Physics Syllabus (JEE Main)
Q. A car travels 100 m in 4 seconds. What is its average speed?
A.
20 m/s
B.
25 m/s
C.
30 m/s
D.
35 m/s
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Solution
Average speed = total distance / total time = 100 m / 4 s = 25 m/s.
Correct Answer: B — 25 m/s
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Q. A car travels 100 m in 5 seconds. What is its average speed?
A.
20 m/s
B.
25 m/s
C.
15 m/s
D.
30 m/s
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Solution
Average speed = total distance / total time = 100 m / 5 s = 20 m/s.
Correct Answer: B — 25 m/s
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Q. A car travels 100 m north and then 100 m east. What is the magnitude of the displacement from the starting point? (2000)
A.
100 m
B.
141.42 m
C.
200 m
D.
50 m
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Solution
Displacement = √(100^2 + 100^2) = √20000 = 141.42 m.
Correct Answer: B — 141.42 m
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Q. A car travels around a circular track of radius 50 m at a speed of 15 m/s. What is the centripetal force acting on the car if its mass is 1000 kg?
A.
450 N
B.
225 N
C.
150 N
D.
75 N
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Solution
Centripetal force (F_c) = mv²/r = 1000 kg * (15 m/s)² / 50 m = 4500 N.
Correct Answer: A — 450 N
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Q. A car travels at 90 km/h and a truck at 60 km/h in opposite directions. What is the relative speed of the car with respect to the truck?
A.
30 km/h
B.
60 km/h
C.
150 km/h
D.
90 km/h
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Solution
Relative speed = Speed of car + Speed of truck = 90 km/h + 60 km/h = 150 km/h.
Correct Answer: C — 150 km/h
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Q. A car travels at a speed of 80 km/h and a bike travels at 60 km/h. If they start from the same point and travel in the same direction, how far apart will they be after 1 hour?
A.
20 km
B.
10 km
C.
30 km
D.
40 km
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Solution
Relative speed = 80 - 60 = 20 km/h. Distance apart after 1 hour = 20 km.
Correct Answer: A — 20 km
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Q. A car's speed is measured as 60 km/h with a relative error of 5%. What is the absolute error?
A.
3 km/h
B.
2 km/h
C.
4 km/h
D.
5 km/h
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Solution
Absolute error = Relative error * Measured value = 0.05 * 60 = 3 km/h.
Correct Answer: A — 3 km/h
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Q. A changing magnetic field induces a current in a closed loop. What is this phenomenon called?
A.
Electromagnetic induction
B.
Magnetic resonance
C.
Electrolysis
D.
Magnetism
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Solution
This phenomenon is known as electromagnetic induction, where a changing magnetic field induces an electromotive force (EMF) in a conductor.
Correct Answer: A — Electromagnetic induction
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Q. A changing magnetic field induces an electric field. This phenomenon is known as?
A.
Electromagnetic induction
B.
Electrostatics
C.
Magnetostatics
D.
Electrodynamics
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Solution
The phenomenon of a changing magnetic field inducing an electric field is known as electromagnetic induction.
Correct Answer: A — Electromagnetic induction
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Q. A charge of +10μC is placed at the origin. What is the electric potential at a point 2m away from the charge?
A.
4500 V
B.
2250 V
C.
5000 V
D.
1000 V
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Solution
V = k * q / r = (9 × 10^9) * (10 × 10^-6) / 2 = 4500 V.
Correct Answer: B — 2250 V
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Q. A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 0.1m in the direction of the field?
A.
0.5 J
B.
1 J
C.
2 J
D.
0.1 J
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Solution
Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 0.1 m = 0.5 J.
Correct Answer: A — 0.5 J
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Q. A charge of +10μC is placed in a uniform electric field of strength 500 N/C. What is the work done in moving the charge 2m in the direction of the field?
A.
10 J
B.
1 J
C.
100 J
D.
0.5 J
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Solution
Work done W = F * d = (E * q) * d = (500 N/C * 10 × 10^-6 C) * 2m = 0.01 J.
Correct Answer: A — 10 J
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Q. A charge of +2μC is placed in an electric field of 1000 N/C. What is the force experienced by the charge? (2000)
A.
2000 N
B.
2 N
C.
0.002 N
D.
1000 N
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Solution
The force F on a charge in an electric field is given by F = qE. Here, F = 2 * 10^-6 C * 1000 N/C = 0.002 N.
Correct Answer: A — 2000 N
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Q. A charge of +3μC is placed at the origin. What is the electric potential at a point 0.5m away?
A.
5400 V
B.
1800 V
C.
7200 V
D.
3600 V
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Solution
V = k * q / r = (9 × 10^9) * (3 × 10^-6) / 0.5 = 54000 V.
Correct Answer: D — 3600 V
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Q. A charge of +3μC is placed at the origin. What is the potential at a point 0.3m away?
A.
9000 V
B.
3000 V
C.
10000 V
D.
15000 V
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Solution
V = k * q / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) / 0.3 m = 9000 V.
Correct Answer: A — 9000 V
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Q. A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
A.
