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. What is the frequency of a wave with a period of 0.5 seconds? (2019)
A.
2 Hz
B.
1 Hz
C.
0.5 Hz
D.
4 Hz
Show solution
Solution
Frequency (f) = 1/Period (T) = 1/0.5 s = 2 Hz.
Correct Answer:
A
— 2 Hz
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Q. What is the frequency of a wave with a speed of 300 m/s and a wavelength of 3 m? (2023)
A.
100 Hz
B.
150 Hz
C.
200 Hz
D.
250 Hz
Show solution
Solution
Frequency (f) = speed (v) / wavelength (λ) = 300 m/s / 3 m = 100 Hz.
Correct Answer:
B
— 150 Hz
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Q. What is the frequency of a wave with a wavelength of 2 m and a speed of 340 m/s?
A.
170 Hz
B.
340 Hz
C.
85 Hz
D.
420 Hz
Show solution
Solution
Frequency (f) is given by the formula f = speed / wavelength. Here, f = 340 m/s / 2 m = 170 Hz.
Correct Answer:
A
— 170 Hz
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Q. What is the functional group of alcohols? (2020)
A.
-OH
B.
-COOH
C.
-CHO
D.
-NH2
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Solution
The functional group of alcohols is the hydroxyl group (-OH).
Correct Answer:
A
— -OH
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Q. What is the functional group of aldehydes?
A.
-OH
B.
-CHO
C.
-C=O
D.
-COOH
Show solution
Solution
The functional group of aldehydes is -CHO.
Correct Answer:
B
— -CHO
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Q. What is the functional group of phenols? (2019) 2019
A.
-OH
B.
-COOH
C.
-CHO
D.
-NH2
Show solution
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic ring.
Correct Answer:
A
— -OH
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Q. What is the functional group of phenols? (2021)
A.
-OH
B.
-COOH
C.
-CHO
D.
-NH2
Show solution
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic hydrocarbon group.
Correct Answer:
A
— -OH
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Q. What is the functional group present in alcohols? (2023) 2023
A.
Aldehyde
B.
Ketone
C.
Hydroxyl
D.
Carboxylic acid
Show solution
Solution
Alcohols contain a hydroxyl (-OH) functional group.
Correct Answer:
C
— Hydroxyl
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Q. What is the functional group present in phenols? (2021)
A.
-OH
B.
-COOH
C.
-CHO
D.
-NH2
Show solution
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic hydrocarbon group.
Correct Answer:
A
— -OH
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Q. What is the functional group present in phenols? (2021) 2021
A.
-OH
B.
-COOH
C.
-CHO
D.
-NH2
Show solution
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic hydrocarbon group.
Correct Answer:
A
— -OH
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Q. What is the general formula for secondary amines? (2021) 2021
A.
RNH2
B.
R2NH
C.
R3N
D.
RNH
Show solution
Solution
The general formula for secondary amines is R2NH, where R represents an alkyl or aryl group.
Correct Answer:
B
— R2NH
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Q. What is the general formula for secondary amines? (2023)
A.
RNH2
B.
R2NH
C.
R3N
D.
RNH
Show solution
Solution
The general formula for secondary amines is R2NH, where R represents an alkyl group.
Correct Answer:
B
— R2NH
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Q. What is the general solution of the equation y' + 4y = 0?
A.
y = Ce^(-4x)
B.
y = Ce^(4x)
C.
y = 4x + C
D.
y = Cx^4
Show solution
Solution
This is a separable equation, and integrating gives y = Ce^(-4x).
Correct Answer:
A
— y = Ce^(-4x)
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Q. What is the general solution of the equation y' = 4y + 3?
A.
y = Ce^(4x) - 3/4
B.
y = Ce^(4x) + 3/4
C.
y = 3e^(4x)
D.
y = Ce^(3x) + 4
Show solution
Solution
The integrating factor is e^(-4x). The solution is y = Ce^(4x) + 3/4.
Correct Answer:
B
— y = Ce^(4x) + 3/4
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Q. What is the general solution of the equation y'' - 3y' + 2y = 0?
A.
y = C1 e^(x) + C2 e^(2x)
B.
y = C1 e^(2x) + C2 e^(x)
C.
y = C1 e^(3x) + C2 e^(0)
D.
y = C1 e^(0) + C2 e^(3x)
Show solution
Solution
The characteristic equation is r^2 - 3r + 2 = 0, which factors to (r - 1)(r - 2) = 0. Thus, the general solution is y = C1 e^(2x) + C2 e^(x).
Correct Answer:
B
— y = C1 e^(2x) + C2 e^(x)
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Q. What is the general solution of the equation y'' - 4y' + 4y = 0?
A.
y = (C1 + C2x)e^(2x)
B.
y = C1 e^(2x) + C2 e^(-2x)
C.
y = C1 e^(4x) + C2 e^(-4x)
D.
y = C1 cos(2x) + C2 sin(2x)
Show solution
Solution
The characteristic equation has a repeated root r = 2. The general solution is y = (C1 + C2x)e^(2x).
