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# To find the value of $g$ using simple pendulum $T=2.00 sec,\;l=50cm.$ was measured. The maximum possible errors in $g$ is (Error in $T$=0.01 sec,$l$=0.1 cm)

$(a)\;1.4 \%\quad (b)\;1.1 \% \quad (c)\;1.5 \%\quad (d)\;1.2 \%$

Can you answer this question?

$T=2 \pi \sqrt {\large\frac{l}{g}}$
$g=\large\frac{4 \pi^2l}{T^2}$
$\large \frac {\Delta g}{g}=\frac {\Delta l}{l}+\frac{ 2 \Delta T}{T}$
$\qquad=\bigg(\large\frac{0.1}{50}+\frac{2 \times 0.01}{2.00}\bigg)$$\times 100$
$\qquad=1.2 \%$
Hence d is the correct answer.
answered Jun 19, 2013 by
edited May 30, 2014