Q. A 10 kg object is dropped from a height. What is the force acting on it just before it hits the ground?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
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Solution
The force acting on the object is its weight, F = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer:
B
— 20 N
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Q. A 10 kg object is hanging at rest from a rope. What is the tension in the rope?
A.
0 N
B.
10 N
C.
100 N
D.
50 N
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Solution
The tension in the rope must balance the weight of the object. T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer:
C
— 100 N
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Q. A 10 kg object is hanging from a rope. What is the tension in the rope when the object is at rest?
A.
0 N
B.
10 N
C.
100 N
D.
50 N
Show solution
Solution
At rest, the tension in the rope equals the weight of the object: T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer:
C
— 100 N
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Q. A 10 kg object is lifted to a height of 10 m. How much work is done against gravity?
A.
0 J
B.
100 J
C.
200 J
D.
1000 J
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Solution
Work done against gravity is equal to the change in potential energy, W = mgh = 10 * 9.8 * 10 = 980 J.
Correct Answer:
D
— 1000 J
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Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity?
A.
20 J
B.
40 J
C.
60 J
D.
80 J
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Solution
Work done = mass × g × height = 10 kg × 9.8 m/s² × 2 m = 196 J.
Correct Answer:
B
— 40 J
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Q. A 10 kg object is lifted to a height of 5 m. How much work is done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Work done = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer:
B
— 100 J
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Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained by the object?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential energy (PE) = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J.
Correct Answer:
B
— 100 J
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Q. A 10 kg object is lifted to a height of 5 m. What is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Work done against gravity = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer:
B
— 100 J
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Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 10 m/s. What is the centripetal force acting on the object?
A.
20 N
B.
10 N
C.
5 N
D.
50 N
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Solution
Centripetal force F = mv²/r = 10 kg * (10 m/s)² / 5 m = 200 N.
Correct Answer:
A
— 20 N
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Q. A 10 kg object is moving in a circular path of radius 5 m with a speed of 4 m/s. What is the centripetal force acting on the object?
A.
8 N
B.
16 N
C.
32 N
D.
40 N
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Solution
Centripetal force F = mv²/r = 10 kg * (4 m/s)² / 5 m = 32 N.
Correct Answer:
B
— 16 N
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Q. A 10 kg object is moving with a speed of 3 m/s. If it comes to a stop, how much work is done by the friction force?
A.
45 J
B.
90 J
C.
135 J
D.
180 J
Show solution
Solution
Work done = Change in Kinetic Energy = 0 - (0.5 × 10 kg × (3 m/s)²) = -45 J.
Correct Answer:
B
— 90 J
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Q. A 10 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
A.
45 J
B.
30 J
C.
60 J
D.
90 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer:
A
— 45 J
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Q. A 10 kg object is moving with a speed of 5 m/s. What is its total mechanical energy?
A.
125 J
B.
250 J
C.
500 J
D.
1000 J
Show solution
Solution
Total mechanical energy = KE + PE. KE = 0.5 * 10 * (5)² = 125 J. Assuming PE = 0, total energy = 125 J.
Correct Answer:
B
— 250 J
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Q. A 10 kg object is moving with a velocity of 2 m/s. If it comes to rest, what is the change in momentum?
A.
20 kg·m/s
B.
10 kg·m/s
C.
5 kg·m/s
D.
0 kg·m/s
Show solution
Solution
Initial momentum = mv = 10 kg * 2 m/s = 20 kg·m/s. Final momentum = 0. Change in momentum = 20 kg·m/s - 0 = 20 kg·m/s.
Correct Answer:
A
— 20 kg·m/s
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
15 J
B.
30 J
C.
45 J
D.
60 J
Show solution
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer:
C
— 45 J
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is its momentum?
A.
10 kg·m/s
B.
30 kg·m/s
C.
3 kg·m/s
D.
0.3 kg·m/s
Show solution
Solution
Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.
Correct Answer:
B
— 30 kg·m/s
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Q. A 10 kg object is moving with a velocity of 3 m/s. What is the total mechanical energy if it is at a height of 5 m?
A.
150 J
B.
180 J
C.
300 J
D.
450 J
Show solution
Solution
Total Mechanical Energy = Kinetic Energy + Potential Energy = 0.5 × 10 kg × (3 m/s)² + 10 kg × 9.8 m/s² × 5 m = 45 J + 490 J = 535 J.
