Q. A particle moves in a circular path with a constant speed of 10 m/s. What is the angular velocity if the radius of the path is 2 m?
A.
2 rad/s
B.
5 rad/s
C.
10 rad/s
D.
20 rad/s
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Solution
Angular velocity (ω) = v/r = 10/2 = 5 rad/s.
Correct Answer: B — 5 rad/s
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Q. A particle moves in a circular path with a constant speed. Which of the following statements is true?
A.
The particle has zero acceleration
B.
The particle has constant acceleration
C.
The particle has centripetal acceleration
D.
The particle's velocity is constant
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Solution
The particle has centripetal acceleration directed towards the center of the circle.
Correct Answer: C — The particle has centripetal acceleration
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Q. A particle moves in a circular path with a radius of 10 m and completes one revolution in 5 seconds. What is the linear speed of the particle?
A.
2π m/s
B.
4π m/s
C.
10 m/s
D.
20 m/s
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Solution
Linear speed (v) = Circumference / Time = (2π * 10 m) / 5 s = 4π m/s.
Correct Answer: A — 2π m/s
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Q. A particle moves in a circular path with a radius of 10 m at a constant speed of 5 m/s. What is the period of the motion?
A.
2π s
B.
4π s
C.
10 s
D.
20 s
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Solution
Circumference = 2πr = 20π m. Time = distance/speed = 20π m / 5 m/s = 4π s.
Correct Answer: B — 4π s
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Q. A particle moves in a circular path with a radius of 10 m at a speed of 5 m/s. What is the time period of the motion?
A.
2π s
B.
4π s
C.
10 s
D.
20 s
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Solution
Time period (T) = (2πr)/v = (2π * 10)/5 = 4π s.
Correct Answer: A — 2π s
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Q. A particle moves in a circular path with a radius of 15 m and completes one revolution in 10 seconds. What is the linear speed of the particle?
A.
1.5 m/s
B.
3 m/s
C.
6 m/s
D.
9 m/s
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Solution
Linear speed (v) = 2πr/T = 2π(15 m)/10 s = 3 m/s.
Correct Answer: B — 3 m/s
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Q. A particle moves in a circular path with a radius of 2 m and completes one revolution in 4 seconds. What is the linear speed of the particle?
A.
1.57 m/s
B.
3.14 m/s
C.
6.28 m/s
D.
12.56 m/s
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Solution
Circumference = 2πr = 2π(2) = 4π m. Speed = Distance/Time = 4π/4 = π m/s ≈ 3.14 m/s.
Correct Answer: C — 6.28 m/s
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Q. A particle moves in a circular path with a radius of 4 m and completes one revolution in 8 seconds. What is the centripetal acceleration?
A.
0.5 m/s²
B.
1 m/s²
C.
2 m/s²
D.
4 m/s²
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Solution
Centripetal acceleration (a_c) = v²/r. First find v = 2πr/T = 2π(4)/8 = π m/s. Then a_c = (π)²/4 = 2.5 m/s².
Correct Answer: B — 1 m/s²
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Q. A particle moves in a circular path with a radius r and a constant speed v. If the speed is doubled, what happens to the angular momentum of the particle?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
Angular momentum L = mvr; if v is doubled, L also doubles.
Correct Answer: B — It doubles
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Q. A particle moves in a straight line under the influence of a constant force of 12 N. If the particle moves 3 m, what is the work done by the force?
A.
12 J
B.
24 J
C.
36 J
D.
48 J
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Solution
Work done = Force × Distance = 12 N × 3 m = 36 J.
Correct Answer: B — 24 J
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Q. A particle moves in a straight line under the influence of a constant force of 3 N. If it moves 6 m, what is the work done by the force?
A.
9 J
B.
12 J
C.
15 J
D.
18 J
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Solution
Work done = Force × Distance = 3 N × 6 m = 18 J.
Correct Answer: D — 18 J
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Q. A particle moves in a straight line under the influence of a constant force. If the initial kinetic energy is 100 J and the work done by the force is 50 J, what is the final kinetic energy?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Final kinetic energy = Initial kinetic energy + Work done = 100 J + 50 J = 150 J.
Correct Answer: C — 150 J
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Q. A particle moves in a straight line with a constant velocity. What is its angular momentum about a point not on the line of motion?
A.
Zero
B.
Constant
C.
Varies with time
D.
Depends on distance
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Solution
Angular momentum is constant as the particle moves with constant velocity.
Correct Answer: B — Constant
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Q. A particle moves in a straight line with a constant velocity. What is the angular momentum of the particle about a point not on the line of motion?
A.
Zero
B.
Depends on the distance from the point
C.
Infinite
D.
Constant
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Solution
The angular momentum is non-zero and depends on the distance from the point to the line of motion.
Correct Answer: B — Depends on the distance from the point
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Q. A particle moves in a straight line with a velocity v. What is its angular momentum about a point P located at a distance d from the line of motion?
