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 :

Question

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 : (A) 92 ± 2 s (B) 92 ± 5.0 s (C) 92 ± 1.8 s (D) 92 ± 3 s

Tardigrade Admin · Accepted Answer

overlinex = (∑ xi/N) = (90 + 91+ 95 + 92/4) = 92 Mean deviation = (∑ | barx - xi|/N) = (2 + 1+ 3 +0/4) = 1.5 L.C. = 1 s. ∴ Required value = 92 ± 2 s

Q. 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 :

$92 \pm 2 s$

$92 \pm 5.0 s$

$92 \pm 1.8 s$

$92 \pm 3 s$

$\overline{x} = \frac{\sum x _{i}}{N} = \frac{90 + 91 + 95 + 92}{4} = 92$
Mean deviation = $\frac{\sum ∣ x ˉ - x _{i} ∣}{N} = \frac{2 + 1 + 3 + 0}{4} = 1.5$
$L . C . = 1 s$ .
$∴$ Required value $= 92 \pm 2 s$

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

92±2s

92±5.0s

92±1.8s

92±3s