Correct Answer:
C
— 300 J
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Q. A 10 kg object is moving with a velocity of 4 m/s. What is its momentum?
A.
40 kg m/s
B.
20 kg m/s
C.
10 kg m/s
D.
30 kg m/s
Show solution
Solution
Momentum (p) = m * v = 10 kg * 4 m/s = 40 kg m/s
Correct Answer:
A
— 40 kg m/s
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Q. A 10 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.
10 kg·m/s
B.
25 kg·m/s
C.
50 kg·m/s
D.
75 kg·m/s
Show solution
Solution
Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.
Correct Answer:
C
— 50 kg·m/s
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Q. A 10 kg object is pushed with a force of 30 N. If the frictional force is 10 N, what is the net force?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
Show solution
Solution
Net force = applied force - friction = 30 N - 10 N = 20 N.
Correct Answer:
B
— 20 N
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Q. A 10 kg object is thrown upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.
11.5 m
B.
22.5 m
C.
15.3 m
D.
10.0 m
Show solution
Solution
Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (15)²/(2*9.8) = 11.5 m.
Correct Answer:
A
— 11.5 m
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Q. A 10 kg object is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
11.25 m
B.
22.5 m
C.
15 m
D.
7.5 m
Show solution
Solution
Using energy conservation: KE_initial = PE_max; 1/2 * m * v^2 = m * g * h; h = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m
Correct Answer:
A
— 11.25 m
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Q. A 10 Ω resistor and a 20 Ω resistor are connected in series. What is the total resistance?
A.
10 Ω
B.
20 Ω
C.
30 Ω
D.
5 Ω
Show solution
Solution
In series, the total resistance R = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.
Correct Answer:
C
— 30 Ω
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Q. A 10-ohm resistor and a 20-ohm resistor are connected in series. What is the total resistance?
A.
10 ohms
B.
20 ohms
C.
30 ohms
D.
5 ohms
Show solution
Solution
In series, the total resistance R = R1 + R2 = 10 + 20 = 30 ohms.
Correct Answer:
C
— 30 ohms
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Q. A 10-ohm resistor has a voltage of 20 volts across it. What is the current flowing through the resistor?
A.
1 A
B.
2 A
C.
3 A
D.
4 A
Show solution
Solution
Using Ohm's Law, I = V / R = 20 V / 10 Ω = 2 A.
Correct Answer:
B
— 2 A
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Q. A 10-ohm resistor has a voltage of 50 volts across it. What is the current flowing through the resistor?
A.
5 A
B.
10 A
C.
15 A
D.
20 A
Show solution
Solution
Using Ohm's Law, I = V / R = 50 V / 10 Ω = 5 A.
Correct Answer:
B
— 10 A
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Q. A 10-ohm resistor is connected to a 20-volt battery. What is the power dissipated by the resistor?
A.
40 W
B.
20 W
C.
10 W
D.
2 W
Show solution
Solution
Power (P) can be calculated using P = V^2 / R = 20^2 / 10 = 400 / 10 = 40 W.
Correct Answer:
A
— 40 W
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Q. A 1000 kg car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the average power exerted by the engine?
A.
2000 W
B.
4000 W
C.
5000 W
D.
6000 W
Show solution
Solution
First, calculate the work done: W = (1/2)mv^2 = (1/2)(1000 kg)(20 m/s)^2 = 200000 J. Then, power is P = W/t = 200000 J / 10 s = 20000 W.
Correct Answer:
C
— 5000 W
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Q. A 1000 W heater operates for 1 hour. How much energy does it consume?
A.
3600 J
B.
1000 J
C.
3600000 J
D.
100000 J
Show solution
Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 1 hour = 3600 seconds. Thus, E = 1000 W * 3600 s = 3600000 J.
Correct Answer:
C
— 3600000 J
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Q. A 1000 W heater operates for 2 hours. How much energy does it consume?
A.
7200000 J
B.
2000000 J
C.
3600000 J
D.
1000000 J
Show solution
Solution
Energy consumed can be calculated using E = P * t. Here, P = 1000 W and t = 2 hours = 7200 seconds. Thus, E = 1000 W * 7200 s = 7200000 J.
Correct Answer:
C
— 3600000 J
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Showing 31 to 60 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!
