To find the value of `g` using simple pendulum , `T = 2.00 s and l = 1.00 m` were measured . Estimate maximum permissible error in `g` . Also find the value of `g`.
Text Solution
AI Generated Solution
To find the value of `g` using a simple pendulum, we start with the formula for the period of a pendulum:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Where:
- \( T \) is the period of the pendulum,
- \( l \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.
...
To find the value of g using simple pendulum. T=2.00 +-0.01 sec , L=1.00+-0.01 m was measured. Estimate maximum permissible error in g . (use pi^(2)=10 )
While determining the value of g using simple pendulum, we plot a graph between l and T^(2) which is
The error in the measurement of the length of the sample pendulum is 0.2% and the error in time period 4% . The maximum possible error in measurement of (L)/(T^(2)) is
In an experiment of simple pendulum the errors in the measurement of length of the pendulum L and time period T are 3% and 2% respectively. Find the maximum percentage error in the value of acceleration due to gravity.
In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) are 3% and 2% respectively. The maximum percentage error in the value of L//T^(2) is
In Ohm's law experiment , the potential drop acros a resistance was as V = 5.0 V and the current was measured as I = 2.00 A . Find the maximum permissible error in resistance.
In an experiment of simple pendulum , the time period measured was 50 s for 25 vibrations when the length of the simple pendulum was taken 100 cm . If the least count of stop watch is 0.1 s and that of meter scale is 0.1 cm . Calculate the maximum possible error in the measurement of value of g . If the actual value of g at the place of experiment is 9.7720 m s^(-2) , Calculate the percentage error.
An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be:
Find the length of seconds pendulum at a place where g = pi^(2) m//s^(2) .
A boy performs an experiment in which he uses a simple pendulum to find the value of 'g' using the formula g=(4pi^(2)l)/(T^(2))= Where l = 1m. Error in measurements of length is Delta l , human error in time measurement is 0.15 sec and least count of stop watch Delta T , for what value of amplitude (A). Delta l and Delta T error in calculation of 'g' is maximum.
