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Archimedes' principle can be used to determine the specific gravity of a solid using a known liquid(Example 10–8). The reverse can be done as well. (a) Asan example, a 3.80-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid. (b) Determine a formula for finding the density of a liquid using this procedure.

Short Answer

Expert verified

The density of the liquid is \(1210\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}\).

The formula for the density of the liquid is \({\rho _{{\rm{liquid}}}} = \left( {\frac{{{m_{{\rm{object}}}} - {m_{{\rm{apparent}}}}}}{{{m_{{\rm{object}}}}}}} \right){\rho _{{\rm{object}}}}\).

Step by step solution

01

Step1:Understanding the density of a liquid

The density of a liquid is the ratio of the mass of liquid and its volume. The density of a liquid decreases with an increase in temperature. It is expressed in kilograms per cubic meter.

02

Step2:Identification of the given data

Mass of the aluminum ballis \({m_{{\rm{ball}}}} = 3.80\;{\rm{kg}}\).

The apparent mass of aluminum ball submerged in a liquidis \({m_{{\rm{apparent}}}} = 2.10\;{\rm{kg}}\).

03

Step3:Part (a) - Determination of the density of liquid

The mass of the liquid displaced by the ball is equal to the difference between the actual mass and the apparent mass of the aluminum ball.

\(\Delta m = {m_{{\rm{actual}}}} - {m_{{\rm{apparent}}}}\)

The mass of the liquid displaced is the product of the ball's volume and the liquid's density.

\(\begin{aligned}{\rho _{{\rm{liquid}}}}{V_{{\rm{ball}}}} &= \Delta m\\{\rho _{{\rm{liquid}}}}\frac{{{m_{{\rm{ball}}}}}}{{{\rho _{{\rm{Al}}}}}} &= \Delta m\\{\rho _{{\rm{liquid}}}} &= \frac{{\Delta m}}{{{m_{{\rm{ball}}}}}}{\rho _{Al}}\\{\rho _{{\rm{liquid}}}} &= \frac{{{m_{{\rm{actual}}}} - {m_{{\rm{apparent}}}}}}{{{m_{{\rm{ball}}}}}}{\rho _{{\rm{Al}}}}\end{aligned}\)

Substitute the values in above formula.

\(\begin{aligned}{\rho _{{\rm{liquid}}}} &= \frac{{\left( {3.80\;{\rm{kg}} - 2.10\;{\rm{kg}}} \right)}}{{3.80\;{\rm{kg}}}}\left( {2.70 \times {{10}^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}} \right)\\{\rho _{{\rm{liquid}}}} &= 1210\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}\end{aligned}\)

Hence, the density of the liquid is \(1210\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}\).

04

Part (b) :Determination of the formula of the density of liquid

Generalizing the relation from above, we have

\({\rho _{{\rm{liquid}}}} = \left( {\frac{{{m_{{\rm{object}}}} - {m_{{\rm{apparent}}}}}}{{{m_{{\rm{object}}}}}}} \right){\rho _{{\rm{object}}}}\).

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