T=2πgℓ⇒g=T24π2ℓ
gΔg=ℓΔℓ+T2ΔT
ΔT=numberofoscillations(n)leastcountoftime(ΔT0)
gΔg=ℓΔℓ+nT2ΔT0
As Δℓ and ΔT0 are same for all observations so
gΔg is minimum for highest value of ℓ,n and T
⇒ Minimum percentage error in g is for student number 1.
Three students S_1,S_2 and S_3 perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different…
JEE Main 2021 — Physics Mechanics
Three students S1,S2 and S3 perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.
| Student No. | Length of pendulum (cm) | Number of oscillations (n) | Total time for n oscillations | Time period (s) |
| 1. | 64.0 | 8 | 128.0 | 16.0 |
| 2. | 64.0 | 4 | 64.0 | 16.0 |
| 3. | 20.0 | 4 | 36.0 | 9.0 |
If E1,E2 and E3 are the percentage errors in g for students 1,2 and 3, respectively, then the minimum percentage error is obtained by student no .
Held on 22 Jul 2021 · Verified 1 Jul 2026.
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