The energy of a system as a function of time `t` is given as `E(t) = A^(2)exp(-alphat)`, `alpha = 0.2 s^(-1)`. The measurement of `A` has an error of `1.25%`. If the error In the measurement of time is `1.50%`, the percentage error in the value of `E(t)` at t = 5 s` is

Answer

D

Explanation

(4) `E = A^(2)e^(-0.2t)`
:. `log_(e) E = 2log_(e)A - 0.2t`
On differentiating we get
`(dE)/(E) = 2(dA)/(A) - 0.2(dt)/t) xx t`
As errors always add up therefore
:.`(dE)/(E) xx 100 = 2((dA)/(A) xx 100)) + 0.2t((dt)/(t) xx 100))`
`(dE)/(E) xx 100 = 2 xx 1.25% + 0.2 xx 5 xx 1.5%`
:. `(dE)/(E) xx 100 = 4%`