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This section contains 5 questions. The answer to each of the questions is a single digit integer ranging from 0 to 9.
The period of oscillation of a simple pendulum is 2p. 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 wrist watch of 1 s resolution. What is the percentage error in the determination of g?
3
g = 4p2L/T2;
Here, T = and DT = . Therefore,
This errors in both L and t are the least count errors. Therefore
(Dg/g) = (DL/L) + 2 (DT/T)
= = 0.027
Thus, the percentage error in g is
100 (Dg/g) = 100(DL/L) + 2 × 100(DT/T) = 3%
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