JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Units & Measurement

The energy of a system as a function of time

t

is given as

E (t) = A^{2} exp (- α t)

, where

α = 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)

t = 5 s

Answer: 4

Solution

Given

E (t) = A^{2} e^{- α t}

with

α = 0.2 s^{- 1}

.

Taking logarithm:

lo g E = 2 lo g A - α t

Differentiating for error propagation:

\frac{d E}{E} = 2 \frac{d A}{A} + α d t

Substituting values:

\frac{d E}{E} = 2 (1.25%) + 0.2 \times (1.50% \times 5)

= 2.5% + 0.2 \times 7.5 = 2.5% + 1.5% = 4%

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer

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