Question

You can determine the volume of fluid that can be contained by a cylindrical tube from the equation V=πr^2L, where r is the radius and L is the length. If r can be measured with a standard deviation of 4%, and L can be measured with a standard deviation of 6%, what is the percent error in the volume? [Report the answer as an integer (i.e., without the % sign and no decimals).]

You can determine the volume of fluid that can be contained by a cylindrical tube from the equation V = arL, where ris the radius and Lis the length. If rcan be measured with a standard deviation of 4%, and I can be measured with a standard deviation of 6%, what is the percent error in the volume? [Report the answer as an integer (i.e., without the % sign and no decimals).]

Answer 1

Volume of a cylinder is given by:

17 = 1

Deviation observed in 'r' = 4%

0 : 04- rܠ ܂

Deviation observed in L = 6%

* - 006 0.06

We have,

17 = 1

Differentiating both sides we get,

ΔV = 2πη LΔr + πη?ΔL

Divide both sides by 'V'

ΔV 2π,LΔ, πιΔL να π2Ιπχ2 )

Simplifying further we get,

$\frac{\Delta V}{V}=2\frac{\Delta r}{r}+\frac{\Delta L}{L}$

Plugging in given values we get,

AV -= 2 * 0.04 + 0.06 = 0.08 + 0.06 = 0.14

Therefore deviation found in volume is 0.14*100 = 14%

Percent error in the volume is 14.