Solve the problem.
-The elasticity $\varepsilon$ of a particular thermoplastic can be modeled approximately by the relation $\varepsilon = \frac { 2.5 \times 10 ^ { 5 } } { \mathrm {~T} ^ { 2.3 } }$, where $\mathrm { T }$ is the Kelvin temperature. If the thermometer used to measure $\mathrm { T }$ is accurate to $1 \%$, and if the measured temperature is $478 \mathrm {~K}$, how should the elasticity be reported?
A) $\varepsilon = \pm 0.004$
B) $\varepsilon = 0.172 \pm 0.002$
C) $\varepsilon = 0.172 \pm 0.004$
D) $\varepsilon = 0.172$

Question

Quizplus · Accepted Answer

To find the uncertainty in $\varepsilon$, we need to use the formula for propagation of uncertainties:
$$\frac{\Delta\varepsilon}{\varepsilon} = \sqrt{\left(\frac{\partial\varepsilon}{\partial T}\right)^2\left(\frac{\Delta T}{T}\right)^2}$$
Taking the derivative of $\varepsilon$ with respect to T, we get:
$$\frac{\partial\varepsilon}{\partial T} = -2.3\frac{2.5\times10^5}{T^{3.3}} = -\frac{5.75\times10^4}{T^{3.3}}$$
Using the given temperature of $478$ K and the accuracy of $1\%$, we can write: 
$$\Delta T = 478\mathrm{~K}\times 0.01 = 4.78\mathrm{~K}$$
Now, substituting the values in the formula for propagation of uncertainties, we get:
$$\frac{\Delta\varepsilon}{\varepsilon} = \sqrt{\left(\frac{-5.75\times10^4}{478^{3.3}}\right)^2\left(\frac{4.78}{478}\right)^2} \approx 0.0235$$
Therefore, the uncertainty in $\varepsilon$ is given by:
$$\Delta \varepsilon \approx 0.0235\times 0.172 \approx 0.004$$
Hence, the elasticity should be reported as $\varepsilon = 0.172 \pm 0.004$ (Option C).

Solve the Problem ε\varepsilonε Of a Particular Thermoplastic Can Be Modeled Approximately by the Can

Solve the Problem $\varepsilon$ Of a Particular Thermoplastic Can Be Modeled Approximately by the Can