Q. A long straight wire carries a uniform linear charge density λ. What is the electric field at a distance r from the wire?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
λ/(2πε₀r²)
D.
0
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Solution
Using Gauss's law for a cylindrical surface around the wire, the electric field E at a distance r is E = λ/(2πε₀r).
Correct Answer: A — λ/(2πε₀r)
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Q. A loop of wire is moved into a magnetic field at a constant speed. What happens to the induced EMF as it enters the field?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It becomes zero
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Solution
As the loop enters the magnetic field, the rate of change of magnetic flux increases, leading to an increase in induced EMF.
Correct Answer: A — It increases
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Q. A loop of wire is moved into a magnetic field at a constant speed. What happens to the induced EMF as the loop enters the field?
A.
Induced EMF increases
B.
Induced EMF decreases
C.
Induced EMF remains constant
D.
Induced EMF becomes zero
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Solution
As the loop enters the magnetic field, the magnetic flux through the loop increases, leading to an increase in induced EMF.
Correct Answer: A — Induced EMF increases
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Q. A loop of wire is moved into a magnetic field at a constant speed. What is the effect on the induced current as the loop enters the field?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It becomes zero
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Solution
As the loop enters the magnetic field, the area of the loop within the field increases, leading to an increase in magnetic flux and thus an increase in the induced current according to Faraday's law.
Correct Answer: A — It increases
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Q. A loop of wire is moved into a uniform magnetic field at a constant speed. What happens to the induced EMF as it enters the field?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It becomes zero
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Solution
As the loop enters the magnetic field, the area exposed to the magnetic field increases, leading to an increase in the induced EMF.
Correct Answer: A — It increases
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Q. A loop of wire is placed in a changing magnetic field. What phenomenon is observed?
A.
Electromagnetic induction
B.
Magnetic resonance
C.
Electrolysis
D.
Thermal conduction
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Solution
A changing magnetic field induces an electromotive force (EMF) in the loop, a phenomenon known as electromagnetic induction.
Correct Answer: A — Electromagnetic induction
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Q. A loop of wire is placed in a changing magnetic field. What phenomenon is this an example of?
A.
Electromagnetic induction
B.
Magnetic resonance
C.
Electrostatics
D.
Magnetic hysteresis
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Solution
This is an example of electromagnetic induction, where a changing magnetic field induces an electromotive force (EMF) in the loop of wire.
Correct Answer: A — Electromagnetic induction
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Q. A loop of wire is placed in a uniform magnetic field. If the angle between the field and the normal to the loop is 60 degrees, what is the effective magnetic flux?
A.
0.5 B A
B.
0.866 B A
C.
0.866 B A²
D.
B A
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Solution
Effective magnetic flux (Φ) = B * A * cos(θ) = B * A * cos(60°) = 0.5 B A.
Correct Answer: B — 0.866 B A
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Q. A loop of wire is placed in a uniform magnetic field. If the field strength is increased, what happens to the induced EMF?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It becomes zero
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Solution
According to Faraday's law, an increase in magnetic field strength leads to an increase in the induced EMF.
Correct Answer: A — It increases
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Q. A loop of wire is placed in a uniform magnetic field. What happens if the magnetic field strength is increased?
A.
Induced current flows in the loop
B.
No effect on the loop
C.
The loop will heat up
D.
The loop will move
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Solution
According to Faraday's law of electromagnetic induction, a change in magnetic field strength induces an electromotive force (EMF) in the loop, causing a current to flow.
Correct Answer: A — Induced current flows in the loop
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Q. A loop of wire is placed in a uniform magnetic field. What happens to the induced EMF if the area of the loop is increased?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Depends on the magnetic field strength
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Solution
According to Faraday's law of electromagnetic induction, the induced EMF is directly proportional to the rate of change of magnetic flux, which increases with an increase in the area of the loop.
Correct Answer: A — Increases
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Q. A machine does 300 J of work in 5 seconds. What is its power?
A.
60 W
B.
30 W
C.
50 W
D.
20 W
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Solution
Power (P) = Work / Time = 300 J / 5 s = 60 W
Correct Answer: A — 60 W
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Q. A machine does 500 J of work in 10 seconds. What is its power output?
A.
50 W
B.
100 W
C.
200 W
D.
500 W
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Solution
Power is calculated using the formula P = W/t. Here, W = 500 J and t = 10 s, so P = 500 J / 10 s = 50 W.
Correct Answer: B — 100 W
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Q. A machine does 500 J of work in 10 seconds. What is its power?
A.
50 W
B.
100 W
C.
200 W
D.
500 W
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Solution
Power = Work / Time = 500 J / 10 s = 50 W.
Correct Answer: B — 100 W
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Q. A machine does 600 J of work in 5 seconds. What is its power?
A.
120 W
B.
100 W
C.
150 W
D.
200 W
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Solution
Power (P) = Work / Time = 600 J / 5 s = 120 W
Correct Answer: A — 120 W
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Q. A magnetic field of 0.3 T is perpendicular to a circular loop of radius 0.1 m. What is the magnetic flux through the loop?
A.
0.03 Wb
B.
0.03 Tm²
C.
0.1 Wb
D.