Correct Answer:
A
— y = (C1 + C2x)e^(2x)
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Q. What is the geometry of a molecule with four bonding pairs and no lone pairs around the central atom? (2023)
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
The geometry is tetrahedral when there are four bonding pairs and no lone pairs around the central atom.
Correct Answer:
C
— Tetrahedral
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Q. What is the geometry of a molecule with four bonding pairs and no lone pairs? (2023)
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
Show solution
Solution
A molecule with four bonding pairs and no lone pairs has a tetrahedral geometry.
Correct Answer:
C
— Tetrahedral
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Q. What is the geometry of a molecule with two bonding pairs and no lone pairs on the central atom? (2023)
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Bent
Show solution
Solution
The geometry is linear when there are two bonding pairs and no lone pairs on the central atom.
Correct Answer:
A
— Linear
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Q. What is the geometry of the complex ion [Ag(NH3)2]+? (2019) 2019
A.
Tetrahedral
B.
Square planar
C.
Octahedral
D.
Linear
Show solution
Solution
The complex [Ag(NH3)2]+ has a linear geometry due to the presence of two ligands around the silver ion.
Correct Answer:
D
— Linear
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Q. What is the gravitational field strength at a distance of 10 m from a mass of 100 kg? (G = 6.67 × 10^-11 N m²/kg²)
A.
6.67 × 10^-10 N/kg
B.
6.67 × 10^-9 N/kg
C.
6.67 × 10^-8 N/kg
D.
6.67 × 10^-7 N/kg
Show solution
Solution
g = G * M / r² = (6.67 × 10^-11) * (100) / (10²) = 6.67 × 10^-10 N/kg
Correct Answer:
A
— 6.67 × 10^-10 N/kg
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Q. What is the gravitational field strength at a distance of 3R from the center of the Earth? (R = radius of the Earth) (2022)
A.
g/3
B.
g/9
C.
g/6
D.
g/12
Show solution
Solution
Gravitational field strength decreases with the square of the distance. At 3R, it is g/9.
Correct Answer:
B
— g/9
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Q. What is the gravitational force acting on a 50 kg object at the surface of the Earth?
A.
490 N
B.
500 N
C.
510 N
D.
520 N
Show solution
Solution
Weight = mass * g = 50 kg * 9.8 m/s² = 490 N
Correct Answer:
A
— 490 N
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Q. What is the gravitational force acting on a 50 kg object on the surface of the Earth?
A.
490 N
B.
50 N
C.
5 N
D.
500 N
Show solution
Solution
Weight W = m * g = 50 kg * 9.8 m/s² = 490 N.
Correct Answer:
A
— 490 N
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Q. What is the gravitational force between two 1 kg masses placed 1 meter apart? (G = 6.67 × 10^-11 N m²/kg²)
A.
6.67 × 10^-11 N
B.
1.67 × 10^-10 N
C.
6.67 × 10^-10 N
D.
0 N
Show solution
Solution
F = G * (m1 * m2) / r² = (6.67 × 10^-11) * (1 * 1) / (1²) = 6.67 × 10^-11 N
Correct Answer:
A
— 6.67 × 10^-11 N
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Q. What is the gravitational force between two 10 kg masses separated by a distance of 2 m?
A.
0.25 N
B.
5 N
C.
10 N
D.
20 N
Show solution
Solution
Using Newton's law of gravitation: F = G(m1*m2)/r² = (6.674 × 10⁻¹¹)(10*10)/(2²) = 0.25 N.
Correct Answer:
A
— 0.25 N
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Q. What is the gravitational force between two objects of mass 5 kg and 10 kg separated by a distance of 2 m? (G = 6.67 × 10^-11 N m²/kg²)
A.
1.67 × 10^-10 N
B.
1.25 × 10^-10 N
C.
1.00 × 10^-10 N
D.
2.00 × 10^-10 N
Show solution
Solution
Using the formula F = G * (m1 * m2) / r², we have F = (6.67 × 10^-11) * (5 * 10) / (2²) = 1.67 × 10^-10 N.
Correct Answer:
A
— 1.67 × 10^-10 N
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Q. What is the gravitational force between two objects of mass m1 and m2 separated by a distance r? (2021)
A.
G * (m1 * m2) / r^2
B.
G * (m1 + m2) / r^2
C.
G * (m1 - m2) / r^2
D.
G * (m1 * m2) * r^2
Show solution
Solution
The gravitational force is given by Newton's law of gravitation: F = G * (m1 * m2) / r^2.
Correct Answer:
A
— G * (m1 * m2) / r^2
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Q. What is the gravitational potential energy of a 10 kg mass at a height of 10 m?
A.
100 J
B.
200 J
C.
500 J
D.
1000 J
Show solution
Solution
Gravitational potential energy (PE) is given by PE = mgh. Here, PE = 10 kg * 9.81 m/s² * 10 m = 981 J.
Correct Answer:
D
— 1000 J
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Q. What is the gravitational potential energy of a 10 kg mass at a height of 5 m? (2019)
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Gravitational potential energy (PE) = m * g * h = 10 kg * 9.81 m/s² * 5 m = 490.5 J.
Correct Answer:
B
— 100 J
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