Question 60: Units , Dimensions and Error Analysis

Question

The energy of a system is given as $\mathrm{E}(\mathrm{t})=\alpha^3 \mathrm{e}^{-\beta t}$, where t is the time and $\beta=0.3 \mathrm{~s}^{-1}$. The errors in the measurement of $\alpha$ and $t$ are $1.2 \%$ and $1.6 \%$, respectively. At $t=5 \mathrm{~s}$, maximum percentage error in the energy is :

Accepted Answer

We start with the energy function:
$ E(t)= \alpha^3 e^{-\beta t}, $
where $\beta=0.3\, 	ext{s}^{-1}$.
Step 1. Take the Logarithm
Taking the logarithm of both sides:
$ \ln E = 3 \ln \alpha - \beta t. $
Step 2. Differentiate to Find the Relative Error
Differentiate both sides:
$ \frac{dE}{E} = 3\,\frac{d\alpha}{\alpha} - \beta\, dt. $
For maximum error estimation, we consider the absolute values and sum the contributions:
$ \left|\frac{\Delta E}{E}ight| \approx 3\,\left|\frac{\Delta \alpha}{\alpha}ight| + \beta\, |\Delta t|. $
Step 3. Plug in the Given Errors
The percentage error in $\alpha$ is $1.2\%$ (i.e., $\Delta \alpha/\alpha = 0.012$).
The percentage error in $t$ is $1.6\%$, so for $t=5\,	ext{s}$:
$   \Delta t = 0.016 	imes 5 = 0.08\, 	ext{s}.   $
Now substitute these values:
$ \left|\frac{\Delta E}{E}ight| \approx 3(0.012) + 0.3(0.08). $
Calculating each term:
$3(0.012) = 0.036$ (or $3.6\%$),
$0.3(0.08) = 0.024$ (or $2.4\%$).
Step 4. Compute the Total Maximum Percentage Error
Add the contributions:
$ \left|\frac{\Delta E}{E}ight| \approx 0.036 + 0.024 = 0.06, $
which is $6\%$.
Thus, the maximum percentage error in the energy at $t = 5\,	ext{s}$ is:
$ \boxed{6\%} $
Answer: Option A (6%).

Answer

$6 \%$

Answer

$11.6 \%$

Answer

$4 \%$

Answer

$8.4 \%$

Units , Dimensions and Error Analysis

The energy of a system is given as \mathrm{E}(\mathrm{t})=\alpha^3 \mathrm{e}^{-\beta t}, where t is the time and \beta=0.3 \mathrm{~s}^{-1}. The errors in the measurement of \alpha and t are 1.2 \% and 1.6 \%, respectively. At t=5 \mathrm{~s}, maximum percentage error in the energy is :

Choose the correct answer: