A simple pendulum is being used to determine th value of gravitational acceleration g at a certain place. Th length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

Solution

Physics sir · 4 steps

Write g in terms of L and T, then add the fractional errors, remembering T comes squared.

1. Formula for g. From T = 2π√(L/g) we get g = 4π² L/T². For a product of powers, the fractional errors add, each multiplied by its power.

   (Δ g)/g = (Δ L)/L + 2 (Δ T)/T

2. Length error. Writing the length as 25.0 cm means it was measured to 0.1 cm. So (Δ L)/L = 0.1/25.0 = 0.004, which is 0.4%.
3. Time error. T = t/40, so (Δ T)/T = (Δ t)/t = 1/50 = 0.02, which is 2%. Because T is squared, it adds 2 × 2% = 4%.
4. So. (Δ g)/g = 0.4% + 4% = 4.4%, which is option (a).

The idea to keep. Turn the formula into fractional errors: each quantity contributes its fractional error times its power. Counting many oscillations does not change Δ t/t, because both T and Δ T shrink by the same factor.

Answer from the NTA final answer key. Solution by PrepWiser.