A student measures the time period of 10...

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:

`92 pm 3s`

`92 pm 2s`

`92 pm 5.0 s`

`92 pm 1.8 s`

Text Solution

Verified by Experts

The correct Answer is:

`T_(AV)=92s`
`(|DeltaT|)_("mean")=1.5s`
since uncertainity is 1.5 s
so digit 2 in 92 uncertain
so reported mean time should be
`92 pm 2`

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