Physics, 30.12.2019 09:31 graceaowen

What fraction of a piece of lead will be submerged when it floats in mercury? the density of lead is 11.3×103kg/m3 and density of mercury is 13.6×103kg/m3?

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Answer from: j015

According to the Archimedes’ Principle, a body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.

In this case we have a piece of lead floating (partially immersed) in liquid mercury. If we observe in the image attached, this object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.

This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero:

$F_{total}=0$ (1)

Here, the piece of lead experiences two kind of forces:

A force downwards because of its weight:

W=mg (2)

Where is the mass of the piece of lead and the acceleration of gravity, which is $9.8\frac{m}{s^{2}}$

And a force applied upwards because of the buoyancy:

$F_{buoyancy}=d_{m}gV_{i}$ (3)

Where;

$d_{m}$ is the density of liquid mercury which is $13.6(10^{3})\frac{kg}{m^{3}}$

And $V_{i}$ is the volume of the immersed part of the piece of lead

Well, if the total net force exerted to the object is zero:

$F_{total}= F_{buoyancy}-W=0$

Note has a negative sign because this force is applied downwards

This means the buoyancy is equal to the weight:

$F_{buoyancy}=W$ (4)

$d_{m}gV_{i}=mg$ (5)

Dividing by in both sides:

$d_{m}V_{i}=m$ (6)

Now, the density of an object or fluid is the relation between the mass and the volume, in the case of the object of this problem (the piece of lead) is:

$d=\frac{m}{V}$ (7)

Where is the Volume of the piece of lead.

From (7) we can find the mass of the piece of lead:

m=dV (8)

and substitute this value in (6), in order to get a relation between the density of the object and the density of the liquid mercury (the fluid), as follows:

$d_{m}V_{i}=dV$

$\frac{V_{i}}{V}=\frac{d}{d_{m}}$ (9)

$\frac{d}{d_{m}}=\frac{11.3(10^{3})kg/m^{3}}{13.6(10^{3})kg/m^{3}}$

Then;

$\frac{11.3(10^{3})kg/m^{3}}{13.6(10^{3})kg/m^{3}}=0.83$

This result is the same as $\frac{83}{100}$ which represents the $83\%$

Therefore:

$83\%$ of the mass of the piece of lead is immersed in mercury

$What fraction of a piece of lead will be submerged when it floats in mercury? the density of lead i$

Answer from: Quest

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Answer from: Quest

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