QJEEM25Phca0e: JEE Main 2025 Physics - Class 11 JEEM25Phca0e Solution: The energy of a system is given as , where t is the time and .…

We start with the energy function:

$E(t)= \alpha^3 e^{-\beta t},$

where $\beta=0.3\, \text{s}^{-1}$ .

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:

$\left|\frac{\Delta E}{E}\right| \approx 3\,\left|\frac{\Delta \alpha}{\alpha}\right| + \beta\, |\Delta t|.$

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 $t=5\,\text{s}$ :

$\Delta t = 0.016 \times 5 = 0.08\, \text{s}.$

Now substitute these values:

$\left|\frac{\Delta E}{E}\right| \approx 3(0.012) + 0.3(0.08).$

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:

$\left|\frac{\Delta E}{E}\right| \approx 0.036 + 0.024 = 0.06,$

which is $6\%$ .

Thus, the maximum percentage error in the energy at $t = 5\,\text{s}$ is:

$\boxed{6\%}$

Answer: Option A (6%).