Engineering Entrance MCQ & Objective Questions
Preparing for Engineering Entrance exams is crucial for aspiring engineers in India. Mastering MCQs and objective questions not only enhances your understanding of key concepts but also boosts your confidence during exams. Regular practice with these questions helps identify important topics and improves your overall exam preparation.
What You Will Practise Here
Fundamental concepts of Physics and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theorems relevant to engineering
Diagrams and graphical representations for better understanding
Conceptual questions that challenge your critical thinking
Previous years' question papers and their analysis
Time management strategies while solving MCQs
Exam Relevance
The Engineering Entrance syllabus is integral to various examinations like CBSE, State Boards, NEET, and JEE. Questions often focus on core subjects such as Physics, Chemistry, and Mathematics, with formats varying from direct MCQs to application-based problems. Understanding the common question patterns can significantly enhance your performance and help you tackle the exams with ease.
Common Mistakes Students Make
Overlooking the importance of units and dimensions in calculations
Misinterpreting questions due to lack of careful reading
Neglecting to review basic concepts before attempting advanced problems
Rushing through practice questions without thorough understanding
FAQs
Question: What are the best ways to prepare for Engineering Entrance MCQs?Answer: Focus on understanding concepts, practice regularly with objective questions, and review previous years' papers.
Question: How can I improve my speed in solving MCQs?Answer: Regular practice, time-bound mock tests, and familiarizing yourself with common question types can help improve your speed.
Start your journey towards success by solving Engineering Entrance MCQ questions today! Test your understanding and build a strong foundation for your exams.
Q. In which direction does the magnetic field line emerge from a bar magnet? (2022)
A.
From North to South
B.
From South to North
C.
In a circular path
D.
Randomly
Show solution
Solution
Magnetic field lines emerge from the north pole and enter the south pole of a bar magnet.
Correct Answer:
A
— From North to South
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Q. In which of the following processes is adsorption utilized? (2023)
A.
Water purification
B.
Sublimation
C.
Distillation
D.
Filtration
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Solution
Adsorption is widely used in water purification processes to remove impurities.
Correct Answer:
A
— Water purification
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Q. In which state of matter do particles have the highest energy? (2023)
A.
Solid
B.
Liquid
C.
Gas
D.
Bose-Einstein Condensate
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Solution
Gas particles have the highest energy compared to solids and liquids, allowing them to move freely and occupy more space.
Correct Answer:
C
— Gas
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Q. In which state of matter do particles have the least amount of energy?
A.
Solid
B.
Liquid
C.
Gas
D.
Plasma
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Solution
Particles in a solid have the least amount of energy compared to liquids, gases, and plasma.
Correct Answer:
A
— Solid
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Q. In which state of matter do particles move freely and are far apart? (2021)
A.
Solid
B.
Liquid
C.
Gas
D.
None of the above
Show solution
Solution
In gases, particles move freely and are far apart compared to solids and liquids.
Correct Answer:
C
— Gas
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Q. Solve the differential equation dy/dx = 2x + 1.
A.
y = x^2 + x + C
B.
y = x^2 + 2x + C
C.
y = 2x^2 + x + C
D.
y = x^2 + C
Show solution
Solution
Integrating both sides, we get y = ∫(2x + 1)dx = x^2 + x + C.
Correct Answer:
A
— y = x^2 + x + C
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Q. Solve the differential equation dy/dx = 2y + 3. (2023)
A.
y = Ce^(2x) - 3/2
B.
y = Ce^(-2x) + 3/2
C.
y = 3e^(2x)
D.
y = 2e^(2x) + C
Show solution
Solution
Using an integrating factor, we find the solution is y = Ce^(2x) - 3/2.
Correct Answer:
A
— y = Ce^(2x) - 3/2
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Q. Solve the differential equation dy/dx = 6x^2y.
A.
y = Ce^(2x^3)
B.
y = Ce^(3x^2)
C.
y = Ce^(6x^2)
D.
y = Ce^(x^6)
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Solution
This is a separable equation. Integrating gives y = Ce^(2x^3).
