Question

The heights of mercury surfaces in the two arms of the manometer shown in figure are 2 cm and 8 cm.
Atmospheric pressure = 1.01 × 105 N−2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.
Figure

Answer 1

Explanation:

Given data,

Atmospheric pressure = 1.01 × 105 N/m−2

h1 = 2cm; h2 = 8cm

(a) The pressure of the gas in the cylinder

• P    = P0 +  ρhg
• [P0 = Atmospheric pressure  
• h = (8cm – 2 cm) = 6 cm = 6× 10-2 m
• ρ = density of mercury = 13600 kg/m3
• g = gravitational field strength = 10ms-²

=  (1.01 × 105) + (13600× 6×10-2 × 10)  

= (1.01 × 105) + (0.08× 105)

= 1.09 × 105 Pa or N/m2.

(b) The pressure of mercury at the bottom of the U tube

• P    = P0 +  ρhg

• [for finding pressure at the bottom of the tube, the total height should be considered in the calculation => h = 8cm = 8× 10-2 m.]

   =  (1.01 × 105) + (13600  8×10-2 × 10)  

= (1.01 × 105) + (0.11× 105)

= 1.12 × 105 Pa or N/m2.