liquid b, it floats, and in liquid c it floats and displaces liquid equal to its own weight? I couldn't tell by looking at the graph. Thank you in advance!
What are your thoughts on that?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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