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Solution
Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point P.
Correct Answer: B — mvd
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Q. A particle moves in a straight line with an acceleration of 2 m/s². If its initial velocity is 3 m/s, what will be its velocity after 5 seconds?
A.
10 m/s
B.
13 m/s
C.
15 m/s
D.
20 m/s
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Solution
Final velocity (v) = u + at = 3 + (2 * 5) = 13 m/s.
Correct Answer: B — 13 m/s
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Q. A particle moves in a straight line with an acceleration of 4 m/s². If its initial velocity is 2 m/s, what will be its velocity after 5 seconds?
A.
22 m/s
B.
20 m/s
C.
18 m/s
D.
16 m/s
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Solution
Using the formula: final velocity = initial velocity + acceleration * time. Final velocity = 2 + 4 * 5 = 22 m/s.
Correct Answer: A — 22 m/s
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Q. A particle moves in a straight line with an initial velocity of 10 m/s and accelerates at 2 m/s². What will be its velocity after 5 seconds?
A.
20 m/s
B.
30 m/s
C.
40 m/s
D.
50 m/s
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Solution
Final velocity = initial velocity + (acceleration * time) = 10 m/s + (2 m/s² * 5 s) = 20 m/s.
Correct Answer: B — 30 m/s
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Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle about the axis?
A.
mr
B.
mr^2
C.
m/r
D.
m/r^2
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Solution
The moment of inertia of a point mass about an axis is given by I = mr^2.
Correct Answer: B — mr^2
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Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle?
A.
mr
B.
mr^2
C.
m/r
D.
m/r^2
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Solution
The moment of inertia of a point mass is given by I = mr^2.
Correct Answer: B — mr^2
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Q. A particle of mass m is moving in a circular path of radius r with a constant speed v. What is the angular momentum of the particle about the center of the circle?
A.
mv
B.
mvr
C.
mr^2
D.
mv^2
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Solution
Angular momentum L = mvr, where v is the linear speed and r is the radius.
Correct Answer: B — mvr
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Q. A particle with charge q moves with velocity v in a magnetic field B. What is the expression for the magnetic force acting on the particle?
A.
F = qvB
B.
F = qvB sin(θ)
C.
F = qB
D.
F = qvB cos(θ)
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Solution
The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvB sin(θ), where θ is the angle between the velocity vector and the magnetic field vector.
Correct Answer: B — F = qvB sin(θ)
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Q. A pendulum is measured to have a length of 2.0 m with an uncertainty of ±0.1 m. What is the relative uncertainty in the length?
A.
5%
B.
10%
C.
2.5%
D.
1%
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Solution
Relative uncertainty = (Uncertainty / Measured value) * 100 = (0.1 / 2.0) * 100 = 5%.
Correct Answer: C — 2.5%
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Q. A pendulum of length 2 m swings from a height of 1 m. What is the speed at the lowest point of the swing? (g = 9.8 m/s²)
A.
4.4 m/s
B.
3.1 m/s
C.
2.8 m/s
D.
5.0 m/s
Show solution
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.4 m/s.
Correct Answer: A — 4.4 m/s
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Q. A pendulum swings back and forth with a period of 1 second. If the length of the pendulum is doubled, what will be the new period?
A.
1 s
B.
1.41 s
C.
2 s
D.
4 s
Show solution
Solution
The period of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T ≈ 1.41 s.
Correct Answer: B — 1.41 s
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Q. A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?
A.
2 m/s
B.
4 m/s
C.
6 m/s
D.
8 m/s
Show solution
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.
Correct Answer: B — 4 m/s
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Q. A pendulum swings from a height of 5 m. What is the speed at the lowest point of the swing?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.
Correct Answer: B — 10 m/s
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Q. A pendulum swings with a maximum angle of 30 degrees. What is the approximate period of the pendulum if its length is 1 m?
A.
1.0 s
B.
1.5 s
C.
2.0 s
D.
2.5 s
Show solution
Solution
The period T of a simple pendulum is given by T = 2π√(L/g). Here, L = 1 m and g = 9.8 m/s². Thus, T = 2π√(1/9.8) ≈ 2.0 s.
Correct Answer: C — 2.0 s
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Q. A pendulum swings with a period of 1 second. If the length of the pendulum is doubled, what will be the new period?
A.
1 s
B.
1.41 s
C.
2 s
D.
4 s
Show solution
Solution
The period T of a simple pendulum is given by T = 2π√(L/g). If L is doubled, T becomes T' = 2π√(2L/g) = √2 * T = 1.41 s.
Correct Answer: B — 1.41 s
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Q. A pendulum swings with a period of 1 second. If the length of the pendulum is increased by a factor of 4, what will be the new period?
A.
1 s
B.
2 s
C.
4 s
D.
√4 s
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Solution
The period T = 2π√(L/g). If L is increased by a factor of 4, T increases by a factor of √4 = 2.
Correct Answer: B — 2 s
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