SOLUTIONS
2024 BOARD EXAM
CBSE Class 12 PCM Physics Board Paper 2024 — Set 1 Solutions
2024
Physics
Class 12 PCM
CBSE EXAMINATION PAPER-2024
PHYSICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 38 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 18 are multiple choice questions
- Section C – questions number 19 to 24 are very short answer
- Section D – questions number 25 to 32 are short answer
- Section E – questions number 33 to 38 are long answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
(1) A double-convex lens, with each face having the same radius of curvature R, is made of glass of refractive index n. Its power is : (A) 2(n - 1)/R (B) 2n-1/R (C) n-1/2R (D) 2n-1/2R
(2) A double-convex lens of power P, with each face having the same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be: (A) 2P (B) P (C) 4P (D) P/2
(3) The above two parts are kept in contact with each other as shown in the figure. The power of the combination will be : (A) P/2 (B) P (C) 2P (D) P/4
[1 Mark]
(4) A double-convex lens of power P, with each face having same radius of curvature, is cut along its principal axis. The two parts are arranged as shown in the figure. The power of the combination will be : (A) Zero (B) P (C) 2P (D) P/2
[1 Mark]
(5) (b) Two convex lenses of focal lengths 60 cm and 20 cm are held coaxially in contact with each other. The power of the combination is : (A) 6·6 D (B) 15 D (C) 1/15 D (D) 1/80 D
Junction Diode as a Rectifier:
The process of conversion of an ac voltage into a dc voltage is called rectification and the device which performs this conversion is called a rectifier. The characteristics of a p-n junction diode reveal that when a p-n junction diode is forward biased, it offers a low resistance and when it is reverse biased, it offers a high resistance. Hence, a p-n junction diode conducts only when it is forward biased. This property of a p-n junction diode makes it suitable for its use as a rectifier.
Thus, when an ac voltage is applied across a p-n junction, it conducts only during those alternate half cycles for which it is forward biased. A rectifier which rectifies only half cycle of an ac voltage is called a half-wave rectifier and one that rectifies both the half cycles is known as a full-wave rectifier.
(1) The root mean square value of an alternating voltage applied to a full-wave rectifier is V₀/√2. Then the root mean square value of the rectified output voltage is: (A) V₀/√2 (B) V₀²/√2 (C) 2V₀/√2 (D) V₀/2√2
(2) In a full-wave rectifier, the current in each of the diodes flows for : (A) Complete cycle of the input signal (B) Half cycle of the input signal (C) Less than half cycle of the input signal (D) Only for the positive half cycle of the input signal
(3) In a full-wave rectifier : (A) Both diodes are forward biased at the same time. (B) Both diodes are reverse biased at the same time. (C) One is forward biased and the other is reverse biased at the same time. (D) Both are forward biased in the first half of the cycle and reverse biased in the second half of the cycle.
(4) A signal, as shown in the figure, is applied to a p-n junction diode. Identify the output across resistance R_L :
[1 Mark]
(5) (a) An alternating voltage of frequency of 50 Hz is applied to a half-wave rectifier. Then the ripple frequency of the output will be : (A) 100 Hz (B) 50 Hz (C) 25 Hz (D) 150 Hz
Section B
A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density . The magnitude of the electric field at its centre is:
Ten capacitors, each of capacitance 1 μF, are connected in parallel to a source of 100 V. The total energy stored in the system is equal to:
The formula for the energy stored in a capacitor is U = (1/2) C V². For capacitors in parallel, the total capacitance C = C1 + C2 + ... + Cn. Here, with 10 capacitors each of 1 μF, the total capacitance is 10 μF. Thus, U = (1/2) * 10*10^-6 F * (100 V)² = 0.5 * 10^-3 J, which corresponds to the option '0.5 × 10-³ J'.
