Question

1. the sides of a small rectangular box are measured to be 1.80 ± 0.01 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long. calculate its volume and uncertainty in cubic centimeters.
2. when non - metric units were used in the united kingdom, a unit of mass called the pound - mass (lbm) was employed, where 1 lbm = 0.4539 kg. (a) if there is an uncertainty of 0.0001 kg in the pound - mass unit, what is its percent uncertainty? (b) based on that percent uncertainty, what mass in pound - mass has an uncertainty of 1 kg when converted to kilograms?
3. the length and width of a rectangular room are measured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m. calculate the area of the room and its uncertainty in square meters.

Explanation:

25.

Step1: Calculate the volume

The volume $V$ of a rectangular box with sides $a = 1.80$ cm, $b=2.05$ cm and $c = 3.1$ cm is given by $V=a\times b\times c$.
$V=1.80\times2.05\times3.1=11.391$ $cm^{3}$

Step2: Calculate the relative uncertainties

The relative uncertainty in a product $y = x_1\times x_2\times\cdots\times x_n$ is $\frac{\Delta y}{y}=\sqrt{(\frac{\Delta x_1}{x_1})^2+(\frac{\Delta x_2}{x_2})^2+\cdots+(\frac{\Delta x_n}{x_n})^2}$
For $a = 1.80\pm0.01$ cm, $\frac{\Delta a}{a}=\frac{0.01}{1.80}$; for $b = 2.05\pm0.02$ cm, $\frac{\Delta b}{b}=\frac{0.02}{2.05}$; for $c=3.1\pm0.1$ cm, $\frac{\Delta c}{c}=\frac{0.1}{3.1}$
$\frac{\Delta a}{a}=\frac{0.01}{1.80}\approx0.00556$, $\frac{\Delta b}{b}=\frac{0.02}{2.05}\approx0.00976$, $\frac{\Delta c}{c}=\frac{0.1}{3.1}\approx0.0323$
$\frac{\Delta V}{V}=\sqrt{(0.00556)^2+(0.00976)^2+(0.0323)^2}$
$=\sqrt{0.0000309+0.0000953+0.0010433}=\sqrt{0.0011695}\approx0.0342$

Step3: Calculate the uncertainty in volume

$\Delta V = V\times\frac{\Delta V}{V}=11.391\times0.0342\approx0.39$ $cm^{3}$

Step1: Calculate the percent - uncertainty

The percent - uncertainty is given by $\text{Percent uncertainty}=\frac{\Delta m}{m}\times100\%$
Given $m = 0.4539$ kg and $\Delta m=0.0001$ kg
$\text{Percent uncertainty}=\frac{0.0001}{0.4539}\times100\%\approx0.022\%$

Step1: Let the mass in pound - mass be $m_{lbm}$

We know that the percent uncertainty is constant. Let the mass in kg be $m_{kg}$ and its uncertainty $\Delta m_{kg}=1$ kg.
Since $\frac{\Delta m_{kg}}{m_{kg}}=\frac{\Delta m_{lbm}}{m_{lbm}}$ and we know the percent uncertainty from part (a) is $\frac{\Delta m_{lbm}}{m_{lbm}}=\frac{0.0001}{0.4539}$
If $\Delta m_{kg} = 1$ kg, then $m_{kg}=\frac{m_{lbm}\times\Delta m_{kg}}{\Delta m_{lbm}}$
Since $\frac{\Delta m_{lbm}}{m_{lbm}}=\frac{0.0001}{0.4539}$, we can rewrite it as $m_{lbm}=\frac{m_{kg}\times\Delta m_{lbm}}{\Delta m_{kg}}$
We know that $1$ lbm $ = 0.4539$ kg. Let the mass in lbm be $x$. Its mass in kg is $0.4539x$.
The percent uncertainty is the same, so $\frac{1}{0.4539x}=\frac{0.0001}{0.4539}$
$x=\frac{1}{0.0001}=10000$ lbm

Answer:

The volume is $V = 11.4\pm0.4$ $cm^{3}$

26.

(a)