Physics Syllabus (JEE Main)
Q. If a solid sphere of mass M and radius R is rotating about an axis through its center, what is its moment of inertia?
A.
2/5 MR^2
B.
3/5 MR^2
C.
1/2 MR^2
D.
1/3 MR^2
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Solution
The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer: A — 2/5 MR^2
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Q. If a solid sphere of radius R and mass M is rotating about an axis through its center, what is its moment of inertia?
A.
2/5 MR^2
B.
3/5 MR^2
C.
1/2 MR^2
D.
1/3 MR^2
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Solution
The moment of inertia of a solid sphere about its center is I = 2/5 MR^2.
Correct Answer: A — 2/5 MR^2
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Q. If a sound wave has a frequency of 440 Hz, what is its period?
A.
0.00227 s
B.
0.0045 s
C.
0.01 s
D.
0.1 s
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Solution
The period (T) is the inverse of frequency (f); T = 1/f = 1/440 ≈ 0.00227 s.
Correct Answer: A — 0.00227 s
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Q. If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string?
A.
2 N
B.
4 N
C.
8 N
D.
16 N
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Solution
Centripetal force (F_c) = mv²/r. Assuming mass (m) = 1 kg, F_c = 1(4²)/2 = 8 N.
Correct Answer: C — 8 N
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Q. If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string if the mass of the stone is 1 kg?
A.
2 N
B.
4 N
C.
8 N
D.
10 N
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Solution
Centripetal force = m(v²/r) = 1(4²/2) = 8 N. Tension in the string = 8 N.
Correct Answer: C — 8 N
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Q. If a system absorbs 100 J of heat and does 40 J of work, what is the change in internal energy?
A.
60 J
B.
40 J
C.
100 J
D.
140 J
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Solution
Using the first law of thermodynamics, ΔU = Q - W = 100 J - 40 J = 60 J.
Correct Answer: A — 60 J
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Q. If a system absorbs 200 J of heat and does 50 J of work, what is the change in internal energy?
A.
150 J
B.
250 J
C.
200 J
D.
50 J
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Solution
Using the First Law of Thermodynamics, ΔU = Q - W = 200 J - 50 J = 150 J.
Correct Answer: A — 150 J
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Q. If a system does 150 J of work on the surroundings and absorbs 100 J of heat, what is the change in internal energy?
A.
-50 J
B.
50 J
C.
100 J
D.
250 J
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Solution
Using the first law, ΔU = Q - W = 100 J - 150 J = -50 J.
Correct Answer: A — -50 J
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Q. If a temperature is recorded as 25°C with an error of ±0.5°C, what is the maximum possible temperature?
A.
25.5°C
B.
24.5°C
C.
26°C
D.
25°C
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Solution
Maximum possible temperature = Measured value + Error = 25 + 0.5 = 25.5°C.
Correct Answer: A — 25.5°C
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Q. If a temperature is recorded as 25°C with an error of ±0.5°C, what is the minimum temperature?
A.
24.5°C
B.
25°C
C.
25.5°C
D.
26°C
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Solution
Minimum temperature = Recorded temperature - Error = 25 - 0.5 = 24.5°C.
Correct Answer: A — 24.5°C
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Q. If a thermometer reads 100°C with an error of ±0.5°C, what is the maximum possible temperature?
A.
100.5°C
B.
99.5°C
C.
101°C
D.
100°C
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Solution
Maximum possible temperature = Measured value + Error = 100 + 0.5 = 100.5°C.
Correct Answer: A — 100.5°C
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Q. If a torque of 15 Nm is produced by a force acting at a distance of 0.3 m from the pivot, what is the magnitude of the force?
A.
50 N
B.
45 N
C.
40 N
D.
30 N
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Solution
Force = Torque / Distance = 15 Nm / 0.3 m = 50 N.
Correct Answer: A — 50 N
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Q. If a torque of 25 Nm is applied and the lever arm is 5 m, what is the angle at which the force is applied if the force is 10 N?
A.
0 degrees
B.
30 degrees
C.
60 degrees
D.
90 degrees
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Solution
Torque = Force × Distance × sin(θ) => 25 Nm = 10 N × 5 m × sin(θ) => sin(θ) = 0.5 => θ = 30 degrees.
Correct Answer: C — 60 degrees
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Q. If a torque of 25 Nm is applied to a wheel with a radius of 0.5 m, what is the force applied at the edge of the wheel?
A.
50 N
B.
40 N
C.
30 N
D.
20 N
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Solution
Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 50 N
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Q. If a torque of 25 Nm is generated by a force acting at a distance of 0.5 m, what is the force applied?
A.
50 N
B.
40 N
C.
30 N
D.
20 N
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Solution
Force = Torque / Distance = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 50 N
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Q. If a torque of 30 Nm is applied to a lever arm of 3 m, what is the force applied?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
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Solution
Force = Torque / Distance = 30 Nm / 3 m = 10 N.
Correct Answer: B — 10 N
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Q. If a torque of 5 Nm is applied to a rotating object with a moment of inertia of 2 kg·m², what is the angular acceleration?
