The radius of a thin wire is 0.16 mm. The area of cross-section of wire in mm2 with correct number of significant figures is

Question

The radius of a thin wire is 0.16 mm. The area of cross-section of wire in mm2 with correct number of significant figures is (A) 0.081 (B) 0.080 (C) 0.0804 (D) 0.080457

Tardigrade Admin · Accepted Answer

Here, R = 0.16 mm Hence, A = π R2 = (22/7) × (0.16)2 = 0.080457 mm2 Since radius has two significant figures so area also will have two significant figures. ∴ : :: : A = 0.080 mm2

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

0.081

0.080

0.0804

0.080457

Here, $R$ = 0.16 mm
Hence, $A = π R^{2}$
$= \frac{22}{7} \times (0.16)^{2} = 0.080457 m m^{2}$
Since radius has two significant figures so area also will have two significant figures.
$∴ A = 0.080 m m^{2}$

Q. The radius of a thin wire is 0.16 mm. The area of cross-section of wire in mm2 with correct number of significant figures is