# Answer to Question #15810 in Mechanics | Relativity for Viola

Question #15810

I'm not asking you without trying to slove it myself, here's the quetion 1st:

If the three quantities; mass, diameter, and height for a cylinder are measured with the same percentage error, which would contribute the most to the error in the density?

::

so, I tried to answer it like this:

since the equation for the cylinder density is p=2*M/(pi*d^2*h)

since the mass and the height aren't squared the error would remain the same, however the diameter is squared which means the error is squared as well.. so it's the diameter!!!!

BUT my friend says since p(density)=M/V , then mass would contribute the most not diameter ;O

HELP PLEASE , I dont know which answer to put in our report D:?!

Expert's answer

Error for indirect measurement is calculated the next way:

Sqrt(((dp/dM)DM)^2+((dp/dh)DH)^2+((dp/dd)Dd)^2)

where DM, DH, Dd - errors of direct measurements, dp/dm,h,d - partial derivatives.

So,

DM*dp/dm=(1/V)DM.

Dh*dp/dh=-(2*M/pi*d^2)(1/h^2)DH.

Dd*dp/dd=-(2*M/pi*h)(2/d^3)Dd.

So, when you put every of last three lines to square, then sum and take square root, you will get error of density.

We don't know the values of direct errors (M, h, d), but by experience, we think that mass contributes the most.

Sqrt(((dp/dM)DM)^2+((dp/dh)DH)^2+((dp/dd)Dd)^2)

where DM, DH, Dd - errors of direct measurements, dp/dm,h,d - partial derivatives.

So,

DM*dp/dm=(1/V)DM.

Dh*dp/dh=-(2*M/pi*d^2)(1/h^2)DH.

Dd*dp/dd=-(2*M/pi*h)(2/d^3)Dd.

So, when you put every of last three lines to square, then sum and take square root, you will get error of density.

We don't know the values of direct errors (M, h, d), but by experience, we think that mass contributes the most.

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