Question

The least count of a stop watch is 0.2s. The time of 20 oscillations of a pendulum is measured to
be 25s. The percentage error in the measurement of time will be:

A. $16\% $
B. $1.8\% $
C. $0.8\% $
D. $0.1\% $

Answer 1

Hint: First, we need to calculate the time period of the pendulum from the number of oscillations
and the time taken for oscillation. The error can be calculated from the least count and the time
period.
The time period of a simple pendulum is defined as the time taken by the pendulum to finish one
full oscillation and is denoted by T. One oscillation of a simple pendulum is one complete cycle
of swinging one way and then returning to its original starting position.
Given,
The number of oscillations, $n = 20$
Time taken for $20$ oscillations, $t = 25{\rm{ s}}$
The least count of stop watch is $0.2{\rm{s}}$
The error in time for $20$ oscillations, $\Delta T = 0.2{\rm{ s}}$
The formula for calculating time period is,
$T = \dfrac{t}{n} = \dfrac{{25}}{{20}} = 1.25{\rm{ s}}$
Now, given that the error in time for $20$oscillations,$\Delta T = 0.2{\rm{ s}}$
Hence, error in time for 1 oscillations, $\Delta T' = \dfrac{{0.2}}{n}{\rm{ s}}$
Therefore, percentage error is calculated as, $\% error = \dfrac{{\Delta T'}}{T} \times {\rm{100
= }}\dfrac{{0.2}}{{20 \times 1.25}} \times 100\% = 0.8\% $
Hence, the correct answer is (C).
Note: In the solution, the students can first calculate the time period of the pendulum from the
given number of oscillations and time taken. From the least count the error in time for 20
oscillations can be identified and for one oscillation the error can be calculated. The error in time
per oscillation divided by the time period and number of oscillation gives the percentage error in
the measurement of time.