Question

state the number of significant figures in 6.37*10^6

Answer 1

5.52×103kgm−35.52×103kgm-3

Solution :

Given

Mass of the earth (M)(M) Radius of the earth (R)(R)

Volume of the earth (V)(V)

=43×πR3m3=43×(3.142)×(6.37×106)3=43×πR3m3=43×(3.142)×(6.37×106)3

Average density (D)(D)

=MassVolume=MV=5.975×102443×(3.142)×(6.37×106)3=MassVolume=MV=5.975×102443×(3.142)×(6.37×106)3

=0.005517×106kgm−3=0.005517×106kgm-3

=5.52×103kgm−3=5.52×103kgm-3 ( to three significant figures)

The density is accurate only up to three significant figures which is the accuracy of the least accurate factor , namely , the radius of the earth .