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

Options

A. 0.01
B. 0.02
C. 0.03
D. 0.04

Answer

(d)

Explanation

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^(-alpha t)`
or `log E = 2 log A - alpha t`
Differentiating both sides, of given relation we get
`(dE)/(E ) = +- (2dA)/(A) +- alpha dt ...... (i)`
`:. (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%`