JEE Main

Conducting Body: National Testing Agency (NTA)
For: Undergraduate engineering, architecture, and planning aspirants
Papers: B.E./B.Tech, B.Arch, B.Plan
Subjects (B.E./B.Tech): Physics, Chemistry, Mathematics
Question Types: MCQs and Numerical Answer Type questions
Duration: 3 hours per paper
Marking Scheme: Positive marks for correct answers; negative marking for incorrect MCQs
Frequency: Multiple sessions per year
Qualifies for: Admission to NITs, IIITs, CFTIs; eligibility for JEE Advanced

The Joint Entrance Examination (JEE) Main is a national-level entrance test in India for students seeking admission to undergraduate engineering, architecture, and planning courses. It is primarily taken by students after completing their Class 12 science education. The exam serves as the admission criterion for institutions like the National Institutes of Technology (NITs) and Indian Institutes of Information Technology (IIITs). It is also the mandatory qualifying step for JEE Advanced, the entrance test for the Indian Institutes of Technology (IITs).

The B.E./B.Tech paper is a computer-based test covering Physics, Chemistry, and Mathematics. It features a mix of multiple-choice questions and questions requiring a numerical answer. The B.Arch and B.Plan papers have their own specific formats, including sections on aptitude and, for architecture, a drawing test.

The primary difficulty of JEE Main lies in its immense scale and competition. A very large number of students compete for a limited number of seats, resulting in low selection rates. The syllabus is extensive, covering the curriculum of both Class 11 and 12. Success demands not only a thorough understanding of concepts but also the ability to solve complex problems quickly and accurately under time pressure. The presence of negative marking for incorrect multiple-choice answers means that students must be confident in their responses.

How a question gets worked here

From a physics paper rather than JEE Main, worked the way every question on this site is. JEE Main questions are being added.

JEE Advanced 2014

A glass capillary tube is of the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height $h$ , where the radius of its cross section is $b$ . If the surface tension of water is $S$ , its density is $\rho$ , and its contact angle with glass is $\theta$ , the value of $h$ will be ( $g$ is the acceleration due to gravity)

Mechanical Properties of Fluids A glass capillary tube is of the shape of a truncated cone with an apex angle

\alpha

so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height

h

, where the radius of its cross section is

b

. If the surface tension of water is

S

, its density is

\rho

, and its contact angle with glass is

\theta

, the value of

h

will be (

g

is the acceleration due to gravity)

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Step 1: Identify the geometry of the capillary tube.

The capillary tube is a truncated cone with apex angle $\alpha$ , meaning the walls are tilted from the vertical by the semi-angle.

\text{Semi-angle of the cone} = \frac{\alpha}{2}

Step 2: Determine the angle of the meniscus tangent with the vertical.

The contact angle of water with the glass wall is $\theta$ , and since the wall is tilted by $\frac{\alpha}{2}$ from the vertical, the tangent to the liquid meniscus at the contact line makes an angle of $\theta + \frac{\alpha}{2}$ with the vertical.

\phi = \theta + \frac{\alpha}{2}

Step 3: Relate the radius of curvature of the meniscus to the tube radius.

Let $R$ be the radius of curvature of the spherical meniscus, and $b$ be the radius of the capillary tube at height $h$ .

b = R \cos\left(\theta + \frac{\alpha}{2}\right)

R = \frac{b}{\cos\left(\theta + \frac{\alpha}{2}\right)}

Step 4: Apply the pressure balance equation.

The pressure difference across the curved meniscus is balanced by the hydrostatic pressure of the water column of height $h$ .

\Delta P = \frac{2S}{R}

\rho g h = \frac{2S}{R}

Step 5: Solve for the height $h$ .

Substitute the expression for $R$ into the pressure balance equation.

\rho g h = \frac{2S \cos\left(\theta + \frac{\alpha}{2}\right)}{b}

h = \frac{2S \cos\left(\theta + \frac{\alpha}{2}\right)}{b \rho g}

Answer:

h = \frac{2S \cos\left(\theta + \frac{\alpha}{2}\right)}{b \rho g}

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