Answer:
0.49 cm
Explanation:
We are given that
[tex]A_1=10 cm^2[/tex]
[tex]A_2=5 cm^2[/tex]
Mass of water,m=100 g
a.[tex]\rho_w=1 g/cm^3[/tex]
Volume,[tex]V=\frac{mass}{density}=\frac{100}{1}=100 cm^3[/tex]
Length of water column in the right arm=[tex]L=\frac{V}{A_2}=\frac{100}{5}=20 cm[/tex]
b.[tex]\rho_m=13.6 g/cm^3[/tex]
In equilibrium condition
Pressure at point A=Pressure at point B
[tex]P+\rho_mg(h+h_2)=P+\rho_wgL[/tex]
[tex]hA_1=h_2A_2[/tex]
[tex]h_2=\frac{hA_1}{A_2}[/tex]
[tex]13.6\times 9.8(h+h(\frac{A_1}{A_2}))=1\times 9.8\times 20[/tex]
[tex]13.6\times h(\frac{A_2+A_1}{A_2})=20[/tex]
[tex]h=\frac{20A_2}{13.6(A_2+A_1)}[/tex]
[tex]h=\frac{20\times 5}{13.6\times (10+5)}=0.49 cm[/tex]
