Question

Two narrow bores of diameter $5.0\, mm$ and $8.0 \,mm$ 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 $T =7.3 \times 10^{-2} Nm ^{-1}$, angle of contact $=0, g =10\, ms ^{-2}$ 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]
$\therefore P _{\text{atm }}-\frac{2 T }{ r _{1}}+\rho g ( x +\Delta h )= P _{ \text{atm }}-\frac{2 T }{ r _{2}}+\rho gx$
$\therefore \rho g \Delta h =2 T \left[\frac{1}{ r _{1}}-\frac{1}{ r _{2}}\right] $
$=2 \times 7.3 \times 10^{-2}\left[\frac{1}{2.5 \times 10^{-3}}-\frac{1}{4 \times 10^{-3}}\right]$
$\therefore \Delta h =\frac{2 \times 7.3 \times 10^{-2} \times 10^{3}}{10^{3} \times 10}\left[\frac{1}{2.5}-\frac{1}{4}\right]$
$=2.19 \times 10^{-3} \,m =2.19 \,mm$