The number of significant figures in 0.06900 is

Rules for Counting Significant Figures

(i) All the non-zero digits are significant. In 1.248, the

number of significant figures is 4.

(ii) All the zeroes between two non-zero digits are

significant, no matter where the decimal point is, if at

all. As examples, 406 and 9.001 have 3 and

4 significant figures respectively.

(iii) If the measurement of number is less than 1, the

zero(es) on the right of decimal point and to the left

of the first non-zero digit are non-significant.

In 0.00606, first three underlined zeroes are

non-significant and the number of significant figures

is only 3.

(iv) The terminal or trailing zero(es) in a number

without a decimal point are not significant. Thus,

12.3 =1230cm=12300 mm has only 3 significant

figures.

(v) The trailing zero(es) in number with a decimal point

are significant. Thus, 3.800 kg has 4 significant

figures.

(vi) A choice of change of units does not change the

number of significant digits or figures in a measurement.

So, here using rule (iii) and (v) we get, in 0.06900

-

0.06900 the underlined part is not significant and the bold part is significant.

Therefore, the number of significant figures is 4.