Consider the circuit shown in the figure. The potential difference between points A and B is:
The problem states that a 10 V source is connected across points A and C with resistors AB=100 Ω, BC=10 Ω, CD=5 Ω, and DA=60 Ω arranged in the circuit, and a galvanometer connected between B and D. By analyzing the circuit using the principles of series and parallel resistors, and applying Ohm's law and potential division rule, the potential difference between points A and B is calculated to be 6 V. The other given values do not align with the voltage drops expected across the 100 Ω resistor segment of the circuit.
A loop carrying a current I clockwise is placed in x-y plane, in a uniform magnetic field directed along z-axis. The tendency of the loop will be to:
A 10 cm long wire lies along y-axis. It carries a current of 1.0 A in positive y-direction. A magnetic field 𝐵⃗=(5 mT)𝑗̂−(8 mT)𝑘̂ exists in the region. The force on the wire is :
To calculate the force on the wire, we use the formula F = I (dl × B). Here, I = 1.0 A, dl is along the y-axis (0, 0.1, 0) in meters, and the magnetic field B = (0, 5 x 10^-3, -8 x 10^-3) T. Using the cross product, F = I * (dl × B) results in F = 1.0 * (0, 0.1, 0) × (0, 5 x 10^-3, -8 x 10^-3), which simplifies to (-0.8 mN) ^ i. Therefore, the correct answer is -(0.8 mN) ^ i.
A galvanometer of resistance G ohm is converted into an ammeter of range 0 to I A. If the current through the galvanometer is 0.1% of I A, the resistance of the ammeter is :
The reactance of a capacitor of capacitance C connected to an ac source of frequency ω is 'X'. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
In the four regions, I, II, III and IV, the electric fields are described as : Region I : Ex = E₀ sin (kz -ωt)
Region II : Ex = E₀
Region III : Ex = E₀ sin kz
Region IV : Ex = E₀cos kz
The displacement current will exist in the region :
The displacement current exists where the electric field changes with time, which is true for Region I where Ex = E0 sin (kz -ωt). The other regions (II, III, IV) either have a constant electric field or only vary with z, but not with time, so they do not contribute to a displacement current.
The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to :
Ge is doped with As. Due to doping,
Two beams, A and B whose photon energies are 3·3 eV and 11·3 eV respectively, illuminate a metallic surface (work function 2·3 eV) successively. The ratio of maximum speed of electrons emitted due to beam A to that due to beam B is :
The waves associated with a moving electron and a moving proton have the same wavelength . It implies that they have the same
Assertion (A) : In photoelectric effect, the kinetic energy of the emitted photoelectrons increases with increase in the intensity of the incident light.
Reason (R) : Photoelectric current depends on the wavelength of the incident light.
Assertion (A) : The mutual inductance between two coils is maximum when the coils are wound on each other.
Reason (R) : The flux linkage between two coils is maximum when they are wound on each other.
Assertion (A) : Two long parallel wires, freely suspended and connected in series to a battery, move apart.
Reason (R) : Two wires carrying current in opposite directions repel each other.
Assertion (A) : Plane and convex mirrors cannot produce real images under any circumstance.
Reason (R) : A virtual image cannot serve as an object to produce a real image.
Section C
Find the temperature at which the resistance of a wire made of silver will be twice its resistance at 20°C. Take 20°C as the reference temperature and temperature coefficient of resistance of silver at 20°C = 4·0*10⁻³ K⁻¹.
Resistance at 20°C = R20, Temperature coefficient of resistance, α = 4.0 * 10-3 K-1.
Resistance R at temperature t is given by R = R20 [1 + α (t - 20)].
We need R = 2 R20.
So, 2 R20 = R20 [1 + α (t - 20)] => 2 = 1 + α (t - 20).
Therefore, α (t - 20) = 1 => t - 20 = 1 / α = 1 / (4.0 * 10-3) = 250°C.
Hence, t = 20 + 250 = 270°C.
Answer: The resistance of silver wire will be twice its resistance at 20°C at 270°C.
Monochromatic light of frequency 5.0 × 10¹⁴ Hz passes from air into a medium of refractive index 1.5. Find the wavelength of the light (i)reflected and (ii) reflected at the interface of the two media.
