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

Given $$E(t) = \alpha^3 e^{-\beta t}$$ with $$\beta = 0.3$$ s⁻¹ and relative errors $$\Delta\alpha/\alpha = 1.2\%$$, $$\Delta t/t = 1.6\%$$, we want to find the maximum percentage error in $$E$$ at $$t = 5\,$$s}.

We begin by taking the natural logarithm of $$E$$ to prepare for error propagation:

$$\ln E = 3\ln\alpha - \beta t$$

By differentiating this relation, the relative error in $$E$$ becomes

$$\frac{\Delta E}{E} = 3\frac{\Delta\alpha}{\alpha} + \beta \Delta t$$

Since $$\Delta t/t = 1.6\%$$ at $$t = 5\,$$s}, we calculate

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

Substituting $$\Delta\alpha/\alpha = 1.2\%$$ and the computed $$\Delta t$$ into the expression for $$\Delta E/E$$ and converting to a percentage gives

$$\frac{\Delta E}{E} \times 100 = 3 \times 1.2\% + \beta \times \Delta t \times 100$$

$$= 3.6\% + 0.3 \times 0.08 \times 100$$

$$= 3.6\% + 2.4\%$$

$$= 6\%$$

Therefore, the maximum percentage error in $$E$$ at $$t=5\,$$s} is 6\%, which corresponds to Option 1.