Question 81: Units , Dimensions and Error Analysis

Question

Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as $$4.62 \mathrm{~s}, 4.632 \mathrm{~s}, 4.6 \mathrm{~s}$$ and $$4.64 \mathrm{~s}$$. The arithmetic mean of these readings in correct significant figure is :

Accepted Answer

To find the arithmetic mean of the time periods recorded, we need to sum up the values and then divide by the number of readings. Let's calculate the sum first:
$$4.62 \mathrm{~s} + 4.632 \mathrm{~s} + 4.6 \mathrm{~s} + 4.64 \mathrm{~s}$$
Adding these values together:
$$4.62 + 4.632 + 4.6 + 4.64 = 18.492 \mathrm{~s}$$
Now, we divide this sum by the number of readings, which is 4:
$$ \frac{18.492 \mathrm{~s}}{4} = 4.623 \mathrm{~s} $$
So, the arithmetic mean of these readings is $$4.623 \mathrm{~s}$$. However, we need to consider the significant figures. The least number of significant figures among the readings is 2 (from 4.6 s). Hence, the mean should also be represented with 2 significant figures.
In this case, the correct answer with proper significant figures is:
Option B
4.6 s

Answer

5 s

Answer

4.6 s

Answer

4.62 s

Answer

4.623 s

Units , Dimensions and Error Analysis

Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as $4.62 \mathrm{~s}, 4.632 \mathrm{~s}, 4.6 \mathrm{~s} and 4.64 \mathrm{~s}$. The arithmetic mean of these readings in correct significant figure is :

Choose the correct answer: