Changing Units — Question 2

Question 2

A hole in the plate, and a missing power in the conversion

A rectangular plate contains the rectangular through-hole shown. Its thickness is uniform. Use $1\ \mathrm{cm}=10^{-2}\ \mathrm{m}$ and $1\ \mathrm{mm}=10^{-3}\ \mathrm{m}$, both exact. Treat all dimensions as exact within this model.

1. Find the area of remaining material in $\mathrm{cm^2}$, then convert it to $\mathrm{m^2}$. Verify by converting the four plan dimensions to metres before computing areas.

2. Find the material volume in $\mathrm{m^3}$ and $\mathrm{mm^3}$. The hole removes material through the entire thickness.

3. A student multiplies the numerical area in $\mathrm{cm^2}$ by $10^{-2}$ and labels it $\mathrm{m^2}$. Explain the error using a square of side $1\ \mathrm{cm}$, and find the factor by which this would overestimate the volume if the thickness were converted correctly.

Question 2 - Solution

A hole in the plate, and a missing power in the conversion

(a) Subtract compatible areas, then square the length factor.

$$A=[(10.0)(6.0)-(4.0)(2.0)]\ \mathrm{cm^2}=52.0\ \mathrm{cm^2}.$$ $$A=52.0(10^{-2})^2\ \mathrm{m^2}=\boxed{0.00520\ \mathrm{m^2}}.$$ Converting lengths first gives $[(0.100)(0.060)-(0.040)(0.020)]\ \mathrm{m^2}=0.00520\ \mathrm{m^2}$, the same result.

(b) Include thickness as a third length.

The thickness is $0.0020\ \mathrm{m}$, so $$V=A d=(0.00520)(0.0020)\ \mathrm{m^3} =\boxed{1.04\times10^{-5}\ \mathrm{m^3}}.$$ Because $1\ \mathrm{m^3}=10^9\ \mathrm{mm^3}$, this is $\boxed{10\,400\ \mathrm{mm^3}}$. Directly, the net plan area is $5200\ \mathrm{mm^2}$, multiplied by $2.0\ \mathrm{mm}$.

(c) A conversion factor acts on both side lengths.

$$\frac{10^{-2}}{10^{-4}}=100,\qquad \frac{V_{\mathrm{wrong}}}{V_{\mathrm{correct}}}=100\quad\text{(same thickness)}.$$

A $1\ \mathrm{cm}$ square is $0.01\ \mathrm{m}$ by $0.01\ \mathrm{m}$, hence has area $10^{-4}\ \mathrm{m^2}$. Using $10^{-2}$ instead of $10^{-4}$ makes the area, and therefore the volume at fixed correct thickness, 100 times too large. The aperture subtraction does not change the required power.