CUET

Conducting Body: National Testing Agency (NTA)
Who it is for: Students seeking undergraduate admission after Class 12
Exam Sections: Language Test, Domain-Specific Subjects, General Test
Question Types: Multiple Choice Questions (MCQs)
Mode: Computer-Based Test (CBT)
Marking Scheme: Marks for correct answers, penalty for incorrect answers
Frequency: Annually
Qualifies you for: Admission to undergraduate courses in participating universities

The Common University Entrance Test, or CUET, is a national-level examination for admission to undergraduate programmes at central, state, and other participating universities across India. It is designed for students who have completed or are about to complete their Class 12 education. The exam's purpose is to provide a single, standardized platform for admissions, creating a more level playing field for candidates from different educational boards.

The paper is a computer-based test consisting entirely of multiple-choice questions. It is structured in three sections: language proficiency tests, domain-specific subject tests based on the Class 12 syllabus, and a General Test that assesses aptitude and general awareness. Candidates select the specific combination of tests they need to take based on the eligibility requirements of the university courses they are applying for.

The difficulty of CUET comes from the vastness of the syllabus and the strategic planning it demands. Students must cover the entire Class 12 curriculum for their chosen domain subjects. The greater challenge, however, is navigating the complex eligibility criteria of various universities to select the correct set of test papers. An error in this selection can make a student ineligible for their desired course, regardless of their score.

How a question gets worked here

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

JEE Advanced 2023

An ideal gas of density $\rho = 0.2\,\text{kg m}^{-3}$ enters a chimney of height $h$ at the rate of $\alpha = 0.8\,\text{kg s}^{-1}$ from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is $A_{1} = 0.1\,\text{m}^{2}$ and the upper end is $A_{2} = 0.4\,\text{m}^{2}$ . The pressure and the temperature of the gas at the lower end are $600\,\text{Pa}$ and $300\,\text{K}$ , respectively, while its temperature at the upper end is $150\,\text{K}$ . The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take $g = 10\,\text{m s}^{-2}$ and the ratio of specific heats of the gas $\gamma = 2$ . Ignore atmospheric pressure. Which of the following statement(s) is(are) correct?

Thermodynamics An ideal gas of density

\rho = 0.2\,\text{kg m}^{-3}

enters a chimney of height

h

at the rate of

\alpha = 0.8\,\text{kg s}^{-1}

from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is

A_{1} = 0.1\,\text{m}^{2}

and the upper end is

A_{2} = 0.4\,\text{m}^{2}

. The pressure and the temperature of the gas at the lower end are

600\,\text{Pa}

and

300\,\text{K}

, respectively, while its temperature at the upper end is

150\,\text{K}

. The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take

g = 10\,\text{m s}^{-2}

and the ratio of specific heats of the gas

\gamma = 2

. Ignore atmospheric pressure. Which of the following statement(s) is(are) correct?

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

Step 1: Calculate the velocity of the gas at the lower end.

The mass flow rate is constant throughout the chimney.

\alpha = \rho_1 A_1 v_1

v_1 = \frac{\alpha}{\rho_1 A_1}

v_1 = \frac{0.8}{0.2 \times 0.1} = 40\,\text{m s}^{-1}

Step 2: Determine the density of the gas at the upper end.

For an adiabatic process of an ideal gas, the relation between temperature and density is given by the adiabatic relation.

T \rho^{1-\gamma} = \text{constant}

\rho \propto T^{\frac{1}{\gamma - 1}}

Substitute $\gamma = 2$ .

\rho_2 = \rho_1 \left(\frac{T_2}{T_1}\right)

\rho_2 = 0.2 \times \frac{150}{300} = 0.1\,\text{kg m}^{-3}

Step 3: Calculate the velocity of the gas at the upper end.

Apply the mass flow rate equation at the upper end.

\alpha = \rho_2 A_2 v_2

v_2 = \frac{\alpha}{\rho_2 A_2}

v_2 = \frac{0.8}{0.1 \times 0.4} = 20\,\text{m s}^{-1}

Step 4: Calculate the pressure of the gas at the upper end.

Use the ideal gas equation of state.

\frac{p}{\rho T} = \text{constant}

p_2 = p_1 \left(\frac{\rho_2 T_2}{\rho_1 T_1}\right)

p_2 = 600 \times \left(\frac{0.1 \times 150}{0.2 \times 300}\right) = 150\,\text{Pa}

Step 5: Calculate the height of the chimney.

Apply the energy conservation equation for steady adiabatic flow.

\frac{v^2}{2} + gh + \frac{\gamma}{\gamma-1}\frac{p}{\rho} = \text{constant}

\frac{v_1^2}{2} + \frac{\gamma}{\gamma-1}\frac{p_1}{\rho_1} = \frac{v_2^2}{2} + gh + \frac{\gamma}{\gamma-1}\frac{p_2}{\rho_2}

h = \frac{\frac{1}{2}(v_1^2 - v_2^2) + \frac{\gamma}{\gamma-1}\left(\frac{p_1}{\rho_1} - \frac{p_2}{\rho_2}\right)}{g}

h = \frac{\frac{1}{2}(40^2 - 20^2) + \frac{2}{2-1}\left(\frac{600}{0.2} - \frac{150}{0.1}\right)}{10}

h = \frac{600 + 2(3000 - 1500)}{10}

h = \frac{600 + 3000}{10} = 360\,\text{m}

Answer: The correct statement is that the velocity of the gas at the lower end is $v_1 = 40\,\text{m s}^{-1}$ and at the upper end is $v_2 = 20\,\text{m s}^{-1}$ .

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