Question

The normal atmospheric pressure can support water column of height L which is, (given density of mercury = $13600\, kg\, m^{-3}$)

• A. 0.76 m
• B. 10.336 m
• C. 103.6 m
• D. 76 m

Answer 1

Answer: (B)

10.336 m

Answer 2

Here, 1 atm = 0.76 m of Hg = L m of water,
$\therefore \:\:\: 0.76 \, \rho_{Hg} g = L \times \rho_w g$
or, $0.76 \times 13600 = L \times 1000$
or, $L = 10.336\, m$