A student measures the distance covered and time taken by a body under free fall which was initially at rest. He uses this data for estimating g , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e 1 and e 2 respectively, then the maximum possible percentage error in the estimation of g is

Question

A student measures the distance covered and time taken by a body under free fall which was initially at rest. He uses this data for estimating g , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e 1 and e 2 respectively, then the maximum possible percentage error in the estimation of g is, e 2 − e 1, e 1 + 2 e 2, e 1 + e 2, e 1 − 2 e 2

MARKS App · Accepted Answer

From the relation h = ut + 1 2 gt 2 h = 1 2 gt 2 ⇒ g = 2 h t 2 ∵ body initially at rest Taking natural logarithms on both sides, we get ln g = ln h - 2 ln t + ln 2 Differentiating, Δ g g = Δ h h - 2 Δ t t For maximum permissible errors, we add both the errors. So Δ g g × 1 0 0 max = Δ h h × 100 + 2 × Δ t t × 100 Δ h h × 100 = e 1 and Δ t t × 100 = e 2 Therefore, Δ g g × 100 max = e 1 + 2 e 2

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