# Calculating % uncertainties

A manufacturer needs to determine the volume of metal used to produce a metal pipe of the following dimensions:
Length, L 40 + 1 m
External diameter, D 12.0 + 0.2 cm
Internal diameter, d 10.0 + 0.2 cm
Estimate the greatest possible percentage error of the volume.

the correct ans is 22.5%.

but no matter how many times i've tried, my ans is always 34%. Help pls thank you! :)

Related Introductory Physics Homework Help News on Phys.org
Could you show us your working?

• Simon Bridge
Could you show us your working?
V=(pi)(R^2)(L) - (pi)(r^2)(L)

delta(R)=(1/2)(0.2)=0.1cm
delta(r)=(1/2)(0.2)=0.1cm

delta(V) = delta(pi*R^2*L) + delta(pi*r^2*L)

Let x=(pi)(R^2)(L)

delta(x)/x = 2(0.1/6) + 1/40= 7/120

delta(x)=7/120 * (pi*R^2*L)=7/120 * 14 400 pi = 840pi

Let y=pi*r^2*L

delta(y)/y = 2(0.1/5) + 1/40=13/200

delta(y)=13/200 * pi*r^2*L=13/200 * 10 000pi= 650pi

delta(V) = delta(x) + delta(y)
=1490pi
delta(V)/V *100% = 1490pi/4400pi *100 = 34%

fyi 4400pi is the value of V
thank you

Drakkith
Staff Emeritus
Hi harvey,

I've moved your thread to the homework forums, but in the future please post all homework questions in the appropriate homework forum and use the template provided.

haruspex
Homework Helper
Gold Member
You have counted the 1/40 twice over. There may be other errors.
Safest way is just to compute the maximum possible volume.

You have counted the 1/40 twice over. There may be other errors.
Safest way is just to compute the maximum possible volume.
i'm sorry, what do you mean by counting 1/40 twice? is my method not the correct approach anyway?

haruspex
Homework Helper
Gold Member
i'm sorry, what do you mean by counting 1/40 twice? is my method not the correct approach anyway?
You counted it here:
delta(x)/x = 2(0.1/6) + 1/40= 7/120
and here:
delta(y)/y = 2(0.1/5) + 1/40=13/200
Your problem starts with this line:
delta(V) = delta(pi*R^2*L) + delta(pi*r^2*L)
Let's back up one step:
##\Delta V = \Delta (\pi*R^2*L-\pi*r^2*L)##
You cannot split that as ##\Delta (\pi*R^2*L)+\Delta(\pi*r^2*L)## because the two terms are not independent.
If the actual length is L-ΔL on the inner radius it will be L-ΔL on the outer radius. Your analysis exaggerates the range by allowing a volume like
##\pi((R-\Delta R)^2(L-\Delta L)-(r+\Delta r)^2(L+\Delta L))##

You counted it here:

and here:

Your problem starts with this line:

Let's back up one step:
##\Delta V = \Delta (\pi*R^2*L-\pi*r^2*L)##
You cannot split that as ##\Delta (\pi*R^2*L)+\Delta(\pi*r^2*L)## because the two terms are not independent.
If the actual length is L-ΔL on the inner radius it will be L-ΔL on the outer radius. Your analysis exaggerates the range by allowing a volume like
##\pi((R-\Delta R)^2(L-\Delta L)-(r+\Delta r)^2(L+\Delta L))##
however, it is true that if R=A+B,
then delta(R) = delta(A) + delta(B).
i don't get what you mean by independent though. i'm sorry but i just want to fully understand this, thank you!