Assuming the density of air to be 1.295 kg m^-3, find the fall in barometric height in mm of Hg at a height of 107 m above the sea level. Take density of mercury = 13.6 x 10³ kg m^-3

As we know,

decrease in pressure = h ρ g

Given,

density of air = 1.295 kg m^-3

h = 107 m

Therefore,

Decrease in pressure = (107) x (1.295) x (g)    [Equation 1]

Let, decrease in mercury height = H

Therefore, decrease in barometric height = (H) x (13.6 x 10³) x(g)   [Equation 2]

Equating 1 and 2 we get,

$107 \times 1.295 \times g = H \times 13.6 \times 10^{3} \times g \\[0.5em] \Rightarrow H = \dfrac{107 \times 1.295}{13.6 \times 10^{3}} \\[0.5em] \Rightarrow H = 0.0101 \text { m of Hg}$

Converting m into mm, we get

1 m = 1000 mm

Therefore, 0.0101 m = 0.0101 m x 1000 = 10.1 mm

Therefore, fall in barometric height = 10 mm of Hg.