A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle ≈ 0°) with 30° inclined with the vertical. The length of water risen in the capillary is _______ cm. (Take g = 9.8 m/s²)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q32 – JEE-MAIN 2025

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Home Physics Mechanical Properties of FluidsQuestion #275

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 the vertical. The length of water risen in the capillary is _______ cm. (Take

g = 9.8 {m/s}^{2}

)

Clue 1 of 3: Capillary Rise in Inclined Tube

The vertical height of liquid rise in a capillary tube is independent of its inclination, but the length of the liquid column along the tube depends on the inclination.

🥷

Ninja Strategy•Relative Magnitude Check

First, calculate the vertical capillary rise ( $h$ ). Then, use the relationship $L = h / \cos α$ to understand that $L$ must be greater than $h$ (since $\cos α < 1$ for $α > 0$ ) and eliminate options that are too small or disproportionately large.

🧠 Full Solution Path

Step 1: Calculate the vertical capillary rise✦ Active

The vertical height of water rise in a capillary tube is given by the formula $h = \frac{2 T \cos θ}{r ρ g}$ . First, convert all given values to CGS units for consistency. Given radius $r = 0.1 mm = 0.01 cm$ , surface tension $T = 70 dyn/cm$ , contact angle $θ = 0^{\circ}$ (so $\cos θ = 1$ ), density of water $ρ = 1 {g/cm}^{3}$ , and acceleration due to gravity $g = 9.8 {m/s}^{2} = 980 {cm/s}^{2}$ .

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} cm

Step 2: Calculate the length of water risen in the inclined tube○ Expand

The capillary tube is inclined at $30^{\circ}$ with the vertical. If $h$ is the vertical rise, the length $L$ of the water column along the inclined tube is given by $L = \frac{h}{\cos α}$ , where $α$ is the angle of inclination with the vertical. Here, $α = 30^{\circ}$ , so $\cos α = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$ .

L = \frac{100 / 7}{\sqrt{3} / 2} = \frac{100}{7} \times \frac{2}{\sqrt{3}} = \frac{200}{7 \sqrt{3}} cm

Step 3: Evaluate the numerical value and match with options○ Expand

Using the approximation $\sqrt{3} \approx 1.732$ , we can calculate the numerical value of $L$ :

L = \frac{200}{7 \times 1.732} = \frac{200}{12.124} \approx 16.495 cm

Comparing this value with the given options: 1) $\frac{71}{5} = 14.2 cm$ 2) $\frac{82}{5} = 16.4 cm$ 3) $\frac{57}{2} = 28.5 cm$ 4) $\frac{68}{5} = 13.6 cm$ The calculated value $16.495 cm$ is closest to $16.4 cm$ , which is option 2.

💡 Teacher's Secret Hint

Be careful with unit conversions and approximations of constants like $\sqrt{3}$ to match the options accurately.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 32 – JEE-MAIN 2025

Clue 1 of 3: Capillary Rise in Inclined Tube

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