First slide

Accuracy; precision of instruments and errors in measurements

Question

In a certain observation made to measure the length of wire and radius of bob of a simple pendulum we got, $l = 23.2 cm, r = 1.32 cm$ and time taken for 10 oscillations was 10.0 s. Find, maximum percentage error in determination of ' $g$ '

Moderate

A

1.4%
B

2.4%
C

3.4%
D

4.4%

Solution

$l = 23.2 cm \Rightarrow Δ l = 0.1 cm$
$r = 1.32 cm \Rightarrow Δ r = 0.01 cm$
$t = 10.0 s \Rightarrow Δ t = 0.1 s$
$g = 4 π^{2} (\frac{L}{T^{2}}) = 4 π^{2} [\frac{1 + r}{(t / n)^{2}}]$
$g = 4 π^{2} n^{2} (\frac{l + r}{t^{2}})$
Maximum percentage error in g will be
$(\frac{Δ g}{g}) \times 100 = [\frac{Δ l + Δ r}{l + r} + 2 (\frac{Δ t}{t})] \times 100$
$= [\frac{0.1 + 0.01}{23.2 + 1.32} + 2 \times \frac{0.1}{10.0}] \times 100$
= 2.4%

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