Calculate percentage error in determination of time period of a pendulum.
`T = 2pi (sqrt(l))/(g)`
where, l and g are measured with `+-1%` and `+-2%`.

Calculate percentage error in the determination of g = 4pi^2 I//t^2, when I and t are measured with +- 2% and +- 3% errors respectively.

What is the percentage error in the measurement of time period of a pendulum if maximum errors in the measurement of l ands g are 1% and 6% respectively?

Calculate the percentage of error in the determination of g=4pi^2l//_t2 when l and t are measured with _-^+2% and _-^+3% errors respectively.

What is the percentage error in the measurement of time period of a pendulum if maximum errors in the measurement of l ands g are 2% and 4% respectively?

What is the percentage error in the measurement of time period of a pendulum if maximum errors in the measurement of l ands g are 2% and 4% respectively?

The percentage error in determination fo g= 4pi^2(I)/(t^2), when I and t are measured with +- 1% and +- 2% errors is

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then find the maximum error in the measurement of acceleration due to gravity.

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is