The diameter and height of a cylinder are measured by a meter scale to be $$12.6 \pm 0.1$$ cm and $$34.2 \pm 0.1$$ cm, respectively. What will be the value of its volume in appropriate significant figures?

We begin with the mathematical expression for the volume of a right circular cylinder. The volume is given by the well-known formula

$$V=\pi r^{2}h,$$

where $$r$$ is the radius and $$h$$ is the height.

The diameter has been measured as $$d = 12.6 \pm 0.1\ {\rm cm}.$$ Since the radius is one-half of the diameter, we have

$$r=\frac{d}{2}= \frac{12.6\ {\rm cm}}{2}=6.3\ {\rm cm},$$

and the corresponding absolute error is also halved:

$$\Delta r=\frac{\Delta d}{2}= \frac{0.1\ {\rm cm}}{2}=0.05\ {\rm cm}.$$

The height has been measured as $$h = 34.2 \pm 0.1\ {\rm cm},$$ so

$$\Delta h = 0.1\ {\rm cm}.$$

Now we calculate the numerical value of the volume using the central (mean) values:

First find $$r^{2}$$:

$$r^{2}= (6.3\ {\rm cm})^{2}=39.69\ {\rm cm^{2}}.$$

Next multiply by the height:

$$r^{2}h = 39.69\ {\rm cm^{2}}\times 34.2\ {\rm cm}=1357.398\ {\rm cm^{3}}.$$

Finally multiply by $$\pi$$ (using $$\pi = 3.1416$$):

$$V = \pi r^{2}h = 3.1416 \times 1357.398\ {\rm cm^{3}} = 4264.4016\ {\rm cm^{3}}.$$

Thus, ignoring errors for the moment, $$V \approx 4264.4\ {\rm cm^{3}}.$$

Next we evaluate the uncertainty in $$V$$. For a quantity of the form $$V=\pi r^{2}h,$$ the fractional (relative) error is obtained by simple addition of the fractional errors of the individual factors, each multiplied by the power with which they occur. Stating the rule explicitly:

$$\frac{\Delta V}{V}=2\left(\frac{\Delta r}{r}\right) + \left(\frac{\Delta h}{h}\right).$$

We already have

$$\frac{\Delta r}{r} = \frac{0.05}{6.3}=0.007936, \qquad \frac{\Delta h}{h} = \frac{0.1}{34.2}=0.002924.$$

Substituting these into the formula,

$$\frac{\Delta V}{V}=2(0.007936)+0.002924 =0.015872+0.002924 =0.018796.$$

The absolute error in $$V$$ is therefore

$$\Delta V = V \left(\frac{\Delta V}{V}\right) = 4264.4\ {\rm cm^{3}}\times 0.018796 \approx 80.16\ {\rm cm^{3}}.$$

In reporting the final result we keep the error to one (or at most two) significant figures; $$80.16$$ rounds to $$80.$$ Once the error is rounded to the nearest ten ($$80\ {\rm cm^{3}}$$), the central value must be rounded to the same place (the tens place). Hence

$$V = 4260 \pm 80\ {\rm cm^{3}}.$$

Hence, the correct answer is Option C.