
When measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $2\% $ in the length of the pendulum and a positive error of $1\% $ in the value of time period. His actual percentage error in the measurement of the value of $g$ will be
A) $3\% $
B) $0\% $
C) $4\% $
D) $5\% $
Answer
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Hint:In this problem we are given that the student makes positive errors in measurement of length and time. The error was made while measuring the acceleration due to gravity. We need to find the actual (combined) error in the measurement. The formula of time period has the term of acceleration due to gravity. Use that formula and using error analysis, find the required percentage error.
Complete step-by-step solution:
We are given that the error in the length of the pendulum is $2\% $ and the error in the measurement of time period is $1\% $ . We need to find the percentage error in the measurement of the value of acceleration due to gravity. Acceleration due to gravity and time period are related by the following formula.
$T = 2\pi \sqrt {\dfrac{L}{g}} $
Here, $T$ is the time period
$g$ is the acceleration due to gravity;
$L$ is the length of the pendulum,
Using this formula, we have
$g = \dfrac{{4{\pi ^2}\theta }}{{{T^2}}}$
The error in the value of acceleration due to gravity will be given as:
$\dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
The percentage error will be given as:
$\dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
We are given that $\dfrac{{\Delta L}}{L} = 2$ and $\dfrac{{\Delta T}}{T} = 1$ , substituting this value in the above equation, we get
$\dfrac{{\Delta g}}{g} = 0.02 + 2(0.01)$
$ \Rightarrow \dfrac{{\Delta g}}{g} = 0.04$
The percentage error will be:
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = 4\% $
actual percentage error in the measurement of the value of $g$ will be $4\% $ .
Thus, option C is the correct option.
Note:Remember the relation between time period and the acceleration due to gravity. In this problem, we are given the percentages, be careful to multiply the error in acceleration due to gravity by $100$ to get the percentage. We are given that the errors are positive, be careful to read the question for negative or positive errors.
Complete step-by-step solution:
We are given that the error in the length of the pendulum is $2\% $ and the error in the measurement of time period is $1\% $ . We need to find the percentage error in the measurement of the value of acceleration due to gravity. Acceleration due to gravity and time period are related by the following formula.
$T = 2\pi \sqrt {\dfrac{L}{g}} $
Here, $T$ is the time period
$g$ is the acceleration due to gravity;
$L$ is the length of the pendulum,
Using this formula, we have
$g = \dfrac{{4{\pi ^2}\theta }}{{{T^2}}}$
The error in the value of acceleration due to gravity will be given as:
$\dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
The percentage error will be given as:
$\dfrac{{\Delta g}}{g} = \dfrac{{\Delta L}}{L} + \dfrac{{2\Delta T}}{T}$
We are given that $\dfrac{{\Delta L}}{L} = 2$ and $\dfrac{{\Delta T}}{T} = 1$ , substituting this value in the above equation, we get
$\dfrac{{\Delta g}}{g} = 0.02 + 2(0.01)$
$ \Rightarrow \dfrac{{\Delta g}}{g} = 0.04$
The percentage error will be:
$ \Rightarrow \dfrac{{\Delta g}}{g} \times 100 = 4\% $
actual percentage error in the measurement of the value of $g$ will be $4\% $ .
Thus, option C is the correct option.
Note:Remember the relation between time period and the acceleration due to gravity. In this problem, we are given the percentages, be careful to multiply the error in acceleration due to gravity by $100$ to get the percentage. We are given that the errors are positive, be careful to read the question for negative or positive errors.
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