The period of oscillation of a simple pendulum is T =2pisqrt(L/g). Measuted value of L is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?

The period of oscillation of a simple pendulum is T =2pisqrt(L/g). Mea

1.	The period of oscillation of a simple pendulum is T =2pisqrt(L/g). Measuted value of L is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?
Answer» <html><body><p></p>Solution :`g= <a href="https://interviewquestions.tuteehub.com/tag/4pi-1882352" style="font-weight:bold;" target="_blank" title="Click to know more about 4PI">4PI</a>^(2) L//T^(2)` Here, `T= (T)/(n) and Delta T = (Delta t)/( n)`. <br/> <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, `(Delta T)/( T) = (Delta t)/( t)`. <br/> The errors in both L and t are the least count errors. <br/> Therefore, `(Deltag //g) = (Delta L//L) + 2(Delta T//T)` <br/> `= (0.1)/( 20.0) +2 ((1)/(90) ) = 0.027` <br/> Thus, the percentage <a href="https://interviewquestions.tuteehub.com/tag/error-25548" style="font-weight:bold;" target="_blank" title="Click to know more about ERROR">ERROR</a> in g is <br/> `<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> (Delta g//g) = 100(Delta L//L) + 2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100 (Delta T //T) = 3`</body></html>