## In This post you will get answer of Q.

The radius of a thin wire is 0.16 mm. The area of cross-section of wire in $mm^2$ with correct number of significant figures is

Solution:

## Here, $R$ = 0.16 mm

Hence, $A = pi R^2$

$ = frac{22}{7} times (0.16)^2 = 0.080457 , mm^2$

Since radius has two significant figures so area also will have two significant figures.

$ therefore :::: A = 0.080, mm^2$