Question

ii)40cm3 of water is poured into an empty measuring cylinder. If the density
of the stone is 8.6g/cm3 Find the new reading of the water in the measuring
cylinder.

Answer 1

FINAL ANSWER:-

$4.65\:cm^{3}$

GIVEN:-

• Initial reading of balance F1 = 40 dg N

• density of the stone D = 8.6g/cm3

TO FIND:-

final reading of balance F2 = x dg N

FORMULAS USED:-

$D=\frac{F1}{F2}$

SOLUTION:-

$D=\frac{F1}{F2}$

$8.6=\frac{40}{x}\\8.6x=40\\x=\frac{40}{8.6}\\x=4.65\: dg\:N$

reading of the water in the measuring  cylinder = 4.65 cm³

Answer 2

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