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Free Practice Questions for TU BSc CSIT Entrance

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Key Facts: TU BSc CSIT Entrance Exam

100

Total Questions

120 min

Exam Time

35%

Min Qualifying Score

1,950 NPR

Application Fee

The TU IOST BSc CSIT entrance test is a 100-mark objective examination conducted by Tribhuvan University for admission to the B.Sc. CSIT programme. It has 100 multiple-choice questions: 25 in Physics, 25 in Chemistry, 25 in Mathematics, 15 in English, and 10 in Computer Science and General Knowledge. Candidates need at least 35% to qualify, and campus seats are allocated in order of entrance merit rank. The application fee is NPR 1,950.

Sample TU BSc CSIT Entrance Practice Questions

Try these sample questions to review concepts for the TU BSc CSIT Entrance exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A projectile is launched from ground level at an angle of 30° to the horizontal with an initial speed of 40 m/s. Assuming g = 10 m/s² and negligible air resistance, what is the maximum height reached by the projectile?
A.10 m
B.20 m
C.40 m
D.80 m
Explanation: The maximum height H of a projectile launched with speed u at angle theta is given by H = (u² * sin²(theta)) / (2g). Here u = 40 m/s, theta = 30°, sin(30°) = 0.5, and g = 10 m/s². Thus, H = (40² * 0.5²) / (2 * 10) = (1600 * 0.25) / 20 = 400 / 20 = 20 m.
2A 2.0 kg block moving with a velocity of 6.0 m/s along a frictionless horizontal surface makes a head-on perfectly elastic collision with a stationary 4.0 kg block. What is the velocity of the 2.0 kg block immediately after the collision?
A.+2.0 m/s in the initial direction
B.-1.0 m/s (rebounding backwards)
C.-2.0 m/s (rebounding backwards)
D.-3.0 m/s (rebounding backwards)
Explanation: In a one-dimensional perfectly elastic collision with target at rest (u2 = 0), the velocity of the first mass after impact is v1 = ((m1 - m2) / (m1 + m2)) * u1. Substituting m1 = 2.0 kg, m2 = 4.0 kg, and u1 = 6.0 m/s gives v1 = ((2.0 - 4.0) / (2.0 + 4.0)) * 6.0 = (-2.0 / 6.0) * 6.0 = -2.0 m/s. The negative sign denotes that the 2.0 kg block rebounds in the opposite direction at 2.0 m/s.
3If the escape velocity from the surface of Earth is v_e = 11.2 km/s, what is the orbital speed of an artificial satellite orbiting in a circular orbit close to the Earth's surface?
A.7.92 km/s
B.5.60 km/s
C.11.20 km/s
D.15.84 km/s
Explanation: The orbital speed near Earth's surface is v_o = sqrt(g * R) and escape velocity is v_e = sqrt(2 * g * R). Therefore, v_o = v_e / sqrt(2) = 11.2 / 1.414 ≈ 7.92 km/s.
4A particle executes simple harmonic motion (SHM) with an amplitude of 10 cm and a period of 4.0 s. At what displacement from the mean position are its kinetic energy and potential energy equal?
A.2.50 cm
B.5.00 cm
C.8.66 cm
D.7.07 cm
Explanation: In SHM, KE = (1/2) * k * (A² - x²) and PE = (1/2) * k * x². Setting KE = PE yields (1/2) * k * (A² - x²) = (1/2) * k * x², which simplifies to A² - x² = x², or 2x² = A². Thus, x = A / sqrt(2) = 10 / sqrt(2) = 5 * sqrt(2) ≈ 7.07 cm.
5Two spherical water droplets of radii r and 2r fall vertically through still air under gravity. Assuming laminar flow and applying Stokes' law, what is the ratio of their terminal velocities v1 / v2?
A.1 : 2
B.1 : 4
C.1 : 8
D.2 : 1
Explanation: By Stokes' law, terminal velocity of a falling sphere is v_t = (2/9) * (r² * (rho - sigma) * g) / eta, where rho is droplet density, sigma is air density, and eta is viscosity. Thus, v_t is directly proportional to r². Therefore, v1 / v2 = (r / 2r)² = 1 / 4.
6A curved railway track of radius 400 m is banked for trains traveling at a speed of 20 m/s without relying on lateral wheel-rail friction. Taking g = 10 m/s², what is the tangent of the banking angle theta?
A.0.10
B.0.05
C.0.20
D.0.50
Explanation: For a banked, friction-free curve, the required centripetal force is provided by the horizontal component of the normal reaction: N * sin(theta) = m * v² / r, and N * cos(theta) = m * g. Dividing these equations gives tan(theta) = v² / (r * g). Substituting v = 20 m/s, r = 400 m, and g = 10 m/s² yields tan(theta) = 20² / (400 * 10) = 400 / 4000 = 0.10.
7A solid uniform cylinder of mass M and radius R rolls without slipping down an inclined plane of inclination angle theta. What is the linear acceleration of its center of mass?
A.(1/2) * g * sin(theta)
B.(5/7) * g * sin(theta)
C.(2/3) * g * sin(theta)
D.g * sin(theta)
Explanation: The linear acceleration of an object rolling without slipping down an incline is a = (g * sin(theta)) / (1 + I / (M * R²)). For a solid cylinder, the moment of inertia about its axis of symmetry is I = (1/2) * M * R². Substituting gives a = (g * sin(theta)) / (1 + 1/2) = (g * sin(theta)) / (3/2) = (2/3) * g * sin(theta).
8An ideal gas occupies a volume of 4.0 L at a temperature of 27 °C and a pressure of 1.0 atm. If the gas pressure is doubled to 2.0 atm and its temperature is raised to 127 °C, what is the new volume occupied by the gas?
A.1.33 L
B.2.00 L
C.3.00 L
D.2.67 L
Explanation: From the combined gas law, (P1 * V1) / T1 = (P2 * V2) / T2. Absolute temperatures are T1 = 27 + 273 = 300 K and T2 = 127 + 273 = 400 K. Rearranging for V2 gives V2 = (P1 * V1 * T2) / (P2 * T1) = (1.0 * 4.0 * 400) / (2.0 * 300) = 1600 / 600 = 8/3 ≈ 2.67 L.
9A Carnot heat engine operates between a heat source at 500 K and a heat sink at 300 K. If the engine absorbs 1500 J of heat from the source in each cycle, how much mechanical work does it deliver per cycle?
A.450 J
B.600 J
C.900 J
D.1000 J
Explanation: The theoretical efficiency of a Carnot engine is eta = 1 - (T_C / T_H) = 1 - (300 / 500) = 1 - 0.60 = 0.40 (40%). The work delivered per cycle is W = eta * Q_H = 0.40 * 1500 J = 600 J.
10At what temperature (in Celsius) is the root-mean-square (rms) speed of nitrogen molecules (N2, molar mass 28 g/mol) equal to the rms speed of oxygen molecules (O2, molar mass 32 g/mol) at 47 °C?
A.7 °C
B.15 °C
C.27 °C
D.41 °C
Explanation: The rms speed is v_rms = sqrt((3 * R * T) / M). Equating speeds yields T_N2 / M_N2 = T_O2 / M_O2. Converting temperature to Kelvin: T_O2 = 47 + 273 = 320 K. Then T_N2 / 28 = 320 / 32 = 10 K/(g/mol), so T_N2 = 10 * 28 = 280 K. In Celsius: 280 - 273 = 7 °C.