0.1 Tm²
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Solution
Magnetic flux (Φ) = B * A = 0.3 T * π(0.1 m)² = 0.03 Wb.
Correct Answer: A — 0.03 Wb
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Q. A man can run at 5 km/h in still air. If he runs in a direction opposite to the wind blowing at 3 km/h, what is his effective speed?
A.
2 km/h
B.
5 km/h
C.
8 km/h
D.
3 km/h
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Solution
Effective speed = speed of man - speed of wind = 5 - 3 = 2 km/h.
Correct Answer: A — 2 km/h
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Q. A man can run at a speed of 5 m/s in still air. If he runs in a direction opposite to the wind blowing at 2 m/s, what is his effective speed?
A.
3 m/s
B.
5 m/s
C.
7 m/s
D.
10 m/s
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Solution
Effective speed = speed of man - speed of wind = 5 - 2 = 3 m/s.
Correct Answer: A — 3 m/s
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Q. A man can walk at a speed of 5 km/h. If he walks in a direction opposite to the wind blowing at 3 km/h, what is his effective speed?
A.
2 km/h
B.
5 km/h
C.
8 km/h
D.
3 km/h
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Solution
Effective speed = speed of man + speed of wind = 5 + 3 = 8 km/h.
Correct Answer: C — 8 km/h
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Q. A mass 'm' is lifted to a height 'h' above the Earth's surface. What is the change in gravitational potential energy?
A.
mgh
B.
mg(h + R)
C.
mgR
D.
0
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Solution
The change in gravitational potential energy when a mass m is lifted to a height h is given by ΔU = mgh.
Correct Answer: A — mgh
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Q. A mass attached to a spring oscillates with a damping coefficient of 0.3 kg/s. If the mass is 1 kg and the spring constant is 4 N/m, what is the damping ratio?
A.
0.1
B.
0.3
C.
0.5
D.
0.75
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Solution
Damping ratio (ζ) = c / (2√(mk)) = 0.3 / (2√(1*4)) = 0.3 / 4 = 0.075.
Correct Answer: B — 0.3
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Q. A mass attached to a spring oscillates with a frequency of 1 Hz. If the mass is increased, what happens to the frequency?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
The frequency of a mass-spring system is inversely proportional to the square root of the mass. Increasing the mass decreases the frequency.
Correct Answer: B — Decreases
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Q. A mass attached to a spring oscillates with a frequency of 2 Hz. What is the spring constant if the mass is 0.5 kg?
A.
8 N/m
B.
16 N/m
C.
32 N/m
D.
64 N/m
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Solution
Using the formula f = (1/2π)√(k/m), we can rearrange to find k = (2πf)²m. Thus, k = (2π(2))²(0.5) = 16 N/m.
Correct Answer: B — 16 N/m
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Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the angular frequency of the motion?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency (ω) = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass attached to a spring oscillates with a frequency of 3 Hz. What is the angular frequency?
A.
3 rad/s
B.
6 rad/s
C.
9 rad/s
D.
12 rad/s
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Solution
Angular frequency ω = 2πf = 2π(3) = 6 rad/s.
Correct Answer: B — 6 rad/s
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Q. A mass attached to a spring oscillates with a frequency of 5 Hz. What is the time period of the oscillation?
A.
0.1 s
B.
0.2 s
C.
0.5 s
D.
1.0 s
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Solution
The time period T is the reciprocal of frequency f. Thus, T = 1/f = 1/5 = 0.2 s.
Correct Answer: C — 0.5 s
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Q. A mass attached to a spring oscillates with a maximum speed of 4 m/s. If the spring constant is 100 N/m, what is the maximum displacement?
A.
0.1 m
B.
0.2 m
C.
0.4 m
D.
0.5 m
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Solution
Maximum speed v_max = ωA, where A is the amplitude. We have ω = √(k/m). Here, k = 100 N/m and m = 1 kg (assuming). Thus, A = v_max/ω = 4/10 = 0.4 m.
Correct Answer: C — 0.4 m
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Q. A mass attached to a spring oscillates with a period of 2 seconds. What is the angular frequency of the motion?
A.
0.5 rad/s
B.
1 rad/s
C.
3.14 rad/s
D.
6.28 rad/s
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Solution
Angular frequency ω = 2π/T = 2π/2 = π rad/s ≈ 3.14 rad/s.
Correct Answer: B — 1 rad/s
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Q. A mass attached to a spring oscillates with a period of 2 seconds. What is the frequency of the oscillation?
A.
0.25 Hz
B.
0.5 Hz
C.
1 Hz
D.
2 Hz
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Solution
Frequency (f) is the reciprocal of the period (T). f = 1/T = 1/2 = 0.5 Hz.
Correct Answer: B — 0.5 Hz
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Q. A mass is measured as 15.0 kg with an uncertainty of ±0.3 kg. If this mass is used to calculate the force (F = ma) with an acceleration of 9.8 m/s², what is the uncertainty in the force?
A.
0.3 N
B.
2.94 N
C.
0.5 N
D.
1.5 N
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Solution
Uncertainty in force = a * (uncertainty in mass) = 9.8 * 0.3 = 2.94 N.
Correct Answer: B — 2.94 N
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