Question #205187

Ben has a gold bar that weighs 72N. He then suspended the gold bar by a light cord. When the gold bar is fully immersed in water (ρwater = 1.00 ×10^3 kg/m^-3), the tension in the cord is 40N. When the gold bar is fully immersed in an unknown liquid, the tension is 25N. What is the density of the unknown liquid?


Expert's answer

The tension in the cord:

T=mgρgVT=mg-\rho gV

where mg is the weight of the bar,

ρ\rho is density of liquid,

V is volume of the bar.


V=mgTρg=724010009.8=3.27103 m3V=\frac{mg-T}{\rho g}=\frac{72-40}{1000\cdot9.8}=3.27\cdot10^{-3}\ m^3


For unknown liquid:

ρ=mgTgV=72259.83.27103=1.467102 kg/m3\rho=\frac{mg-T}{gV}=\frac{72-25}{9.8\cdot3.27\cdot10^{-3}}=1.467\cdot10^2 \ kg/m^3


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