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 pm 2s`
B. `92 pm 5.0s`
C. `92 pm 1.8s`
D. `92 pm 3s`

Answer

C

Explanation

Maan time `T= (90+91+95+92)/(4)=92s`
The absolute errors are as follows :
`T_(1)= 90 implies |triangleT_(1)|= |T_(0)-T_(1)|=2`
`T_(2)= 91 implies triangleT_(2)= |T-T_(2)|=1`
`T_(3)= 95 implies triangleT_(3) = |T-T_(3)|=3`
`T_(4)= 92 implies triangleT_(4) = |T -T_(4)|=0`
Mean error `triangleT=(2+1+3+0)/(4)=1.5s`
Least count of measuring clock is 1s
The reported mean time
`=T pm triangleT= (92 pm 1.5)s`
Minimum division in the measuring clock is 1s, thus the reported mean time should be `(92 pm 2)s`.