Q. A tuning fork produces a sound wave of frequency 440 Hz. What is the wavelength of the sound wave in air at a temperature of 20°C?
A.
0.78 m
B.
0.34 m
C.
0.50 m
D.
0.25 m
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Solution
The speed of sound in air at 20°C is approximately 343 m/s. The wavelength λ is given by λ = v/f. Thus, λ = 343/440 ≈ 0.78 m.
Correct Answer: A — 0.78 m
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Q. A tuning fork produces a sound wave with a frequency of 440 Hz. What is the wavelength of the sound wave in air, given that the speed of sound in air is approximately 340 m/s?
A.
0.77 m
B.
0.85 m
C.
0.90 m
D.
1.00 m
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Solution
Wavelength λ is given by the formula λ = v/f. Here, v = 340 m/s and f = 440 Hz. Thus, λ = 340/440 = 0.7727 m, approximately 0.77 m.
Correct Answer: A — 0.77 m
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Q. A uniform electric field of 200 N/C is present. What is the potential difference between two points 3 m apart?
A.
600 V
B.
400 V
C.
200 V
D.
800 V
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Solution
The potential difference V = E * d = 200 N/C * 3 m = 600 V.
Correct Answer: A — 600 V
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Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod when it makes an angle θ with the vertical?
A.
√(g/L)(1-cosθ)
B.
√(2g/L)(1-cosθ)
C.
√(g/L)(1+cosθ)
D.
√(2g/L)(1+cosθ)
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Solution
Using conservation of energy, the potential energy lost equals the rotational kinetic energy gained. The angular velocity ω can be derived as ω = √(2g/L)(1-cosθ).
Correct Answer: B — √(2g/L)(1-cosθ)
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Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity just before it hits the ground?
A.
√(3g/L)
B.
√(2g/L)
C.
√(g/L)
D.
√(4g/L)
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Solution
Using conservation of energy, potential energy at the top = rotational kinetic energy at the bottom. mgh = (1/2)Iω^2. For a rod, I = (1/3)ML^2, h = L/2. Solving gives ω = √(3g/L).
Correct Answer: B — √(2g/L)
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Q. A uniform rod of length L and mass M is rotated about its center. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
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Solution
The moment of inertia of a uniform rod about its center is I = 1/12 ML^2.
Correct Answer: C — 1/2 ML^2
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Q. A uniform rod of length L is pivoted at one end. If it is allowed to fall freely, what is its angular acceleration just after it is released?
A.
g/L
B.
2g/L
C.
g/2L
D.
3g/2L
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Solution
The angular acceleration α = τ/I = (MgL/2)/(1/3 ML^2) = 3g/2L.
Correct Answer: B — 2g/L
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Q. A uniform thin circular ring of mass M and radius R is rotated about an axis through its center. What is its moment of inertia?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
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Solution
The moment of inertia of a thin circular ring about an axis through its center is I = MR^2.
Correct Answer: A — MR^2
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Q. A uniformly charged sphere of radius R has a total charge Q. What is the electric field at a point outside the sphere (r > R)?
A.
0
B.
Q/(4πε₀r²)
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center, thus E = Q/(4πε₀r²).
Correct Answer: B — Q/(4πε₀r²)
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Q. A vehicle moving with a speed of 60 km/h applies brakes and comes to a stop in 5 seconds. What is the deceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
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Solution
Convert speed to m/s: 60 km/h = 16.67 m/s. Deceleration = (final velocity - initial velocity) / time = (0 - 16.67) / 5 = -3.33 m/s², approximately 3 m/s².
Correct Answer: B — 3 m/s²
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Q. A voltage is measured as 12.0 V with an uncertainty of ±0.1 V. What is the absolute error?
A.
0.1 V
B.
0.2 V
C.
0.3 V
D.
0.4 V
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Solution
Absolute error is the same as the uncertainty, which is ±0.1 V.
Correct Answer: A — 0.1 V
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Q. A voltage is measured as 12.0 V with an uncertainty of ±0.2 V. What is the maximum possible voltage?
A.
11.8 V
B.
12.0 V
C.
12.2 V
D.
12.4 V
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Solution
Maximum possible voltage = Measured value + Uncertainty = 12.0 + 0.2 = 12.2 V.
Correct Answer: C — 12.2 V
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Q. A voltage is measured as 15.0 V with an uncertainty of ±0.3 V. What is the fractional error in the voltage measurement?
A.
0.02
B.
0.03
C.
0.01
D.
0.05
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Solution
Fractional error = (absolute error / measured value) = 0.3 / 15.0 = 0.02.
Correct Answer: B — 0.03
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Q. A voltage measurement is taken as 220 V with an uncertainty of ±5 V. What is the absolute error?
A.
5 V
B.
10 V
C.
2.5 V
D.
0.5 V
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Solution
The absolute error is given directly as ±5 V.
Correct Answer: A — 5 V
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Q. A volume is measured as 1000 mL with an error of 10 mL. What is the relative error?
A.
0.01
B.
0.1
C.
0.001
D.
0.05
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Solution
Relative error = Absolute error / Measured value = 10 / 1000 = 0.01.
Correct Answer: B — 0.1
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Q. A volume is measured as 2.0 L with an uncertainty of ±0.1 L. If this volume is used to calculate density, what is the uncertainty in density if mass is measured as 4.0 kg with an uncertainty of ±0.2 kg?
A.
