Question

Two narrow bores of diameters $3\text{mm}$ and $6\text{mm}$ are joined together to form a $\text{u-shaped}$ tube open at both ends. If the u - tube contains water, the difference in it's level of water in the two limbs will be $($Take surface tension of water $=7.3\times {10}^{-2}\text{N/m}$, angle of contact $=0^{\circ }$ and density of water = $1000{\text{kg/m}}^3)$

• A. $6\text{mm}$
• B. $5\text{mm}$
• C. $2\text{cm}$
• D. $3\text{mm}$

Answer 1

The correct option is B $5\text{mm}$

We know that capillary rise,

$h=\frac{4T\cos \theta }{\rho gd}$

where, $d$ is diameter of capillary tube, $T$ is surface tension and $\theta$ is angle of contact.

Capillary rise in each bore,

$h_1=\frac{4T\cos \theta }{\rho g{d}_1}$ and $h_2=\frac{4T\cos \theta }{\rho g{d}_2}$

Given,
$T=7.3\times {10}^{-2}\text{N/m}$
$\theta =0^{\circ }$
diameter of first bore, $d_1=3\text{mm}$
diameter of second bore, $d_2=6\text{mm}$

Now, difference in water level of the two limbs,

$\Delta h=h_1-h_2$

$\Rightarrow \Delta h=\frac{4T\cos \theta }{\rho g}[\frac{1}{d_1}-\frac{1}{d_2}]$

Substituting the values,
$\Rightarrow \Delta h=\frac{4\times 7.3\times {10}^{-2}\times 1}{1000\times 10}[\frac{1}{3\times {10}^{-3}}-\frac{1}{6\times {10}^{-3}}]$

$\Rightarrow \Delta h=4.87\text{mm}\approx 5\text{mm}$

Hence option $\left(b\right)$ is correct answer.