Question #217616
A U-tube manometer has Mercury of relative density off 13.68 as its manometric liquid. Kerosene of relative density of 0.6 was poured into one of the arms and so a drop of 2.0 cm results in that arm. Also, water with a relative density of 1g/cm3 was poured into the other arm and so the two arms became the same again. Calculate 1 the height of the kerosene to the height of the water column in details.
1
Expert's answer
2021-07-16T10:19:48-0400

As per the given question,

Density of mercury (ρ)=13.68(\rho)=13.68

Relative density of kerosene (ρk)=0.6(\rho_k)=0.6

Drop in height (ΔH)=2cm(\Delta H)= 2 cm

Relative density of water (ρw)=1g/cm3(\rho_w)=1g/cm^3

Height of kerosene =?

Height of water = ?

ρ1h1=ρ2h2\rho_1 h_1=\rho_2h_2

Now, substituting the values,

h1=ρ2h2ρ1h_1=\frac{\rho_2 h_2}{\rho_1}


=0.6×213.68cm=\frac{0.6\times 2}{13.68}cm

=0.0877cm=0.0877cm


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