0.1 kg/L
B.
0.2 kg/L
C.
0.05 kg/L
D.
0.4 kg/L
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Solution
Using the formula for density (density = mass/volume), the uncertainty in density can be calculated using the formula for propagation of uncertainty.
Correct Answer: A — 0.1 kg/L
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Q. A volume is measured as 200 L with an error of 5 L. What is the relative error in percentage?
A.
2.5%
B.
5%
C.
1%
D.
0.5%
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Solution
Relative error = (Absolute error / Measured value) * 100 = (5 / 200) * 100 = 2.5%
Correct Answer: A — 2.5%
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Q. A wave has a frequency of 50 Hz and a wavelength of 2 m. What is its speed?
A.
25 m/s
B.
50 m/s
C.
100 m/s
D.
75 m/s
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Solution
Speed v = f × λ = 50 Hz × 2 m = 100 m/s.
Correct Answer: B — 50 m/s
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Q. A wave has a frequency of 50 Hz and a wavelength of 3 m. What is its speed?
A.
150 m/s
B.
100 m/s
C.
50 m/s
D.
200 m/s
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Solution
Speed v = f × λ = 50 Hz × 3 m = 150 m/s.
Correct Answer: B — 100 m/s
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Q. A wave has a wavelength of 0.5 m and a frequency of 600 Hz. What is the speed of the wave?
A.
300 m/s
B.
600 m/s
C.
1200 m/s
D.
1500 m/s
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Solution
The speed v of a wave is given by v = fλ. Here, v = 600 Hz * 0.5 m = 300 m/s.
Correct Answer: A — 300 m/s
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Q. A wave has an amplitude of 0.5 m and a frequency of 10 Hz. What is the maximum speed of a particle in the wave?
A.
3.14 m/s
B.
6.28 m/s
C.
9.42 m/s
D.
12.56 m/s
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Solution
The maximum speed v_max is given by v_max = ωA, where ω = 2πf. Thus, ω = 2π(10) = 20π and v_max = 20π(0.5) = 10π ≈ 31.42 m/s.
Correct Answer: B — 6.28 m/s
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Q. A wave is described by the equation y(x, t) = 0.2 cos(4x - 5t). What is the amplitude of the wave?
A.
0.1 m
B.
0.2 m
C.
0.3 m
D.
0.4 m
Show solution
Solution
The amplitude of the wave is the coefficient of the cosine function, which is 0.2 m.
Correct Answer: B — 0.2 m
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Q. A wave on a string is described by the equation y(x, t) = 0.05 cos(4x - 3t). What is the amplitude of the wave?
A.
0.05 m
B.
0.1 m
C.
0.2 m
D.
0.3 m
Show solution
Solution
The amplitude of the wave is the coefficient of the cosine function, which is 0.05 m.
Correct Answer: A — 0.05 m
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Q. A wave on a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 10t)). What is the speed of the wave?
A.
5 m/s
B.
10 m/s
C.
20 m/s
D.
25 m/s
Show solution
Solution
The speed of the wave v is given by v = fλ. Here, f = 10 Hz and λ = 2 m (from k = 2π/λ). Thus, v = 10 * 2 = 20 m/s.
Correct Answer: B — 10 m/s
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Q. A wave on a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 2t)). What is the speed of the wave?
A.
1 m/s
B.
2 m/s
C.
0.5 m/s
D.
4 m/s
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Solution
The speed of the wave v is given by v = ω/k. Here, ω = 2π(2) = 4π rad/s and k = 2π(0.5) = π rad/m. Thus, v = (4π)/(π) = 4 m/s.
Correct Answer: B — 2 m/s
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Q. A wave on a string is described by the equation y(x, t) = 0.1 sin(2πx - 4πt). What is the amplitude of the wave?
A.
0.1 m
B.
0.2 m
C.
0.4 m
D.
1 m
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Solution
The amplitude is the coefficient of the sine function, which is 0.1 m.
Correct Answer: A — 0.1 m
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Q. A wave on a string is described by the equation y(x, t) = A sin(kx - ωt). What does 'k' represent?
A.
Angular frequency
B.
Wave number
C.
Amplitude
D.
Phase constant
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Solution
'k' is the wave number, which is related to the wavelength of the wave.
Correct Answer: B — Wave number
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Q. A wave on a string is described by the equation y(x, t) = A sin(kx - ωt). What does 'A' represent?
A.
Wavelength
B.
Frequency
C.
Amplitude
D.
Wave number
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Solution
'A' represents the amplitude of the wave, which is the maximum displacement from the equilibrium position.
Correct Answer: C — Amplitude
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Q. A wave traveling along a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 4t)). What is the wave speed?
A.
2 m/s
B.
4 m/s
C.
8 m/s
D.
1 m/s
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Solution
The wave speed v can be found from the equation v = fλ. Here, f = 4 Hz and λ = 2 m (from k = 2π/λ, k = 1 m⁻¹). Thus, v = 4 Hz * 2 m = 8 m/s.
Correct Answer: B — 4 m/s
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Q. A wave traveling along a string is described by the equation y(x, t) = 0.1 sin(2π(0.5x - 2t)). What is the speed of the wave?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
4 m/s
Show solution
Solution
The speed of the wave v is given by v = fλ. Here, f = 2 Hz and λ = 1 m (from the wave equation). Thus, v = 2 * 1 = 2 m/s.
Correct Answer: B — 2 m/s
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