The period of oscillation of a simple pendulum is measured, in successive measurement the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s.
Then, the relative error or percentage error is

detailed solution

Correct option is B

The mean period of oscillation of the pendulum isT=(2.63+2.56+2.42+2.71+2.80)s5T=13.125s=2.624  s=2.62  sAs the periods are measured to a resolution of 0.01 s, all times are to the second decimal, it is proper to put this mean period also to the second decimal. The errors in the measurement are 2.63 s –2.62 s = 0.01s2.56 s –2.62 s = –0.06 s2.42 s –2.62 s = –0.20 s2.71s –2.62 s = 0.09 s2.80 s –2.62 s = 0.18 sNote that the errors have the same units as the quantity to be measured. The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) isΔTmean=[(0.01+0.06+0.20+0.09+0.18)s]/5= 0.54 s / 5 = 0.11sThat means, the period of oscillation of the simple pendulum is (2.62±0.11)s  i.e.,  it  lies  between(2.62+0.11)s  and (2.62−0.11)s or between 2.73 s and 2.51 s. As the arithmetic mean of all the absolute errors in 0.11 s, there is already an error in the tenth of a second. Hence, there is no point in giving the period to a hundredth. A more correct way will to be write, T=2.6±0.1 sNote that the last numeral 6 is unreliable, since it may be anything between 5 and 7. We indicate this by saying that the measurement has two significant figures. In this case, the two significant figures are 2, which is reliable and 6, which has an error associated with it. Hence, the relative error or percentage error isSa=0.12.6×100=4%