Physics Syllabus (JEE Main)
Q. In a series circuit, if one resistor has a resistance of 5 Ω and another has 10 Ω, what is the total resistance?
A.
5 Ω
B.
10 Ω
C.
15 Ω
D.
20 Ω
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Solution
In a series circuit, total resistance R_total = R1 + R2 = 5 Ω + 10 Ω = 15 Ω.
Correct Answer: C — 15 Ω
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Q. In a series circuit, if one resistor has a resistivity of 5 Ω·m and another has 10 Ω·m, what is the total resistance?
A.
5 Ω
B.
10 Ω
C.
15 Ω
D.
20 Ω
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Solution
In series, total resistance R_total = R1 + R2 = 5 + 10 = 15 Ω.
Correct Answer: C — 15 Ω
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Q. In a series circuit, if one resistor is removed, what happens to the total resistance?
A.
It increases.
B.
It decreases.
C.
It remains the same.
D.
It becomes zero.
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Solution
In a series circuit, removing a resistor decreases the total resistance because the total resistance is the sum of all resistances.
Correct Answer: B — It decreases.
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Q. In a series RLC circuit, at resonance, what is the relationship between inductive reactance and capacitive reactance?
A.
X_L > X_C
B.
X_L < X_C
C.
X_L = X_C
D.
X_L + X_C = 0
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Solution
At resonance in a series RLC circuit, the inductive reactance (X_L) equals the capacitive reactance (X_C), hence X_L = X_C.
Correct Answer: C — X_L = X_C
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Q. In a series RLC circuit, if the resistance is increased, what happens to the bandwidth of the resonance?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Increasing the resistance in a series RLC circuit decreases the bandwidth of the resonance because the quality factor (Q) is inversely proportional to resistance.
Correct Answer: B — Decreases
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Q. In a series RLC circuit, if the resistance is increased, what happens to the bandwidth?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
In a series RLC circuit, increasing the resistance decreases the bandwidth because the quality factor (Q) is inversely proportional to resistance.
Correct Answer: B — Decreases
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Q. In a series RLC circuit, if the resistance is increased, what happens to the bandwidth of the resonance peak?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
Increasing the resistance in a series RLC circuit decreases the bandwidth of the resonance peak.
Correct Answer: B — Decreases
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Q. In a series RLC circuit, what happens to the current at resonance?
A.
Maximum
B.
Minimum
C.
Zero
D.
Constant
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Solution
At resonance, the current is maximum in a series RLC circuit.
Correct Answer: A — Maximum
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Q. In a series RLC circuit, what happens to the current when the frequency is increased beyond the resonant frequency?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
Beyond the resonant frequency, the circuit becomes more inductive, causing the current to decrease.
Correct Answer: B — Decreases
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Q. In a series RLC circuit, what is the condition for resonance?
A.
R = 0
B.
L = C
C.
ωL = 1/ωC
D.
V = I
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Solution
The condition for resonance in a series RLC circuit is ωL = 1/ωC.
Correct Answer: C — ωL = 1/ωC
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Q. In a shear stress-strain relationship, what does the slope of the linear portion represent?
A.
Young's modulus
B.
Shear modulus
C.
Bulk modulus
D.
Tensile strength
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Solution
The slope of the linear portion of the shear stress-strain curve represents the shear modulus of the material.
Correct Answer: B — Shear modulus
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Q. In a simple harmonic motion, if the amplitude is halved, how does the total energy change?
A.
Remains the same
B.
Halves
C.
Doubles
D.
Quadruples
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Solution
Total energy E is proportional to the square of the amplitude. If amplitude is halved, energy is reduced to 1/4, hence it halves.
Correct Answer: B — Halves
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Q. In a simple harmonic motion, if the amplitude is increased, what happens to the total energy of the system?
A.
It decreases
B.
It remains the same
C.
It increases
D.
It becomes zero
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Solution
The total energy in simple harmonic motion is proportional to the square of the amplitude. If the amplitude increases, the total energy increases.
Correct Answer: C — It increases
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Q. In a simple harmonic motion, if the mass is increased while keeping the spring constant constant, what happens to the period?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Doubles
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Solution
The period (T) increases with mass (T = 2π√(m/k)).
Correct Answer: A — Increases
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Q. In a simple harmonic motion, if the mass is increased, what happens to the period?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Depends on the spring constant
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Solution
The period T is given by T = 2π√(m/k). If m increases, T increases.
Correct Answer: A — Increases
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Q. In a simple harmonic motion, the maximum displacement from the mean position is called?
A.
Amplitude
B.
Frequency
C.
Wavelength
D.
Period
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Solution
The maximum displacement from the mean position in simple harmonic motion is called Amplitude.
Correct Answer: A — Amplitude
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Q. In a simple harmonic motion, the maximum displacement from the mean position is called what?
