Animated Solution for Physics - Physics and Measurement: The diameter and height of a cylinder are measured by a meter scale to be $12.6\pm 0.1\text{ cm}$ and $34.2\pm 0.1\text{ cm}$, respectively. What will be the value of its volume in appropriate significant figures ?

Imagine you are in a laboratory, holding a solid cylinder. You take out a standard meter scale and carefully measure its dimensions. You find the diameter $D$ to be $12.6\text{ cm}$ and the height $h$ to be $34.2\text{ cm}$. Because you are using a meter scale, your measurements are limited by its least count, which is $0.1\text{ cm}$. Therefore, your readings are recorded as $D=12.6\pm 0.1\text{ cm}$ and $h=34.2\pm 0.1\text{ cm}$.

Our goal is to calculate the volume of this cylinder and, more importantly, to express it with the correct significant figures and uncertainty. This is a classic problem that tests not just your ability to plug numbers into a formula, but your understanding of how precision propagates through mathematical operations.

The volume of a cylinder is given by the well-known geometric formula $V=\pi r^{2}h$. Since our measurement is in terms of diameter $D$, we substitute $r=\frac{D}{2}$ to get:

Let's first calculate the central value of the volume by substituting the measured values without their uncertainties:

When you punch this into a calculator, you get a highly precise-looking number: $V=4262.22{\text{ cm}}^3$. But wait! Is our result really that precise?

This is where the rules of significant figures come into play. A chain is only as strong as its weakest link, and a calculated result is only as precise as its least precise input.

Look at our measured values: $12.6$ has exactly three significant figures, and $34.2$ also has exactly three significant figures. According to the rules of multiplication, our final result must be restricted to the same number of significant figures as the input with the fewest significant figures.

Therefore, we must round $4262.22$ to three significant figures. The first three digits are $4,2,$ and $6$. The next digit is $2$, which is less than $5$, so we round down. This gives us our properly rounded central volume:

Now, we need to figure out the uncertainty, or absolute error, in our volume. When physical quantities are multiplied or divided, their fractional (or relative) errors add up. Furthermore, if a quantity is raised to a power, its fractional error is multiplied by that power.

Applying this rule to our volume formula $V=\frac{\pi }{4}{D}^{2}h$, we get the error propagation equation:

Notice that the constant $\frac{\pi }{4}$ disappears because exact constants have zero uncertainty. Let's substitute our values into this equation:

Calculating the individual fractions gives:

To find the absolute error $\Delta V$, we multiply this total fractional error by the unrounded volume. It is crucial to use the unrounded volume ($4262.22$) here to prevent intermediate rounding errors from skewing the final uncertainty.

In experimental physics, absolute errors represent the bounds of our uncertainty. It doesn't make sense to state an uncertainty with high precision. By convention, we round the absolute error to one significant figure, or to match the decimal place of the least precise measurement.

Rounding $79.7$ gives us an absolute error of $80{\text{ cm}}^3$. Finally, we combine our rounded central volume with our rounded absolute error to state the final result:

This perfectly matches option (b). The beauty of this problem lies in the seamless integration of geometry, error analysis, and the strict discipline of significant figures.