Question

In the determination of acceleration due to gravity ( $ g $ ) using the formula $ T = 2\pi \sqrt {\dfrac{L}{g}} $ , the errors in measurement of L and T are 1% and 2% respectively. The maximum percentage error in the value of $ g $ is
(A) 5%
(B) 4%
(C) 3%
(D) $ 1.5\% $

Answer 1

Hint Percentage error can be given as the error divided by the true or used value (usually the mean value) multiplied by one hundred per cent. We need to derive an expression for error in $ g $ as a function of error in $ L $ , and $ T $ .
Formula used: In this solution we will be using the following formula;
 $ \Rightarrow T = 2\pi \sqrt {\dfrac{L}{g}} $ where $ T $ is the period, $ L $ is the length, $ g $ is the acceleration due to gravity.
 $ \Rightarrow P.{E_v} = \dfrac{{\Delta V}}{V} \times 100\% $ , where $ V $ is a variable, $ \Delta V $ is the error in the variable, and $ P.{E_v} $ is the percentage error of the variable

Complete step by step answer
From the question, we have that
 $ \Rightarrow T = 2\pi \sqrt {\dfrac{L}{g}} $ where $ T $ is the period, $ L $ is the length, $ g $ is the acceleration due to gravity. We are to look for the percentage error in the determination of $ g $ based on the error in the measurement of the period $ T $ and length $ L $ .
First, we must make $ g $ subject of the formula.
Hence, squaring both sides we have
 $ \Rightarrow {T^2} = 4{\pi ^2}\left( {\dfrac{L}{g}} \right) $ . Multiplying by both sides by $ g $ and dividing by $ {T^2} $ , we have
 $ \Rightarrow g = 4{\pi ^2}\left( {\dfrac{L}{{{T^2}}}} \right) $
From mathematical principles, it can be proven that
 $ \Rightarrow \dfrac{{\Delta g}}{g} = \left( {\dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta T}}{T}} \right) $
Multiplying all through by one hundred per cent we have
 $ \Rightarrow \dfrac{{\Delta g}}{g} \times 100\% = \left( {\dfrac{{\Delta L}}{L} \times 100\% + 2\dfrac{{\Delta T}}{T} \times 100\% } \right) $
From the expression of percentage error which is given by
 $ \Rightarrow P.{E_v} = \dfrac{{\Delta V}}{V} \times 100\% $ , where $ V $ is a variable, $ \Delta V $ is the error in the variable, and $ P.{E_v} $ is the percentage error of the variable. We have that
 $ \Rightarrow P.{E_g} = \left( {P.{E_L} + 2 \times P.{E_T}} \right) $ ,
According to the question $ P.{E_L} = 1\% $ and $ P.{E_T} = 2\% $ hence, replacing into formula above, we have
 $ \Rightarrow P.{E_g} = (1\% + 2 \times 2\% ) = 5\% $ .
 $ \therefore P.{E_g} = 5\% $
Hence the correct answer is option A.

Note
To avoid confusions, we shall prove that $ \dfrac{{\Delta g}}{g} = \left( {\dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta T}}{T}} \right) $ as follows:
Mathematically,
 $ \Rightarrow dg = \dfrac{{\partial g}}{{\partial l}}dl + \dfrac{{\partial g}}{{\partial T}}dT $
Hence, from $ g = 4{\pi ^2}\left( {\dfrac{L}{{{T^2}}}} \right) $ , differentiating $ g $ we have,
 $ \Rightarrow dg = \dfrac{{4{\pi ^2}}}{{{T^2}}}dl + \left( { - 2\dfrac{{4{\pi ^2}}}{{{T^3}}}dT} \right) $ , factorising out $ 4{\pi ^2} $ , we have
 $ \Rightarrow dg = 4{\pi ^2}\left[ {\dfrac{1}{{{T^2}}}dl + \left( { - 2\dfrac{L}{{{T^3}}}dT} \right)} \right] $
Now, observe from $ g = 4{\pi ^2}\left( {\dfrac{L}{{{T^2}}}} \right) $ that
 $ \Rightarrow \dfrac{1}{{{T^2}}} = \dfrac{g}{{4{\pi ^2}L}} $ and $ \dfrac{L}{{{T^2}}} = \dfrac{g}{{4{\pi ^2}}} $ . Hence, replacing these in $ g $ , we have
 $ dg = 4{\pi ^2}\left[ {\dfrac{g}{{4{\pi ^2}L}}dL + \left( { - 2\dfrac{g}{{4{\pi ^2}}}dT} \right)} \right] $ .
Dividing through by $ g $ and cancelling $ 4{\pi ^2} $
We have that
 $ \Rightarrow \dfrac{{dg}}{g} = 4{\pi ^2}\left[ {\dfrac{{dL}}{L} + \left( { - 2\dfrac{{dT}}{T}} \right)} \right] $ .
Hence,
 $ \Rightarrow \dfrac{{\Delta g}}{g} = \left[ {\dfrac{{\Delta L}}{L} + \left( { - 2\dfrac{{( - \Delta T)}}{T}} \right)} \right] $ ( $ dT = - \Delta T $ because, in this case, for increase in $ \Delta g $ there’s a corresponding decrease in $ \Delta T $ as obvious in the expression for $ g $ )
Hence, finally,
 $ \Rightarrow \dfrac{{\Delta g}}{g} = \left( {\dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta T}}{T}} \right) $ .