152,277 views
8 votes
8 votes
40 cm³ of water is poured into an empty measuring cylinder. A stone of mass 129g is put into the cylinder. If the density of the mixture of the stone is 8.6g/cm³, find the new reading of the cylinder.​

User Fridojet
by
2.6k points

2 Answers

25 votes
25 votes

Explanation:

I assume the measuring cylinder measures the cm³ of its normally liquid content.

because for anything else we would need more information about the dimensions of the cylinder.

and so, originally the cylinder shows 40 cm³.

after dropping the stone in, how many cm³ of water is the cylinder then showing ?

let's first mention some facts we are going to use :

water weighs 1 kg (1000 g) per liter.

and 1 liter fits exactly into a cube of

10 cm × 10 cm × 10 cm = 1000 cm³

so, 1 cm³ water weighs exactly 1 g and has therefore a density of 1 g / cm³.

the stone has a density of 8.6 g / cm³, is therefore heavier than water and sinks (and replaces water correspondingly).

how many cm³ does the stone have (and replaces water) ?

well, it has 129 g, and 8.6 g of the stone fill a cm³.

so, it has

129 / 8.6 = 15 cm³

therefore, as these 15 cm³ of stone replace 15 cm³ of water, this is the same as putting 40 + 15 = 55 cm³ of water into the measuring cylinder.

and the cylinder reads now 55 cm³.

FYI : but there is still only 40 cm³ of water in there.

this is actually used to calculate the density of objects (by first weighing and then dropping them into the water to see how much water they replace).

User Macbernie
by
2.7k points
22 votes
22 votes

Answer: 55 cm³.

Explanation:


\displaystyle\\m_(stone)=129\ g\ \ \ \ \ \rho_(stone)=8,6\ (g)/(cm^3) \ \ \ \ \ V_(water)=40\ cm^3\ \ \ \ \ V=?.\\ \boxed {\rho=(m)/(V) }\\V=(m)/(\rho) \\V=V_(water)+V_(stone)\\V_(stone)=(m_(stone))/(\rho_(stone)) \\V_(stone)=(129)/(8,6)\\V_(stone )=15\ cm^3.\\V=40+15\\V=55\ cm^3.

User Overthink
by
2.5k points