In a U-tube, two immiscible liquids of density ` rho _ 1 and rho _ 2 ` are arranged as shown in figure. The right part of the tube is closed at the top and some air is entrapped above liquid of density ` rho _2 `. The gauge pressure of this air is :

Options

A. Zero
B. ` (rho_2 - rho_1 ) gh `
C. ` (rho_1 - rho_2) gh`
D. Not possible

Answer

C

Explanation

The pressure of entrapped air with respect to atmospheric pressure is called the gauge pressure. Writing pressure relation starting from the open end, we get ` P _ 0 + rho _ 1 gh - rho _ 2 gh = P_(air) `
Therefore `, P_(air ) - P _ 0 = (rho _1 - rho _2 ) gh` (gauge)