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`)

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 .

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

While determining the value of g using simple pendulum, we plot a graph between l and T^(2) which is

Time is measured using a stop watch of least count 0.1 second In 10 oscilation time taken is 20.0 second Find maximum permissible error in time period .

Find the area of the triangle formed by (0,0),(1,0),(0,1) .

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.

If L = 2.01 m pm 0.01 m , what is 3 L ?

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 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.

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.