Physics Syllabus (JEE Main)
Q. A copper wire has a resistivity of 1.68 x 10^-8 Ω·m. What is the resistance of a 100 m long wire with a cross-sectional area of 1 mm²?
A.
1.68 Ω
B.
0.168 Ω
C.
0.0168 Ω
D.
16.8 Ω
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Solution
Resistance (R) = ρ * (L / A) = 1.68 x 10^-8 * (100 / 1 x 10^-6) = 1.68 Ω.
Correct Answer: A — 1.68 Ω
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Q. A cube encloses a charge Q at its center. What is the electric flux through one face of the cube?
A.
Q/ε₀
B.
Q/6ε₀
C.
Q/3ε₀
D.
Zero
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Solution
The total flux through the cube is Q/ε₀, and since there are 6 faces, the flux through one face is Q/6ε₀.
Correct Answer: B — Q/6ε₀
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Q. A cube of side length a has a charge Q at one of its corners. What is the electric flux through one face of the cube?
A.
Q/(6ε₀)
B.
Q/(12ε₀)
C.
Q/(8ε₀)
D.
Q/(4ε₀)
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Solution
The total flux through the cube is Q/ε₀, and since the charge is at one corner, the flux through one face is Q/(12ε₀).
Correct Answer: B — Q/(12ε₀)
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Q. A cube of side length a has a charge Q at one of its corners. What is the total electric flux through the cube?
A.
Q/ε₀
B.
Q/(6ε₀)
C.
Q/(12ε₀)
D.
0
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Solution
The charge at one corner contributes 1/8 of its flux to the cube. Thus, total flux = (1/8)Q/ε₀ = Q/(12ε₀).
Correct Answer: C — Q/(12ε₀)
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Q. A current-carrying conductor experiences a force in a magnetic field. This phenomenon is known as?
A.
Electromagnetic induction
B.
Lorentz force
C.
Faraday's law
D.
Ampere's law
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Solution
The force experienced by a current-carrying conductor in a magnetic field is known as the Lorentz force.
Correct Answer: B — Lorentz force
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Q. A current-carrying conductor experiences a force in a magnetic field. What is the direction of this force given by?
A.
Right-hand rule
B.
Left-hand rule
C.
Ampere's law
D.
Faraday's law
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Solution
The direction of the force on a current-carrying conductor in a magnetic field is given by the right-hand rule.
Correct Answer: A — Right-hand rule
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Q. A current-carrying wire is placed in a magnetic field. What is the direction of the force acting on the wire?
A.
Parallel to the wire
B.
Perpendicular to the wire and field
C.
Opposite to the field
D.
In the direction of the field
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Solution
The force on a current-carrying wire in a magnetic field is given by the right-hand rule and is perpendicular to both the wire and the magnetic field.
Correct Answer: B — Perpendicular to the wire and field
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Q. A cyclist accelerates from rest at a rate of 1 m/s². How far will he travel in 10 seconds?
A.
50 m
B.
100 m
C.
150 m
D.
200 m
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Solution
Using the formula: distance = initial velocity * time + 0.5 * acceleration * time². Distance = 0 + 0.5 * 1 * 10² = 50 m.
Correct Answer: B — 100 m
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Q. A cyclist accelerates from rest at a rate of 1 m/s². How far will he travel in 8 seconds?
A.
32 m
B.
40 m
C.
48 m
D.
64 m
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Solution
Using the formula: distance = initial velocity * time + 0.5 * acceleration * time². Distance = 0 + 0.5 * 1 * 8² = 32 m.
Correct Answer: C — 48 m
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Q. A cyclist accelerates from rest at a rate of 2 m/s². How far will he travel in 10 seconds?
A.
100 m
B.
50 m
C.
200 m
D.
150 m
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Solution
Using the formula: distance = ut + 0.5at². Here, u = 0, a = 2 m/s², t = 10 s. Distance = 0 + 0.5 * 2 * 10² = 100 m.
Correct Answer: A — 100 m
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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the average power output if the cyclist has a mass of 70 kg?
A.
140 W
B.
280 W
C.
560 W
D.
700 W
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Solution
Kinetic Energy = 0.5 × m × v² = 0.5 × 70 kg × (10 m/s)² = 3500 J. Power = Work / Time = 3500 J / 5 s = 700 W.
Correct Answer: C — 560 W
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Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the work done by the cyclist?
A.
800 J
B.
900 J
C.
1000 J
D.
1200 J
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Solution
Work done = Change in Kinetic Energy = 0.5 × mass × (final velocity² - initial velocity²) = 0.5 × 75 kg × (15 m/s)² = 843.75 J.
Correct Answer: C — 1000 J
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Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy at that speed?
A.
500 J
B.
750 J
C.
1000 J
D.
1250 J
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Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 75 kg × (15 m/s)² = 8437.5 J.
Correct Answer: C — 1000 J
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Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the effective weight of the cyclist?
A.
30 kg
B.
60 kg
C.
90 kg
D.
120 kg
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Solution
Weight = Work / Height = 300 J / 5 m = 60 N; mass = Weight / g = 60 N / 9.8 m/s² ≈ 6.12 kg.
Correct Answer: B — 60 kg
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Q. A cyclist exerts a force of 100 N to maintain a speed of 5 m/s. What is the power output of the cyclist?
A.
200 W
B.
500 W
C.
