Physics Syllabus (JEE Main)
Q. What is the wavelength of light if the first-order maximum occurs at an angle of 30° in a double-slit experiment with slit separation of 0.1 mm and a screen distance of 1 m?
A.
300 nm
B.
600 nm
C.
450 nm
D.
750 nm
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Solution
Using d sin θ = mλ, λ = d sin θ/m. For m=1, d=0.1 mm, θ=30°, λ = 0.1 mm * sin(30°) = 0.1 mm * 0.5 = 0.05 mm = 500 nm.
Correct Answer: B — 600 nm
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Q. What is the wavelength of light if the first-order maximum occurs at an angle of 30° in a double-slit experiment with slit separation of 0.1 mm and distance to screen of 1 m?
A.
0.5 mm
B.
0.3 mm
C.
0.2 mm
D.
0.1 mm
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Solution
Using d sin θ = mλ, λ = d sin θ/m. For m=1, d=0.1 mm, θ=30°, λ = 0.1 mm * sin(30°) = 0.1 mm * 0.5 = 0.05 mm = 0.2 mm.
Correct Answer: C — 0.2 mm
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Q. What is the wavelength of light if the first-order maximum occurs at an angle of 30° in a double-slit experiment with slit separation of 0.1 mm and screen distance of 1 m?
A.
0.5 mm
B.
0.6 mm
C.
0.7 mm
D.
0.8 mm
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Solution
Using d sin θ = mλ, λ = d sin θ/m. For m=1, d=0.1 mm, θ=30°, λ = 0.1 mm * sin(30°) = 0.1 mm * 0.5 = 0.05 mm = 0.5 mm.
Correct Answer: A — 0.5 mm
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Q. What is the wavelength of light with a frequency of 5 x 10^14 Hz?
A.
6 x 10^-7 m
B.
3 x 10^8 m
C.
5 x 10^-7 m
D.
1 x 10^-6 m
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Solution
Using the formula λ = c/f, where c = 3 x 10^8 m/s, λ = 3 x 10^8 / (5 x 10^14) = 6 x 10^-7 m.
Correct Answer: A — 6 x 10^-7 m
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Q. What is the wavelength of light with a frequency of 5 x 10^14 Hz? (Speed of light = 3 x 10^8 m/s)
A.
0.6 m
B.
0.5 m
C.
0.4 m
D.
0.3 m
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Solution
Wavelength λ = c / f = (3 x 10^8 m/s) / (5 x 10^14 Hz) = 0.6 x 10^-6 m = 0.6 µm.
Correct Answer: B — 0.5 m
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Q. What is the weight of a 10 kg mass on Earth (g = 9.8 m/s²)?
A.
9.8 N
B.
10 N
C.
98 N
D.
100 N
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Solution
Weight W = mg = 10 kg * 9.8 m/s² = 98 N.
Correct Answer: C — 98 N
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Q. What is the weight of a 10 kg object on Earth (g = 9.8 m/s²)?
A.
9.8 N
B.
10 N
C.
98 N
D.
100 N
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Solution
Weight W = mg = 10 kg * 9.8 m/s² = 98 N.
Correct Answer: C — 98 N
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Q. What is the weight of a 10 kg object on the surface of a planet where g = 12 m/s²?
A.
120 N
B.
100 N
C.
80 N
D.
60 N
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Solution
Weight W = mg = 10 * 12 = 120 N.
Correct Answer: A — 120 N
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Q. What is the weight of a 10 kg object on the surface of the Earth?
A.
10 N
B.
100 N
C.
1000 N
D.
1 N
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Solution
Weight W = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: B — 100 N
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Q. What is the weight of a 10 kg object on the surface of the Earth? (g = 9.8 m/s²)
A.
9.8 N
B.
19.6 N
C.
98 N
D.
0.98 N
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Solution
Weight W = m * g = 10 kg * 9.8 m/s² = 98 N
Correct Answer: B — 19.6 N
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Q. What is the weight of a 12 kg object on Earth?
A.
12 N
B.
120 N
C.
1.2 N
D.
0.12 N
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Solution
Weight W = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: B — 120 N
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Q. What is the work done against friction if a box is moved 5 m on a surface with a coefficient of friction of 0.2 and a normal force of 200 N?
A.
10 J
B.
20 J
C.
30 J
D.
40 J
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Solution
Frictional force = μ * N = 0.2 * 200 N = 40 N. Work done = frictional force * distance = 40 N * 5 m = 200 J.
Correct Answer: B — 20 J
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Q. What is the work done against friction when a 5 kg box is pushed 10 m across a surface with a coefficient of kinetic friction of 0.4?
A.
20 J
B.
40 J
C.
50 J
D.
60 J
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Solution
Work done against friction = frictional force * distance = (μk * m * g) * d = (0.4 * 5 kg * 9.8 m/s²) * 10 m = 196 J, approximately 40 J.
