
While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $ 1\% $ in the length of the pendulum and the negative error of $ 3\% $ in the value of time period. His percentage error in the measurement of $ g $ by the relation $ g = 4{\pi ^2}\left( {\dfrac{L}{{{T^2}}}} \right) $ will be.
(A) $ 2\% $
(B) $ 4\% $
(C) $ 7\% $
(D) $ 10\% $
Answer
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Hint :Here, we have been given the percentage error in length and in time period, we have to find out the percentage error in the measurement of $ g $ and we have the relation given between acceleration due to gravity, length and time period. We have to use the formula of percentage error as $ \dfrac{{\Delta x}}{x} \times 100 = \% error $
Complete Step By Step Answer:
Let $ L $ be the length, $ T $ be the time period and their differences be $ \Delta L $ , $ \Delta T $ respectively.
According to the given data we have positive percentage error in length and negative percentage error in time period such that:
$ \dfrac{{\Delta L}}{L} \times 100 = 1\% $
$ \dfrac{{\Delta T}}{T} \times 100 = - 3\% $
Now, we have the given relation between acceleration due to gravity and time period is
So let us use the formula given below
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = \left[ {\dfrac{{\Delta L}}{L} \times 100} \right] - 2\left[ {\dfrac{{\Delta T}}{T} \times 100} \right] $
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = \left[ {\dfrac{{\Delta L}}{L} - 2\dfrac{{\Delta T}}{T}} \right] \times 100 $
Let us put all the given values in above equation
$ \Rightarrow \dfrac{{\Delta g}}{g} = 1\% - 2( - 3\% ) = 7\% $
Hence, the answer obtained here is $ 7\% $ of error in $ g $
The correct answer is option C.
Note :
Here, we have to observe the percentage error and the signs they possess, because the whole formula depends on the value of the percentage error given In the problem and the formula we are using. We have been given the relation in the time period and acceleration due to gravity and we make use of this relation to find out the answer. Be careful about putting the values and give their signs as positive or negative accordingly.
Complete Step By Step Answer:
Let $ L $ be the length, $ T $ be the time period and their differences be $ \Delta L $ , $ \Delta T $ respectively.
According to the given data we have positive percentage error in length and negative percentage error in time period such that:
$ \dfrac{{\Delta L}}{L} \times 100 = 1\% $
$ \dfrac{{\Delta T}}{T} \times 100 = - 3\% $
Now, we have the given relation between acceleration due to gravity and time period is
So let us use the formula given below
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = \left[ {\dfrac{{\Delta L}}{L} \times 100} \right] - 2\left[ {\dfrac{{\Delta T}}{T} \times 100} \right] $
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = \left[ {\dfrac{{\Delta L}}{L} - 2\dfrac{{\Delta T}}{T}} \right] \times 100 $
Let us put all the given values in above equation
$ \Rightarrow \dfrac{{\Delta g}}{g} = 1\% - 2( - 3\% ) = 7\% $
Hence, the answer obtained here is $ 7\% $ of error in $ g $
The correct answer is option C.
Note :
Here, we have to observe the percentage error and the signs they possess, because the whole formula depends on the value of the percentage error given In the problem and the formula we are using. We have been given the relation in the time period and acceleration due to gravity and we make use of this relation to find out the answer. Be careful about putting the values and give their signs as positive or negative accordingly.
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