Question

A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle $ \approx 0^\circ $) with $ 30^\circ $ inclined with vertical. The length of water risen in the capillary is ____ cm. (Take $ g = 9.8 $ m/s$^2 $)

• A. \( \frac{82}{5} \)
• B. \( \frac{57}{2} \)
• C. \( \frac{71}{5} \)
• D. \( \frac{68}{5} \)

Hint:

Ensure consistent units throughout the calculation. The vertical height of the liquid column in the capillary is determined by the Jurin's law. When the capillary is inclined, the length of the liquid column along the tube is related to the vertical height through trigonometric relations involving the angle of inclination.

Answer 1

The Correct Option is A

Step 1: Identify the given parameters and convert them to consistent units.
Radius of the capillary tube, \( r = 0.1 \, \text{mm} = 0.01 \, \text{cm} \) Surface tension of water, \( T = 70 \, \text{dyn/cm} \) Contact angle, \( \theta \approx 0^\circ \) Angle of inclination of the capillary tube with the vertical, \( \alpha = 30^\circ \) Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 = 980 \, \text{cm/s}^2 \) Density of water, \( \rho = 1 \, \text{g/cm}^3 \)
Step 2: Determine the vertical height \( h \) of the water risen in the capillary tube.
The formula for the height of the liquid risen in a capillary tube is given by: \[ h = \frac{2T \cos \theta}{r \rho g} \] Substituting the given values: \[ h = \frac{2 \times 70 \times \cos(0^\circ)}{0.01 \times 1 \times 980} = \frac{140 \times 1}{9.8} = \frac{1400}{98} = \frac{100}{7} \, \text{cm} \]
Step 3: Relate the vertical height \( h \) to the length \( l \) of the water risen along the inclined capillary tube.
Let \( l \) be the length of the water risen in the capillary tube. From the geometry of the situation, we have: \[ h = l \cos \alpha \] where \( \alpha \) is the angle of inclination of the capillary tube with the vertical. Given \( \alpha = 30^\circ \), we have \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \). So, \[ l = \frac{h}{\cos \alpha} = \frac{h}{\cos(30^\circ)} = \frac{100/7}{\sqrt{3}/2} = \frac{200}{7\sqrt{3}} \, \text{cm} \] There seems to be a mistake in the calculation or the provided options, as the derived value does not match any of them. Let's recheck the steps. Revisiting
Step 2:
\[ h = \frac{2 \times 70 \times \cos(0^\circ)}{0.01 \times 1 \times 980} = \frac{140}{9.8} = \frac{1400}{98} = \frac{100}{7} \, \text{cm} \] Revisiting
Step 3:
The angle of inclination with the vertical is \( 30^\circ \). The vertical height \( h \) is related to the length along the tube \( l \) by \( h = l \cos(30^\circ) \). So, \( l = \frac{h}{\cos(30^\circ)} = \frac{100/7}{\sqrt{3}/2} = \frac{200}{7\sqrt{3}} \approx \frac{200}{7 \times 1.732} \approx \frac{200}{12.124} \approx 16.5 \, \text{cm} \). This still does not match the options. Let's assume there might be a slight misinterpretation of the question or a potential error in the options. Let's reconsider the problem. The vertical height risen is \( h = \frac{100}{7} \) cm. The length risen along the inclined tube \( l \) is such that its vertical component is \( h \).
Therefore, \( h = l \cos(30^\circ) \). \( l = \frac{h}{\cos(30^\circ)} = \frac{100/7}{\sqrt{3}/2} = \frac{200}{7\sqrt{3}} \). If the question meant the angle with the horizontal, then \( h = l \sin(30^\circ) \), and \( l = \frac{h}{\sin(30^\circ)} = \frac{100/7}{1/2} = \frac{200}{7} \approx 28.57 \). This also doesn't match. Let's check the units carefully. Radius in cm, surface tension in dyn/cm, density in g/cm\(^3\), g in cm/s\(^2\). The height \( h \) will be in cm. Let's assume there was a calculation error in the options and try to match the closest value. \( \frac{200}{7\sqrt{3}} \approx 16.5 \). None of the options are close to this. Let's re-read the question carefully: "with \( 30^\circ \) inclined with vertical." We have \( h = \frac{100}{7} \) cm. The length along the tube \( l \) is given by \( h = l \cos(30^\circ) \). \( l = \frac{h}{\cos(30^\circ)} = \frac{100/7}{\sqrt{3}/2} = \frac{200}{7\sqrt{3}} \). There might be an error in the question or the provided options. However, if we were to choose the closest option based on a potential approximation or a different interpretation, it would be challenging. Let's double-check the calculation for \( h \). \( h = \frac{2 \times 70 \times 1}{0.01 \times 1 \times 980} = \frac{140}{9.8} = \frac{1400}{98} = \frac{100}{7} \) cm. Let's check the options' decimal values: (1) \( \frac{82}{5} = 16.4 \) (2) \( \frac{57}{2} = 28.5 \) (3) \( \frac{71}{5} = 14.2 \) (4) \( \frac{68}{5} = 13.6 \) The value \( \frac{200}{7\sqrt{3}} \approx 16.5 \) is closest to option (1) \( \frac{82}{5} = 16.4 \). There might be a rounding error or a slight difference in the values used.