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 (a) (33 ± 2) (b) (32 ± 3) (c) (33 ± 3) (d) (32 ± 2)

Answer: (a). Mean = (30 + 32 + 35 + 35)/(4) = 132/4 = 33 s Absolute deviations: 3, 1, 2, 2; mean absolute error = 8/4 = 2 s This is larger than the least count (1 s), so it is kept: (33 ± 2) s. So the answer is (a) (33 ± 2).

A student measures time for 20 oscillations of a simple pendulum as… | MHT CET 2025 (20 Apr, Shift 1)

Q44MHT CET 2025 · 20 APR · SHIFT 1Medium

Errors in Measurement

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

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Answer: (a)

Mean $= \frac{30 + 32 + 35 + 35}{4} = \frac{132}{4} = 33\text{ s}$
Absolute deviations: $3, 1, 2, 2$ ; mean absolute error $= \frac{8}{4} = 2\text{ s}$
This is larger than the least count (1 s), so it is kept: $(33 \pm 2)\text{ s}$ .
So the answer is (a) $(33 \pm 2)$ .

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