JEE Advanced Physics questions with solutions

Q1. Rain is falling at an angle α with respect to the vertical at a velocity of 10 m/s. A girl runs in the opposite direction to the rain at a speed of 8 m/s and observes that the rain appears to make an angle β with the vertical. What is the relationship between α and β?

1. tan α = (8 + 10sinβ) / 10cosβ
2. tan β = (8 + 10sinα) / 10cosα
3. tan α = tan β
4. tan α = cot β

Answer: tan β = (8 + 10sinα) / 10cosα

The relative motion of the girl and the rain alters the apparent angle of the rain. Using vector addition, the relationship between α and β is given by tan β = (8 + 10sinα) / 10cosα.

Q2. A vector A is turned by a very small angle Δθ radians (where Δθ is much less than 1) to form a new vector B. What is the magnitude of the difference |B − A|?

1. |A| Δθ
2. |B| Δθ − |A|
3. |A| Δθ² / 2
4. 0

Answer: |A| Δθ

For a small angle Δθ, the magnitude of the difference |B − A| is approximately |A| Δθ, as the arc length of the rotation is proportional to the angle and the magnitude of the vector.

Q3. A swimmer moves through water at a speed of v relative to the water. To cross a river of width d, which flows at a speed u (where u < v), the downstream displacement caused by the river's flow is x. Which of the following is accurate?

1. When crossing the river in the shortest time, x equals du/v.
2. The value of x cannot be smaller than du/v.
3. To minimize x, the swimmer must move at an angle of π/2 + sin⁻¹(v/u) relative to the river's current.
4. The largest value of x occurs when the swimmer moves at an angle of π/2 − sin⁻¹(v/u) relative to the river's current.

Answer: When crossing the river in the shortest time, x equals du/v.

When crossing the river in the shortest time, the swimmer moves perpendicular to the river's flow. The downstream displacement x is caused solely by the river's velocity and is given by x = du/v, derived from the ratio of the river's speed to the swimmer's speed.

Q4. A particle is launched from a point with an initial speed u at an angle θ above the horizontal. At a specific point during its motion, its velocity becomes perpendicular to its original direction. What is the time taken to reach this point?

1. The speed of the particle at this point is u sin θ
2. The speed of the particle at this point is u cos θ
3. The time taken to reach this point is (u/g) sec θ
4. The time taken to reach this point is (u/g) cosec θ

Answer: The time taken to reach this point is (u/g) cosec θ

For velocity to become perpendicular to the launch direction, the component of velocity along the original direction must vanish: u - g t sin(theta) = 0, giving t = u/(g sin theta) = (u/g) cosec theta. The stored 'sec theta' is incorrect; the correct option is (u/g) cosec theta.

Q5. Two objects, each with a mass of M, are positioned at a fixed distance of 2L apart. A smaller particle of mass m is launched from the midpoint of the line connecting their centers, moving in a direction perpendicular to this line. Given the gravitational constant G, which of the following statements is accurate?

1. The least velocity required for the particle of mass m to overcome the gravitational pull of the two objects is √(4GM / L)
2. The least velocity required for the particle of mass m to overcome the gravitational pull of the two objects is √(2GM / L)
3. The gravitational field strength due to the two objects is 2GM / L
4. The total energy of the particle of mass m remains unchanged

Answer: The least velocity required for the particle of mass m to overcome the gravitational pull of the two objects is √(4GM / L)

The least velocity required for the particle to escape the gravitational pull of the two objects is derived from energy conservation. The escape velocity is √(4GM / L), accounting for the combined gravitational potential of both masses.

Q6. A central force is given as F(r) = −k / rⁿ, where k is a constant. What condition must n satisfy for a circular orbit to remain stable?

1. n must equal 2
2. n must be less than 3
3. n must be greater than 3
4. n must equal −1

Answer: n must be less than 3

For a circular orbit to remain stable under a central force F(r) = −k / rⁿ, the effective potential must have a minimum. This condition is satisfied when n < 3, as higher values lead to instability due to excessive inward force.

Q7. A capillary tube with a square-shaped internal cross-section, each side of length 'a', is inserted vertically into a liquid of density ρ and surface tension σ. If the liquid adheres to the tube walls with a contact angle θ, approximately how high will the liquid rise inside the tube? (Ignore the surface tension effects at the corners of the tube.)

1. 2σ cosθ / (aρg)
2. 4σ cosθ / (aρg)
3. 8σ cosθ / (aρg)
4. 4σ secθ / (aρg)

Answer: 4σ cosθ / (aρg)

Surface-tension force pulling the column up equals sigma times the wetted perimeter times cos(theta) = sigma*4a*cos(theta). Setting this equal to the column weight rho*g*a^2*h gives h = 4 sigma cos(theta)/(a rho g). The stored answer used 2 sigma instead of 4 sigma.

