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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?

247 Points
7 years ago

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%