- #1
inno87
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I'm not sure if I am doing this right, can anyone help me out?
A cylindrical cookie has a diameter of 5.0 \pm 0.1 cm, and a thickness of 1.00 \pm 0.01 cm.
A. Assuming the uncertainities are normally distrubited, what is the most likely value of the volume of the cookie?
V=pi*(d/2)^2*h=pi*(1 \pm .01 cm * [(5.0 \pm .1 cm)/2]^2=pi*[1 \pm .01 cm * (2.5 \pm .05 cm)^2=pi*[1 \pm .01 cm * 6.25 \pm .1 cm^2]=pi* 6.25 \pm .11 cm^3 = 19.634 \pm .11 cm^3
Most likely volume is 19.634 cm^3
B. What is the percent uncertainty in the volume?
.11/19.634=.5603%
C. What is the absolute uncertainty in the volume?
.11 cm^3 (taken from question 1)
D. Assuming the thickness uncertainity remains \pm .01 cm, to what value would the diameter's uncertainty (in cm) have to be reduced in order to make the uncertainty in the volume \pm 3%?
I tried setting up something like (.01+D_unc)/(19.634)=.03 but that would mean you'd have to increase the uncertainty so it seems I may have done something wrong here!
A cylindrical cookie has a diameter of 5.0 \pm 0.1 cm, and a thickness of 1.00 \pm 0.01 cm.
A. Assuming the uncertainities are normally distrubited, what is the most likely value of the volume of the cookie?
V=pi*(d/2)^2*h=pi*(1 \pm .01 cm * [(5.0 \pm .1 cm)/2]^2=pi*[1 \pm .01 cm * (2.5 \pm .05 cm)^2=pi*[1 \pm .01 cm * 6.25 \pm .1 cm^2]=pi* 6.25 \pm .11 cm^3 = 19.634 \pm .11 cm^3
Most likely volume is 19.634 cm^3
B. What is the percent uncertainty in the volume?
.11/19.634=.5603%
C. What is the absolute uncertainty in the volume?
.11 cm^3 (taken from question 1)
D. Assuming the thickness uncertainity remains \pm .01 cm, to what value would the diameter's uncertainty (in cm) have to be reduced in order to make the uncertainty in the volume \pm 3%?
I tried setting up something like (.01+D_unc)/(19.634)=.03 but that would mean you'd have to increase the uncertainty so it seems I may have done something wrong here!