Q8. A Carnot engine takes in 1000 J of thermal energy from a source at 127°C and expels 600 J of energy to the surroundings in each cycle. What is the engine's efficiency and the temperature of the cold reservoir?

1. 20% and −43°C
2. 40% and −33°C
3. 50% and −20°C
4. 70% and −10°C

Answer: 40% and −33°C

Efficiency = 1 - 600/1000 = 40%. For a Carnot engine Qc/Qh = Tc/Th, so Tc = (600/1000)*(400 K) = 240 K = -33 C. The correct pairing is 40% and -33 C, option 1; the stored answer (20%, -43 C) is wrong.

Q9. Consider an ideal diatomic gas with molecules of mass 'm' at an absolute temperature T. Let V represent the root mean square speed of the molecules, ignoring vibrational energy contributions. Which statement is incorrect?

1. It is possible for a molecule to have a speed exceeding √2 V.
2. The value of V scales with the square root of T.
3. The mean rotational kinetic energy per molecule equals mV²/4.
4. The mean kinetic energy per molecule is 5mV²/6.

Answer: The mean rotational kinetic energy per molecule equals mV²/4.

With V^2 = 3kT/m, we get kT = mV^2/3. A diatomic molecule has 2 rotational degrees of freedom, so mean rotational KE = 2*(1/2)kT = kT = mV^2/3, not mV^2/4. The mean total KE = (5/2)kT = 5mV^2/6 is correct, so the incorrect statement is the rotational-KE one (mV^2/4).

Q10. A particle constrained to move along the x-axis has a potential energy described by U(x) = k[1 - exp(-x²)], where k is a positive constant and -∞ ≤ x ≤ +∞. Which of the following statements is true?

1. The particle is in an unstable equilibrium at positions away from the origin.
2. For any finite, nonzero x, the force on the particle points outward from the origin.
3. If the total mechanical energy is k/2, the particle's kinetic energy is least at the origin.
4. The motion near x = 0 is simple harmonic for small displacements.

Answer: The motion near x = 0 is simple harmonic for small displacements.

U(x)=k[1-exp(-x^2)] has U'(x)=2kx*exp(-x^2), giving a stable minimum at x=0 where U=0. For small x, U approximately k*x^2, so the motion is simple harmonic. The stored claim that KE is least at the origin is false (KE is maximum there since U is minimum).

Q11. A steel wire is used to suspend an object with a specific gravity of ρ, and the fundamental frequency of transverse standing waves in the wire is 300 Hz. When the object is partially submerged in water such that half of its volume is underwater, what will the new fundamental frequency (in Hz) become?

1. 300√((2ρ−1)/2ρ)
2. 300√(2ρ−1)
3. 300(2ρ−1)/2ρ
4. 300(2ρ−1)/2

Answer: 300√((2ρ−1)/2ρ)

Fundamental frequency f is proportional to sqrt(T). In air T = V*rho*g; with half the volume submerged T' = V*rho*g - (V/2)*1*g = Vg(rho - 1/2). Thus f' = 300*sqrt((rho-1/2)/rho) = 300*sqrt((2rho-1)/(2rho)), which is option 0, not the stored option 2.

Q12. A wave pulse traveling to the right along the x-axis is described by the equation y(x, t) = 2.0 / ((x − 3.0t)² + 1), where x and y are measured in centimeters and t is in seconds. The peak height of the pulse is defined as the largest displacement along the y-axis. Which of the following is true?

1. The peak height of the pulse decreases over time.
2. The peak height of the pulse remains unchanged over time.
3. The pulse moves at a speed of 3.0 cm/s.
4. The pulse moves at a speed of 0.33 cm/s.

Answer: The peak height of the pulse remains unchanged over time.

The equation of the pulse shows that its peak height depends only on the denominator, which remains constant over time. Therefore, the peak height of the pulse does not change as it propagates.

Q13. When a person blows into one end of a long pipe, creating a high-pressure air pulse that moves through it, what happens when this pulse reaches the opposite end of the pipe?

1. A high-pressure pulse begins moving back through the pipe if the far end is open.
2. A low-pressure pulse begins moving back through the pipe if the far end is open.
3. A low-pressure pulse begins moving back through the pipe if the far end is closed.
4. A high-pressure pulse begins moving back through the pipe if the far end is closed.

Answer: A low-pressure pulse begins moving back through the pipe if the far end is open.

