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?


1
Expert's answer
2021-06-10T10:22:50-0400

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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