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The time period of a simple pendulum is given by T = 2*pi*sqrt(l/g). The length of the pendulum is measured as l = 10.0 +/- 0.1 cm and the time period as T = 0.5 +/- 0.02 s. What is the percentage error in the experimentally determined value of g?

1. 1%
2. 5%
3. 9%
4. 18%

Correct answer: 9%

Solution

From g = 4*pi² * l / T², using error propagation for products and powers: delta_g / g = deltaₗ / l + 2 * delta_T / T. Substituting: deltaₗ/l = 0.1/10 = 0.01 (1%), and 2 * delta_T/T = 2 * 0.02/0.5 = 0.08 (8%). Total relative error = 0.01 + 0.08 = 0.09, so percentage error in g = 9%.

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