Question

A tall cylinder with a cross-sectional area 12.0 cm² is partially filled with mercury; the surface of the mercury is 5.00 cm above the bottome of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure?

Answer 1

Answer:

$V = 816 cm^3$

Explanation:

As we know that gauge pressure of the fluid at the bottom of the cylinder is given as

$P = \rho g h$

now we know that pressure at the bottom is double when water is poured on the mercury

So we have

$\rho_{hg} g h_1 = \rho_w g h_2$

so we will have

$13.6 \times 5 = 1 \times h$

so we have

$h = 68 cm$

now the volume of the water added to it is given as

$V = h A$

$V = (68 cm)(12 cm^2)$

$V = 816 cm^3$