90 J
B.
60 J
C.
30 J
D.
45 J
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Solution
Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.
Correct Answer: B — 60 J
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Q. A charge of +4μC is placed at the origin. What is the electric field at a point 2m away on the x-axis?
A.
0 N/C
B.
450 N/C
C.
900 N/C
D.
1800 N/C
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Solution
E = k * |q| / r^2 = (9 × 10^9) * 4 × 10^-6 / (2)^2 = 450 N/C.
Correct Answer: C — 900 N/C
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Q. A charge of +5μC is placed at the origin. What is the electric field at a point 2m away along the x-axis?
A.
0 N/C
B.
1125 N/C
C.
2250 N/C
D.
4500 N/C
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Solution
E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m)² = 1125 N/C.
Correct Answer: C — 2250 N/C
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Q. A charge of +5μC is placed in an electric field of strength 2000 N/C. What is the force experienced by the charge?
A.
10N
B.
1N
C.
100N
D.
200N
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Solution
Force F = qE = (5 × 10^-6 C) * (2000 N/C) = 10 N.
Correct Answer: A — 10N
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Q. A charge of -2 μC is placed in an electric field of 1000 N/C. What is the potential energy of the charge? (2000)
A.
-2000 J
B.
2000 J
C.
0 J
D.
-1000 J
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Solution
Potential energy U = q * V = -2 × 10^-6 C * 1000 N/C = -0.002 J.
Correct Answer: A — -2000 J
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Q. A charge of -2μC is placed in an electric field of 500 N/C. What is the force acting on the charge?
A.
-1 N
B.
1 N
C.
-0.5 N
D.
0.5 N
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Solution
Force F = E * q = 500 N/C * -2 × 10^-6 C = -1 N.
Correct Answer: A — -1 N
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Q. A charge of -4 μC is placed in an electric field of 200 N/C. What is the potential energy of the charge?
A.
-800 μJ
B.
800 μJ
C.
400 μJ
D.
0 μJ
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Solution
Potential energy U = q * V = -4 × 10^-6 C * 200 N/C = -800 μJ.
Correct Answer: A — -800 μJ
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Q. A charge of 4 μC is placed at the origin. What is the electric potential at a point 3 m away?
A.
3000 V
B.
1200 V
C.
4000 V
D.
None of the above
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Solution
V = k * q / r = (9 × 10^9 N m²/C²) * (4 × 10^-6 C) / (3 m) = 1200 V.
Correct Answer: B — 1200 V
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Q. A charge of 4 μC is placed at the origin. What is the electric potential at a point (3, 4) m?
A.
300 V
B.
200 V
C.
100 V
D.
0 V
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Solution
Distance r = √(3² + 4²) = 5 m. V = k * q / r = (9 × 10^9) * (4 × 10^-6) / 5 = 720 V.
Correct Answer: B — 200 V
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Q. A charge of 5 μC is placed in an electric field of 2000 N/C. What is the electric potential energy of the charge?
A.
10 mJ
B.
1 mJ
C.
0.5 mJ
D.
2 mJ
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Solution
Electric potential energy (U) = Charge × Electric Field = 5 μC × 2000 N/C = 10 mJ.
Correct Answer: A — 10 mJ
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Q. A charge of 5 μC is placed in an electric field of 2000 N/C. What is the potential energy of the charge?
A.
10 mJ
B.
1 mJ
C.
0.5 mJ
D.
2 mJ
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Solution
Potential energy (U) = Charge × Electric Field × Distance. Assuming distance = 1 m, U = 5 μC × 2000 N/C = 10 mJ.
Correct Answer: A — 10 mJ
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Q. A charge Q is uniformly distributed over a spherical surface of radius R. What is the electric field at a point outside the sphere at distance r from the center?
A.
0
B.
Q/4πε₀r²
C.
Q/4πε₀R²
D.
Q/4πε₀R
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Solution
For points outside the sphere, the electric field behaves as if all the charge were concentrated at the center, so E = Q/4πε₀r².
Correct Answer: B — Q/4πε₀r²
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Q. A charged capacitor has a potential difference of 12 V across its plates. If the capacitance is 4 µF, what is the charge stored in the capacitor?
A.
48 µC
B.
12 µC
C.
3 µC
D.
24 µC
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Solution
Charge Q = C × V = 4 µF × 12 V = 48 µC.
Correct Answer: A — 48 µC
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Q. A charged particle enters a magnetic field perpendicularly. What is the path it will follow?
A.
Straight line
B.
Circular path
C.
Elliptical path
D.
Parabolic path
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Solution
A charged particle entering a magnetic field perpendicularly will follow a circular path due to the magnetic force acting as a centripetal force.
Correct Answer: B — Circular path
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Q. A charged particle moves from a point of higher electric potential to a point of lower electric potential. What happens to its kinetic energy?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Cannot be determined
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Solution
As the charged particle moves to a lower potential, it loses potential energy, which is converted into kinetic energy, thus increasing its kinetic energy.
Correct Answer: A — Increases
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