Q. If a disc rolls without slipping on a flat surface, what is the relationship between its linear velocity v and angular velocity ω?
A.
v = rω
B.
v = 2rω
C.
v = 1/2 rω
D.
v = 3rω
Show solution
Solution
For rolling without slipping, the linear velocity v is equal to the radius r times the angular velocity ω, hence v = rω.
Correct Answer:
A
— v = rω
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Q. If a drop of liquid is placed on a flat surface, what shape will it take due to surface tension?
A.
Square
B.
Flat
C.
Sphere
D.
Cylinder
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Solution
Due to surface tension, a drop of liquid will take the shape of a sphere, as this shape has the minimum surface area for a given volume.
Correct Answer:
C
— Sphere
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Q. If a fluid has a viscosity of 0.5 Pa·s, what does this indicate about its flow characteristics?
A.
It flows easily
B.
It is very thick
C.
It is a gas
D.
It is a low-density fluid
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Solution
A viscosity of 0.5 Pa·s indicates that the fluid is relatively thick and flows less easily compared to fluids with lower viscosity.
Correct Answer:
B
— It is very thick
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Q. If a force of 10 N is applied to move an object 5 m in the direction of the force, what is the work done?
A.
50 J
B.
30 J
C.
20 J
D.
10 J
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Solution
Work Done (W) = Force * Distance = 10 N * 5 m = 50 J
Correct Answer:
A
— 50 J
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Q. If a force of 10 N is applied to move an object 5 m, what is the work done?
A.
50 J
B.
25 J
C.
10 J
D.
5 J
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Solution
Work Done (W) = Force * Distance = 10 N * 5 m = 50 J
Correct Answer:
A
— 50 J
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Q. If a force of 12 N is applied at an angle of 30 degrees to a lever arm of 1 m, what is the torque about the pivot?
A.
6 Nm
B.
10 Nm
C.
12 Nm
D.
15 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 12 N × 1 m × sin(30°) = 12 N × 1 m × 0.5 = 6 Nm.
Correct Answer:
A
— 6 Nm
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Q. If a force of 12 N is applied at an angle of 30 degrees to the horizontal while moving an object 3 m, what is the work done by the force?
A.
18 J
B.
24 J
C.
30 J
D.
36 J
Show solution
Solution
Work done = F × d × cos(θ) = 12 N × 3 m × cos(30°) = 12 N × 3 m × (√3/2) = 18 J.
Correct Answer:
B
— 24 J
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Q. If a force of 12 N is applied to a 4 kg object, what is the resulting acceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
Using F = ma, we find a = F/m = 12 N / 4 kg = 3 m/s².
Correct Answer:
B
— 3 m/s²
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Q. If a force of 15 N acts on an object and moves it 4 m in the direction of the force, what is the work done?
A.
30 J
B.
60 J
C.
75 J
D.
90 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
B
— 60 J
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Q. If a force of 15 N is applied at a distance of 0.4 m from the pivot, what is the torque?
A.
6 Nm
B.
12 Nm
C.
15 Nm
D.
20 Nm
Show solution
Solution
Torque = Force × Distance = 15 N × 0.4 m = 6 Nm.
Correct Answer:
B
— 12 Nm
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Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1.5 m, what is the torque about the pivot?
A.
3.75 Nm
B.
7.5 Nm
C.
11.25 Nm
D.
12.99 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1.5 m × sin(30°) = 15 × 1.5 × 0.5 = 11.25 Nm.
Correct Answer:
B
— 7.5 Nm
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Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1 m, what is the torque about the pivot?
A.
7.5 Nm
B.
15 Nm
C.
12.99 Nm
D.
10 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1 m × sin(30°) = 15 N × 1 m × 0.5 = 7.5 Nm.
Correct Answer:
C
— 12.99 Nm
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Q. If a force of 15 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done by the force?
A.
30 J
B.
60 J
C.
120 J
D.
