AP Physics 2

AP Physics 2 Practice Test: Practice Test 1

Practice Test 1 for AP Physics 2: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

In a single closed loop, a $12\text{ V}$ battery (positive terminal on the left) is in series with $R_1=3.0\,\Omega$ and $R_2=5.0\,\Omega$ . Conventional current $I$ is clockwise. Traverse the loop clockwise starting at the battery’s negative terminal. Use the sign convention: crossing a battery from $-$ to $+$ is $+\mathcal{E}$ , from $+$ to $-$ is $-\mathcal{E}$ ; crossing a resistor in the direction of current is $-IR$ , opposite is $+IR$ . Which equation correctly represents Kirchhoff’s loop rule for this loop (energy conservation)?

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Question 1

$+12 - 3.0I = 0$
$-12 - 3.0I - 5.0I = 0$
$+12 - 3.0I - 5.0I = 0$ (correct answer)
$+12 + 3.0I + 5.0I = 0$

Explanation: This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around any closed loop must equal zero, reflecting energy conservation. Starting at the battery's negative terminal and traversing clockwise: we cross the battery from negative to positive (+12V), then cross R₁ in the direction of current (-3.0I), then cross R₂ in the direction of current (-5.0I). The equation becomes +12 - 3.0I - 5.0I = 0. Choice B (+12 + 3.0I + 5.0I = 0) incorrectly adds the resistor terms, suggesting the misconception that voltage increases across resistors. To solve loop problems systematically, choose a consistent traversal direction and apply sign conventions carefully: batteries give +ℰ when crossed from - to +, and resistors give -IR when crossed in the current direction.

Question 2

A $9\,\text{V}$ ideal battery drives a circuit where $R_1=3\,\Omega$ is in series with a parallel network between junctions J and K. In Branch 1, $R_2=2\,\Omega$ ; in Branch 2, $R_3=5\,\Omega$ . The branches rejoin at K and return to the battery. Which statement correctly compares the voltage drops across $R_2$ and $R_3$ ?

The voltage across $R_2$ is less than the voltage across $R_3$ .
The voltage across $R_2$ is zero because current splits at J.
The voltage across $R_2$ equals the voltage across $R_3$ . (correct answer)
The voltage across $R_2$ is greater than the voltage across $R_3$ .

Explanation: This problem tests understanding of compound DC circuits. The circuit has R₁ in series with a parallel network of R₂ and R₃ between junctions J and K. In parallel branches, the key principle is that voltage across parallel elements is always equal, regardless of their individual resistances. Both R₂ and R₃ experience the same voltage drop between junctions J and K, even though R₂ = 2Ω and R₃ = 5Ω have different resistances. Choice D incorrectly assumes voltage becomes zero when current splits, reflecting a fundamental misunderstanding of how voltage behaves in parallel circuits. Always remember: parallel elements share the same voltage, while series elements share the same current.

Question 3

A student says an electromagnetic wave in vacuum consists of an oscillating electric field that creates a magnetic field, and the magnetic field creates the electric field, with both fields perpendicular to the direction of travel. Which statement best describes electromagnetic waves?

They are transverse waves with perpendicular $E$ and $B$ fields that can propagate through vacuum. (correct answer)
They propagate by moving electrons back and forth, producing a net flow of charge.
They require a medium because the fields cannot sustain each other in empty space.
They are longitudinal waves because the fields oscillate along the direction of travel.

Explanation: This question tests understanding of electromagnetic waves. Electromagnetic waves are transverse waves consisting of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation. The student correctly describes how the changing electric field induces the magnetic field and vice versa, allowing self-propagation through vacuum without requiring a material medium. This mutual induction between E and B fields is what enables EM waves to travel through empty space. Choice A incorrectly claims EM waves are longitudinal, confusing them with sound waves where oscillations occur along the propagation direction. The key insight is that electromagnetic waves are self-sustaining transverse oscillations that can propagate through vacuum.

Question 4

Water waves in a ripple tank travel at a constant speed set by the water depth. The period of the wave is changed from $T$ to $3T$ by adjusting the source. Which statement correctly relates the new wavelength to the original wavelength?

