Question

In a U-shaped tube the radius of one limb is \(2\;mm\) and that of other limb is \(4\;mm\). A liquid of surface tension \(0.03\;N m^{-1}\), density \(1500\;kg m^{-3}\) and angle of contact zero is taken in the tube. The difference in the heights of the levels of the liquid in the two limbs is
Take \(g=10\;\text{m s}^{-2}\).

• A. \(3\;mm\)
• B. \(2.5\;mm\)
• C. \(1\;mm\)
• D. \(1.5\;mm\)

Hint:

In capillary rise problems, smaller radius gives greater rise. For two limbs, use \[ \Delta h=\frac{2T\cos\theta}{\rho g}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \]

Answer 1

The Correct Option is C

Step 1: Use capillary rise formula.
The capillary rise in a tube of radius \(r\) is \[ h=\frac{2T\cos\theta}{\rho gr} \] Since the angle of contact is zero, \[ \theta=0^\circ \] So, \[ \cos0^\circ=1 \] Hence, \[ h=\frac{2T}{\rho gr} \]

Step 2: Write the height difference formula.
For two limbs of radii \(r_1\) and \(r_2\), the difference in liquid levels is \[ \Delta h=\frac{2T}{\rho g}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \] Given, \[ T=0.03\;N m^{-1} \] \[ \rho=1500\;kg m^{-3} \] \[ g=10\;\text{m s}^{-2} \] \[ r_1=2\;mm=2\times 10^{-3}\;m \] \[ r_2=4\;mm=4\times 10^{-3}\;m \]

Step 3: Substitute the values.
\[ \Delta h=\frac{2(0.03)}{1500\times 10} \left(\frac{1}{2\times 10^{-3}}-\frac{1}{4\times 10^{-3}}\right) \] \[ =\frac{0.06}{15000}(500-250) \] \[ =\frac{0.06}{15000}\times 250 \] \[ =0.001\;m \]

Step 4: Convert into millimetres.
\[ 0.001\;m=1\;mm \]

Step 5: Final conclusion.
Hence, the difference in heights is \[ \boxed{1\;mm} \]