The period of oscillating simple pendulum is T=2 g where length ' ' is 100 cm with error 1 mm. Period is 2 second. The time of 100 oscillations is measured by a stopwatch of least count 0.1 s. The percentage error in gravitational acceleration ' g ' is

Step-by-step solution

The period of a simple pendulum is governed by the relation T = 2 g , which rearranged gives g = 4 ^2 T^2 . Using error propagation for multiplicative relationships, the fractional error in g becomes g g = + 2 T T . The percentage error in length = 1 m with = 0.001 m is 0.001 1 100 % = 0.1 % . The error in period is derived from timing 100 oscillations with a stopwatch of least count 0.1 s, giving T = 0.1 100 = 0.001 s for T = 2 s, so T T = 0.05 % . The percentage error in g is 0.1 % + 2 0.05 % = 0.2 % . Final answ