T=2πgℓ
g=T24π2ℓ
gΔg=ℓΔℓ+T2ΔT
gΔg=11×10−3+2×1.950.01
gΔg=0.0113 or 1.13
JEE Main 2021 — Physics Mechanics
The period of oscillation of a simple pendulum is T=2πgL. 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:
Held on 24 Feb 2021 · Verified 6 Jul 2026.
1.03
1.33
1.30
1.13
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