+0  
 
0
197
4
avatar

So for part (i)I found the volume is equal to v= 6283.19 cm^3 , now for part (ii) does anyone know how to do it? Or any hints? Thanks 

 Apr 30, 2019
 #1
avatar+28353 
+3

You could calculate the volume of a cylinder of radius 10+0.03 cm and height 20+0.05 cm, then determine the difference of the result from your previously calculated volume.

 

You should also do the same for radius 10-0.03 and height 20-0.05 cm.

 

Take the larger of the two differences.

 Apr 30, 2019
 #2
avatar
0

thanks for your answer, this is how I have done it now can you please correct me if I'm wrong? thanks

first I calculated the new volume of the cylinder with radius 10.03 and height 20.05;

v= pi * (10.03)2 * ( 20.05) = 1263.56

The initial volume of the cylinder - the new volume =

6283.19 - 1263.56 = 5019.63

 

then I calculated the volume of the cylinder with radius 9.97 and height 19.95,

V= pi * (9.97)2 * (19.95) = 6229.93

The initial volume of the cylinder - the new volume =

6283.19 - 6229.93 = 53.26 

 

so the approximate maximum error is 5019.63

 

Is that right? Please let me know. 

Guest Apr 30, 2019
 #3
avatar+28353 
+1

You need to check your calculation of pi*10.032*20.05.  !!!

Alan  Apr 30, 2019
 #4
avatar+1821 
+2

Solution:

 

\(V = \pi * r^2 * h \pm \text {(error)}\\ \small \text {For this equation, the easiest method is to calculate each error as a decimal and }\\ \small \text {then use the sum of the decimals to calculate the error range.}\\ \large \text { }\\ \text {Gauss Error Function } \small \text { (uncertainties in decimal).}\\ r= 10 cm \pm 0.03\\ r_{error} = (2)\frac{0.03}{10} = 0.006\\ h= 20 cm \pm 0.05\\ h_{error} = \frac{0.05}{20} = 0.003\\ \small \text{sum of errors } = 0.006+0.003 = 0.009\\ V_{error} = 0.009 (6283.18) = \pm 56.54cm^3 \\ \large \text { }\\ V= \pi * 10^2 * 20 = 6283cm^3 \pm 56.54cm^3 \, | \,\small \text{68% confidence interval}\\ 6226.6cm^3 \leq V \leq 6339.7cm^3 \, | \,\small \text{68% confidence interval}\\ \)

For more information see:

https://web2.0calc.com/questions/physics_81#r2

 

 

 

GA

 Apr 30, 2019
edited by GingerAle  Apr 30, 2019
edited by GingerAle  Apr 30, 2019

10 Online Users

avatar