Question

A student measures times for 20 oscillations of a simple pendulum as 30s, 32s, 35s and 31s. If the minimum division in the measuring clock is 1s, then correct mean time in second , within error limits , is

• A. \((32\pm 5)\text{s}\)
• B. \((32\pm 3)\text{s}\)
• C. \((32\pm 2)\text{s}\)
• D. \((32\pm 1)\text{s}\)

Hint:

Find the mean of the readings, then the mean absolute deviation, and round the error to the least count of the clock.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The result of repeated readings is written as mean \(\pm\) mean absolute error. The error cannot be quoted finer than the least count of the instrument.

Step 2: Calculation of the mean
\(\bar{t}=\dfrac{30+32+35+31}{4}=32\text{ s}\)

Step 3: Calculation of the error
Absolute deviations from the mean are \(2,\ 0,\ 3,\ 1\) seconds.
\(\overline{\Delta t}=\dfrac{2+0+3+1}{4}=1.5\text{ s}\)

Step 4: Round to the least count
The least count is \(1\text{ s}\), so the error \(1.5\text{ s}\) is quoted as \(2\text{ s}\). The mean time is \((32\pm 2)\text{ s}\).

Final Answer:
The mean time with error limits is \((32\pm 2)\text{ s}\). \[ \boxed{\text{(C) }(32\pm 2)\text{ s}} \]