The period of oscillation of a simple pendulum is T = 2 π L g .  Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be:

Option 1 - <p>1.33%</p>
Option 2 - <p>1.13%</p>
Option 3 - <p>1.03%</p>
Option 4 - <p>1.30%</p>
5 Views|Posted 4 months ago
Asked by Shiksha User
1 Answer
V
4 months ago
Correct Option - 2
Detailed Solution:

T = 2 π L g o r T 2 = 4 π 2 ( L g )

g = 4 π 2 ( L T 2 )

Δ g g % = ( Δ L L + 2 Δ T T ) % = [ 1 m m 1 m + 2 * 0 . 0 1 1 . 9 5 ] * 1 0 0 % = ( 0 . 0 0 1 + 0 . 0 1 0 2 ) * 1 0 0 = 1 . 1 3 %

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Physics Units and Measurement 2025

Physics Units and Measurement 2025

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