Percent uncertainty help (physics)

Question

i know you have to divide somehow. but only one number is really given.

some help would be nice.

What is the percent uncertainty for the measurement 5.2?

1% 2% 3% or 4%

What is the percent uncertainty in the measurement 2.58 +- .15 cm

2.9%, 5.8%, 8.7% or 12%

What is the percent uncertainty in the area of a circle whose radius is 1.8x10^4 c?

1.1%, 5.6%, 11%, 56%

What is the volume, and its approximate uncertainty, of a sphere of radius 1.96 +- 0.01m?

31.5 +- .2 m^

31.5 +- .3 m^

31.5 +- .4 m^

31.5 +- .5 m^

please show a small explanation for future questions.

the most answered correctly wins 10 points. thanks.

Answer 1

> percent uncertainty for the measurement 5.2?

I disagree with Joel on this one, based on my experience in physics labs.  When the amount of uncertainty is not given, you should assume that the last digit in the measurement can vary by +/- 1.  This means "5.2" has an uncertainty of "0.1", because the last digit is in the tenths position.  If the measurement were presented as "5.20", the uncertainty (unless otherwise stated) would be "0.01".  That's why the number of trailing zeros in your measurement is important.

That means the measurement and its uncertainty are: 5.2 +/- 0.1.  To determine the "percent" (or "relative") uncertainty, divide the absolute uncertainty (0.1) by the measurement (5.2).  This gives 0.01923..., which is 1.923%, or about 2%.

> percent uncertainty in the measurement 2.58 +- .15 cm

Divide the absolute uncertainty (.15) by the measurement (2.58).

> percent uncertainty in the area of a circle whose radius is 1.8x10^4 cm?

Here the absolute uncertainty is not given, so assume the last reported digit can vary by 1.  That means the absolute uncertainty is 0.1 × 10^4 cm.

Now, when you perform math operations on the measurements, you have to adjust the absolute uncertainty of the final result.  In particular:

1. If you multiply the measurement by some number "n" (which has zero uncertainty), then you must also multiply the uncertainty by "n".

2. If you square the measurement, then you must multiply the uncertainty by 2M (where "M" is the amount of the measurement)

3. If you cube the measurement, then you must multiply the uncertainty by 3M²

So, if the measured radius is "R" (1.8x10^4 cm); and the uncertainty is "Δ" (0.1 × 10^4 cm), then the area is:

Area = πR² +/- π(2R)Δ

(Notice I multiplied the Δ by π (according to Rule 1 above) and then by (2R) (according to Rule 2) above.)

Relative uncertainty = absolute uncertainty / measured value, so:

Relative uncertainty in area = π(2R)Δ / (πR²)

= 2Δ/R

= 0.2 × 10^4 cm / (1.8 times; 10^4 cm)

(which is about 11%)

> volume, and its approximate uncertainty, of a sphere of radius 1.96 +- 0.01m?

volume = (4/3)πR^3 ≈ 31.5 m^3

uncertainty (by Rules 1 and 3) = (4/3)π(3)(1.96m)²(0.01m) ≈ 0.48 m

Answer 2

Assuming the degree of uncertainty is indicated by the number of significant digits,  5.2 has uncertainty +/- .05.  .05/5.2 = .0096 = 1 pct

.15/2.58 = .0581 = 5.8 pct

For the circle radius 1.8*10^4 uncertainty is .05*10^4 = 5*10^2 = .028

Area = pi*r^2 = p*(1.8^10^4)^2.

With maximum uncertainty it is pi*(1.8+.05)^2*10^4^2

= pi*(1.8^2 + 2*.05*1.8 + .05^2)*10^8

Ignoring the .05^2 term, uncertainty is 2*.05*1.8 / 1.8^2 = 2*.05/1.8 = .056 = 5.6pct

For the sphere, one can follow similar logic.  First, the volume is

V = 4/3*pi*r^3 = 31.54 m^3, or 31.5 to 3 significant figures.

Analogous to the circle problem, the uncertainty of the volume will be 3 times the uncertainty of the linear dimension because you are cubing that dimension.

.01/1.96 * 3 = .0153 or about 1.5 pct

.0153 * 31.5 = .48, or about .5 m^3 uncertainty; last choice in the problem.