A high-pressure (compression) pulse reaching the OPEN end of a pipe reflects as a low-pressure (rarefaction) pulse, since an open end is a pressure node. So a low-pressure pulse travels back if the far end is open (option index 1). The stored answer (index 2) wrongly requires a closed end.

Q14. Under which condition can stationary waves form?

1. On a string fixed at both ends.
2. On a string fixed at one end and loose at the other.
3. When a wave reflects back from a rigid surface.
4. When two waves with the same frequency and a phase difference of π travel in the same direction.

Answer: On a string fixed at both ends.

Stationary waves form when two waves of the same frequency and amplitude travel in opposite directions and interfere. This condition is naturally satisfied on a string fixed at both ends.

Q15. A particle with charge -q and mass m revolves in a circular path of radius r around an infinitely long charged wire with linear charge density +λ. What is the time period of its motion?

1. T = 2π√(m/2kλq)
2. T = √(4π²m³/2kλq)
3. T = 2π√(m/kλq)
4. T = √(m/2kλq)

Answer: T = 2π√(m/2kλq)

The charged particle experiences a centripetal force due to the electric field created by the charged wire. Equating the centripetal force to the electrostatic force and solving for the time period, we find that it depends on the mass, charge, and the linear charge density of the wire, leading to the given expression.

Q16. A spherical conductor has two hollow cavities inside it. The conductor itself carries no net charge, but a charge q₁ is placed at the center of one cavity, and a charge q₂ is placed at the center of the other cavity. A third charge q₃ is located at a large distance r from the conductor. If the forces experienced by q₃, q₁, and q₂ are F₁, F₂, and F₃ respectively, which of the following is true? (Assume all charges are positive)

1. F₁ is smaller than F₂, which is smaller than F₃
2. F₁ equals F₂, but both are smaller than F₃
3. F₁ equals F₂, but both are greater than F₃
4. F₁ is greater than F₂, which is greater than F₃

Answer: F₁ is greater than F₂, which is greater than F₃

The forces on q₁ and q₂ are due to the charges they induce on the inner walls of the cavities, which are equal in magnitude but opposite in direction. The force on q₃ is weaker because it is at a large distance and experiences the net field of the entire conductor, which is smaller.

Q17. A dipole is positioned at the center of a spherical surface. Which of the following statements is accurate?

1. The net electric flux passing through the sphere is zero.
2. The electric field vanishes at all points on the sphere.
3. The electric field is non-zero at every location on the sphere.
4. The electric field becomes zero along a circular path on the sphere.

Answer: The net electric flux passing through the sphere is zero.

A dipole has equal and opposite charges, so the total charge enclosed by the spherical surface is zero. By Gauss's law, the net electric flux through the sphere is therefore zero, even though the electric field at individual points on the sphere is non-zero.

Q18. What is the value of the electrostatic potential at point H?

1. σ / ε₀ [(a² + H²)¹/² - H]
2. σ / ε₀ [(a² + H²)¹/² + H]
3. σ / 2ε₀ [(a² + H²)¹/² - H]
4. σ / 2ε₀ [(a² + H²)¹/² + H]

Answer: σ / 2ε₀ [(a² + H²)¹/² - H]

The electrostatic potential at a point is derived by integrating the electric field due to a charged surface. For a point at height H, the potential depends on the geometry of the system, leading to the given expression involving σ, ε₀, and the distance terms.

Q19. The electric current in a conductor varies with time according to i = (20 + 4t). Determine the value of x if the total charge passing through a section of the conductor in 10 seconds is x × 10² C.

1. 2
2. 4
3. 6
4. 8

Answer: 4

Charge Q = integral over 0..10 of (20+4t) dt = 20*10 + 2*10^2 = 200 + 200 = 400 C = 4 x 10^2 C. Thus x = 4, which is option index 1, not the stored index 2.

Q20. Two parallel conductors lie in the plane of the paper, separated by a distance X₀. A charged particle travels with velocity v between these wires, maintaining a distance X₁ from one of them. When both wires carry identical currents I flowing in the same direction, the particle's trajectory has a curvature radius of R₁. Conversely, if the currents in the wires flow in opposite directions, the curvature radius becomes R₂. Given that X₁/X₀ equals 3, determine the ratio R₁/R₂.

1. 1
2. 2
3. 3
4. 4

Answer: 3

When the currents in the wires flow in the same direction, the magnetic forces on the particle add up, resulting in a smaller radius of curvature (R₁). When the currents flow in opposite directions, the forces partially cancel, leading to a larger radius (R₂). The ratio R₁/R₂ is determined to be 3 based on the geometry and current configuration.

Q21. What is the final velocity attained by the loop?

