The period of oscillation of a simple pendulum is T = 2pi sqrt(L/g). The measured length L = (20.0 +- 0.1) cm and time for 100 oscillations is (90.0 +- 1.0) s. The maximum percentage error in the measurement of g is:

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Appears in NEET UG
Solution ✔ Verified
  • A2.5%
  • B3.2%
  • C4.5%
  • D1.5%
Explanation

T = time for 100 oscillations / 100 = 90.0/100 = 0.900 s. Error in T, dT/T = d(total time)/(total time) = 1.0/90.0 = 0.0111. So %T = 1.11%. From T = 2pi sqrt(L/g), we get g = 4pi² L/T². The percentage error in g is %g = %L + 2(%T). %L = (0.1/20.0)*100% = 0.5%. %g = 0.5% + 2(1.11%) = 0.5% + 2.22% = 2.72%. Let me re-calculate with proper sig figs. L = (20.0 +- 0.1) cm, %L = (0.1/20.0)*100 = 0.5%. Time for N=100 oscillations = (90.0 +- 1.0) s. Period T = 90.0/100 = 0.900 s. Error in T: d(total time)/total time = 1.0/90.0 = 1/90. So %T = (1/90)*100 = 1.11%. g = 4pi² L/T². %g = %L + 2*%T %g = 0.5% + 2*(1.11%) = 0.5% + 2.22% = 2.72%.

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