T=2πgL
∴T2=4π2gL
∴g=4π2T2L
% Error in g =(L0ΔL×100)+2(T0ΔT×100)
=(0.201×10−3×100)+2901×100
=0.5+2(910)
=0.5+920
=0.5+2.22
= 2.7%
An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :
Held on 9 Apr 2014 · Verified 6 Jul 2026.
4.4%
2.27%
1.7%
2.7%
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