Time periods of oscillation of the same simpl | vedclass

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(C) The given readings are $4.62 \,s, 4.632 \,s, 4.6 \,s$,and $4.64 \,s$.To find the arithmetic mean,we first calculate the sum: $4.62 + 4.632 + 4.6 +

Vedclass · Accepted Answer

$4.6$

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(C) The given readings are $4.62 \,s, 4.632 \,s, 4.6 \,s$,and $4.64 \,s$.To find the arithmetic mean,we first calculate the sum: $4.62 + 4.632 + 4.6 + 4.64 = 18.492 \,s$.According to the rules of significant figures for addition,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places. Here,$4.6 \,s$ has only one decimal place.Therefore,the sum should be rounded to one decimal place: $18.5 \,s$.Now,calculate the arithmetic mean: $	ext{Mean} = \frac{18.5}{4} = 4.625 \,s$.Finally,applying the rules of significant figures for division,the result must have the same number of significant figures as the measurement with the fewest significant figures. The value $4.6 \,s$ has two significant figures.Rounding $4.625 \,s$ to two significant figures gives $4.6 \,s$.

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

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