The water would stand at a height of 1033.6 cm
If the mass of the mercury, m equals the mass of the water, m' at its height,h', then the weight of the mercury W equals the weight of the water, W'
So, W = W'
mg = m'g
ρV = ρ'V' where ρ = density of mercury = 13.60 g/cm³, V = volume of mercury = Ah where A = cross-sectional area of tube = 1.00 cm and h = height of mercury = 76.0 cm, ρ' = density of water = 1.00 g/cm³, V = volume of water = Ah' where A = cross-sectional area of tube = 1.00 cm and h' = height of water.
So, ρV = ρ'V'
ρAh = ρ'Ah'
h' = ρAh/ρ'A
h' = ρh/ρ'
Substituting the values of the variables into the equation, we have
h' = ρh/ρ'
h' = 13.60 g/cm³ × 76.0 cm/1.00 g/cm³
h' = 13.60 × 76.0 cm
h' = 1033.6 cm
So, the water would stand at a height of 1033.6 cm
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