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1.Units, Dimensions and Measurement
easy
In a simple pendulum experiment, the maximum percentage error in the measurement of length is $2\%$ and that in acceleration due to gravity $g$ is $4\%$. Then the maximum percentage error in determination of the time-period is
A
$\pm 6 \%$
B
$\pm 5 \%$
C
$\pm 4 \%$
D
$\pm 3 \%$
Solution
Time period $T\, = \,\,2\pi \sqrt {\frac{1}{g}} \,\,\, = \,2\pi {\left( {\frac{1}{g}} \right)^{\frac{1}{2}}}\,\, = 2\pi \frac{{{{({1})}^{\frac{1}{2}}}}}{{{{(g)}^{\frac{1}{2}}}}}$
$\left( {\frac{{\Delta T}}{t}\, \times \,100} \right)\,\% \, = \pm \left[ {\frac{1}{2} \times \frac{{\Delta {l}}}{{l}}\, \times \,100 + \,\frac{1}{{2\,}} \times \,\frac{{\Delta g}}{g}\, \times \,100} \right]$
$ = \pm \left[ {1\,\% \, + \,2\,\% } \right]\,$
$\therefore T\,\, = \pm \,\,3\,\% $
Standard 11
Physics
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