Physics Syllabus (JEE Main)
Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
8.66 Nm
D.
17.32 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer: C — 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
17.32 Nm
D.
34.64 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 17.32 Nm
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Q. A force of 25 N is applied at an angle of 30 degrees to the horizontal while moving an object 10 m. What is the work done?
A.
100 J
B.
125 J
C.
150 J
D.
175 J
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Solution
Work done = Force × Distance × cos(θ) = 25 N × 10 m × cos(30°) = 216.5 J.
Correct Answer: B — 125 J
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Q. A force of 25 N is applied at an angle of 60 degrees to the horizontal while moving an object 3 m. What is the work done?
A.
37.5 J
B.
50 J
C.
75 J
D.
100 J
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Solution
Work done = Force × Distance × cos(θ) = 25 N × 3 m × cos(60°) = 37.5 J.
Correct Answer: A — 37.5 J
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
A.
15 Nm
B.
30 Nm
C.
60 Nm
D.
52 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer: D — 52 Nm
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Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?
A.
3 m/s²
B.
6 m/s²
C.
9 m/s²
D.
12 m/s²
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Solution
Using F = ma, a = F/m = 30 N / 5 kg = 6 m/s².
Correct Answer: B — 6 m/s²
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
A.
10 Nm
B.
15 Nm
C.
20 Nm
D.
25 Nm
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Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
20 Nm
B.
40 Nm
C.
34.64 Nm
D.
69.28 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 34.64 Nm
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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
86.6 Nm
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Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 N·m
B.
50 N·m
C.
86.6 N·m
D.
100 N·m
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer: C — 86.6 N·m
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 Nm
B.
50 Nm
C.
43.3 Nm
D.
100 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.
Correct Answer: C — 43.3 Nm
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Q. A forced oscillator has a mass of 3 kg and is driven by a force of 12 N at a frequency of 2 Hz. What is the amplitude of the oscillation if the damping coefficient is 0.1 kg/s?
A.
0.1 m
B.
0.2 m
C.
0.3 m
D.
0.4 m
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Solution
Using F = mAω², we find A = F / (mω²) = 12 / (3*(2π*2)²) ≈ 0.2 m.
Correct Answer: B — 0.2 m
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Q. A gas at 300 K has an RMS speed of 400 m/s. What will be its RMS speed at 600 K?
A.
400 m/s
B.
400 sqrt(2) m/s
C.
800 m/s
D.
200 m/s
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Solution
The RMS speed is proportional to the square root of the temperature. Therefore, at 600 K, the RMS speed will be 400 * sqrt(600/300) = 400 * sqrt(2) m/s.
Correct Answer: B — 400 sqrt(2) m/s
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Q. A gas at 300 K has an RMS speed of 500 m/s. What will be its RMS speed at 600 K?
A.
500 m/s
B.
707 m/s
C.
1000 m/s
D.
250 m/s
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Solution
The RMS speed is proportional to the square root of the temperature. Therefore, v_rms at 600 K = 500 * sqrt(600/300) = 500 * sqrt(2) ≈ 707 m/s.
Correct Answer: B — 707 m/s
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Q. A gas expands isothermally at 300 K from a volume of 1 m³ to 2 m³. What is the work done by the gas?
A.
0 J
B.
300 J
C.
600 J
D.
150 J
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Solution
The work done in an isothermal expansion is W = nRT ln(Vf/Vi). Assuming 1 mole of gas, W = 1 * 8.31 * 300 * ln(2) ≈ 600 J.
Correct Answer: C — 600 J
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Q. A gas has an RMS speed of 500 m/s. If the molar mass of the gas is 0.02 kg/mol, what is the temperature of the gas?
A.
250 K
B.
500 K
C.
1000 K
D.
2000 K
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Solution
Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Substituting v_rms = 500 m/s and M = 0.02 kg/mol gives T = 500 K.
Correct Answer: B — 500 K
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Q. A height is measured as 180 cm with an error of 1 cm. What is the upper limit of the measurement?
A.
181 cm
B.
179 cm
C.
180 cm
D.
182 cm
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Solution
Upper limit = Measured value + Error = 180 + 1 = 181 cm
Correct Answer: A — 181 cm
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Q. A hollow cylinder with charge density λ is placed along the z-axis. What is the electric field at a point outside the cylinder?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
Zero
D.
λ/(ε₀r)
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Solution
For a hollow cylinder, the electric field at a point outside is E = λ/(2πε₀r) using Gauss's law.
Correct Answer: A — λ/(2πε₀r)
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Q. A hollow sphere has a charge +Q distributed uniformly on its surface. What is the electric field at a point inside the sphere?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere has a charge Q distributed uniformly over its surface. What is the electric field inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀)
D.
Q/(4πε₀R)
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Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere has a charge Q uniformly distributed on its surface. What is the electric field inside the sphere?
A.
Q/4πε₀R²
B.
0
C.
Q/ε₀
D.
Q/4πε₀
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Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere rolls down a slope of height h. What fraction of its potential energy is converted into translational kinetic energy at the bottom?
A.
1/3
B.
1/2
C.
2/3
D.
1
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Solution
For a hollow sphere, I = (2/3)mr^2. Using energy conservation, the translational kinetic energy is 2/3 of the potential energy at the top.
Correct Answer: C — 2/3
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Q. A hollow sphere rolls down an incline. If it starts from rest, what fraction of its total energy is translational at the bottom?
A.
1/3
B.
2/3
C.
1/2
D.
1/4
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Solution
For a hollow sphere, the translational kinetic energy at the bottom is 2/3 of the total energy, hence the fraction is 2/3.
Correct Answer: B — 2/3
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Q. A hollow sphere rolls down an incline. If its mass is m and radius is R, what is its moment of inertia?
A.
(2/5)mR^2
B.
(1/2)mR^2
C.
(2/3)mR^2
D.
(3/5)mR^2
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Solution
The moment of inertia of a hollow sphere about its center is I = (2/3)mR^2.
Correct Answer: C — (2/3)mR^2
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Q. A hollow sphere with charge Q has a radius R. What is the electric field at a point inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
D.
Q/(2πε₀R²)
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Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry, as per Gauss's law.
Correct Answer: B — 0
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Q. A hollow spherical conductor carries a charge Q. What is the electric field inside the cavity?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
Inside the cavity of a hollow conductor, the electric field is zero due to the shielding effect of the conductor.
Correct Answer: B — 0
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Q. A length is measured as 100 m with a possible error of 1 m. What is the percentage error?
A.
1%
B.
0.5%
C.
2%
D.
0.1%
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Solution
Percentage error = (Absolute error / True value) * 100 = (1 / 100) * 100 = 1%.
Correct Answer: A — 1%
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Q. A length is measured as 100.0 m with an uncertainty of ±0.5 m. If this length is used to calculate the area of a square, what is the uncertainty in the area?
A.
1 m²
B.
0.5 m²
C.
2 m²
D.
0.25 m²
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Solution
Area = L², so uncertainty in area = 2 * L * (uncertainty in L) = 2 * 100 * 0.5 = 100 m².
Correct Answer: A — 1 m²
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