Q. What is the formula for the equivalent resistance of two resistors R1 and R2 in series?
A.
R_eq = R1 + R2
B.
R_eq = R1 * R2 / (R1 + R2)
C.
R_eq = R1 - R2
D.
R_eq = R1 / R2
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Solution
The equivalent resistance of two resistors in series is simply the sum of their resistances: R_eq = R1 + R2.
Correct Answer: A — R_eq = R1 + R2
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Q. What is the formula for the force experienced by a charged particle moving in a magnetic field?
A.
F = qvB sin(θ)
B.
F = qvB cos(θ)
C.
F = qB
D.
F = qE
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Solution
The force on a charged particle moving in a magnetic field is given by F = qvB sin(θ), where θ is the angle between the velocity and the magnetic field.
Correct Answer: A — F = qvB sin(θ)
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Q. What is the formula for the fringe separation in a double-slit experiment?
A.
λD/d
B.
d/λD
C.
D/λd
D.
λd/D
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Solution
The fringe separation (β) is given by the formula β = λD/d, where λ is the wavelength, D is the distance to the screen, and d is the distance between the slits.
Correct Answer: A — λD/d
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Q. What is the formula for the fringe width in a double-slit experiment?
A.
λD/d
B.
d/λD
C.
D/λd
D.
λd/D
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Solution
The fringe width (β) is given by the formula β = λD/d, where D is the distance from the slits to the screen and d is the distance between the slits.
Correct Answer: A — λD/d
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Q. What is the formula for the magnetic field (B) inside a solenoid?
A.
B = μ₀(nI)
B.
B = μ₀I
C.
B = μ₀(nI)/L
D.
B = μ₀I²
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Solution
The magnetic field inside a solenoid is given by B = μ₀(nI), where n is the number of turns per unit length.
Correct Answer: A — B = μ₀(nI)
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Q. What is the formula for the magnetic force experienced by a charged particle moving in a magnetic field?
A.
F = qE
B.
F = qvBsin(θ)
C.
F = mv^2/r
D.
F = BIL
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Solution
The magnetic force on a charged particle is given by F = qvBsin(θ), where q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
Correct Answer: B — F = qvBsin(θ)
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Q. What is the formula for the magnetic force on a charge q moving with velocity v in a magnetic field B?
A.
F = qvB
B.
F = qvBsinθ
C.
F = qB
D.
F = qvBcosθ
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Solution
The magnetic force on a charge moving in a magnetic field is given by F = qvBsinθ, where θ is the angle between v and B.
Correct Answer: B — F = qvBsinθ
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Q. What is the formula for the magnetic force on a charged particle moving in a magnetic field?
A.
F = qE
B.
F = qvBsinθ
C.
F = mv^2/r
D.
F = qvE
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Solution
The magnetic force on a charged particle is given by F = qvBsinθ, where q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between v and B.
Correct Answer: B — F = qvBsinθ
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Q. What is the formula for the modulus of resilience?
A.
U = 1/2 * σ * ε
B.
U = σ * ε
C.
U = 1/2 * ε^2
D.
U = σ^2 / 2
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Solution
The modulus of resilience is given by U = 1/2 * σ * ε, where σ is the yield stress and ε is the yield strain.
Correct Answer: A — U = 1/2 * σ * ε
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Q. What is the formula for the total power in a series RLC circuit?
A.
P = VI
B.
P = I^2R
C.
P = V^2/R
D.
P = VI cos(φ)
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Solution
The total power in a series RLC circuit is given by P = VI cos(φ), where φ is the phase angle.
Correct Answer: D — P = VI cos(φ)
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Q. What is the formula for Young's modulus?
A.
Stress/Strain
B.
Strain/Stress
C.
Force/Area
D.
Area/Force
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Solution
Young's modulus (E) is defined as the ratio of stress to strain.
Correct Answer: A — Stress/Strain
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Q. What is the frequency of a sound wave with a wavelength of 0.5 m in air (speed of sound = 343 m/s)?
A.
686 Hz
B.
343 Hz
C.
171.5 Hz
D.
1500 Hz
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Solution
Frequency (f) = speed/wavelength = 343 m/s / 0.5 m = 686 Hz.
Correct Answer: A — 686 Hz
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Q. What is the frequency of a sound wave with a wavelength of 0.5 m traveling at 340 m/s?
A.
680 Hz
B.
340 Hz
C.
170 Hz
D.
850 Hz
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Solution
Frequency (f) = Speed (v) / Wavelength (λ) = 340 m/s / 0.5 m = 680 Hz.
Correct Answer: A — 680 Hz
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Q. What is the frequency of a wave if its period is 0.01 seconds?
A.
100 Hz
B.
10 Hz
C.
1 Hz
D.
0.1 Hz
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Solution
Frequency is the reciprocal of the period. Therefore, frequency = 1/period = 1/0.01 = 100 Hz.
Correct Answer: A — 100 Hz
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Q. What is the frequency of a wave if its period is 0.02 seconds?
A.
50 Hz
B.
100 Hz
C.
200 Hz
D.
25 Hz
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Solution
Frequency f is the reciprocal of the period T, given by f = 1/T. Therefore, f = 1/0.02 s = 50 Hz.
Correct Answer: B — 100 Hz
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Q. What is the frequency of a wave with a period of 0.01 seconds?
A.
100 Hz
B.
50 Hz
C.
200 Hz
D.
10 Hz
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Solution
Frequency f is the reciprocal of the period T. Therefore, f = 1/T = 1/0.01 s = 100 Hz.
Correct Answer: A — 100 Hz
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Q. What is the frequency of a wave with a wavelength of 2 m traveling at a speed of 340 m/s?
A.
170 Hz
B.
340 Hz
C.
680 Hz
D.
85 Hz
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Solution
Frequency f = Speed / Wavelength = 340 m/s / 2 m = 170 Hz.
Correct Answer: A — 170 Hz
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Q. What is the fringe width in a double-slit experiment if the wavelength of light is 500 nm, the distance between the slits is 0.2 mm, and the distance to the screen is 1 m?
A.
0.5 mm
B.
1 mm
C.
2 mm
D.
0.25 mm
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Solution
Fringe width β = λD/d = (500 x 10^-9 m)(1 m)/(0.2 x 10^-3 m) = 2.5 mm.
Correct Answer: A — 0.5 mm
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Q. What is the fundamental frequency of a pipe open at both ends if its length is 2 m?
A.
85 Hz
B.
170 Hz
C.
340 Hz
D.
425 Hz
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Solution
The fundamental frequency is given by f = v/λ. For a pipe open at both ends, λ = 2L = 4 m. Thus, f = 343 m/s / 4 m = 85.75 Hz, approximately 85 Hz.
Correct Answer: B — 170 Hz
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Q. What is the fundamental frequency of a pipe open at both ends that is 2 meters long?
A.
85 Hz
B.
170 Hz
C.
340 Hz
D.
425 Hz
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Solution
The fundamental frequency is given by f = v/λ; for a pipe open at both ends, λ = 2L, so f = v/(2L) = 343/(2*2) = 42.875 Hz.
Correct Answer: B — 170 Hz
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Q. What is the general form of the equation for a damped harmonic oscillator?
A.
x(t) = A cos(ωt)
B.
x(t) = A e^(-bt) cos(ωt)
C.
x(t) = A sin(ωt)
D.
x(t) = A e^(bt) cos(ωt)
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Solution
The equation x(t) = A e^(-bt) cos(ωt) describes the motion of a damped harmonic oscillator.
Correct Answer: B — x(t) = A e^(-bt) cos(ωt)
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Q. What is the general form of the equation of motion for a damped harmonic oscillator?
A.
m d²x/dt² + b dx/dt + kx = 0
B.
m d²x/dt² + kx = 0
C.
m d²x/dt² + b dx/dt = 0
D.
m d²x/dt² + b dx/dt + kx = F(t)
Show solution
Solution
The equation of motion for a damped harmonic oscillator includes a damping term and is given by m d²x/dt² + b dx/dt + kx = 0.
Correct Answer: A — m d²x/dt² + b dx/dt + kx = 0
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Q. What is the general form of the equation of motion for a damped oscillator?
A.
m d²x/dt² + b dx/dt + kx = 0
B.
m d²x/dt² + kx = 0
C.
m d²x/dt² + b dx/dt = 0
D.
m d²x/dt² + b dx/dt + kx = F(t)
Show solution
Solution
The equation of motion for a damped oscillator includes a damping term (b dx/dt) along with the restoring force (kx).
Correct Answer: A — m d²x/dt² + b dx/dt + kx = 0
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Q. What is the gravitational field strength at a distance 'R' from the center of a planet of radius 'R' and uniform density?
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Solution
Inside a uniform sphere, the gravitational field strength varies linearly with distance from the center, so at R/2 it is g/2.
Correct Answer: C — g/2
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Q. What is the gravitational field strength at a distance of 3R from the center of a planet of mass M and radius R?
A.
G*M/R²
B.
G*M/(3R)²
C.
G*M/(9R²)
D.
G*M/(6R²)
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Solution
g = GM/r², at 3R, g = GM/(3R)² = GM/9R².
Correct Answer: C — G*M/(9R²)
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Q. What is the gravitational field strength at a distance of 3R from the center of the Earth? (R = radius of Earth)
A.
g/9
B.
g/3
C.
g/6
D.
g/12
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Solution
Gravitational field strength (g') = g / (distance^2) = g / (3R)^2 = g / 9.
Correct Answer: A — g/9
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Q. What is the gravitational field strength at a distance of 3R from the center of a planet of radius R?
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Solution
g = GM/r²; at 3R, g = GM/(3R)² = G/9.
Correct Answer: A — G/9
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Q. What is the gravitational field strength at a distance of 4R from the center of a planet of mass M and radius R?
A.
GM/R²
B.
GM/4R²
C.
GM/16R²
D.
GM/8R²
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Solution
g = GM/r²; at 4R, g = GM/(4R)² = GM/16R².
Correct Answer: C — GM/16R²
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Q. What is the gravitational field strength at a distance of 4R from the center of a planet of radius R?
A.
G/16
B.
G/4
C.
G/2
D.
G
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Solution
Gravitational field strength g = GM/r². At 4R, g = GM/(4R)² = GM/16R² = G/16.
Correct Answer: A — G/16
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Q. What is the gravitational field strength at the surface of a planet of mass M and radius R?
A.
GM/R^2
B.
2GM/R^2
C.
GM/R
D.
G/R^2
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Solution
The gravitational field strength g at the surface of a planet is given by g = GM/R^2.
Correct Answer: A — GM/R^2
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