The wavelength stays $\lambda$ because period does not affect wavelength.
The wavelength becomes $9\lambda$ because both period and speed triple.
The wavelength becomes $\tfrac{1}{3}\lambda$ because frequency increases.
The wavelength becomes $3\lambda$ because the wave speed stays constant. (correct answer)

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ connects wave speed, frequency, and wavelength, where frequency equals 1/period (f = 1/T). In this ripple tank, wave speed is constant because it depends only on water depth, not on the source. When period increases from T to 3T, frequency decreases to f/3, since frequency and period are inversely related. To maintain constant wave speed with one-third the frequency, wavelength must triple to 3λ. Choice A incorrectly assumes wavelength decreases when frequency decreases, missing that constant speed requires them to change oppositely. Remember: when wave speed is fixed, frequency and wavelength change inversely.

Question 5

Two waves travel in different media. Wave 1 has speed $v_1$ and frequency $f_1$ ; wave 2 has speed $2v_1$ and the same frequency $f_1$ . Which statement best describes the wavelengths?

$\lambda_2=\lambda_1$ because wavelength depends only on amplitude.
$\lambda_2=2\lambda_1$ because $\lambda=v/f$ at fixed frequency. (correct answer)
$\lambda_2=4\lambda_1$ because doubling speed doubles frequency as well.
$\lambda_2=\tfrac{1}{2}\lambda_1$ because higher speed implies shorter wavelength.

Explanation: This question tests understanding of properties of wave pulses and waves. The fundamental relationship v = fλ shows that at constant frequency, wavelength is directly proportional to wave speed. Since wave 2 has twice the speed of wave 1 but the same frequency, its wavelength must be twice as large: λ₂ = v₂/f = 2v₁/f = 2(v₁/f) = 2λ₁. Choice A incorrectly suggests higher speed means shorter wavelength, reflecting the misconception that speed and wavelength are inversely related. Remember that at fixed frequency, wavelength scales directly with wave speed.

Question 6

A train horn emits a constant tone as the train moves away from a person standing still. Compared to when the train is not moving, the person hears a frequency that is

the same because only source motion matters, not relative motion
higher because the speed of sound increases behind the train
lower because wavefronts are stretched as the source recedes (correct answer)
higher because the horn is louder when the train is moving

Explanation: This question tests understanding of the Doppler effect. When a sound source moves away from a stationary observer, each successive wavefront is emitted from a position farther from the observer, causing the wavefronts to spread out or stretch. This increased spacing between wave crests means the observer encounters them less frequently, resulting in a lower observed frequency compared to the source frequency. Choice D incorrectly confuses loudness (amplitude) with frequency (pitch), which are independent properties of sound waves. The transferable principle is that relative motion changes frequency, not wave speed or amplitude.

Question 7

A $1.5\,\text{V}$ ideal battery charges a capacitor $C=2.0\,\mu\text{F}$ , then is removed. Define the system as the capacitor only. The capacitor is then connected to an identical uncharged capacitor in parallel. Which statement correctly describes what happens to the electric potential energy stored in the capacitors?

It stays the same because total charge is conserved when the capacitors are connected.
It decreases because some energy must leave the capacitor system during charge redistribution. (correct answer)
It increases because the equivalent capacitance doubles when connected in parallel.
It becomes equal to the battery’s power output because the battery set the initial voltage.

Explanation: This problem tests conservation of electric energy. Initially, the charged 2.0μF capacitor stores U₁ = ½(2.0μF)(1.5V)² = 2.25μJ. When connected to an identical uncharged capacitor, charge redistributes until both reach the same voltage. The total charge Q = (2.0μF)(1.5V) = 3.0μC is conserved, giving final voltage Vf = Q/Ctotal = 3.0μC/4.0μF = 0.75V. The final total energy is Uf = ½(4.0μF)(0.75V)² = 1.125μJ, which is half the initial energy. Choice A incorrectly assumes charge conservation implies energy conservation. Choice C wrongly thinks parallel connection increases energy, missing that energy is lost during charge redistribution. The strategy is to recognize that connecting capacitors at different voltages always dissipates energy, even though charge is conserved.

Question 8

A beam travels in air and enters water at $35^\circ$ to the normal. Water has the higher refractive index ( $n_{water} > n_{air}$ ). Which option correctly describes the refraction direction at the boundary?

