# finding the density of a liquid

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terr123_78787

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#1

A long wooden cylinder is placed into a liquid and it floats. The length of the cylinder below the liquid level is 15cm.

The pressure exerted by the liquid alone on the bottom of the cylinder is 1.9 x 10^3 Pa.

Calculate the density of the liquid.

Hi can someone please help me with this question, my working and thoughts are below!

i thought you could use the equation: pressure = height x density x g

so 1.9 x10^3 = 0.15 x density x 9.81, density thus = 1291. 19 = 1300 kgm^-3

or i thought you could use this way below:

FB = PA

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi = density x pi x 0.15 x 9.81

5969.026046 = density x 4.62285359

density =5969.026046/4.62285359

= 1291.199456 = 1291.2 = 1300 kgm^-3

is this correct?

The pressure exerted by the liquid alone on the bottom of the cylinder is 1.9 x 10^3 Pa.

Calculate the density of the liquid.

Hi can someone please help me with this question, my working and thoughts are below!

i thought you could use the equation: pressure = height x density x g

so 1.9 x10^3 = 0.15 x density x 9.81, density thus = 1291. 19 = 1300 kgm^-3

or i thought you could use this way below:

FB = PA

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi = density x pi x 0.15 x 9.81

5969.026046 = density x 4.62285359

density =5969.026046/4.62285359

= 1291.199456 = 1291.2 = 1300 kgm^-3

is this correct?

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Joinedup

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#2

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#2

Yeah good method, result seems sensible (density of pure water is 1000 kg per m

^{3})
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terr123_78787

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#3

(Original post by

Yeah good method, result seems sensible (density of pure water is 1000 kg per m

**Joinedup**)Yeah good method, result seems sensible (density of pure water is 1000 kg per m

^{3})
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terr123_78787

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#4

can someone help me with these questions, they carry on from the question above!

the cylinder is pushed down into the liquid and then allowed to oscillate freely. the graph of displacement x against time is shown below:

the cylinder oscillates with simple harmonic motion with frequency of 1.4 Hz.

1. calculate the displacement, in cm, at time t = 0.60.

2. calculate the maximum speed of the oscillating cylinder.

3. the cylinder is now pushed down further into the liquid before being. As before, the cylinder oscillates with simple harmonic motion. state the effect this has on both the amplitude and the period.

if anyone can help me with these or explain to me how to start, id appreciate it!!

the cylinder is pushed down into the liquid and then allowed to oscillate freely. the graph of displacement x against time is shown below:

the cylinder oscillates with simple harmonic motion with frequency of 1.4 Hz.

1. calculate the displacement, in cm, at time t = 0.60.

2. calculate the maximum speed of the oscillating cylinder.

3. the cylinder is now pushed down further into the liquid before being. As before, the cylinder oscillates with simple harmonic motion. state the effect this has on both the amplitude and the period.

if anyone can help me with these or explain to me how to start, id appreciate it!!

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0le

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#5

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#5

(Original post by

A long wooden cylinder is placed into a liquid and it floats. The length of the cylinder below the liquid level is 15cm.

The pressure exerted by the liquid alone on the bottom of the cylinder is 1.9 x 10^3 Pa.

Calculate the density of the liquid.

Hi can someone please help me with this question, my working and thoughts are below!

i thought you could use the equation: pressure = height x density x g

so 1.9 x10^3 = 0.15 x density x 9.81, density thus = 1291. 19 = 1300 kgm^-3

or i thought you could use this way below:

FB = PA

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi = density x pi x 0.15 x 9.81

5969.026046 = density x 4.62285359

density =5969.026046/4.62285359

= 1291.199456 = 1291.2 = 1300 kgm^-3

is this correct?

**terr123_78787**)A long wooden cylinder is placed into a liquid and it floats. The length of the cylinder below the liquid level is 15cm.

The pressure exerted by the liquid alone on the bottom of the cylinder is 1.9 x 10^3 Pa.

Calculate the density of the liquid.

