Question

Two narrow bores of diameter $5.0mm$ and $8.0mm$ are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube.
[Take surface tension of water $T=7.3\times 10^{-2}N{m}^{-1}$, angle of contact $=0,g=10m{s}^{-2}$ and density of water $=1.0\times 10^{3}kg{m}^{-3}$ ]

• A. 3.62 mm
• B. 2.19 mm
• C. 5.34 mm
• D. 4.97 mm

Answer 1

Answer: (B)

We have $P_A=P_B$.
[Points $A$ & $B$ at same horizontal level]
$\therefore P_{\text{atm }}-\frac{2T}{r_1}+\rho g(x+\Delta h)=P_{\text{atm }}-\frac{2T}{r_2}+\rho gx$
$\therefore \rho g\Delta h=2T\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$
$=2\times 7.3\times 10^{-2}\left[\frac{1}{2.5\times 10^{-3}}-\frac{1}{4\times 10^{-3}}\right]$
$\therefore \Delta h=\frac{2\times 7.3\times 10^{-2}\times 10^3}{10^3\times 10}\left[\frac{1}{2.5}-\frac{1}{4}\right]$
$=2.19\times 10^{-3}m=2.19mm$