Answer to Question #205187 in Mechanics | Relativity for :-)

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=mg-\\rho gV"

where mg is the weight of the bar,

"\\rho" is density of liquid,

V is volume of the bar.


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


For unknown liquid:

"\\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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