Browse Questions

# 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 $0.67%$

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

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.