1000 W
D.
250 W
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Solution
Power can be calculated using P = F * v. Here, F = 100 N and v = 5 m/s. Thus, P = 100 N * 5 m/s = 500 W.
Correct Answer: B — 500 W
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Q. A cyclist is moving around a circular track of radius 100 m. If he completes one lap in 40 seconds, what is his average speed?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Circumference = 2πr = 2π(100 m). Average speed = distance/time = (2π(100 m))/40 s ≈ 15.7 m/s.
Correct Answer: B — 10 m/s
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Q. A cyclist is moving at 15 m/s and a pedestrian at 5 m/s in the same direction. How fast does the cyclist appear to be moving to the pedestrian?
A.
10 m/s
B.
15 m/s
C.
20 m/s
D.
5 m/s
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Solution
Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.
Correct Answer: A — 10 m/s
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Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the speed of the cyclist relative to the pedestrian?
A.
10 m/s
B.
15 m/s
C.
5 m/s
D.
20 m/s
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Solution
Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.
Correct Answer: A — 10 m/s
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Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the relative speed of the pedestrian with respect to the cyclist?
A.
10 m/s
B.
5 m/s
C.
20 m/s
D.
15 m/s
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Solution
Relative speed = Speed of pedestrian - Speed of cyclist = 5 m/s - 15 m/s = -10 m/s (10 m/s behind).
Correct Answer: A — 10 m/s
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Q. A cyclist is moving at 15 m/s and passes a stationary observer. If the observer starts moving at 5 m/s in the same direction, what is the speed of the cyclist relative to the observer?
A.
10 m/s
B.
15 m/s
C.
20 m/s
D.
5 m/s
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Solution
Relative speed = Speed of cyclist - Speed of observer = 15 m/s - 5 m/s = 10 m/s.
Correct Answer: A — 10 m/s
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Q. A cyclist is moving at 15 m/s towards the east while a car is moving at 25 m/s towards the west. What is the relative speed of the cyclist with respect to the car?
A.
10 m/s
B.
15 m/s
C.
40 m/s
D.
25 m/s
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Solution
Relative speed = Speed of cyclist + Speed of car = 15 m/s + 25 m/s = 40 m/s.
Correct Answer: C — 40 m/s
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Q. A cyclist is moving at 15 m/s while a car is moving at 25 m/s in the same direction. What is the speed of the cyclist relative to the car?
A.
10 m/s
B.
15 m/s
C.
25 m/s
D.
40 m/s
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Solution
Relative speed = Speed of car - Speed of cyclist = 25 m/s - 15 m/s = 10 m/s.
Correct Answer: A — 10 m/s
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Q. A cyclist is moving at a speed of 15 km/h. If a car is moving in the same direction at 30 km/h, what is the relative speed of the car with respect to the cyclist?
A.
15 km/h
B.
30 km/h
C.
45 km/h
D.
0 km/h
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Solution
Relative speed = speed of car - speed of cyclist = 30 - 15 = 15 km/h.
Correct Answer: A — 15 km/h
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Q. A cyclist is moving in a circular path of radius 10 m at a speed of 5 m/s. What is the angle of banking required to prevent slipping?
A.
30°
B.
45°
C.
60°
D.
90°
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Solution
tan(θ) = v²/(rg) = (5²)/(10*9.8) = 0.25. θ = tan⁻¹(0.25) ≈ 14°.
Correct Answer: A — 30°
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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 5 m/s. What is the angular velocity of the cyclist?
A.
0.2 rad/s
B.
0.5 rad/s
C.
1 rad/s
D.
2 rad/s
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Solution
Angular velocity (ω) = v/r = 5 m/s / 15 m = 1/3 rad/s.
Correct Answer: B — 0.5 rad/s
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Q. A cyclist is moving in a circular path of radius 15 m with a speed of 6 m/s. What is the angular velocity of the cyclist?
A.
0.4 rad/s
B.
0.6 rad/s
C.
0.8 rad/s
D.
1.0 rad/s
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Solution
Angular velocity (ω) = v/r = 6 m/s / 15 m = 0.4 rad/s.
Correct Answer: B — 0.6 rad/s
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Q. A cyclist is moving in a circular track of radius 30 m with a speed of 15 m/s. What is the net force acting on the cyclist if the mass of the cyclist is 60 kg?
A.
180 N
B.
120 N
C.
90 N
D.
60 N
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Solution
Centripetal force F = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.
Correct Answer: A — 180 N
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Q. A cyclist is moving in a circular track of radius 30 m. If he completes one round in 12 seconds, what is his average speed?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Circumference = 2πr = 2π(30) = 60π m. Average speed = distance/time = 60π m / 12 s = 5 m/s.
Correct Answer: B — 10 m/s
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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the angular velocity of the cyclist?
A.
π/6 rad/s
B.
π/3 rad/s
C.
2π/6 rad/s
D.
2π/3 rad/s
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Solution
Angular velocity (ω) = 2π/T = 2π/12 = π/6 rad/s.
Correct Answer: B — π/3 rad/s
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Q. A cyclist is moving in a circular track of radius 30 m. If the cyclist completes one round in 12 seconds, what is the average speed of the cyclist?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
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Solution
Circumference = 2πr = 2π(30 m). Average speed = distance/time = (2π(30 m))/12 s ≈ 5 m/s.
Correct Answer: B — 10 m/s
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