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` isA. 0.01B. 0.02C. 0.03D. 0.04

The energy of a system as a function of time `t` is given as `E(t) = A

1.	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` isA. 0.01B. 0.02C. 0.03D. 0.04
Answer» Correct Answer - (d) Given `E (t) =A^2 exp (- alpha t), alpha = 0.2 s^(-1),` `(dA)/(A) = 1.25%, (dt)/(t) = 1.50%, t =5 sec., (dE)/(E ) = ?` Taking log of both sides, we get `log E = log A^2 e^(-alpha t)` `log E = log A^2 + log e^(-alphat)` or `log E =2 log A - alpha t` Differentiating both sides, we get `(dE)/(E ) = +- (2dA)/(A) +- alpha dt ........ (i)` `:gt (dt)/(t) =1.50%, t= 5 sec.` `:. (dt)/(5) = 1.50%, dt = 7.50%` From (i), `(dE)/(E ) = +- 2(1.25) + -0.2xx7.5` `= +- 2.5 +- 1.5 = +- 4%`