Question

Two narrow bores of diameter $5.0\text{mm}$ and $8.0\text{mm}$ 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.

$\text{Take surface tension of water,}T=7.3\times {10}^{-2}{\text{Nm}}^{-1}$
$\text{angle of contact}=0$
$g=10{\text{ms}}^{-2}$
$\text{density of water}=1.0\times {10}^3{\text{kgm}}^{-3}$

• A. $4.97\text{mm}$
• B. $3.62\text{mm}$
• C. $5.34\text{mm}$
• D. $2.19\text{mm}$

Answer 1

The correct option is D $2.19\text{mm}$

$d_1=5\text{mm},$ $d_2=8\text{mm}$
$T=7.3\times {10}^{-2}{\text{Nm}}^{-1}$
$\theta =0^{\circ }$
$g=10{\text{ms}}^{-2}$
$\rho =1.0\times {10}^3{\text{kgm}}^{-3}$

The level difference between the two limbs of the U-tube will be equal to the difference between the capillary rise in the two limbs.

Now, the capillary rise is given by, $h=\frac{2T\cos \theta }{r\rho g}$
$\therefore h_1=\frac{2\times 7.3\times {10}^{-2}\times 1}{2.5\times {10}^{-3}\times {10}^3\times 10}=5.84\text{mm}$

And, $h_2=\frac{2\times 7.3\times {10}^{-2}\times 1}{4\times {10}^{-3}\times {10}^3\times 10}=3.65\text{mm}$

$\therefore$ The level difference is,
$\Delta h=h_1-h_2=5.84-3.65=2.19\text{mm}$

Hence, $\left(D\right)$ is the correct answer