
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. the length of pendulum is 55.0 cm. the percentage error in the determination of g is close to :-
$\begin{align}
&\text{A}\text{. 0}\text{.7 }\!\!\%\!\!\text{ } \\
&\text{B}\text{. 0}\text{.2 }\!\!\%\!\!\text{ } \\
&\text{C}\text{. 3}\text{.5 }\!\!\%\!\!\text{ } \\
&\text{D}\text{. 6}\text{.8 }\!\!\%\!\!\text{ } \\
\end{align}$
Answer
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Hint: Obtain the mathematical expression for the time period of simple pendulum. From this equation express acceleration due to gravity g in terms the length of the pendulum and the time period of the pendulum. Find the least count of the measurements and apply the error formula to find the percentage error of the value of g.
Complete answer:
Given, for the simple pendulum, the length of the pendulum is $l=55.0cm$
Time taken for 20 oscillations is 30 sec.
So, the time period of the pendulum is given as,
$T=\dfrac{30}{20}s$
Time period of pendulum is given as,
$T=2\pi \sqrt{\dfrac{l}{g}}$
Where, l is the length of the pendulum and g is the acceleration due to gravity.
We can write from the above equation that,
$g=\dfrac{4{{\pi }^{2}}l}{{{T}^{2}}}$
Now we can find the percentage error of g from the above equation.
$\dfrac{\Delta g}{g}\times 100=\left( \dfrac{\Delta L}{L}+\dfrac{2\Delta T}{T} \right)\times 100$
Now the length of the pendulum is given as 55.0 cm.
So, the value of $\Delta L$ will be 0.1cm
Again, the value of $\Delta T$ can be written as,
$\Delta T=\dfrac{1}{20}s$
Putting the values on the above equation,
$\dfrac{\Delta g}{g}\times 100=\left( \dfrac{\Delta L}{L}+\dfrac{2\Delta T}{T} \right)\times 100$
$~\dfrac{\Delta g}{g}\times 100=\left( \dfrac{0.1}{55}+\dfrac{2\times \dfrac{1}{20}}{\dfrac{30}{20}} \right)\times 100$
$\dfrac{\Delta g}{g}\times 100=6.8$
So, the percentage error in the value of g is $6.8\%$
So, the correct answer is “Option D”.
Note:
Least count of an instrument is the smallest value that can be measured by the instrument. When we measure the error in an instrument, we use the least count.
Absolute error is the difference between the actual and measured value. Percentage error is the relative error shown as percentage.
Complete answer:
Given, for the simple pendulum, the length of the pendulum is $l=55.0cm$
Time taken for 20 oscillations is 30 sec.
So, the time period of the pendulum is given as,
$T=\dfrac{30}{20}s$
Time period of pendulum is given as,
$T=2\pi \sqrt{\dfrac{l}{g}}$
Where, l is the length of the pendulum and g is the acceleration due to gravity.
We can write from the above equation that,
$g=\dfrac{4{{\pi }^{2}}l}{{{T}^{2}}}$
Now we can find the percentage error of g from the above equation.
$\dfrac{\Delta g}{g}\times 100=\left( \dfrac{\Delta L}{L}+\dfrac{2\Delta T}{T} \right)\times 100$
Now the length of the pendulum is given as 55.0 cm.
So, the value of $\Delta L$ will be 0.1cm
Again, the value of $\Delta T$ can be written as,
$\Delta T=\dfrac{1}{20}s$
Putting the values on the above equation,
$\dfrac{\Delta g}{g}\times 100=\left( \dfrac{\Delta L}{L}+\dfrac{2\Delta T}{T} \right)\times 100$
$~\dfrac{\Delta g}{g}\times 100=\left( \dfrac{0.1}{55}+\dfrac{2\times \dfrac{1}{20}}{\dfrac{30}{20}} \right)\times 100$
$\dfrac{\Delta g}{g}\times 100=6.8$
So, the percentage error in the value of g is $6.8\%$
So, the correct answer is “Option D”.
Note:
Least count of an instrument is the smallest value that can be measured by the instrument. When we measure the error in an instrument, we use the least count.
Absolute error is the difference between the actual and measured value. Percentage error is the relative error shown as percentage.
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