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JEE Advanced · Physics · 1. Math in Physics

The energy of a system as a function of time t is given as Et=A2-αt , 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) at t=5 s is

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(B) 4

Step-by-step Solution

Detailed explanation

Energy E=A2e-αt
For small % errors, we can, do differentiation
dE=2AdAe-αt+A2-αe-αtdt
Fractional error \(=\frac{ dE }{ E }=\frac{2 Ae ^{-\alpha t }( dA )+\left(-\alpha A ^2 e ^{-\alpha t }\right) dt }{ A ^2 e ^{-\alpha t }}=2\left(\frac{ dA }{ A }\right)+\) \(\left(-\alpha \frac{ dt }{ t }\right) t\)
% error =21.25%+0.2×1.5%×5
=4% (errors always add up)
Alternate solution:
E=A2e-αt
Taking natural logarithm on both sides,
lnE=lnA2+-αt
Differentiating
dEE=2dAA+-αdt
For small fractional erros, errors always add up
dEE=2dAA+αdtt×t
=21.25%+0.21.5%5
=4%
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