It bends away from the normal in water
It bends toward the normal in water (correct answer)
It bends away from the normal because the speed decreases
It bends toward the normal because the frequency increases

Explanation: This problem addresses refraction. When light travels from air (lower refractive index) to water (higher refractive index), the light wave slows down as it enters the denser medium. This speed reduction at the boundary causes the light ray to bend toward the normal. The refracted ray in water makes a smaller angle with the normal than the 35° incident angle. Choice C incorrectly states the ray bends away from the normal because speed decreases, revealing confusion about the relationship between speed change and bending direction. To predict refraction: when light enters a denser medium (higher n), it slows down and bends toward the normal.

Question 9

A polarized electromagnetic wave travels in vacuum. The electric field oscillates only along $\hat y$ , while the magnetic field oscillates only along $\hat z$ . Which statement is correct?

The wave is transverse because both fields are perpendicular to the direction of travel. (correct answer)
The wave is longitudinal because the electric field points along the direction of travel.
The wave must be in a medium because polarization cannot occur in vacuum.
The wave transports net charge because the electric field has a single direction.

Explanation: This question tests understanding of electromagnetic waves. Electromagnetic waves are transverse waves because both the electric and magnetic fields oscillate perpendicular to the direction of propagation. With E along ŷ and B along ẑ, both fields are perpendicular to the propagation direction (along x̂), confirming the transverse nature. Choice B incorrectly claims the wave is longitudinal—this misconception confuses EM waves with sound waves or assumes fields must oscillate along the propagation direction. Remember that EM waves are always transverse, with E ⊥ B ⊥ propagation direction.

Question 10

A student shines red light of very high intensity on a metal surface and observes no photoelectrons. Switching to dim blue light causes immediate emission of electrons. The student keeps the metal and setup unchanged. Which reasoning best accounts for the change?

Blue light has higher frequency, so each photon can exceed the work function (correct answer)
Blue light is dimmer, so electrons are less likely to be trapped
Red light needs more time for electrons to store energy from the wave
Any light will eject electrons if the intensity is high enough

Explanation: The photoelectric effect. Blue light has higher frequency than red light, meaning each blue photon carries more energy according to E = hf. Even though the red light has very high intensity (many photons per second), each red photon lacks sufficient energy to overcome the metal's work function. The dim blue light has fewer photons per second, but each blue photon has enough energy to eject an electron immediately upon absorption. Choice C incorrectly invokes the classical wave model where electrons could accumulate energy over time. The key principle is that photoelectric emission depends on individual photon energy (frequency), not the total energy delivered (intensity).

Question 11

A wire segment lies along the +y direction and carries $I$ upward. It is in a uniform magnetic field that points in the +y direction. Which statement best describes the magnetic force on the wire?

The force is in the +y direction.
The force is zero. (correct answer)
The force is in the +x direction.
The force is in the -y direction.

Explanation: This problem tests magnetism and current-carrying wires. The wire carries current in the +y direction (upward), and the magnetic field also points in the +y direction. When current and magnetic field are parallel (or antiparallel), the cross product I × B equals zero. Therefore, there is no magnetic force on the wire. Choice A incorrectly suggests a force in the +y direction, which represents the misconception that parallel vectors produce a force in their common direction. Remember that magnetic force requires current and field to have some perpendicular component.

Question 12

X-rays of wavelength $\lambda$ scatter from electrons initially at rest. A student finds $\Delta\lambda$ depends on scattering angle but not on the target material. Which conclusion about light is supported by this wavelength change behavior?

The shift comes from photon–electron collisions, set by momentum conservation. (correct answer)
The shift comes from wave refraction, so it should depend strongly on material index.
The shift is an interference artifact that depends mainly on crystal spacing.
The shift occurs because electrons absorb photons and later emit longer wavelengths.

Explanation: This question examines Compton scattering. The wavelength shift depends only on scattering angle and not on target material because it results from photon-electron collisions governed by momentum conservation. The shift Δλ = λc(1 - cos θ) depends only on the scattering angle θ and the Compton wavelength λc = h/(mec), which involves only fundamental constants and electron mass. This material-independence proves the interaction is between individual photons and electrons, not a bulk material property. Choice A incorrectly suggests material-dependent refraction, which would vary with different targets. Momentum exchange reveals particle-like behavior of light.

Question 13

Two point charges $+2q$ and $-q$ are separated by $d$ . If the distance is reduced to $\tfrac{1}{2}d$ with charges unchanged, the force magnitude becomes

four times as large (correct answer)
twice as large
one-fourth as large
eight times as large

Explanation: This problem tests understanding of electric charge and electric force. Initially, the force between charges +2q and -q separated by distance d is F = k(2q)(q)/d² = 2kq²/d². When the distance is reduced to d/2, the new force becomes F = k(2q)(q)/(d/2)² = 2kq²/(d²/4) = 8kq²/d². Comparing the new force to the original: (8kq²/d²)/(2kq²/d²) = 4, so the new force is four times as large. A common misconception is thinking that halving the distance doubles the force, forgetting that force depends on 1/r², so halving r quadruples the force. Always square the distance ratio when calculating how force changes with distance.

Question 14

A gas in a sealed cylinder with a movable piston is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). During a process, the gas does $W=+80\ \text{J}$ , and its internal energy increases by $\Delta U=+20\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?

Heat enters the gas: $Q=+100\ \text{J}$ . (correct answer)
Heat enters the gas: $Q=+60\ \text{J}$ .
Heat leaves the gas: $Q=-60\ \text{J}$ .
No heat is transferred: $Q=0\ \text{J}$ .

Explanation: This problem tests the first law of thermodynamics. The first law states ΔU = Q - W, which rearranges to Q = ΔU + W for finding heat transfer. Given W = +80 J (work done by gas) and ΔU = +20 J (internal energy increases), we calculate Q = 20 + 80 = +100 J. Heat enters the gas. Choice C gives only 60 J, perhaps by subtracting W from ΔU instead of adding, a sign error when rearranging the equation. Always solve for the unknown by properly rearranging the first law equation and maintaining sign consistency.

Question 15

A wave pulse on a string reaches a boundary where the string is fixed to a wall. The boundary condition changes the end’s displacement to zero. Which statement best describes the reflected pulse?

It reflects inverted, with the same speed in the original string. (correct answer)
It reflects inverted, with a higher speed because it hits a wall.
It is not reflected because fixed ends absorb all wave energy.
It reflects upright, with a lower frequency than the incident pulse.

Explanation: This question tests understanding of boundary behavior of waves and polarization. At a fixed boundary (hard boundary), the string cannot move, creating a boundary condition where displacement must be zero. To satisfy this condition, the reflected wave must have opposite polarity (inverted) to cancel the incident wave's displacement at the boundary. The wave speed in the string depends only on string properties (v = √(T/μ)) and doesn't change upon reflection. Choice B incorrectly suggests the pulse reflects upright, which would violate the zero-displacement boundary condition at a fixed end. At fixed boundaries, waves always reflect with inversion to maintain zero displacement at the boundary.

Question 16

A diverging lens with focal length magnitude $|f|=20\ \text{cm}$ is placed in air. An upright object is located $60\ \text{cm}$ to the left of the lens. Ray tracing shows refracted rays diverge and their backward extensions intersect on the left side of the lens. Which statement best describes the image?

Real, inverted, and larger than the object
Virtual, upright, and smaller than the object (correct answer)
Real, upright, and on the far side of the lens
Virtual, inverted, and located on the lens surface

Explanation: This question tests understanding of images formed by lenses. Diverging lenses always produce virtual, upright, and reduced images for real objects, regardless of object distance. With the object at 60 cm from a diverging lens of focal length magnitude 20 cm, the parallel ray refracts as if coming from the near focal point while the center ray passes straight through. These refracted rays diverge, so only their backward extensions intersect on the object's side of the lens, confirming a virtual image. The image is upright (as all virtual images are) and smaller than the object because diverging lenses always demagnify. Choice A incorrectly suggests a diverging lens can produce a real, inverted image, revealing a fundamental misconception—diverging lenses cannot converge light from real objects to form real images. When encountering a diverging lens problem, immediately recognize that the image will be virtual, upright, and smaller, then use ray tracing to find its exact position.

Question 17

A composite bar has two equal-length sections in series, X then Y, same area. $k_X$ is larger than $k_Y$ . Ends are held at fixed temperatures. Where is the temperature gradient larger?

In section Y, because lower conductivity requires a larger gradient to drive the same heat flow (correct answer)
In section X, because higher specific heat creates a larger gradient
Equal in both sections, because fixed end temperatures force equal gradients
In section X, because higher conductivity requires a larger gradient to drive the same heat flow

Explanation: This question tests understanding of specific heat and thermal conductivity. In steady-state heat conduction through materials in series, the heat current must be the same through both sections. From Fourier's law, heat current = kA(dT/dx), where dT/dx is the temperature gradient. Since the heat current and area A are the same for both sections, the temperature gradient is inversely proportional to thermal conductivity: lower k requires larger gradient. Section Y, with lower thermal conductivity than section X, must have a larger temperature gradient to maintain the same heat flow. Choice A incorrectly claims higher conductivity requires larger gradient, reversing the inverse relationship. In series thermal conduction, remember: lower conductivity means steeper temperature gradient for the same heat flow.

Question 18

A capacitor is connected to an ideal $5.0\,\text{V}$ battery and reaches steady state. If the plate separation is doubled while still connected, which statement best describes the charge on the capacitor?

It decreases because the capacitance decreases at fixed voltage (correct answer)
It increases because the electric field must increase
It remains the same because the battery keeps charge constant
It becomes zero because capacitors do not store charge at steady state

Explanation: This question tests understanding of Capacitors. For a parallel-plate capacitor, capacitance is given by C = εA/d, where d is the plate separation. When d doubles, capacitance halves. Since the capacitor remains connected to the battery, the voltage stays constant at 5.0 V. Using Q = CV, if C halves while V remains constant, then Q must also halve, so the charge decreases. Choice C incorrectly assumes the battery keeps charge constant rather than voltage—batteries are constant voltage sources, not constant charge sources. Always identify whether the capacitor stays connected (constant V) or is isolated (constant Q) during changes.

Question 19

A nucleus of $^{40}_{19}\text{K}$ becomes $^{40}_{20}\text{Ca}$ with no change in mass number. Which particle is emitted?

A beta-plus particle (positron), increasing atomic number by 2
A beta-minus particle (electron), increasing atomic number by 1 (correct answer)
A gamma photon, increasing atomic number by 1
An alpha particle, increasing atomic number by 1

Explanation: This question tests understanding of types of radioactive decay. In beta-minus decay, a neutron converts to a proton by emitting an electron (beta-minus particle), increasing atomic number by 1 while mass number remains unchanged. Potassium-40 (Z=19, A=40) becomes calcium-40 (Z=20, A=40), showing atomic number increases by 1 (19→20) with constant mass number (40→40). Alpha decay would change both A and Z, while gamma decay changes neither. Choice D incorrectly claims positron emission increases atomic number by 2, confusing both the particle type and magnitude of change in beta decays. Remember: beta-minus increases Z by 1 (neutron→proton), beta-plus decreases Z by 1 (proton→neutron).

Question 20

The magnetic flux through a loop decreases linearly from $+3\,\text{mWb}$ to $0$ over $2\,\text{s}$ , while the loop’s area and orientation are constant. Which statement correctly describes the induced current’s magnetic field direction during the decrease?

It points in the same direction as the original flux to oppose the decrease. (correct answer)
It points opposite the original flux direction to make the flux decrease faster.
It points perpendicular to the original flux direction because the loop is stationary.
It is zero because the flux remains positive during the interval.

Explanation: This question tests understanding of electromagnetic induction and Faraday's law. As flux decreases from +3 mWb to 0, the rate of change dΦ/dt is negative, inducing an emf by Faraday's law: ε = -dΦ/dt. Lenz's law states that induced effects oppose the change causing them—here, the induced current must oppose the flux decrease. To oppose a decreasing positive flux, the induced current creates a magnetic field in the same direction as the original flux, trying to maintain it, making B correct. Choice A incorrectly suggests the induced field speeds up the decrease, violating Lenz's law. Always remember: induced currents create fields that oppose flux changes, not flux itself.

Question 21

A circuit has an ideal $18\text{ V}$ battery, then $R_1=6\,\Omega$ to junction J. From J, two branches run to junction K: branch A has $R_2=6\,\Omega$ ; branch B has $R_3=12\,\Omega$ . The branches rejoin at K and return to the battery. Which statement correctly compares the power dissipated in $R_2$ and $R_3$ ?

$R_2$ dissipates more power than $R_3$ . (correct answer)
$R_2$ dissipates the same power as $R_3$ .
$R_2$ dissipates less power than $R_3$ .
Both dissipate zero power because the current is consumed in $R_1$ .

Explanation: This problem tests understanding of compound DC circuits. R₁ (6Ω) is in series with parallel branches containing R₂ (6Ω) and R₃ (12Ω). In parallel branches, voltage is the same across each resistor, but current divides inversely with resistance—R₂ gets twice the current of R₃ since it has half the resistance. Power dissipation follows P = V²/R for parallel resistors with the same voltage, so lower resistance means higher power. Since R₂ (6Ω) has half the resistance of R₃ (12Ω), it dissipates twice the power. Choice D incorrectly assumes current is "consumed" in R₁, violating conservation of charge. To analyze power in compound circuits, first determine whether resistors are in series (same current) or parallel (same voltage), then apply the appropriate power formula.

Question 22

An ideal gas in a cylinder with a movable piston is compressed slowly from $4.0\,\text{L}$ to $2.0\,\text{L}$ while the temperature and amount of gas are held constant. Initially the pressure is $100\,\text{kPa}$ . Compared to the initial pressure, the final pressure is

$50\,\text{kPa}$ because pressure decreases when volume decreases.
$100\,\text{kPa}$ because temperature is constant.
$150\,\text{kPa}$ because pressure increases by $50\,\text{kPa}$ .
$200\,\text{kPa}$ because pressure is inversely proportional to volume. (correct answer)

Explanation: This question tests the ideal gas law. The ideal gas law (PV = nRT) shows that when temperature and amount of gas are constant, pressure and volume are inversely proportional (P₁V₁ = P₂V₂). As volume decreases from 4.0 L to 2.0 L (halved), pressure must double to maintain the same product PV. Therefore, pressure increases from 100 kPa to 200 kPa. Choice A incorrectly states that pressure decreases when volume decreases, revealing a fundamental misconception about the inverse relationship between P and V. To solve ideal gas problems correctly, identify which variables are constant and apply the appropriate proportionality relationship.

Question 23

A long straight wire carries current downward (− $y$ ). At a point to the east of the wire (+ $x$ ), which statement best describes the magnetic field direction there?

Into the page (− $z$ ) by the right-hand rule.
Downward (− $y$ ), in the same direction as the current.
Out of the page (+ $z$ ) by the right-hand rule. (correct answer)
Zero unless a moving charge passes through the point.

Explanation: This question tests understanding of magnetic fields. Magnetic fields circulate around current-carrying wires in a pattern determined by the right-hand rule: point your thumb in the current direction, and your fingers show how field lines wrap around the wire. For current flowing downward (-y), point your thumb down; at a point east of the wire (+x direction), your fingers curl from east toward north, then west, then south, meaning the field points out of the page (+z direction) at the eastern point. The field forms concentric circles centered on the wire, perpendicular to the current direction. Choice C incorrectly assumes magnetic fields point in the same direction as current flow, confusing field orientation with charge motion. Apply the right-hand rule systematically: thumb along current, fingers show field circulation, determining field direction at any point around the wire.

Question 24

A 0.15 m wire segment is vertical and carries $I=5.0\ \text{A}$ upward. A uniform magnetic field of $0.30\ \text{T}$ points into the page. Which statement best describes the direction of the magnetic force on the segment?

The force is upward along the wire.
The force is to the left.
The force is into the page.
The force is to the right. (correct answer)

Explanation: This problem tests magnetism and current-carrying wires. The wire carries current upward (vertical), and the magnetic field points into the page. Using the right-hand rule: point fingers upward (current direction), curl them into the page (field direction), and the thumb points to the right. Therefore, the magnetic force on the wire segment is to the right. Choice C incorrectly suggests the force is into the page, which represents the misconception of confusing the field direction with the force direction. Always apply the right-hand rule carefully: fingers along current, curl toward field, thumb shows force.

Question 25

A pin is placed closer to a converging lens than its focal length. Which statement best describes the image?

Real, inverted, and larger than the object, formed on the opposite side of the lens
Virtual, upright, and larger than the object, formed on the same side as the object (correct answer)
Real, upright, and smaller than the object, formed on the lens surface
Virtual, inverted, and smaller than the object, formed on the opposite side of the lens

Explanation: This question tests understanding of images formed by lenses. When an object is placed inside the focal length of a converging lens, the refracted rays diverge on the opposite side and never actually meet. However, when traced backward, these rays appear to originate from a point on the same side as the object, creating a virtual image. This virtual image is upright (maintains object orientation) and larger than the object, which is why magnifying glasses work when held close to objects. Choice A incorrectly suggests a real image forms, but this is impossible when the object is inside f because rays diverge rather than converge. To determine image type, check if the object is inside or outside the focal length: inside f always produces virtual images for converging lenses.

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