The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of thier 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 :

Properties of Matter PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q64 with Solution

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Correct answer: A

For a liquid in a capillary tube, $h = \frac{2T\cos\theta}{\rho g r}$ where:

For water, $\theta_w \approx 0^\circ$ , so $\cos 0^\circ = 1$ Hence rise of water is $h = \frac{2T_w}{\rho_w g r_2}$

For mercury, $\theta_m \approx 135^\circ$ , so $\cos 135^\circ = -\frac{1}{\sqrt{2}}$ Since mercury is depressed, we take magnitude of depression:

= \frac{2T_m}{\rho_m g r_1}\cdot \frac{1}{\sqrt{2}}$$ 2. **Given both magnitudes are equal** So, $$\frac{2T_m}{\rho_m g r_1}\cdot \frac{1}{\sqrt{2}} = \frac{2T_w}{\rho_w g r_2}$$ Cancel $2$ and $g$: $$\frac{T_m}{\rho_m r_1}\cdot \frac{1}{\sqrt{2}} = \frac{T_w}{\rho_w r_2}$$ Rearrange for $\dfrac{r_1}{r_2}$: $$\frac{r_1}{r_2} = \frac{T_m}{T_w}\cdot \frac{\rho_w}{\rho_m}\cdot \frac{1}{\sqrt{2}}$$ 3. **Substitute given ratios** Given: $$\frac{T_m}{T_w} = 7.5, \qquad \frac{\rho_m}{\rho_w} = 13.6$$ Thus, $$\frac{\rho_w}{\rho_m} = \frac{1}{13.6}$$ So, $$\frac{r_1}{r_2} = 7.5\cdot \frac{1}{13.6}\cdot \frac{1}{\sqrt{2}}$$ Now, $$\frac{7.5}{13.6} \approx 0.5515, \qquad \frac{1}{\sqrt{2}} \approx 0.707$$ Therefore, $$\frac{r_1}{r_2} \approx 0.5515 \times 0.707 \approx 0.39$$ 4. **Match with options** $$0.39 \approx \frac{2}{5} = 0.4$$ So the closest option is **A**. 5. **Final answer** $$\boxed{\frac{r_1}{r_2} \approx \frac{2}{5}}$$

View written solutionFree

Correct answer: A

For a liquid in a capillary tube, $h = \frac{2T\cos\theta}{\rho g r}$ where:

For water, $\theta_w \approx 0^\circ$ , so $\cos 0^\circ = 1$ Hence rise of water is $h = \frac{2T_w}{\rho_w g r_2}$

For mercury, $\theta_m \approx 135^\circ$ , so $\cos 135^\circ = -\frac{1}{\sqrt{2}}$ Since mercury is depressed, we take magnitude of depression:

= \frac{2T_m}{\rho_m g r_1}\cdot \frac{1}{\sqrt{2}}$$ 2. **Given both magnitudes are equal** So, $$\frac{2T_m}{\rho_m g r_1}\cdot \frac{1}{\sqrt{2}} = \frac{2T_w}{\rho_w g r_2}$$ Cancel $2$ and $g$: $$\frac{T_m}{\rho_m r_1}\cdot \frac{1}{\sqrt{2}} = \frac{T_w}{\rho_w r_2}$$ Rearrange for $\dfrac{r_1}{r_2}$: $$\frac{r_1}{r_2} = \frac{T_m}{T_w}\cdot \frac{\rho_w}{\rho_m}\cdot \frac{1}{\sqrt{2}}$$ 3. **Substitute given ratios** Given: $$\frac{T_m}{T_w} = 7.5, \qquad \frac{\rho_m}{\rho_w} = 13.6$$ Thus, $$\frac{\rho_w}{\rho_m} = \frac{1}{13.6}$$ So, $$\frac{r_1}{r_2} = 7.5\cdot \frac{1}{13.6}\cdot \frac{1}{\sqrt{2}}$$ Now, $$\frac{7.5}{13.6} \approx 0.5515, \qquad \frac{1}{\sqrt{2}} \approx 0.707$$ Therefore, $$\frac{r_1}{r_2} \approx 0.5515 \times 0.707 \approx 0.39$$ 4. **Match with options** $$0.39 \approx \frac{2}{5} = 0.4$$ So the closest option is **A**. 5. **Final answer** $$\boxed{\frac{r_1}{r_2} \approx \frac{2}{5}}$$