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JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 17)

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 :
5 s
4.6 s
4.62 s
4.623 s

설명

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

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