The ratio of surface tensions of mercury and water is given to be 7.5… | JEE Main 2019 (10 Apr, Shift 1)

Q61JEE Main 2019 · 10 APR · SHIFT 1Medium

Angle of Contact & Capillarity

The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2. The ratio, (r1/r2), is then close to:

Show answer and solution

Answer: (a)

$h = \frac{2T|\cos\theta|}{r\rho g}$ for both, so $r \propto \frac{T|\cos\theta|}{\rho}$ for the same $h$ .
$\frac{r_1}{r_2} = \frac{T_m}{T_w} \times \frac{\rho_w}{\rho_m} \times \frac{|\cos 135^\circ|}{\cos 0^\circ} = \frac{7.5}{13.6} \times 0.707$
$\frac{r_1}{r_2} \approx 0.39 \approx \frac{2}{5}$
So the answer is (a) $\frac{2}{5}$ .

Report a mistake

Angle of Contact & Capillarity question from JEE Main 2019 (10 Apr, Shift 1)

More questions on Angle of Contact & Capillarity