Practice portal › Experimental Skills › Simple Pendulum and g
Asked in JEE Main 28th July 2nd Shift 2022 · Error and accuracy in g
Given: L=10 cm with Δ L=1 mm=0.1 cm; 100 oscillations timed with a 1 s watch.
Total time =100×0.5=50 s, so Δ t=1 s on 50 s.
g=(4π²L)/(T²), so (Δ g)/g=(Δ L)/L+2(Δ t)/t.
=(0.1)/(10)+2×1/(50)=0.01+0.04=0.05.
x=5.
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