At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in barometer is replaced by water ? Take density of mercury to be 13.6 x 10³ kg m^-3

We know that,

Pressure = h ρ g

Given,

ρ = 13.6 x 10³ kg m^-3

g = 9.8 m s ^-2

h = 0.70 m

Substituting the values in the formula above we get,

$P = (0.70) \times (13.6 \times 10^{3}) \times (9.8) \\[0.5em] \Rightarrow P = 93.3 \times 10^{3} \text { Pa} \\[0.5em]$

Hence, P = 93.3 x 10³Pa

Now, if we use a water barometer,

P = 93.3 x 10³ Pa

ρ = 1 x 10³ kg m^-3

g = 9.8 m s ^-2

Substituting the values in the formula above we get,

$93.3 \times 10^{3} = (h) \times (1 \times 10^{3}) \times (9.8) \\[0.5em] h = \dfrac{93.3 \times 10^{3}}{1 \times 10^{3} \times 9.8} \\[0.5em] h = 9.52 \text { m} \\[0.5em]$

Hence, height of water colum = 9.52 m