Hi can someone please help me with this question, my working and thoughts are below!

i thought you could use the equation: pressure = height x density x g

so 1.9 x10^3 = 0.15 x density x 9.81, density thus = 1291. 19 = 1300 kgm^-3

or i thought you could use this way below:

FB = PA

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi = density x pi x 0.15 x 9.81

5969.026046 = density x 4.62285359

density =5969.026046/4.62285359

= 1291.199456 = 1291.2 = 1300 kgm^-3

is this correct?

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caffeine--

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#6

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#7

(Original post by

You just copied the other user (or you all three are the same or friends) and you made the same mistake that they did.

**0le**)You just copied the other user (or you all three are the same or friends) and you made the same mistake that they did.

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#8

(Original post by

No, it isn't correct. They've ignored the pressure acting on the sides of the cylinder which also contributes to the buoyant force.

**0le**)No, it isn't correct. They've ignored the pressure acting on the sides of the cylinder which also contributes to the buoyant force.

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0le

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#9

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#9

(Original post by

sorry what other user? what mistake did i make?

**terr123_78787**)sorry what other user? what mistake did i make?

https://www.thestudentroom.co.uk/sho...47&postcount=4

FB = PA

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x A = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

5969.026042 x r^2 = density x r^2 x 4.62285359

5969.026042 = density x 4.62285359

Density = 5969.026042/4.62285359 = 1291.199456 = 1291.2 kgm^-3

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x A = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

5969.026042 x r^2 = density x r^2 x 4.62285359

5969.026042 = density x 4.62285359

Density = 5969.026042/4.62285359 = 1291.199456 = 1291.2 kgm^-3

FB = PA

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi = density x pi x 0.15 x 9.8

5969.026046 = density x 4.62285359

density =5969.026046/4.62285359

= 1291.199456 = 1291.2 = 1300 kgm^-3

PA = density x volume x g

1.9 x 10^3 x A = density x volume x 9.81

1.9 x 10^3 x pi x r^2 = density x pi x r^2 x 0.15 x 9.81

1.9 x 10^3 x pi = density x pi x 0.15 x 9.8

5969.026046 = density x 4.62285359

density =5969.026046/4.62285359

= 1291.199456 = 1291.2 = 1300 kgm^-3

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#10

(Original post by

Come on, stop playing dumb. You copied even the same symbols. Here is the post in their thread:

https://www.thestudentroom.co.uk/sho...47&postcount=4

Here is your post without the spacing:

Looks familiar doesn't it?

**0le**)Come on, stop playing dumb. You copied even the same symbols. Here is the post in their thread:

https://www.thestudentroom.co.uk/sho...47&postcount=4

Here is your post without the spacing:

Looks familiar doesn't it?

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0le

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#11

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#11

(Original post by

ohh i think that person is doing the same paper as me as ive seen a few of their posts but i didn't copy their working, the ocr a physics revision guide has an example of a similar question to this i worked from there? i am so confused lmao

**terr123_78787**)ohh i think that person is doing the same paper as me as ive seen a few of their posts but i didn't copy their working, the ocr a physics revision guide has an example of a similar question to this i worked from there? i am so confused lmao

*every*line

*and*write the same symbols. Anyway I've removed my other post. Stick with the approach you've used in the OP. It is the correct approach. Your answer for the density is correct.

I understand now what they've done:

The buoyant force is calculated by integrating the pressure acting on the total surface area of an immersed object. This is always true. The pressure also varies with depth. If you go deeper into the ocean, the pressure is higher.

However what I glaringly omitted is that the pressures on either side of the column part of the cylinder just cancel out. On that part of the cylinder they are acting on the horizontal direction on each side of the cylinder. On the left side of the cylinder the pressure acts inwards and on the right side of the cylinder it acts inwards.

So what you and the other user did was numerically correct, even if you got there by misunderstanding my post about total surface area. My formula and logic were correct, but my mistake was not realizing the simple fact that the pressures on that part of the cylinder just cancel out. All you then have is the pressure acting on the base. This means you don't need to know the radius of the cylinder.