180 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 120 J
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Q. If a force of 15 N is applied at an angle of 60° to the horizontal while moving an object 4 m, what is the work done by the force?
A.
30 J
B.
60 J
C.
120 J
D.
180 J
Show solution
Solution
Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.
Correct Answer:
C
— 120 J
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Q. If a force of 15 N is applied to a mass of 3 kg, what is the net force acting on the mass if there is a frictional force of 5 N opposing the motion?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
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Q. If a force of 15 N is applied to a mass of 3 kg, what is the net force acting on the mass if it is also experiencing a frictional force of 5 N?
A.
10 N
B.
15 N
C.
20 N
D.
5 N
Show solution
Solution
Net force = applied force - frictional force = 15 N - 5 N = 10 N.
Correct Answer:
A
— 10 N
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Q. If a force of 15 N is applied to move an object 3 m in the direction of the force, what is the work done?
A.
45 J
B.
30 J
C.
15 J
D.
60 J
Show solution
Solution
Work done = Force × Distance = 15 N × 3 m = 45 J.
Correct Answer:
A
— 45 J
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Q. If a force of 15 N is applied to move an object 4 m in the direction of the force, what is the work done?
A.
30 J
B.
60 J
C.
45 J
D.
75 J
Show solution
Solution
Work done = Force × Distance = 15 N × 4 m = 60 J.
Correct Answer:
B
— 60 J
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Q. If a force of 30 N is applied to a mass of 10 kg, what is the net force acting on the mass if there is a frictional force of 10 N opposing the motion?
A.
20 N
B.
30 N
C.
40 N
D.
10 N
Show solution
Solution
Net force = applied force - frictional force = 30 N - 10 N = 20 N.
Correct Answer:
A
— 20 N
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Q. If a force of 5 N acts on an object and moves it 4 m in the direction of the force, what is the work done?
A.
10 J
B.
15 J
C.
20 J
D.
25 J
Show solution
Solution
Work done = Force × Distance = 5 N × 4 m = 20 J.
Correct Answer:
C
— 20 J
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Q. If a force of 5 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m, what is the work done by the force?
A.
7.5 J
B.
15 J
C.
12.99 J
D.
10 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 5 N × 3 m × cos(60°) = 5 N × 3 m × 0.5 = 7.5 J.
Correct Answer:
C
— 12.99 J
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Q. If a force of 5 N is applied at an angle of 60 degrees to the horizontal while moving an object 4 m, what is the work done by the force?
A.
10 J
B.
20 J
C.
5 J
D.
15 J
Show solution
Solution
Work done = Force × Distance × cos(θ) = 5 N × 4 m × cos(60°) = 5 N × 4 m × 0.5 = 10 J.
Correct Answer:
A
— 10 J
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Q. If a forced oscillator is driven at a frequency much lower than its natural frequency, what happens to the amplitude?
A.
Increases significantly
B.
Decreases
C.
Remains constant
D.
Fluctuates
Show solution
Solution
At frequencies much lower than the natural frequency, the amplitude of the forced oscillator increases significantly.
Correct Answer:
B
— Decreases
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Q. If a galvanometer shows a deflection when connected to a potentiometer, what does it indicate?
A.
The circuit is open.
B.
The potential difference is zero.
C.
The potential difference is equal to the reference voltage.
D.
The current is flowing through the galvanometer.
Show solution
Solution
A deflection in the galvanometer indicates that the potential difference across the galvanometer is equal to the reference voltage.
Correct Answer:
C
— The potential difference is equal to the reference voltage.
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Q. If a gas expands against a constant external pressure, the work done by the gas is given by:
A.
W = P_ext * ΔV
B.
W = ΔU + Q
C.
W = Q - ΔU
D.
W = P_ext / ΔV
Show solution
Solution
The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.
Correct Answer:
A
— W = P_ext * ΔV
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Q. If a gas expands and does 150 J of work while absorbing 100 J of heat, what is the change in internal energy?
