91Ó°ÊÓ

The given data can be listed below,

• The length of the rectangular piece is,$\mathrm{l}=7.60Â\pm 0.01\mathrm{cm}$
• The width of the rectangular piece is,$\mathrm{W}=1.90Â\pm 0.01\mathrm{cm}$

The area of an object is the total space occupied by that object anywhere in a 2D plane. It is a two-dimensional concept and therefore has two parameters to work with.

The area of a rectangle with uncertainty is given by,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{total}}=\mathrm{A}Â\pm \mathrm{δ´¡} \\ =\mathrm{I}×\mathrm{W}Â\pm \left[\left(\mathrm{δ\pm õ}\right)\mathrm{W}+\mathrm{I}\left(\mathrm{δ°Â}\right)\right] \end{gathered}$$

Here, l is the length of the rectangle and w is the width of the rectangle,role="math" localid="1655803396714" $\mathrm{δ}I$ is the uncertainty in length, and$\mathrm{δ}W$ is the uncertainty in width.

Substitute all the values in the above equation,

role="math" localid="1655803656925" $$\begin{gathered} \mathrm{A}=(7.60\mathrm{cm})(1.90\mathrm{cm})Â\pm \left[\left(0.01\mathrm{cm}\right)1.90\mathrm{cm}Â\pm (0.01\mathrm{cm})7.60\mathrm{cm}\right] \\ =(14.44Â\pm 0.095){\mathrm{cm}}^2 \end{gathered}$$

Thus, the area and uncertainty in the area are$(14.44Â\pm 0.095){\mathrm{cm}}^2$

For area, the percentage of fractional uncertainty (F) is,

$\mathrm{F}=\frac{\mathrm{uncertain}\mathrm{value}}{\mathrm{absolute}\mathrm{value}}$

Substitute values in the above expression,

$$\begin{gathered} \mathrm{F}=\frac{0.095{\mathrm{cm}}^2}{14.44{\mathrm{cm}}^2}×100\% \\ =0.66\% \end{gathered}$$

For length, using the same formula stated above, the percentage uncertainty (L) is given by,

$\mathrm{L}=\frac{\mathrm{uncertain}\mathrm{value}\mathrm{of}\mathrm{length}}{\mathrm{absolute}\mathrm{value}\mathrm{of}\mathrm{length}}$

Substitute the values in the above,

$$\begin{gathered} \mathrm{L}=\frac{0.01\mathrm{cm}}{7.60\mathrm{cm}}×100\% \\ =0.13\% \end{gathered}$$

For width, using the same formula the percentage uncertainty (W) is given by,

role="math" localid="1655806553656" $\mathrm{W}=\frac{\mathrm{uncertain}\mathrm{value}\mathrm{of}\mathrm{width}}{\mathrm{absolute}\mathrm{value}\mathrm{of}\mathrm{width}}$

Substitute the values in the above,

$$\begin{gathered} \mathrm{W}=\frac{0.01\mathrm{cm}}{1.9\mathrm{cm}}×100\% \\ =0.53\% \end{gathered}$$

The sum of the uncertainty values of length and width is given by,

$$\begin{gathered} \mathrm{S}=0.13\%+0.53\% \\ =0.66\% \end{gathered}$$

So, the sum of the uncertainties is 0.66%

Therefore, we can see that, $S=F$.

Thus, it is verified that the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width.