Question

14. The liquid in the tube has a density of 13,600 kilograms per cubicmeter. The atmospheric pressure is 1.013 x 10^5 newtons persquare meter. The pressure P is most nearly(a) 1.05 x 10^5 pa(b) 1.25 x 10^5 Pa(c) 1.33 x 10^5 pa(d) 1.47 x 10^5 Pa

Answer 1

Given,

The density of the liquid,ρ=13600 kg/m^3

The atmospheric pressure, P₁=1.013×10⁵

The difference between the height of the liquid in the column, h=0.240 m

The pressure P is given by,

$P=P_1+g\rho h$

Where g is the acceleration due to gravity.

On substituting the known values,

$P=1.013\times10^5+9.8\times13600\times0.240=1.33\times10^5\text{ Pa}$

Hence pressure, P=1.33×10⁵ Pa