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$
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$