The period of oscillation of a simple pendulum is given by T = 2pisqrt((L)/(g)), where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has 500 divisions. If the measured value L is 50 cm, the accuracy in the determination of g is 1.1% and the time taken for 100 oscillations is 100 seconds, what should be the possible error in measurement of the clock in one minute (in milliseconds) ?

The period of oscillation of a simple pendulum is given by T = 2pisqrt

1.	The period of oscillation of a simple pendulum is given by T = 2pisqrt((L)/(g)), where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has 500 divisions. If the measured value L is 50 cm, the accuracy in the determination of g is 1.1% and the time taken for 100 oscillations is 100 seconds, what should be the possible error in measurement of the clock in one minute (in milliseconds) ?
Answer» 1 2 5 0.25 Answer :C

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