Last edited by 0le; 1 year ago

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#12

(Original post by

Come on, you both happen to write to the exact same digits for

I understand now what they've done:

The buoyant force is calculated by integrating the pressure acting on the total surface area of an immersed object. This is always true. The pressure also varies with depth. If you go deeper into the ocean, the pressure is higher.

However what I glaringly omitted is that the pressures on either side of the column part of the cylinder just cancel out. On that part of the cylinder they are acting on the horizontal direction on each side of the cylinder. On the left side of the cylinder the pressure acts inwards and on the right side of the cylinder it acts inwards.

So what you and the other user did was numerically correct, even if you got there by misunderstanding my post about total surface area. My formula and logic were correct, but my mistake was not realizing the simple fact that the pressures on that part of the cylinder just cancel out. All you then have is the pressure acting on the base. This means you don't need to know the radius of the cylinder

**0le**)Come on, you both happen to write to the exact same digits for

*every*line*and*write the same symbols. Anyway I've removed my other post. Stick with the approach you've used in the OP. It is the correct approach. Your answer for the density is correct.I understand now what they've done:

The buoyant force is calculated by integrating the pressure acting on the total surface area of an immersed object. This is always true. The pressure also varies with depth. If you go deeper into the ocean, the pressure is higher.

However what I glaringly omitted is that the pressures on either side of the column part of the cylinder just cancel out. On that part of the cylinder they are acting on the horizontal direction on each side of the cylinder. On the left side of the cylinder the pressure acts inwards and on the right side of the cylinder it acts inwards.

So what you and the other user did was numerically correct, even if you got there by misunderstanding my post about total surface area. My formula and logic were correct, but my mistake was not realizing the simple fact that the pressures on that part of the cylinder just cancel out. All you then have is the pressure acting on the base. This means you don't need to know the radius of the cylinder

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#13

can anyone guide me through part 1, 2 and 3 ?

1. calculate the displacement, in cm, at time t = 0.60.

i know that displacement is max amplitude so i thought it was 2.0cm?

2. calculate the maximum speed of the oscillating cylinder.

im not sure how to do this

3. the cylinder is now pushed down further into the liquid before being. As before, the cylinder oscillates with simple harmonic motion. state the effect this has on both the amplitude and the period.

not sure on this either?

1. calculate the displacement, in cm, at time t = 0.60.

i know that displacement is max amplitude so i thought it was 2.0cm?

2. calculate the maximum speed of the oscillating cylinder.

im not sure how to do this

3. the cylinder is now pushed down further into the liquid before being. As before, the cylinder oscillates with simple harmonic motion. state the effect this has on both the amplitude and the period.

not sure on this either?

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reply

Callicious

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#14

(Original post by

can anyone guide me through part 1, 2 and 3 ?

1. calculate the displacement, in cm, at time t = 0.60.

i know that displacement is max amplitude so i thought it was 2.0cm?

2. calculate the maximum speed of the oscillating cylinder.

im not sure how to do this

3. the cylinder is now pushed down further into the liquid before being. As before, the cylinder oscillates with simple harmonic motion. state the effect this has on both the amplitude and the period.

not sure on this either?

**terr123_78787**)can anyone guide me through part 1, 2 and 3 ?

1. calculate the displacement, in cm, at time t = 0.60.

i know that displacement is max amplitude so i thought it was 2.0cm?

2. calculate the maximum speed of the oscillating cylinder.

im not sure how to do this

3. the cylinder is now pushed down further into the liquid before being. As before, the cylinder oscillates with simple harmonic motion. state the effect this has on both the amplitude and the period.

not sure on this either?

2) SHM equation for x(t): . Find max from this.

3) Pushed further. Consider equation I gave in (2): is higher clearly. Consider perfect vs. imperfect oscillator (the response of will depend on this, i.e. drag/no drag/etc).

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