[" The time period of oscillation of a simple "],[" pendulum is given by "T=2 pi sqrt((ell)/(g))" .The length of "],[" the pendulum is measured as "ell=10+-0.1cm],[" and the time period as "T=0.5+-0.02s" .The "],[" percentage error in the value of g is - "]

The time period of oscillation of a simple pendulum is given by T=2pisqrt(l//g) The length of the pendulum is measured as 1=10+-0.1 cm and the time period as T=0.5+-0.02s . Determine percentage error in te value of g.

The time period of oscillation of a simple pendulum is given by T=2pisqrt(l//g) The length of the pendulum is measured as 1=10+-0.1 cm and the time period as T=0.5+-0.02s . Determine percentage error in te value of g.

The period of oscillation of a simple pendulum is given by T=2pi sqrt((l)/(g)) . The length l of the pendulum is about 0.5s. The time of 100 oscillations is measured with a watch of 1 s resolution. Calcualte percentage error in measurment of g.

The time period T of oscillation of a simple pendulum of length l is given by T=2pi. sqrt((l)/(g)) . Find the percentage error in T corresponding to 2% in the value of l.

The time T of oscillation of as simple pendulum of length l is given by T=2pi sqrt(l/g) . The percentage error in T corresponding to an error of 2% in the value of l is

The time period of a pendulum is given by T = 2 pi sqrt((L)/(g)) . The length of pendulum is 20 cm and is measured up to 1 mm accuracy. The time period is about 0.6 s . The time of 100 oscillations is measured with a watch of 1//10 s resolution. What is the accuracy in the determination of g ?

The time period of a pendulum is given by T = 2 pi sqrt((L)/(g)) . The length of pendulum is 20 cm and is measured up to 1 mm accuracy. The time period is about 0.6 s . The time of 100 oscillations is measured with a watch of 1//10 s resolution. What is the accuracy in the determination of g ?