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Units and Measurements — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements4 Marks
Question
The acceleration due to gravity is determined by using a simple pendulum of length $l = (100 ± 0.1) \ cm$. If its time period is $T = (2 ± 0.01) s$, find the maximum percentage error in the measurement of g.
✓

Answer

Given: $\Delta I =0.1 \ cm, I =100 \ cm, \Delta T =0.01 s$,
$T = 2s$
To find: Percentage error
Formulae:
$i. T =2 \pi \sqrt{\frac{l}{g}}$
$ii.$ Percentage error $=\frac{\Delta g \times 100}{g}$
Calculation: From formula $(i),$
$T ^2=4 \pi^2 l / g \ldots .($ Squaring both sides$)$
$\therefore g =4 \pi^2 \frac{l}{T^2} $
$\therefore \frac{\Delta g }{ g }=\frac{\Delta l}{l}+\frac{2 \Delta T }{ T }$
$....($Using result from error in division$)$
From formula $(ii),$
$\text { Percentage error }=\left(\frac{\Delta l}{l}+\frac{2 \Delta T }{ T }\right) \times 100$
$=\left(\frac{0.1}{100}+\frac{2 \times 0.01}{2}\right) \times 100$
$=(0.001+0.01) \times 100=1.1 \%$
Percentage error in measurement of $g$ is $1.1 \%.$

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