Q. A box is pushed across a floor with a force of 50 N. If the coefficient of kinetic friction is 0.4, what is the net force acting on the box if the normal force is 100 N?
A.
10 N
B.
20 N
C.
30 N
D.
40 N
Solution
Frictional force = μk * N = 0.4 * 100 N = 40 N. Net force = applied force - frictional force = 50 N - 40 N = 10 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.3. If the normal force is 100 N, what is the net force acting on the box?
A.
20 N
B.
30 N
C.
50 N
D.
70 N
Solution
Frictional force = μk * N = 0.3 * 100 N = 30 N. Net force = applied force - frictional force = 50 N - 30 N = 20 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.4. What is the acceleration of the box if its mass is 10 kg?
A.
1 m/s²
B.
2 m/s²
C.
3 m/s²
D.
4 m/s²
Solution
Net force = applied force - frictional force. Frictional force = μ_k * N = 0.4 * 10 kg * 9.8 m/s² = 39.2 N. Net force = 50 N - 39.2 N = 10.8 N. Acceleration = F/m = 10.8 N / 10 kg = 1.08 m/s², approximately 1 m/s².
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. If the frequency of the AC source is increased, what happens to the capacitive reactance?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes infinite
Solution
Capacitive reactance (X_C) is given by X_C = 1/(2πfC). If the frequency (f) increases, X_C decreases.
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. What is the frequency if the capacitance is 10 microfarads?
A.
1 kHz
B.
10 kHz
C.
100 Hz
D.
1000 Hz
Solution
Capacitive reactance (X_C) is given by X_C = 1 / (2πfC). Rearranging gives f = 1 / (2πX_CC). Substituting X_C = 50 ohms and C = 10 x 10^-6 F gives f = 318.31 Hz, approximately 1 kHz.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is doubled, what happens to the voltage across the capacitor?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
Solution
When the distance is doubled, the capacitance decreases, leading to an increase in voltage since Q = CV is constant.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is increased, what happens to the charge?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not affect the charge.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the voltage is doubled?
A.
Charge doubles
B.
Charge halves
C.
Charge remains the same
D.
Charge quadruples
Solution
The charge on a capacitor is given by Q = C * V. If the voltage is doubled, the charge also doubles, assuming capacitance remains constant.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the distance between the plates is increased?
A.
Charge increases
B.
Charge decreases
C.
Charge remains the same
D.
Charge becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not change the charge.
Q. A capacitor of capacitance C is charged to a voltage V and then connected in parallel with another uncharged capacitor of capacitance C. What is the final voltage across the capacitors?
A.
V/2
B.
V
C.
2V
D.
0
Solution
When connected in parallel, the total charge is conserved. The final voltage across both capacitors is V/2.
Q. A capacitor of capacitance C is charged to a voltage V and then connected to another uncharged capacitor of capacitance C. What is the final voltage across both capacitors?
A.
V/2
B.
V
C.
2V
D.
0
Solution
When connected, charge redistributes between the two capacitors, resulting in a final voltage of V/2 across each.
Q. A capacitor of capacitance C is connected to a battery of voltage V. If the battery is removed and the capacitor is connected to another capacitor of capacitance 2C, what is the final voltage across the combination?
A.
V/3
B.
V/2
C.
V
D.
2V
Solution
When the charged capacitor C is connected to an uncharged capacitor 2C, the final voltage is V_final = Q_total / C_eq = V/(1 + 1/2) = V/3.
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!
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