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The period of oscillation of a simple pendulum is given by

$T\ =2\pi\sqrt{\frac{l}{g}}$ , Where $l$ is about $100\ cm$ known with an accuracy of $1\ mm$ . This time is about $2\ seconds$ . The time up to $100$ oscillations is measured with a stopwatch of at least $0.1\ seconds.$ The percentage error in g is:
Option: 1
$0.1\ %$

Option: 2
$1\ %$

Option: 3
$0.2\ %$

Option: 4
$0.8\ %$

Answers (1)

best_answer

$T\ =2\pi\sqrt{\frac{l}{g}}$

Or, $T^2=\frac{4\pi^2l}{g}$

Or, $g=\frac{4\pi^2l}{T^2}$

Here percentage error in $l$ is:

$\frac{1\ mm}{100\ cm\ }\times100$

Or, $\frac{0.1}{100}\times100\ =\ 0.1\$

Again,

The Percentage error in T:

$\frac{0.1}{2\times100}=100\ =0.5$

Now,

Percentage error in g = Percentage error in $l$ + 2(Percentage error in T)

$0.1+2\times0.05\ =\ 0.2%$

Posted by

himanshu.meshram

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