Question

A student measures time for 20 oscillations of a simple pendulum as 30 s, 32 s, 35 s and 35 s. If the minimum division in the measuring clock is 1 s, then correct mean time (in second) is ______.

Options

• (33 ± 2)

• (32 ± 3)

• (33 ± 3)

• (32 ± 2)

Answer 1

A student measures time for 20 oscillations of a simple pendulum as 30 s, 32 s, 35 s and 35 s. If the minimum division in the measuring clock is 1 s, then correct mean time (in second) is (33 ± 2).

Explaation:

Mean time, t = \[\frac{30+32+35+35}{4}=\frac{132}{4}\] = 33s

Δ mean time, Δt = \[\frac{\Delta\mathrm{t}_{1}+\Delta\mathrm{t}_{2}+\Delta\mathrm{t}_{3}+\Delta\mathrm{t}_{4}}{4}\] = \[\frac{3+1+2+2}{4}\] = \[\frac {8}{4}\] = 2

Δt = 2s

\[\therefore\] Correct mean time = t ± Δt = (33 ± 2) s.