Question

A physical quantity $ Z $ as a function of time is given as $ Z\left( t \right) = {A^{\dfrac{3}{2}}}{e^{ - kt}} $ , where $ k = 0.1{s^{ - 1}} $ . The measurement of $ A $ is $ 2.00\% $ . If the error in the measurement of time is $ 1.25\% $ , then the percentage error in the value of $ Z\left( t \right) $ at $ t = 10s $ would be
(1) $ 4.25\% $
(2) $ 2.50\% $
(3) $ 3.75\% $
(4) $ 3.50\% $

Answer 1

Hint : On performing any mathematical operation error always increases hence whether the quantity is added, subtracted, multiplied, divided the error in the quantities always gets added. Percentage error is the difference between the estimated value and the actual value in comparison to the actual value and is expressed as a percentage.

Complete step by step answer
From question,
 $\Rightarrow Z\left( t \right) = {A^{\dfrac{3}{2}}}{e^{ - kt}} $
Taking logarithm to the base $ e $ on both sides,
 $\Rightarrow \ln (Z\left( t \right){\text{)}} = \dfrac{3}{2}\ln A - kt $
Differentiating both sides,
 $\Rightarrow d(\ln (Z\left( t \right){\text{)) }} = d(\dfrac{3}{2}\ln A - kt) $
Percentage error is the difference between the calculated value and the actual value with respect to the actual value and is expressed in a percentage format. In calculus $ dx $ represents the infinitesimal difference between the calculated value and the actual value.
 $\Rightarrow \dfrac{{d(Z\left( t \right){\text{)}}}}{{Z\left( t \right)}} = \dfrac{3}{2} \times \dfrac{{dA}}{A} - kdt $
We know that performing any mathematical operation error always increases hence whether the quantity is added, subtracted, multiplied, divided the error in the quantities always gets added.
Hence,
 $\Rightarrow \dfrac{{d(Z\left( t \right){\text{)}}}}{{Z\left( t \right)}} \times 100 = \dfrac{3}{2} \times \dfrac{{dA}}{A} \times 100 + k\dfrac{{dt}}{t}t $
This is the percentage error in $ Z $ is the sum of percentage error in $ A $ multiplied by $ \dfrac{3}{2} $ and percentage error in $ t $ multiplied by $ kt $ .
 $\Rightarrow \dfrac{{d(Z\left( t \right){\text{)}}}}{{Z\left( t \right)}} \times 100 = \dfrac{3}{2} \times 2 \times 100 + 0.1 \times 1.25 \times 10 $
As in the question, it is given that the percentage error in $ A $ is $ 2.00\% $ , the error in the measurement of time is $ 1.25\% $ , and as we have to find percentage error in the value of $ Z\left( t \right) $ when $ 10s $ has elapsed so $ t = 10s $ .
 $\Rightarrow \dfrac{{d(Z\left( t \right){\text{)}}}}{{Z\left( t \right)}} \times 100 = 3 + 1.25 $
 $\Rightarrow \dfrac{{d(Z\left( t \right){\text{)}}}}{{Z\left( t \right)}} \times 100 = 4.25 $
Hence the percentage error in calculating $ Z $ is $ 4.25\% $ .
Therefore, the correct answer to the given question is (A) $ 4.25\% $ .

Note
For a mathematical equation, $ Z = \dfrac{{{a^x}{b^y}}}{{{c^z}}} $ the relative error in calculating $ Z $ will be $ x\dfrac{{da}}{a} + y\dfrac{{db}}{b} + z\dfrac{{dc}}{c} $ but in this question we also proved how this equation came for your conceptual understanding. In physics calculus is an indispensable part so a student must know basic calculus to solve physics numerically.