Question

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

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

Answer 1

Answer: (B)

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, $L=10.336m$