Question

As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury barometer drops by 21.6 mm from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013 ✕ 105 Pa and the density of mercury is 13.6 g/cm3.

Answer 1

Answer:

new atmospheric pressure is 0.9838 × $10^{5}$  Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × $10^{5}$  = 13.6 × 10³ × 9.81 × h

h =  0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × $10^{5}$  Pa