Correct Answer:
A
— y = Ce^(2x^3)
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Q. Solve the differential equation dy/dx = y/x. (2023)
A.
y = Cx
B.
y = Cx^2
C.
y = C/x
D.
y = C ln(x)
Show solution
Solution
This is a separable equation. Integrating gives y = Cx.
Correct Answer:
A
— y = Cx
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Q. Solve the differential equation y' = 5 - 2y.
A.
y = 5/2 + Ce^(-2x)
B.
y = 5 + Ce^(-2x)
C.
y = 2 + Ce^(2x)
D.
y = 5/2 - Ce^(-2x)
Show solution
Solution
This is a linear first-order equation. The solution is y = 5/2 + Ce^(-2x).
Correct Answer:
A
— y = 5/2 + Ce^(-2x)
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Q. Solve the differential equation y' = 5y + 3.
A.
y = (3/5) + Ce^(5x)
B.
y = (5/3) + Ce^(5x)
C.
y = Ce^(5x) - 3
D.
y = Ce^(3x) + 5
Show solution
Solution
Using the integrating factor method, we find the solution y = (3/5) + Ce^(5x).
Correct Answer:
A
— y = (3/5) + Ce^(5x)
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Q. Solve the differential equation y'' - 3y' + 2y = 0.
A.
y = C1e^(2x) + C2e^(x)
B.
y = C1e^(x) + C2e^(2x)
C.
y = C1e^(-x) + C2e^(-2x)
D.
y = C1e^(3x) + C2e^(x)
Show solution
Solution
The characteristic equation is r^2 - 3r + 2 = 0, which factors to (r - 1)(r - 2) = 0. The general solution is y = C1e^(x) + C2e^(2x).
Correct Answer:
B
— y = C1e^(x) + C2e^(2x)
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Q. Solve the equation y' = 6y + 12.
A.
y = 2 - Ce^(-6x)
B.
y = Ce^(6x) - 2
C.
y = 2 + Ce^(6x)
D.
y = 6Ce^(-x)
Show solution
Solution
This is a first-order linear equation. The integrating factor method gives the solution y = 2 - Ce^(-6x).
Correct Answer:
A
— y = 2 - Ce^(-6x)
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Q. Solve the first-order differential equation dy/dx = y/x.
A.
y = Cx
B.
y = Cx^2
C.
y = C/x
D.
y = C ln(x)
Show solution
Solution
This is a separable equation. Separating variables and integrating gives y = Cx.
Correct Answer:
A
— y = Cx
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Q. Solve the first-order linear differential equation dy/dx + 2y = 6.
A.
y = 3 - Ce^(-2x)
B.
y = 3 + Ce^(-2x)
C.
y = 6 - Ce^(-2x)
D.
y = 6 + Ce^(-2x)
Show solution
Solution
Using an integrating factor e^(2x), we solve to get y = 3 - Ce^(-2x).
Correct Answer:
A
— y = 3 - Ce^(-2x)
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Q. Solve the first-order linear differential equation dy/dx + y/x = 1.
A.
y = x + C/x
B.
y = Cx - x
C.
y = Cx + x
D.
y = C/x + x
Show solution
Solution
Using the integrating factor e^(∫(1/x)dx) = x, we solve to get y = x + C/x.
Correct Answer:
A
— y = x + C/x
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Q. Solve the first-order linear differential equation dy/dx = y/x.
A.
y = Cx
B.
y = Cx^2
C.
y = C/x
D.
y = C ln(x)
Show solution
Solution
This is separable: dy/y = dx/x. Integrating gives ln|y| = ln|x| + C, thus y = Cx.
Correct Answer:
A
— y = Cx
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Q. The Arrhenius equation relates which of the following? (2023)
A.
Rate constant and temperature
B.
Concentration and time
C.
Pressure and volume
D.
Energy and temperature
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Solution
The Arrhenius equation relates the rate constant (k) to temperature (T) and activation energy (Ea).
Correct Answer:
A
— Rate constant and temperature
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Q. The dimensional formula for energy is: (2020)
A.
[M^1 L^2 T^-2]
B.
[M^2 L^1 T^-2]
C.
[M^0 L^2 T^0]
D.
[M^1 L^0 T^-1]
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Solution
Energy is defined as work done, which has the dimensional formula [M^1 L^2 T^-2].