Correct Answer: B — 40 J
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Q. What is the work done by a constant force of 10 N acting on an object that moves 5 m in the direction of the force?
A.
10 J
B.
50 J
C.
5 J
D.
25 J
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Solution
Work done (W) = Force (F) * Distance (d) = 10 N * 5 m = 50 J.
Correct Answer: B — 50 J
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Q. What is the work done by a constant force of 10 N acting over a distance of 5 m in the direction of the force?
A.
50 J
B.
10 J
C.
5 J
D.
25 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. What is the work done by a constant force of 10 N moving an object 5 m in the direction of the force?
A.
10 J
B.
25 J
C.
50 J
D.
100 J
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Solution
Work done (W) = Force (F) × Distance (d) = 10 N × 5 m = 50 J.
Correct Answer: C — 50 J
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Q. What is the work done by a force of 10 N moving an object 5 m in the direction of the force?
A.
50 J
B.
10 J
C.
5 J
D.
0 J
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Solution
Work done W = Force x Distance = 10 N x 5 m = 50 J.
Correct Answer: A — 50 J
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Q. What is the work done by a force of 5 N moving an object 3 m in the direction of the force?
A.
15 J
B.
5 J
C.
3 J
D.
0 J
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Solution
Work done W = Force × Distance = 5 N × 3 m = 15 J.
Correct Answer: A — 15 J
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Q. What is the work done by a gas during an isobaric expansion?
A.
Zero
B.
PΔV
C.
ΔU
D.
Q
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Solution
In an isobaric process, the work done by the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume.
Correct Answer: B — PΔV
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Q. What is the work done by a gas during expansion against a constant external pressure?
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Solution
The work done by a gas during expansion against a constant external pressure is given by W = PΔV, where P is the pressure and ΔV is the change in volume.
Correct Answer: A — PΔV
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Q. What is the work done by a gas during expansion at constant pressure?
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Solution
The work done by a gas during expansion at constant pressure is given by W = PΔV.
Correct Answer: A — PΔV
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Q. What is the work done by a gas during isothermal expansion?
A.
Zero
B.
Depends on the temperature
C.
Is equal to the heat absorbed
D.
Is equal to the change in internal energy
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Solution
During isothermal expansion, the work done by the gas is equal to the heat absorbed, as the internal energy remains constant.
Correct Answer: C — Is equal to the heat absorbed
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Q. What is the work done by an ideal gas during an isobaric process?
A.
PΔV
B.
nRT
C.
0
D.
nRΔT
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Solution
The work done by an ideal gas during an isobaric process is given by W = PΔV.
Correct Answer: A — PΔV
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Q. What is the work done by an ideal gas during an isothermal expansion?
A.
Zero
B.
nRT ln(Vf/Vi)
C.
nRT (Vf - Vi)
D.
nR (Tf - Ti)
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Solution
The work done by an ideal gas during an isothermal expansion is given by W = nRT ln(Vf/Vi).
Correct Answer: B — nRT ln(Vf/Vi)
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Q. What is the work done by friction when a 5 kg block slides 2 m on a surface with a coefficient of kinetic friction of 0.4?
A.
-4 N·m
B.
-8 N·m
C.
-10 N·m
D.
-20 N·m
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Solution
Frictional force F_friction = μk * N = 0.4 * 5 kg * 9.8 m/s² = 19.6 N. Work done by friction = -F_friction * distance = -19.6 N * 2 m = -39.2 N·m, approximately -40 N·m.
Correct Answer: B — -8 N·m
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Q. What is the work done by the system during an isochoric process?
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the temperature
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Solution
In an isochoric process, the volume remains constant, so no work is done by the system (W = PΔV = 0).
Correct Answer: C — Zero
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Q. What is the work done by the system in an isochoric process?
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the temperature
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Solution
In an isochoric process, the volume remains constant, so no work is done by the system (W = PΔV = 0).
Correct Answer: C — Zero
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Q. What is the work done by the system when it expands against a constant external pressure?
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Solution
The work done by the system during expansion against a constant external pressure is given by W = PΔV.
Correct Answer: A — PΔV
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Q. What is the work done in moving a charge of +2μC from a point A to B in a uniform electric field of 500 N/C over a distance of 0.4m?
A.
400 J
B.
200 J
C.
100 J
D.
80 J
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Solution
Work done W = F * d = q * E * d = (2 × 10^-6 C) * (500 N/C) * (0.4 m) = 0.4 J = 80 J.
Correct Answer: D — 80 J
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Q. What is the work done in moving a charge of +2μC through a potential difference of 10V?
A.
20 μJ
B.
200 μJ
C.
2 μJ
D.
0.2 μJ
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Solution
W = q * V = 2 × 10^-6 C * 10 V = 20 × 10^-6 J = 20 μJ.
Correct Answer: B — 200 μJ
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