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Question

# Explain how Archimedes' principle may be applied to find the relative density of a solid.

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Solution

## We know that,$RelativeDensity=\frac{Densityofbody}{Densityofwaterat{4}^{\circ }C}\phantom{\rule{0ex}{0ex}}=\frac{Massofthebody}{Massofwaterat{4}^{\circ }Cofvolumeequaltothatofthebody}$Using Archimedes' principle, the mass of water of volume equal to that of the body is obtained by finding the mass of water displaced by that body when it is completely immersed in water since a body when immersed in water, displaces water equal to its own volume. Therefore,$RelativeDensity\left(R.D\right)=\frac{Massofthebody}{Massofwaterdiplacedbythebody}=\frac{Weightofthebody}{Massofwaterdiplacedbythebody}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{Weightofthebody}{Lossofweightofthebodyinwater\left(orupthrust\right)}=\frac{Weightofbodyinair}{Weightofbodyinair-Weightofbodyinwater}\phantom{\rule{0ex}{0ex}}$Hence, to find the relative density of a solid body using Archimedes' principle, we can use this formula $R.D.=\frac{{W}_{1}}{{W}_{1}-{W}_{2}}$.

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