Question

A tall cylinder with a cross-sectional area 13.0 cm2 is partially filled with mercury; the surface of the mercury is 6.00 cm above the bottom 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 = 1060.8 cm³

Explanation:

we know that the pressure,

  P = density  x  gravity  x  depth

ρm is the density of mercury

ρw is the density of water

the pressure due to mercury P₁= (ρm) g h₁

the pressure due to water P₂ = (ρw) g h₂

the total pressure

P = P₁ + P₂

P =  (ρm) g h₁  + (ρw) g h₂

but the total pressure is double the pressure due to mercury.  

(ρm) g h₁ =(ρw) g h₂

$h_2 = \dfrac{\rho_m\times h_1}{\rho_w}$

$h_2 =13.6\times 6$

h₂ = 81.6 cm

the height of the water is 81.6 cm

the volume

V = height  x area

V = 81.6 x 13

V = 1060.8 cm³

the volume of water must be added to double the gauge pressure is V = 1060.8 cm³