MCQMediumJEE 2025Surface Tension & Capillarity

JEE Physics 2025 Question with Solution

A capillary tube of radius 0.1mm0.1 \, \text{mm} is partly dipped in water (surface tension 70dyn/cm70 \, \text{dyn/cm} and glass water contact angle 0\approx 0^\circ) with 3030^\circ inclined with vertical. The length of water risen in the capillary is _____ cm. (Take g=9.8m/s2g = 9.8 \, \text{m/s}^2)

  • A

    825\frac{82}{5}

  • B

    572\frac{57}{2}

  • C

    715\frac{71}{5}

  • D

    685\frac{68}{5}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: surface tension T=70dyn/cm=0.07N/mT = 70 \, \text{dyn/cm} = 0.07 \, \text{N/m}, contact angle θ0\theta \approx 0^\circ so cosθ=1\cos \theta = 1, radius r=0.1mm=0.0001mr = 0.1 \, \text{mm} = 0.0001 \, \text{m}, density of water ρ=1000kg/m3\rho = 1000 \, \text{kg/m}^3, and g=9.8m/s2g = 9.8 \, \text{m/s}^2.

Find: the length of water risen along the capillary tube when the tube is inclined at 3030^\circ with the vertical.

Using Jurin's law, the vertical rise is

h=2Tcosθrρgh = \frac{2T \cos \theta}{r \rho g}

Substituting the given values,

h=2×0.07×10.0001×1000×9.8h = \frac{2 \times 0.07 \times 1}{0.0001 \times 1000 \times 9.8} =0.140.98= \frac{0.14}{0.98} =0.142857m=14.2857cm= 0.142857 \, \text{m} = 14.2857 \, \text{cm}

Since the capillary is inclined at 3030^\circ with the vertical, if ll is the length of the liquid column along the tube, then

h=lcos30h = l \cos 30^\circ

So,

l=hcos30l = \frac{h}{\cos 30^\circ} l=14.28573/216.5cml = \frac{14.2857}{\sqrt{3}/2} \approx 16.5 \, \text{cm}

This is closest to

825=16.4cm\frac{82}{5} = 16.4 \, \text{cm}

Therefore, the correct option is A.

Unit conversion and inclination relation

Given: the tube radius is 0.1mm0.1 \, \text{mm}, the surface tension is 70dyn/cm70 \, \text{dyn/cm}, the contact angle is 00^\circ, and the tube makes 3030^\circ with the vertical.

Find: the length of rise measured along the inclined tube.

The key step is to keep units consistent. Convert

70dyn/cm=0.07N/m70 \, \text{dyn/cm} = 0.07 \, \text{N/m}

and

0.1mm=0.01cm=104m0.1 \, \text{mm} = 0.01 \, \text{cm} = 10^{-4} \, \text{m}

Now apply capillary rise formula for vertical height:

h=2Tcosθrρgh = \frac{2T \cos \theta}{r \rho g}

Because θ=0\theta = 0^\circ,

cos0=1\cos 0^\circ = 1

Hence,

h=2×0.07104×1000×9.8h = \frac{2 \times 0.07}{10^{-4} \times 1000 \times 9.8} =0.140.98= \frac{0.14}{0.98} =0.142857m= 0.142857 \, \text{m} =14.2857cm= 14.2857 \, \text{cm}

This is the vertical rise. The tube is inclined, so the actual liquid length inside the tube is larger. Since the inclination is given with the vertical,

lcos30=hl \cos 30^\circ = h

Therefore,

l=14.28570.86616.5cml = \frac{14.2857}{0.866} \approx 16.5 \, \text{cm}

Among the given options, this matches 825\frac{82}{5} most closely.

Therefore, the length of water risen in the capillary is 825cm\frac{82}{5} \, \text{cm}, so the correct option is A.

The other extracted approach on the page uses an inconsistent numerical evaluation and the relation hsin30\frac{h}{\sin 30^\circ}, which disagrees with the geometry for an angle measured with the vertical.

Common mistakes

  • Using l=hsin30l = \frac{h}{\sin 30^\circ} is incorrect here because the tube is inclined with the vertical, not with the horizontal. The vertical component of the liquid length is lcos30l \cos 30^\circ, so use l=hcos30l = \frac{h}{\cos 30^\circ} instead.

  • Converting the radius wrongly can change the answer by a factor of 1010. Here 0.1mm=104m=0.01cm0.1 \, \text{mm} = 10^{-4} \, \text{m} = 0.01 \, \text{cm}, not 103m10^{-3} \, \text{m}.

  • For water in a clean glass capillary, the contact angle is approximately 00^\circ, so cosθ=1\cos \theta = 1. Replacing it with another trigonometric value without justification gives an incorrect rise.

Practice more Surface Tension & Capillarity questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions