To linearize pendulum data, a class plots the square of the period, T², on the vertical axis against the length L on the horizontal axis. Small-angle swings were used. Pendulum data L (m) T (s) T² (s²) 0.25 1.00 1.00 0.50 1.41 2.00 1.00 2.00 4.00 Based on these data, what value of g does the slope of the graph imply?

g = 4π²/slope = π² ≈ 9.9 m/s²

g = 4π² × slope ≈ 158 m/s²

g = 2π/√slope = π ≈ 3.1 m/s²

Students drop a bar magnet, north pole first, straight down through a vertical coil connected to a voltage sensor. The trace shows a positive pulse as the magnet enters the coil and then a negative pulse as it leaves. The negative pulse has a larger peak and a shorter duration. One student says: "The second pulse should be positive too, since the magnet is still moving in the same direction." Which explanation best accounts for the trace?

Flux rises on entry and falls on exit, so the EMF reverses; the magnet exits faster, so that pulse is larger and shorter

The leading south pole reverses the current as the magnet leaves; the exit pulse is larger because a south pole is stronger

Lenz's law slows the magnet inside the coil, so the exit pulse is reversed and larger because the magnet takes longer to leave

Flux is greatest as the magnet leaves, so the EMF reverses there; the exit pulse is larger because EMF is proportional to flux

A student measures the intensity I of light from a small bulb at several distances r. Intensity data r (m) I (W/m²) 1.0 16 2.0 4.0 4.0 1.0 Which graph of these data would be a straight line through the origin, and what would its slope be?

I versus 1/r², with a slope of 16 W

√I versus r, with a slope of 4 W/m²

I versus 1/r, with a slope of 16 W/m

I versus r², with a slope of 16 W/m⁴

Melting temperature of ice Pressure (atm) Melting temperature (°C) 1 0.00 100 −0.74 200 −1.5 A block of ice is held at −1.0°C while the pressure on it is raised from 1 atm to 200 atm. What happens to the ice?

It melts, because at 200 atm the melting temperature (−1.5°C) is below −1.0°C.

It stays solid, because higher pressure raises the melting temperature above −1.0°C.

It stays solid, because −1.0°C is below the 1 atm melting temperature of 0°C.

It sublimes, because −1.0°C is below the temperature of the triple point of water.

Students drive a mass–spring system with a motor at various frequencies and record the steady-state amplitude. Driven oscillator data Driving frequency (Hz) Amplitude (cm) 1.0 0.6 1.5 1.4 1.8 3.5 2.0 8.0 2.2 3.2 2.5 1.2 3.0 0.5 The students attach a light vane that dips into water to add damping and repeat the run. Which prediction is best supported by the physics of driven oscillators?

The peak amplitude will be smaller and the curve broader, with the peak still near 2.0 Hz

The peak will shift to a much higher frequency, with about the same peak amplitude

The peak will stay the same height but become narrower, centered at 2.0 Hz

The peak amplitude will be larger, because the vane adds mass that stores more energy

A simple AC generator has a 100-turn rectangular coil of area 0.020 m² rotating at an angular speed of 30 rad/s in a uniform 0.50 T magnetic field. What is the peak EMF of the generator, and what happens if the rotation rate is doubled?

30 V; both the peak EMF and the frequency double

1.0 V; both the peak EMF and the frequency double

30 V; the frequency doubles but the peak EMF stays the same

30 V; the peak EMF doubles but the frequency stays the same

Praxis Physics (5266)

Free Praxis Physics 5266 Practice Questions

10 sample questions for the Praxis Physics (5266), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Nature and Impact of Science and Engineering, Principles and Models of Matter and Energy, Mechanics, Electricity and Magnetism, Waves.

Question 1 · Nature and Impact of Science and Engineering
To linearize pendulum data, a class plots the square of the period, T², on the vertical axis against the length L on the horizontal axis. Small-angle swings were used.
Pendulum data
L (m) T (s) T² (s²)
0.25 1.00 1.00
0.50 1.41 2.00
1.00 2.00 4.00
Based on these data, what value of g does the slope of the graph imply?
- A.g = 4π² × slope ≈ 158 m/s²
- B.g = 4π²/slope = π² ≈ 9.9 m/s²
- C.g = slope = 4.0 m/s²
- D.g = 2π/√slope = π ≈ 3.1 m/s²
Show answer and explanation
Answer: B. g = 4π²/slope = π² ≈ 9.9 m/s²
Since T = 2π√(L/g), squaring gives T² = (4π²/g)L, so a plot of T² versus L is a line through the origin with slope 4π²/g. The slope is 4.00 s² ÷ 1.00 m = 4.0 s²/m, so g = 4π²/4.0 = π² ≈ 9.9 m/s². Reading g directly as the slope ignores the 4π² factor, multiplying by the slope inverts the relationship, and 2π/√slope comes from mixing the T-versus-L form with the linearized form.
Question 2 · Principles and Models of Matter and Energy
A 0.50 kg sample of an unknown solid is heated by a 500 W immersion heater. Assume all of the heater's output goes into the sample.
Heating data
Time (min) Temperature (°C)
0 −20
1 −10
2 0
4 0
6 0
7 0
8 5
9 10
What is the latent heat of fusion of the substance?
- A.5.0 × 10³ J/kg
- B.4.2 × 10⁵ J/kg
- C.7.5 × 10⁴ J/kg
- D.3.0 × 10⁵ J/kg
Show answer and explanation
Answer: D. 3.0 × 10⁵ J/kg
The temperature stays at 0°C from 2 min to 7 min, so melting takes 5 min = 300 s. Q = Pt = (500 W)(300 s) = 1.5 × 10⁵ J, and L = Q/m = 1.5 × 10⁵ J ÷ 0.50 kg = 3.0 × 10⁵ J/kg. 7.5 × 10⁴ J/kg multiplies by the mass instead of dividing, 5.0 × 10³ J/kg uses minutes instead of seconds, and 4.2 × 10⁵ J/kg uses the full 7 min instead of only the plateau.
Question 3 · Mechanics
Planet X has twice Earth's mass and twice Earth's radius. An astronaut has a mass of 50 kg.
What would the astronaut weigh on planet X? (Use g = 10 m/s² on Earth.)
- A.500 N
- B.1000 N
- C.250 N
- D.125 N
Show answer and explanation
Answer: C. 250 N
Surface gravity g = GM/R², so on planet X g = 10 × 2/2² = 5 m/s², and the 50 kg astronaut weighs 50 × 5 = 250 N. Considering only the doubled mass gives 1000 N; applying only the 1/R² factor to the 500 N Earth weight gives 125 N; 500 N assumes weight does not change between planets.
Question 4 · Mechanics
A block on a frictionless horizontal surface oscillates on a spring with amplitude A.
I. The block's velocity is zero. II. The net force on the block is zero. III. The block's acceleration has its maximum magnitude and points toward the equilibrium position.
Which statements are true at the instant the block is at its maximum displacement x = +A?
- A.II and III only
- B.I, II, and III
- C.I and III only
- D.I only
Show answer and explanation
Answer: C. I and III only
At x = ±A the block momentarily stops, so v = 0 (I true). The restoring force F = −kx has its largest magnitude there, so the net force is not zero (II false) and the acceleration a = −(k/m)x is maximum and points toward equilibrium (III true). Position and acceleration are 180° out of phase, and velocity is 90° out of phase with both; a common error is to assume zero velocity means zero force.
Question 5 · Electricity and Magnetism
Students drop a bar magnet, north pole first, straight down through a vertical coil connected to a voltage sensor. The trace shows a positive pulse as the magnet enters the coil and then a negative pulse as it leaves. The negative pulse has a larger peak and a shorter duration.
One student says: "The second pulse should be positive too, since the magnet is still moving in the same direction."
Which explanation best accounts for the trace?
- A.The leading south pole reverses the current as the magnet leaves; the exit pulse is larger because a south pole is stronger
- B.Lenz's law slows the magnet inside the coil, so the exit pulse is reversed and larger because the magnet takes longer to leave
- C.Flux rises on entry and falls on exit, so the EMF reverses; the magnet exits faster, so that pulse is larger and shorter
- D.Flux is greatest as the magnet leaves, so the EMF reverses there; the exit pulse is larger because EMF is proportional to flux
Show answer and explanation
Answer: C. Flux rises on entry and falls on exit, so the EMF reverses; the magnet exits faster, so that pulse is larger and shorter
By Faraday's law the EMF depends on the rate of change of flux. Flux through the coil increases as the north pole approaches and decreases as the magnet leaves, so the sign of the EMF reverses even though the magnet keeps moving down. The magnet gains speed while falling, so the flux changes faster on exit, producing a taller, narrower pulse. EMF is not proportional to the flux itself, the two poles of a bar magnet are equally strong, and a magnet that took longer to leave would give a wider, smaller pulse, not a narrower, larger one.
Question 6 · Nature and Impact of Science and Engineering
A student measures the intensity I of light from a small bulb at several distances r.
Intensity data
r (m) I (W/m²)
1.0 16
2.0 4.0
4.0 1.0
Which graph of these data would be a straight line through the origin, and what would its slope be?
- A.√I versus r, with a slope of 4 W/m²
- B.I versus 1/r², with a slope of 16 W
- C.I versus 1/r, with a slope of 16 W/m
- D.I versus r², with a slope of 16 W/m⁴
Show answer and explanation
Answer: B. I versus 1/r², with a slope of 16 W
Each doubling of r cuts I by a factor of 4 (16 → 4 → 1), so I ∝ 1/r², i.e. I = k/r². Plotting I against 1/r² gives a line through the origin with slope k = I·r² = (16 W/m²)(1 m²) = 16 W (check: 4 × 4 = 16 and 1 × 16 = 16). A plot of I versus 1/r would curve (I·r is 16, 8, 4, not constant), I versus r² decreases nonlinearly, and √I = 4/r is inversely, not linearly, related to r.
Question 7 · Principles and Models of Matter and Energy
Melting temperature of ice
Pressure (atm) Melting temperature (°C)
1 0.00
100 −0.74
200 −1.5
A block of ice is held at −1.0°C while the pressure on it is raised from 1 atm to 200 atm. What happens to the ice?
- A.It stays solid, because higher pressure raises the melting temperature above −1.0°C.
- B.It melts, because at 200 atm the melting temperature (−1.5°C) is below −1.0°C.
- C.It stays solid, because −1.0°C is below the 1 atm melting temperature of 0°C.
- D.It sublimes, because −1.0°C is below the temperature of the triple point of water.
Show answer and explanation
Answer: B. It melts, because at 200 atm the melting temperature (−1.5°C) is below −1.0°C.
The data show that water's solid–liquid boundary has a negative slope: raising the pressure lowers the melting temperature, because liquid water is denser than ice. At 200 atm ice melts at −1.5°C, so a sample held at −1.0°C is above its melting point and melts. The idea that pressure raises the melting point holds for most substances but contradicts these data, sublimation requires pressures below the triple point (0.006 atm), not above 1 atm, and the 1 atm melting point no longer applies at 200 atm.
Question 8 · Mechanics
The escape speed from Earth's surface is about 11 km/s. Planet Z has twice Earth's mass and half Earth's radius.
What is the escape speed from the surface of planet Z?
- A.44 km/s
- B.16 km/s
- C.11 km/s
- D.22 km/s
Show answer and explanation
Answer: D. 22 km/s
Escape speed is v = √(2GM/R), so it scales as √(M/R). For planet Z, M/R is 2 ÷ ½ = 4 times Earth's, so v = √4 × 11 = 22 km/s. 44 km/s forgets the square root, 11 km/s ignores the changes, and 16 km/s (≈ √2 × 11) accounts for only the doubled mass.
Question 9 · Mechanics
Students drive a mass–spring system with a motor at various frequencies and record the steady-state amplitude.
Driven oscillator data
Driving frequency (Hz) Amplitude (cm)
1.0 0.6
1.5 1.4
1.8 3.5
2.0 8.0
2.2 3.2
2.5 1.2
3.0 0.5
The students attach a light vane that dips into water to add damping and repeat the run. Which prediction is best supported by the physics of driven oscillators?
- A.The peak will shift to a much higher frequency, with about the same peak amplitude
- B.The peak amplitude will be smaller and the curve broader, with the peak still near 2.0 Hz
- C.The peak will stay the same height but become narrower, centered at 2.0 Hz
- D.The peak amplitude will be larger, because the vane adds mass that stores more energy
Show answer and explanation
Answer: B. The peak amplitude will be smaller and the curve broader, with the peak still near 2.0 Hz
Resonance occurs when the driving frequency matches the natural frequency, here about 2.0 Hz. Increased damping removes energy each cycle, so the peak amplitude drops and the response curve broadens; for light-to-moderate damping the peak frequency shifts only slightly (slightly lower, not higher). Damping cannot raise or sharpen the peak, and a light vane does not add significant mass.
Question 10 · Electricity and Magnetism
A simple AC generator has a 100-turn rectangular coil of area 0.020 m² rotating at an angular speed of 30 rad/s in a uniform 0.50 T magnetic field.
What is the peak EMF of the generator, and what happens if the rotation rate is doubled?
- A.1.0 V; both the peak EMF and the frequency double
- B.30 V; the frequency doubles but the peak EMF stays the same
- C.30 V; the peak EMF doubles but the frequency stays the same
- D.30 V; both the peak EMF and the frequency double
Show answer and explanation
Answer: D. 30 V; both the peak EMF and the frequency double
For a coil rotating in a uniform field, ε_max = NBAω = (100)(0.50 T)(0.020 m²)(30 rad/s) = 30 V. Doubling ω makes the flux change twice as fast, doubling the peak EMF, and the output completes twice as many cycles per second, doubling the frequency. The value 1.0 V is NBA, omitting ω, and the other options treat only one of the two effects of faster rotation.

Pendulum data
L (m)	T (s)	T² (s²)
0.25	1.00	1.00
0.50	1.41	2.00
1.00	2.00	4.00

Heating data
Time (min)	Temperature (°C)
0	−20
1	−10
2	0
4	0
6	0
7	0
8	5
9	10

Intensity data
r (m)	I (W/m²)
1.0	16
2.0	4.0
4.0	1.0

Melting temperature of ice
Pressure (atm)	Melting temperature (°C)
1	0.00
100	−0.74
200	−1.5

Driven oscillator data
Driving frequency (Hz)	Amplitude (cm)
1.0	0.6
1.5	1.4
1.8	3.5
2.0	8.0
2.2	3.2
2.5	1.2
3.0	0.5

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