A.
2.5 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
1 rad/s²
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Solution
Angular acceleration α = Torque / Moment of inertia = 5 Nm / 2 kg·m² = 2.5 rad/s².
Correct Answer: A — 2.5 rad/s²
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Q. If a torque of 8 Nm is produced by a force acting at a distance of 0.2 m, what is the force?
A.
20 N
B.
30 N
C.
40 N
D.
50 N
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Solution
Force = Torque / Distance = 8 Nm / 0.2 m = 40 N.
Correct Answer: A — 20 N
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Q. If a train travels at a speed of 72 km/h, how long will it take to cover a distance of 180 km?
A.
2 h
B.
2.5 h
C.
3 h
D.
3.5 h
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Solution
Time = distance / speed = 180 km / (72 km/h) = 2.5 h.
Correct Answer: C — 3 h
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Q. If a volume is measured as 200 L with an error of 5 L, what is the range of possible volumes?
A.
195 L to 205 L
B.
200 L to 205 L
C.
195 L to 210 L
D.
200 L to 210 L
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Solution
Range of possible volumes = 200 - 5 to 200 + 5 = 195 L to 205 L.
Correct Answer: A — 195 L to 205 L
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Q. If a weight is measured as 10 kg with a 0.2 kg error, what is the maximum possible true weight?
A.
10.2 kg
B.
9.8 kg
C.
10 kg
D.
10.4 kg
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Solution
Maximum true weight = Measured weight + Error = 10 kg + 0.2 kg = 10.2 kg.
Correct Answer: A — 10.2 kg
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Q. If a wheel of radius R rolls without slipping, what is the distance traveled by the center of mass after one complete rotation?
A.
2πR
B.
πR
C.
4R
D.
R/2
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Solution
The distance traveled by the center of mass after one complete rotation is equal to the circumference of the wheel, which is 2πR.
Correct Answer: A — 2πR
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Q. If a wire carries a current of 5 A and is placed in a magnetic field of 0.1 T, what is the force per unit length on the wire?
A.
0.5 N/m
B.
1 N/m
C.
0.2 N/m
D.
0.1 N/m
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Solution
Using the formula F = BIL, where B is the magnetic field, I is the current, and L is the length of the wire (1 m), F = 0.1 T * 5 A * 1 m = 0.5 N.
Correct Answer: B — 1 N/m
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Q. If a wire carries a current of 5 A, what is the magnetic field at a distance of 0.1 m from the wire?
A.
0.1 T
B.
0.2 T
C.
0.5 T
D.
1.0 T
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Solution
Using Ampere's Law, B = μ₀I/2πr, where μ₀ = 4π × 10^-7 Tm/A, we find B = (4π × 10^-7)(5)/(2π(0.1)) = 0.1 T.
Correct Answer: B — 0.2 T
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Q. If a wire carrying current I is bent into a semicircular arc of radius R, what is the magnetic field at the center of the arc?
A.
μ₀I/(4R)
B.
μ₀I/(2R)
C.
μ₀I/(8R)
D.
μ₀I/(πR)
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Solution
The magnetic field at the center of a semicircular arc of radius R carrying current I is given by B = μ₀I/(4R).
Correct Answer: A — μ₀I/(4R)
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Q. If a wire is bent into a semicircular shape, what is the magnetic field at the center of the semicircle due to current I?
A.
μ₀I/(4R)
B.
μ₀I/(2R)
C.
μ₀I/(πR)
D.
μ₀I/(8R)
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Solution
The magnetic field at the center of a semicircular wire carrying current I is given by B = μ₀I/(4R) for the semicircle, which is half of the full circular loop.
Correct Answer: C — μ₀I/(πR)
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Q. If a wire is bent into a semicircular shape, what is the magnetic field at the center of the semicircle due to the current I?
A.
μ₀I/(4R)
B.
μ₀I/(2R)
C.
μ₀I/(8R)
D.
μ₀I/(6R)
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Solution
The magnetic field at the center of a semicircular wire carrying current I is given by B = μ₀I/(4R).
Correct Answer: A — μ₀I/(4R)
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Q. If a wire is stretched and its length increases by 2%, what is the strain?
A.
0.02
B.
0.2
C.
2
D.
200
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Solution
Strain is defined as the change in length divided by the original length, so 2% = 0.02.
Correct Answer: A — 0.02
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Q. If a wire of length L and cross-sectional area A is stretched by a force F, what is the expression for the elongation?
A.
F * L / (A * E)
B.
A * F / (L * E)
C.
E * F / (A * L)
D.
L * E / (A * F)
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Solution
The elongation (ΔL) can be expressed as ΔL = F * L / (A * E).
Correct Answer: A — F * L / (A * E)
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Q. If a wire's length is doubled while keeping its cross-sectional area constant, how does its resistance change?
A.
Remains the same
B.
Doubles
C.
Halves
D.
Quadruples
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Solution
Resistance is directly proportional to length; doubling the length doubles the resistance.
Correct Answer: B — Doubles
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