Answer
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Hint: We need to find the relation between the percentage errors involved in each physical quantity in the given physical quantity with the percentage error involved in the physical quantity. We can find the solution for this problem using this relation.
Complete answer:
The dimensional formula of a physical phenomenon gives the idea on the physical quantities involved or related to them. They also give the operations involved in the relation, thus giving us the idea to calculate the errors involved with the calculation of the physical phenomenon.
Let us recap the error analysis method for different mathematical operations.
For, \[z=x-y\], the error involved is
\[\pm \partial z=\pm \partial x\pm \partial y\]
For,\[d=\dfrac{ab}{c}\], the error involved is
\[\dfrac{\partial d}{d}=\dfrac{\partial a}{a}+\dfrac{\partial b}{b}+\dfrac{\partial c}{c}\]
For, \[d=\dfrac{{{a}^{p}}{{b}^{q}}}{{{c}^{r}}}\], the error involved is
\[\dfrac{\partial d}{d}=\dfrac{p\partial a}{a}+\dfrac{q\partial b}{b}+\dfrac{r\partial c}{c}\text{ --(1)}\]
Now, let us consider the given physical quantity. It is given that the acceleration due to gravity can be related to a number of other physical quantities as –
\[g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}}\]
We are also given the percentage errors involved in the measurement of length ‘L’ and the measurement of time ‘T’. We can use the above relations to find the percentage error of the acceleration due to gravity as –
\[\begin{align}
& g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}} \\
& \Rightarrow \dfrac{\partial g}{g}\times 100=\dfrac{\partial L}{L}\times 100+\dfrac{2\partial T}{T}\times 100 \\
& \pm \dfrac{\partial g}{g}\times 100=2\%+(2)(3)\% \\
& \therefore \dfrac{\partial g}{g}\times 100=8\% \\
\end{align}\]
The percentage error in acceleration due to gravity is \[8\%\].
The correct answer is option A.
Note:
The error analysis is not just a method to find the errors involved in the calculation or physical setup of the experiment, it also gives ideas in the different aspects where mathematical approach is limited to accuracy in the practical approach.
Complete answer:
The dimensional formula of a physical phenomenon gives the idea on the physical quantities involved or related to them. They also give the operations involved in the relation, thus giving us the idea to calculate the errors involved with the calculation of the physical phenomenon.
Let us recap the error analysis method for different mathematical operations.
For, \[z=x-y\], the error involved is
\[\pm \partial z=\pm \partial x\pm \partial y\]
For,\[d=\dfrac{ab}{c}\], the error involved is
\[\dfrac{\partial d}{d}=\dfrac{\partial a}{a}+\dfrac{\partial b}{b}+\dfrac{\partial c}{c}\]
For, \[d=\dfrac{{{a}^{p}}{{b}^{q}}}{{{c}^{r}}}\], the error involved is
\[\dfrac{\partial d}{d}=\dfrac{p\partial a}{a}+\dfrac{q\partial b}{b}+\dfrac{r\partial c}{c}\text{ --(1)}\]
Now, let us consider the given physical quantity. It is given that the acceleration due to gravity can be related to a number of other physical quantities as –
\[g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}}\]
We are also given the percentage errors involved in the measurement of length ‘L’ and the measurement of time ‘T’. We can use the above relations to find the percentage error of the acceleration due to gravity as –
\[\begin{align}
& g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}} \\
& \Rightarrow \dfrac{\partial g}{g}\times 100=\dfrac{\partial L}{L}\times 100+\dfrac{2\partial T}{T}\times 100 \\
& \pm \dfrac{\partial g}{g}\times 100=2\%+(2)(3)\% \\
& \therefore \dfrac{\partial g}{g}\times 100=8\% \\
\end{align}\]
The percentage error in acceleration due to gravity is \[8\%\].
The correct answer is option A.
Note:
The error analysis is not just a method to find the errors involved in the calculation or physical setup of the experiment, it also gives ideas in the different aspects where mathematical approach is limited to accuracy in the practical approach.
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