Q54P - University Physics with Modern Physics

Question

A rectangular piece of aluminum is $\mathbf{7}\mathbf{.}\mathbf{60}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{cm}$ long and $\mathbf{1}\mathbf{.}\mathbf{90}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{cm}$wide (a) Find the area of the rectangle and the uncertainty in the area. (b) Verify that the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width.

Step-by-Step Solution

Answer

a)  The area and uncertainty in the area are $(14.44\pm 0.095) {\mathrm{cm}}^2$.

b)  It is verified that the values of uncertainty in the area and the sum of fractional uncertainties in width and length values are equal to each other.

Step 1: Identification of given data

The given data can be listed below,

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

Step 2: Concept/Significance of the area of an object

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.

Step 3: (a) Determination of the area of the rectangle and the uncertainty in the area.

The area of a rectangle with uncertainty is given by,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{total}}=\mathrm{A}\pm \mathrm{\delta A} \\ =\mathrm{I}\times \mathrm{W}\pm \left[\left(\mathrm{\delta I}\right)\mathrm{W}+\mathrm{I}\left(\mathrm{\delta W}\right)\right] \end{gathered}$$

Here, l is the length of the rectangle and w is the width of the rectangle,  $\delta I$ is the uncertainty in length, and $\delta W$ is the uncertainty in width.

Substitute all the values in the above equation, 

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

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

Step 4: (b) Verification of the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width.

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,

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

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,

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

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

  $\mathrm{W}=\frac{\mathrm{uncertain} \mathrm{value} \mathrm{of} \mathrm{width}}{\mathrm{absolute} \mathrm{value} \mathrm{of} \mathrm{width} }$

Substitute the values in the above,

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

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.