Answer
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Hint : We need to take logarithm on both sides for the physical quantity Z as a function of time. Then we need to assign the values of percent error in the equation, we get the percentage error in the value of $ Z\left( t \right) $ at $ t = 10s $ .
Complete step by step answer
Given that, $ Z\left( t \right) = {A^{\dfrac{3}{2}}}{e^{ - kt}} $ where $ k = 0.1{s^{ - 1}} $ and measurement of A has an error of 2.00%.
Taking logarithm on both sides for the equation, $ Z\left( t \right) = {A^{\dfrac{3}{2}}}{e^{ - kt}} $ ,
$ \log Z = \dfrac{3}{2}\log A - kt $
Percent error formula is the absolute value of the difference of the measured value and the actual value divided by the actual value and multiplied by 100. Percent error calculation helps to know how close a measured value is to a true value.
$ {\text{Percentage Error = }}\left[ {{{\left( {ApproximateValue - Exact{\text{ }}Value} \right)} \mathord{\left/
{\vphantom {{\left( {ApproximateValue - Exact{\text{ }}Value} \right)} {Exact{\text{ }}Value}}} \right.} {Exact{\text{ }}Value}}} \right] \times 100 $
Given that, the measurement of A has an error of 2.00%. The error in the measurement of time is 1.25%, and $ k = 0.1{s^{ - 1}} $ .
For calculating the percentage error, we assign the error values, in the equation, . $ \log Z = \dfrac{3}{2}\log A - kt $ .
We get, the percentage error in Z,
$ \log Z = \dfrac{3}{2} \times 2\% - \left( {0.1} \right) \times 10 \times 1.25\% $
$ \Rightarrow \log Z = 3\% + 1.25\% = 4.25\% $
Hence, the percentage error in the value of $ Z\left( t \right) $ at $ t = 10s $ would be $ 4.25\% $ .
Hence, the correct answer is Option A.
Note
In mathematics, the logarithm is the inverse function to exponentiation. A logarithm is defined as the power to which number must be raised to get some other values. It is the most convenient way to express large numbers. A logarithm has various important properties that prove multiplication and division of logarithms can also be written in the form of logarithm of addition and subtraction.
In a general way, the logarithmic function can be written or denoted as,
$ {\log _b}x = n $
$ \Rightarrow {b^n} = x $
Some basic rules of logarithm,
$ {\log _b}\left( {mn} \right) = {\log _b}m + {\log _b}n $
$ {\log _b}\left( {{m \mathord{\left/
{\vphantom {m n}} \right.} n}} \right) = {\log _b}m - {\log _b}n $
$ {\log _b}{m^n} = n{\log _b}m $
$ {\log _b}m = \dfrac{{{{\log }_a}m}}{{{{\log }_a}b}} $ .
Complete step by step answer
Given that, $ Z\left( t \right) = {A^{\dfrac{3}{2}}}{e^{ - kt}} $ where $ k = 0.1{s^{ - 1}} $ and measurement of A has an error of 2.00%.
Taking logarithm on both sides for the equation, $ Z\left( t \right) = {A^{\dfrac{3}{2}}}{e^{ - kt}} $ ,
$ \log Z = \dfrac{3}{2}\log A - kt $
Percent error formula is the absolute value of the difference of the measured value and the actual value divided by the actual value and multiplied by 100. Percent error calculation helps to know how close a measured value is to a true value.
$ {\text{Percentage Error = }}\left[ {{{\left( {ApproximateValue - Exact{\text{ }}Value} \right)} \mathord{\left/
{\vphantom {{\left( {ApproximateValue - Exact{\text{ }}Value} \right)} {Exact{\text{ }}Value}}} \right.} {Exact{\text{ }}Value}}} \right] \times 100 $
Given that, the measurement of A has an error of 2.00%. The error in the measurement of time is 1.25%, and $ k = 0.1{s^{ - 1}} $ .
For calculating the percentage error, we assign the error values, in the equation, . $ \log Z = \dfrac{3}{2}\log A - kt $ .
We get, the percentage error in Z,
$ \log Z = \dfrac{3}{2} \times 2\% - \left( {0.1} \right) \times 10 \times 1.25\% $
$ \Rightarrow \log Z = 3\% + 1.25\% = 4.25\% $
Hence, the percentage error in the value of $ Z\left( t \right) $ at $ t = 10s $ would be $ 4.25\% $ .
Hence, the correct answer is Option A.
Note
In mathematics, the logarithm is the inverse function to exponentiation. A logarithm is defined as the power to which number must be raised to get some other values. It is the most convenient way to express large numbers. A logarithm has various important properties that prove multiplication and division of logarithms can also be written in the form of logarithm of addition and subtraction.
In a general way, the logarithmic function can be written or denoted as,
$ {\log _b}x = n $
$ \Rightarrow {b^n} = x $
Some basic rules of logarithm,
$ {\log _b}\left( {mn} \right) = {\log _b}m + {\log _b}n $
$ {\log _b}\left( {{m \mathord{\left/
{\vphantom {m n}} \right.} n}} \right) = {\log _b}m - {\log _b}n $
$ {\log _b}{m^n} = n{\log _b}m $
$ {\log _b}m = \dfrac{{{{\log }_a}m}}{{{{\log }_a}b}} $ .
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