Correct Answer:
A
— [M^1 L^2 T^-2]
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Q. The dimensional formula for frequency is: (2023)
A.
[T^-1]
B.
[M^0 L^0 T^-1]
C.
[M^1 L^1 T^-1]
D.
[M^0 L^1 T^0]
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Solution
Frequency is defined as the number of cycles per unit time, hence its dimensional formula is [T^-1].
Correct Answer:
A
— [T^-1]
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Q. The dimensional formula for velocity is: (2022)
A.
[M^0 L^1 T^-1]
B.
[M^1 L^1 T^0]
C.
[M^1 L^0 T^-1]
D.
[M^0 L^0 T^1]
Show solution
Solution
Velocity is defined as displacement per unit time, hence its dimensional formula is [M^0 L^1 T^-1].
Correct Answer:
A
— [M^0 L^1 T^-1]
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Q. The direction of induced current can be determined by which law? (2023)
A.
Lenz's Law
B.
Faraday's Law
C.
Ohm's Law
D.
Ampere's Law
Show solution
Solution
Lenz's Law states that the direction of induced current is such that it opposes the change in magnetic flux that produced it.
Correct Answer:
A
— Lenz's Law
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Q. The force on a charged particle in a magnetic field is maximum when the angle between the velocity and the magnetic field is: (2021)
A.
0 degrees
B.
90 degrees
C.
180 degrees
D.
45 degrees
Show solution
Solution
The force on a charged particle in a magnetic field is maximum when the angle between the velocity and the magnetic field is 90 degrees.
Correct Answer:
B
— 90 degrees
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Q. The force on a charged particle moving in a magnetic field is maximum when the angle between the velocity and the magnetic field is: (2021)
A.
0 degrees
B.
90 degrees
C.
180 degrees
D.
45 degrees
Show solution
Solution
The force on a charged particle moving in a magnetic field is maximum when the angle between the velocity and the magnetic field is 90 degrees.
Correct Answer:
B
— 90 degrees
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Q. The force on a charged particle moving in a magnetic field is maximum when: (2021)
A.
The particle moves parallel to the field
B.
The particle moves perpendicular to the field
C.
The particle is at rest
D.
The particle moves in a circular path
Show solution
Solution
The force on a charged particle moving in a magnetic field is maximum when the particle moves perpendicular to the field.
Correct Answer:
B
— The particle moves perpendicular to the field
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Q. The frequency of a tuning fork is 440 Hz. What is the time period of the fork? (2019)
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.005 s
Show solution
Solution
Time period (T) = 1 / frequency (f) = 1 / 440 Hz ≈ 0.00227 s.
Correct Answer:
A
— 0.00227 s
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Q. The frequency of a tuning fork is 440 Hz. What is the time period of the sound wave produced? (2022)
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.005 s
Show solution
Solution
Time period (T) = 1 / frequency (f) = 1 / 440 Hz ≈ 0.00227 s.
Correct Answer:
A
— 0.00227 s
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Q. The frequency of a wave is doubled. How does this affect its wavelength if the speed of the wave remains constant? (2022)
A.
Wavelength doubles
B.
Wavelength halves
C.
Wavelength remains the same
D.
Wavelength quadruples
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Solution
According to the wave equation v = fλ, if frequency (f) is doubled and speed (v) remains constant, wavelength (λ) must halve.
Correct Answer:
B
— Wavelength halves
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Q. The half-life of a first-order reaction is dependent on which of the following? (2020) 2020
A.
Initial concentration of reactants
B.
Rate constant
C.
Temperature
D.
All of the above
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Solution
The half-life of a first-order reaction is independent of the initial concentration and is directly proportional to the rate constant.
Correct Answer:
B
— Rate constant
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Q. The half-life of a first-order reaction is given by which of the following expressions? (2020)
A.
t1/2 = 0.693/k
B.
t1/2 = k/0.693
C.
t1/2 = 1/k
D.
t1/2 = k/1
Show solution
Solution
For a first-order reaction, the half-life is constant and is given by the formula t1/2 = 0.693/k.
Correct Answer:
A
— t1/2 = 0.693/k
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