Question

use the rules for working with measurements to give your answer to the appropriate accuracy and/or precision. find the volume (in mm³) of the cylinder. 16.0 mm 30.0 mm × mm³

Explanation:

Step1: Recall the formula for the volume of a cylinder

The formula for the volume \( V \) of a cylinder is \( V=\pi r^{2}h \), where \( r \) is the radius and \( h \) is the height.

Step2: Identify the values of \( r \) and \( h \)

From the diagram, the radius \( r = 16.0\space mm \) and the height \( h=30.0\space mm \).

Step3: Substitute the values into the formula

Substitute \( r = 16.0 \) and \( h = 30.0 \) into the formula:
\( V=\pi\times(16.0)^{2}\times30.0 \)
First, calculate \( (16.0)^{2}=16.0\times16.0 = 256.0 \)
Then, multiply by \( 30.0 \): \( 256.0\times30.0=7680.0 \)
Then, multiply by \( \pi \): \( V=\pi\times7680.0\approx3.14159\times7680.0 \)
\( 3.14159\times7680.0 = 24127.4304 \)
Now, consider the significant figures. The given values \( 16.0 \) and \( 30.0 \) have three significant figures. So, the result should be rounded to three significant figures.
\( 24127.4304\approx24100 \) (or in scientific notation \( 2.41\times 10^{4} \)) but more accurately, when we calculate \( \pi\times16^{2}\times30 \):
\( 16^{2}=256 \), \( 256\times30 = 7680 \), \( 7680\times\pi\approx7680\times3.1416 = 24127.68 \). Rounding to three significant figures, we get \( 24100 \space mm^{3} \) or \( 2.41\times 10^{4}\space mm^{3} \), but a more precise calculation with the given precision (since \( 16.0 \) and \( 30.0 \) have one decimal place in the context of precision, but significant figures are three) gives \( V=\pi r^{2}h=\pi\times(16.0)^{2}\times30.0 = \pi\times256.0\times30.0=\pi\times7680.0\approx24127.43\space mm^{3} \). But considering significant figures (three from \( 16.0 \) and \( 30.0 \)), the answer is \( 24100\space mm^{3} \) (or \( 2.41\times 10^{4}\space mm^{3} \)) or more accurately, if we use the exact multiplication:

Wait, maybe I made a mistake in significant figures. \( 16.0 \) has three significant figures, \( 30.0 \) has three significant figures. So the product should have three significant figures.

Calculating \( \pi\times16.0^{2}\times30.0 \):

\( 16.0^{2}=256.0 \) (four significant figures, but since \( 16.0 \) has three, it's \( 256 \) (three? Wait, \( 16.0 \) is three significant figures, so \( 16.0^{2}=256. \) (three significant figures? Wait, no: \( 16.0 \) has three significant figures, so when we square it, \( 16.0\times16.0 = 256.0 \) (four significant figures? Wait, no, the number of significant figures is determined by the least number? No, when multiplying or dividing, the result has the same number of significant figures as the least precise measurement. Wait, \( 16.0 \) has three, \( 30.0 \) has three. So \( r^{2} \) is \( 16.0\times16.0 = 256.0 \) (four, but since the original has three, we can consider it as three? Wait, no, the rule is that when you have a measurement with \( n \) significant figures, any operation on it preserves the number of significant figures in the context of multiplication/division. So \( 16.0 \) (three sig figs) squared is \( 256. \) (three sig figs? Wait, \( 16.0 \times 16.0 = 256.0 \), which is four, but the uncertainty in \( 16.0 \) is \( \pm0.1 \), so \( 16.0^{2} \) is \( (16.0\pm0.1)^{2}=256.0\pm3.2 + 0.01\approx256.0\pm3.2 \), so the result has three significant figures (since the first digit of the uncertainty is in the tenths place? No, maybe I'm overcomplicating. Let's just calculate the value:

\( V=\pi r^{2}h=\pi\times(16.0)^{2}\times30.0=\pi\times256.0\times30.0 = 7680.0\pi\approx7680.0\times3.1415926535 = 24127.4304\space mm^{3} \)

Now, \( 16.0 \) and \( 30.0 \) have three significant figures, so we round \( 24127.4304 \) to three significant figures, which is \( 24100\space mm^{3} \) (or \( 2.41\times 10^{4}\space mm^{3} \)). But maybe the problem expects the use of \( \pi\approx3.14 \) for simplicity. Let's try that:

\( 3.14\times256\times30 \)

\( 256\times30 = 7680 \)

\( 3.14\times7680 = 24115.2 \), which rounds to \( 24100\space mm^{3} \) (three significant figures) or \( 2.41\times 10^{4}\space mm^{3} \).

Alternatively, if we consider the precision of the measurements: \( 16.0 \) mm is precise to the tenths place, \( 30.0 \) mm is precise to the tenths place. The volume formula is \( V = \pi r^{2}h \). The radius is \( 16.0 \), height \( 30.0 \). So \( r^{2}=256.0 \) (precise to the tenths place? No, \( 16.0\times16.0 = 256.0 \), which is precise to the tenths place? Wait, \( 16.0 \) has a precision of \( 0.1 \) mm, so \( r^{2} \) has a precision of \( 2\times16.0\times0.1=3.2 \), so \( r^{2}=256.0\pm3.2 \). Then \( h = 30.0\pm0.1 \), so \( V=\pi r^{2}h \) has a precision that we can calculate, but maybe the problem just wants the numerical answer with the appropriate precision, so using \( \pi\approx3.1416 \), we get \( V\approx24127\space mm^{3} \), which can be written as \( 2.41\times 10^{4}\space mm^{3} \) or \( 24100\space mm^{3} \). But the most accurate calculation with the given numbers is:

\( V=\pi\times(16.0)^{2}\times30.0 = \pi\times256.0\times30.0 = 7680.0\pi\approx24127.43\space mm^{3} \)

So, depending on the required precision, but since the given values have three significant figures, the answer should be \( 24100\space mm^{3} \) (or \( 2.41\times 10^{4}\space mm^{3} \)) or more precisely, \( 24100\space mm^{3} \) (rounded to three significant figures).

Answer:

\( 24100\space mm^{3} \) (or \( 2.41\times 10^{4}\space mm^{3} \) or \( 24127\space mm^{3} \) depending on precision, but the most appropriate with three significant figures is \( 24100\space mm^{3} \))