The period of oscillation of a simple pendulum is T=2π √ g. Measured value of l is 20.0 cm, known to 1 mm accuracy and time for 100oscillations of…
JEE Main 2015 — Physics Mechanics
2015mcqeasy
The period of oscillation of a simple pendulum is T=2πgl. Measured value of l is 20.0cm, known to 1mm accuracy and time for 100oscillations of the pendulum is found to be 90s using a wristwatch of 1s resolution. The accuracy in the determination of g is
Official previous-year question
Held on 4 Apr 2015 · Verified 6 Jul 2026.
Options
A
5
B
4
C
3
D
1
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Solution
∴T=2πgl⇒g=4π2T2l
∴ Error in g can be calculated as
gΔg=lΔl+T2ΔT.
∴ Total time for noscillation is t=nT where T= time for oscillation.
⇒tΔt=TΔT
⇒gΔg=lΔl+t2Δt
Given that Δl=1mm=10−3m,l=20×10−2m
Δt=1s,t=90s.
gΔg×100=(lΔl+t2Δt)×100
=(20×10−210−3+902×1)×100=21+920=0.5+2.22=2.72
≅3
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