A.
Amplitude
B.
Frequency
C.
Wavelength
D.
Period
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Solution
The maximum displacement from the mean position in simple harmonic motion is called the amplitude.
Correct Answer: A — Amplitude
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Q. In a simple harmonic motion, the phase difference between displacement and acceleration is:
A.
0 degrees
B.
90 degrees
C.
180 degrees
D.
270 degrees
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Solution
Acceleration is always opposite to displacement in SHM, hence 180 degrees.
Correct Answer: C — 180 degrees
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Q. In a simple harmonic motion, the restoring force is directly proportional to what?
A.
Displacement
B.
Velocity
C.
Acceleration
D.
Mass
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Solution
In SHM, the restoring force is directly proportional to the displacement from the mean position.
Correct Answer: A — Displacement
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Q. In a simple harmonic motion, the restoring force is directly proportional to which of the following?
A.
Displacement
B.
Velocity
C.
Acceleration
D.
Mass
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Solution
In simple harmonic motion, the restoring force is directly proportional to the displacement from the mean position.
Correct Answer: A — Displacement
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Q. In a simple harmonic motion, the velocity is maximum at which point?
A.
Mean position
B.
Amplitude
C.
Equilibrium position
D.
None of the above
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Solution
The velocity is maximum at the mean position where the displacement is zero.
Correct Answer: A — Mean position
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Q. In a simple harmonic oscillator, if the maximum speed is 4 m/s and the amplitude is 2 m, what is the angular frequency?
A.
2 rad/s
B.
4 rad/s
C.
6 rad/s
D.
8 rad/s
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Solution
Maximum speed (v_max) = ωA. Thus, ω = v_max/A = 4/2 = 2 rad/s.
Correct Answer: B — 4 rad/s
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Q. In a single-slit diffraction experiment, what happens to the width of the central maximum as the slit width decreases?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
As the slit width decreases, the central maximum becomes wider due to increased diffraction.
Correct Answer: A — It increases
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Q. In a single-slit diffraction pattern, how does the intensity of the central maximum compare to the first minimum?
A.
Equal
B.
Twice
C.
Four times
D.
Half
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Solution
The intensity of the central maximum is four times that of the first minimum in a single-slit diffraction pattern.
Correct Answer: C — Four times
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Q. In a single-slit diffraction pattern, how does the intensity of the first minimum compare to the intensity of the central maximum?
A.
It is equal
B.
It is half
C.
It is zero
D.
It is one-fourth
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Solution
The intensity at the first minimum is zero, while the central maximum has maximum intensity.
Correct Answer: C — It is zero
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Q. In a single-slit diffraction pattern, the width of the central maximum is 4 mm. If the slit width is halved, what will be the new width of the central maximum?
A.
2 mm
B.
4 mm
C.
8 mm
D.
16 mm
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Solution
The width of the central maximum is inversely proportional to the slit width. Halving the slit width doubles the width of the central maximum to 8 mm.
Correct Answer: C — 8 mm
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Q. In a single-slit diffraction pattern, what is the angle for the first minimum if the slit width is 0.5 mm and the wavelength of light is 600 nm?
A.
30°
B.
60°
C.
45°
D.
15°
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Solution
For single-slit diffraction, the first minimum occurs at sin θ = λ/a. Here, sin θ = 600 x 10^-9 m / 0.5 x 10^-3 m = 0.0012, θ ≈ 0.0698 rad ≈ 4°.
Correct Answer: C — 45°
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Q. In a single-slit diffraction pattern, what is the angle for the first minimum?
A.
sin(θ) = λ/a
B.
sin(θ) = 2λ/a
C.
sin(θ) = 3λ/a
D.
sin(θ) = 0
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Solution
The angle for the first minimum in single-slit diffraction is given by sin(θ) = λ/a, where a is the slit width.
Correct Answer: A — sin(θ) = λ/a
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Q. In a single-slit diffraction pattern, what is the angular position of the first minimum?
A.
sin(θ) = λ/a
B.
sin(θ) = 2λ/a
C.
sin(θ) = 3λ/a
D.
sin(θ) = 0
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Solution
The first minimum in a single-slit diffraction pattern occurs at sin(θ) = λ/a, where a is the width of the slit.
Correct Answer: A — sin(θ) = λ/a
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Q. In a single-slit diffraction pattern, what is the angular width of the central maximum if the slit width is 0.5 mm and the wavelength of light is 500 nm?
A.
0.1 rad
B.
0.2 rad
C.
0.3 rad
D.
0.4 rad
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Solution
Angular width = 2λ/a. Here, a = 0.5 mm = 500 μm, so angular width = 2 * 500 nm / 500 μm = 0.002 rad = 0.2 rad.
Correct Answer: B — 0.2 rad
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