1. mgR/(B₀²a²)
2. mgR/(B₀a)
3. 2mgR/(B₀²a²)
4. mgR/(2B₀²a²)

Answer: mgR/(B₀²a²)

The final velocity of the loop is determined by balancing the magnetic force and gravitational force. Using the given parameters, the velocity is proportional to mgR and inversely proportional to B₀²a².

Q22. What factor influences the visibility of fringes in an interference pattern?

1. The spacing between successive fringes
2. The wavelength of light used
3. The relative brightness of the light sources
4. The separation between the two slits

Answer: The relative brightness of the light sources

The visibility of fringes in an interference pattern depends on the relative brightness of the light sources, as it affects the contrast between bright and dark fringes.

Q23. When the screen is moved further along the x-axis away from the source, which of the following statements is false?

1. The central bright fringe moves along the x-axis.
2. The positions of all bright fringes except the central one are altered.
3. The spacing between consecutive fringes does not change.
4. The angular separation between fringes is affected by the shift.

Answer: The central bright fringe moves along the x-axis.

The central bright fringe does not move along the x-axis when the screen is shifted; it remains at the same position because it corresponds to the point of zero path difference.

Q24. The photoelectric effect provides evidence for the quantum behavior of light because:

1. photoelectrons are not emitted if the light's frequency is below a certain threshold value
2. the highest kinetic energy of photoelectrons is determined solely by the light's frequency, not its brightness
3. photoelectrons are ejected instantly even when the light is very dim
4. the charge carried by photoelectrons is discrete and quantized

Answer: photoelectrons are not emitted if the light's frequency is below a certain threshold value

The photoelectric effect demonstrates that light behaves as discrete packets of energy (photons). Electrons are only emitted if the light's frequency exceeds a threshold, proving that energy transfer is quantized.

Q25. What is the kinetic energy of an electron in the nth orbit?

1. nhα / 4π
2. 4π² / nh²α
3. nhα / 2π
4. nh²α / π

Answer: nhα / 4π

The kinetic energy of an electron in the nth orbit is derived from centripetal force and energy quantization principles, resulting in the formula nhα / 4π.

Q26. What is the increase in potential energy of a spring when it is stretched or compressed by a distance x from its equilibrium position?

1. 1/2 kx
2. kx
3. 1/2 kx²
4. kx²

Answer: 1/2 kx²

The potential energy stored in a spring follows Hooke's law and is proportional to the square of the displacement, giving the formula 1/2 kx².

Q27. The average kinetic energy of a single atom is calculated as (3/2)kT, where k = 1.38 × 10⁻²³ J/K and T = 160 K. What is the resulting value?

1. 3.312 × 10⁻²¹ J
2. 3.412 × 10⁻²¹ J
3. 3.212 × 10⁻²¹ J
4. 3.512 × 10⁻²¹ J

Answer: 3.312 × 10⁻²¹ J

Average KE = (3/2)kT = 1.5 x 1.38e-23 x 160 = 3.312e-21 J, which is option index 0. The stored answer 3.212e-21 J (index 2) is an arithmetic error.

Q28. A long coaxial cable is made up of two cylindrical conductors with radii a and b. The inner conductor carries a constant current i, while the outer conductor acts as the return path for the current. What is the self-inductance of a segment of the cable of length l?

1. μ₀l ln(b/a) / 2π
2. μ₀l ln(a/b) / π
3. μ₀l ln(a/b) / 2π
4. μ₀l ln(b/a) / π

Answer: μ₀l ln(b/a) / 2π

The self-inductance of a coaxial cable depends on the logarithmic ratio of the radii of the conductors and the length of the cable. The correct expression is μ₀l ln(b/a) / 2π.

Q29. Determine the inductance per unit length for a loop created by connecting the ends of two infinitely long parallel wires, each with radius r, and separated by a distance d between their centers. Ignore the magnetic flux inside the wires and the effects at the ends.

1. μ₀ / (2π) * logₑ((d − r) / r)
2. μ₀ / (2π) * logₑ((d + r) / r)
3. μ₀ / π * logₑ((d − r) / r)
4. None of the above

Answer: μ₀ / π * logₑ((d − r) / r)

The inductance per unit length for the loop is derived by considering the magnetic flux outside the wires and the separation distance. The correct formula is μ₀ / π * logₑ((d − r) / r).

Q30. If the electromotive force is 10 V and the rate of change of current is calculated as 2.5 A/s, what is the self-inductance of the coil, given by L = e / (dl/dt)?

1. 4 H
2. 2 H
3. 6 H
4. 8 H

Answer: 4 H

The self-inductance is calculated using L = e / (di/dt). Substituting the given values, L = 10 V / 2.5 A/s = 4 H.