Question

A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90s$, $91s$, $95s$ and $92s$. If the minimum division in the measuring clock is $1s$, then the reported mean time should be

• A. $92\pm 5.0s$
• B. $92\pm 3s$
• C. $92\pm 2s$
• D. $92\pm 1.8s$

Answer 1

The correct option is C $92\pm 2s$

Find mean value of time.
Measured time period of $100$ oscillations by the student are $90s$,$91s$, $95s$ and $92s$.
Mean value of time,
$t_m=\frac{90+91+95+92}{4}=92s$

Find mean absolute error in measurement.
Formula Used: $|\Delta t|=|T_m-t|$
The absolute error in measurement can be calculated as
$|\Delta t_1|=|t_m-t_1|=2s$
$|\Delta t_1|=|t_m-t_1|=1s$
$|\Delta t_3|=|t_m-t_3|=3s$
$|\Delta t_4|=|t_m-t_4|=0s$
Mean absolute error
$\Delta t_{\text{mean}}=\frac{2+1+3+0}{4}$$=1.5s$
But the least count of the measuring clock is $1s$, so it can not measure up to $0.5$ second, so we have to round it off. So mean error will be $2s$.
Hence mean time $(92\pm 2s)$.
Final Answer: (a)