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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
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The time period of a sample pendulum is to be determined with the help of a stop clock having a least count of $ \large\frac{1}{3}$ seconds. If the estimated time period is about 2 second time for how many oscillations must be recorded so that the periodic time is determined with an error less than $0.67%$

$\begin{array}{1 1}(A)\;6 \\(B)\;12 \\(C)\;18 \\(D)\;25 \end{array} $

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If t is the time for N oscillations, the $T =\large\frac{t}{N}$ and $\Delta T =\large\frac{\Delta t}{N}$
Here , $\large\frac{\Delta T}{T} =\frac{0.67 }{100} =\frac{2}{300}$
$\Delta t = \large\frac{1}{3}$$s$
and $T= 2s$
$N= \large\frac{\Delta t}{\Delta T}= \frac{\Large\frac{1}{3}}{(2/300)(2)} $
$\qquad= 25$
Hence D is the correct answer.
answered Jul 10, 2014 by meena.p

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