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Suppose to a scientist was able to construct a barometer with a liquid being twice denser than mercury, then how would the liquid raise at standard pressure?​

Sagot :

Answer:

           h = h₀/2,  the liquid column rises half the height

Explanation:

Pressure is defined as force per area,  

         P = F / A

in a barometer the force is the weights of the liquid column

         F = W = mg

         P = m g/A

if we use the definition of density

         ρ = [tex]\frac{m}{A h}[/tex]

         [tex]\frac{m}{A}[/tex] = ρ h

we substitute

         P = ρ g h

Let's use the index o for the mercury

         P₀ = ρ₀ g h₀

when we change the liquid for another with

          ρ = 2ρ₀

the pressure expression is

         P = 2ρ₀ g h

in the problem they indicate that the pressure is equal to the initial P = P₀

we substitute

           ρ₀ g h₀ = 2ρ₀ g h

           h₀ = 2h

           h = h₀/2

the liquid column rises half the height