The period of oscillation of a simple pendulum is T = 2 π l g . Measured value of l is 20 . 0 cm , known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of 1 s resolution. The accuracy in the determination of g is

Step-by-step solution

∴ T = 2 π l g ⇒ g = 4 π 2 l T 2 ∴ Error in g can be calculated as Δ g g = Δ l l + 2 Δ T T . ∴ Total time for n   oscillation is t = n T where T = time for oscillation. ⇒ Δ t t = Δ T T ⇒ Δ g g = Δ l l + 2 Δ t t Given that Δ l = 1   mm = 10 - 3   m ,   l = 20 × 10 - 2   m Δ t = 1   s ,  t = 90   s . %   e r r o r   i n   g  is Δ g g × 100 = Δ l