A student measures the time period of `100` ocillations 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 men time should be:

Options

A. `92 +- 2s`
B. `92 +- 5.0s`
C. `92 +- 1.8s`
D. `92 +- 3s`

Answer

a

Explanation

Measured time period of `100` oscillation are `90 sec, 91sec , 95 sec and 92 sec`
Mean value of time `t_(m) = (90 + 91 + 95 +92)/(4) = 92 sec`
Absolute error in meansurement
`|Delta I_(1)| = |t_(m) - t_(1)| = 2 sec`
`|Delta I_(2)| = |t_(m) - t_(2)| = 1 sec`
`|Delta I_(3)| = |t_(m) - t_(3)| = 3 sec`
`|Delta I_(4)| = |t_(m) - t_(4)| = 0 sec`
Mean absolate error `EDeltat_(mean) = (2 + 1 + 3+ 0)/(4) = 1.5sec`
But the least count of thge measuring clock is `1 sec` so it connot measure up to `0.5` second so we have to round it off mean error will be `2` second Hence mean time `92 +- 2 sec`