About the TU BSc CSIT Entrance Exam

The TU IOST BSc CSIT Entrance Examination is the university-wide undergraduate entrance examination administered annually by Tribhuvan University's Institute of Science and Technology. It serves as the gateway for admission into the four-year Bachelor of Science in Computer Science and Information Technology program across TU constituent campuses and affiliated private colleges throughout Nepal.

Exam sponsor: Tribhuvan University, Institute of Science and Technology (IOST). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Single paper of 100 multiple-choice questions (1 mark each, 100 marks) in 120 minutes: Physics 25, Chemistry 25, Mathematics 25, English 15, and Computer & GK 10.

Time Limit

120 minutes

Passing Score

35% (35 out of 100 marks minimum)

Exam / Certification Fees

1,950 NPR

Exam sponsor website

Reported exam pass rate: Not published; campus seats are allocated by entrance merit rank. Scoring at least 35% is required for qualification, but competitive top ranks are needed to secure seats in constituent campuses such as ASCOL, Patan Multiple Campus, and Central Department. Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

25%

Physics

Mechanics, heat and thermodynamics, wave and optics, electrostatics, electromagnetism, and modern physics based on +2 science curricula.

25%

Chemistry

Physical chemistry stoichiometry, states of matter, thermodynamics, chemical bonding, inorganic elements, and organic reactions.

25%

Mathematics

Algebra, matrices, trigonometry, coordinate geometry, calculus limits and integrals, vectors, and probability.

15%

English

Grammatical structures, prepositions, idiomatic phrases, vocabulary, and reading comprehension passage analysis.

10%

Computer Science & General Knowledge

Computer architecture, binary representations, programming concepts, networking, and contemporary technology GK.

Preparing for the TU BSc CSIT Entrance Exam

What You Need to Know

  • Passing score: 35% (35 out of 100 marks minimum)
  • Assessment: Single paper of 100 multiple-choice questions (1 mark each, 100 marks) in 120 minutes: Physics 25, Chemistry 25, Mathematics 25, English 15, and Computer & GK 10.
  • Time limit: 120 minutes
  • Exam / certification fees: 1,950 NPR Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

TU BSc CSIT Entrance: Suggested Study Strategy

1Prioritize fundamental formulas in calculus, coordinate geometry, and kinematics for rapid mental calculation without a calculator.
2Review high-frequency organic chemistry conversion reactions and functional group identification.
3Understand logic gate truth tables, binary and hexadecimal arithmetic, and basic memory hierarchies for the computer section.
4Practice timed pacing across all five sections to ensure all 100 questions are attempted within the 120-minute limit.

Frequently Asked Questions

What is the format and duration of the TU BSc CSIT entrance examination?

The examination consists of 100 objective multiple-choice questions with four options each, totaling 100 marks. The test duration is 2 hours (120 minutes) and is completed on OMR sheets.

What is the minimum qualifying score for admission?

Candidates must achieve at least 35% (35 marks out of 100) to qualify for merit listing and participate in college admission rounds.

What subjects are included in the 100-mark entrance exam?

The paper includes 25 marks of Physics, 25 marks of Chemistry, 25 marks of Mathematics, 15 marks of English, and 10 marks of Computer Science and General Knowledge.

Is OpenExamPrep affiliated with Tribhuvan University or IOST?

No. OpenExamPrep is an independent study resource and is not affiliated with or endorsed by Tribhuvan University or the Institute of Science and Technology. Its practice questions are written independently.