A.
-50 J
B.
50 J
C.
250 J
D.
100 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.
Correct Answer:
A
— -50 J
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Q. If a gas expands and does 50 J of work while absorbing 30 J of heat, what is the change in internal energy?
A.
-20 J
B.
20 J
C.
80 J
D.
30 J
Show solution
Solution
Using the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 30 J - 50 J = -20 J.
Correct Answer:
B
— 20 J
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Q. If a gas occupies 10 L at 1 atm, what will be its volume at 2 atm if the temperature remains constant?
A.
5 L
B.
10 L
C.
20 L
D.
15 L
Show solution
Solution
Using Boyle's Law (P1V1 = P2V2), if P1 = 1 atm, V1 = 10 L, and P2 = 2 atm, then V2 = (P1V1)/P2 = (1 atm * 10 L) / 2 atm = 5 L.
Correct Answer:
A
— 5 L
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Q. If a gas occupies a volume of 10 L at 1 atm, what will be its volume at 2 atm if the temperature remains constant?
A.
5 L
B.
10 L
C.
20 L
D.
15 L
Show solution
Solution
Using Boyle's Law (P1V1 = P2V2), if the pressure doubles from 1 atm to 2 atm, the volume will halve from 10 L to 5 L.
Correct Answer:
A
— 5 L
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Q. If a hollow cylinder and a solid cylinder of the same mass and radius roll down the same incline, which one reaches the bottom first?
A.
Hollow cylinder
B.
Solid cylinder
C.
Both reach at the same time
D.
Depends on the angle of incline
Show solution
Solution
The solid cylinder reaches the bottom first because it has a smaller moment of inertia compared to the hollow cylinder.
Correct Answer:
B
— Solid cylinder
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Showing 1651 to 1680 of 5000 (167 Pages)
Physics Syllabus (JEE Main) MCQ & Objective Questions
The Physics Syllabus for JEE Main is crucial for students aiming to excel in their exams. Understanding this syllabus not only helps in grasping fundamental concepts but also enhances problem-solving skills through practice. Engaging with MCQs and objective questions is essential for effective exam preparation, as it allows students to identify important questions and strengthen their knowledge base.
What You Will Practise Here
Mechanics: Laws of Motion, Work, Energy, and Power
Thermodynamics: Laws of Thermodynamics, Heat Transfer
Waves and Oscillations: Simple Harmonic Motion, Wave Properties
Electromagnetism: Electric Fields, Magnetic Fields, and Circuits
Optics: Reflection, Refraction, and Optical Instruments
Modern Physics: Quantum Theory, Atomic Models, and Nuclear Physics
Fluid Mechanics: Properties of Fluids, Bernoulli's Principle
Exam Relevance
The Physics Syllabus (JEE Main) is integral to various examinations, including CBSE, State Boards, and competitive exams like NEET and JEE. Questions often focus on conceptual understanding and application of theories. Common patterns include numerical problems, conceptual MCQs, and assertion-reason type questions, which test both knowledge and analytical skills.
Common Mistakes Students Make
Misinterpreting the question stem, leading to incorrect answers.
Neglecting units and dimensions in calculations.
Overlooking the significance of diagrams in understanding concepts.
Confusing similar concepts, such as velocity and acceleration.
Failing to apply formulas correctly in different contexts.
FAQs
Question: What are the key topics in the Physics Syllabus for JEE Main?Answer: Key topics include Mechanics, Thermodynamics, Waves, Electromagnetism, Optics, Modern Physics, and Fluid Mechanics.
Question: How can I improve my performance in Physics MCQs?Answer: Regular practice of MCQs, understanding concepts deeply, and revising important formulas can significantly enhance your performance.
Start solving practice MCQs today to test your understanding of the Physics Syllabus (JEE Main). This will not only boost your confidence but also prepare you effectively for your upcoming exams. Remember, consistent practice is the key to success!
