A student measures the time period of 100 osc | vedclass

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(A) Step $1$: Calculate the mean time period $(T_{mean})$: $T_{mean} = \frac{90 + 91 + 95 + 92}{4} = \frac{368}{4} = 92\;s$. Step $2$: Calculate the a

Vedclass · Accepted Answer

$92\pm 2\;s$

Vedclass · Answer

(A) Step $1$: Calculate the mean time period $(T_{mean})$: $T_{mean} = \frac{90 + 91 + 95 + 92}{4} = \frac{368}{4} = 92\;s$. Step $2$: Calculate the absolute errors for each measurement: $|\Delta T_1| = |92 - 90| = 2\;s$ $|\Delta T_2| = |92 - 91| = 1\;s$ $|\Delta T_3| = |92 - 95| = 3\;s$ $|\Delta T_4| = |92 - 92| = 0\;s$ Step $3$: Calculate the mean absolute error $(\Delta T_{mean})$: $\Delta T_{mean} = \frac{2 + 1 + 3 + 0}{4} = \frac{6}{4} = 1.5\;s$. Step $4$: Rounding off based on the least count: The least count of the clock is $1\;s$. Therefore,the mean absolute error should be rounded to the nearest integer,which is $2\;s$. Thus,the reported mean time is $92 \pm 2\;s$.

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

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