An object is placed 30 cm in front of a concave mirror of radius of curvature 40 cm. Find the
(i) position of the image formed and
(ii) magnification of the image.
Consider a neutron (mass m) of kinetic energy E and a photon of the same energy. Let λₙ and λₚ be the de Broglie wavelength of neutron and the wavelength of photon respectively. Obtain an expression for λₙ/λₚ.
For neutron, kinetic energy E = pn2 / (2m), where pn is its momentum.
So, pn = sqrt(2mE).
De Broglie wavelength of neutron, λn = h / pn = h / sqrt(2mE).
For photon, energy E = pp * c, where pp is momentum, c is speed of light.
So, pp = E / c.
Wavelength of photon, λp = h / pp = h c / E.
Therefore, the ratio λn / λp = (h / sqrt(2mE)) / (h c / E) = E / (c sqrt(2mE)) = sqrt(E) / (c sqrt(2m)) = 1 / c * sqrt(E / 2m).
Plot a graph showing the variation of current with voltage for the material GaAs. On the graph, mark the region where :
a) resistance is negative, and
b) Ohm's law is obeyed.
A plano-convex lens of focal length 16 cm is made of a material of refractive index 1.4. Calculate the radius of the curved surface of the lens.
Section D
A cube of side 0.1 m is placed, as shown in the figure, in a region where electric field 𝐄⃗ = 500 x𝑖̂ exists. Here x is in meters and E in N/C⁻¹.
Calculate:
(a) the flux passing through the cube, and
(b) the charge within the cube.
The flux through the two faces perpendicular to the x-axis:
Flux_in = E(x) * A = 500 * x * a2
Flux_out = E(x + a) * A = 500 * (x + 0.1) * a2
Net flux, Φ = Flux_out - Flux_in = 500 * (x + 0.1) * (0.1)2 - 500 * x * (0.1)2 = 500 * 0.12 * 0.1 = 500 * 0.01 * 0.1 = 0.5 N m2/C.
Flux through other faces is zero as E_y = 0 and E_z = 0.
(b) Using Gauss's law, charge inside cube q = ε_0 * Φ, where ε_0 = 8.854 * 10-12 C2/N m2.
q = 8.854 * 10-12 * 0.5 = 4.427 * 10-12 C.
Therefore,
(a) Flux through the cube = 0.5 N m2/C
(b) Charge inside the cube = 4.427 * 10-12 C.
Define current density. Is it a scalar or a vector? An electric field 𝐄⃗ is maintained in a metallic conductor. If n be the number of electrons (mass m, charge -e) per unit volume in the conductor and ζ its relaxation time, show that the current density 𝐣⃗ = α𝐄⃗, where α=(ne²/m)ζ
(b) Scalar or vector: Current density is a vector quantity because it has both magnitude and direction. Its direction is the same as the flow of positive charge.
(c) Derivation of j = αE: Consider a metallic conductor with n electrons per unit volume, each electron having mass m and charge -e. An electric field E is applied. Due to E, electrons gain an average drift velocity vd. The force on each electron is F = -eE, and acceleration a = F/m = -eE/m.
The electrons collide with atoms and lose velocity after an average relaxation time ζ. Using the equation of motion, drift velocity vd = acceleration * relaxation time = (-eE/m)*ζ = (-eζ/m)E.
The current density j = number of electrons per unit volume * charge * drift velocity = n * (-e) * vd.
Substituting vd, j = n*(-e)*(-eζ/m)E = (ne2ζ/m)E.
Since E and j are vectors in the same direction, we write j = αE, where α = (ne2/m) ζ.
This shows current density is proportional to electric field with proportionality constant α.
A proton with kinetic energy 1.3384 × 10⁻¹⁴ J moving horizontally from north to south, enters a uniform magnetic field B of 2.0 mT directed eastward. Calculate:
(a) the speed of the proton,
(b) the magnitude of acceleration of the proton,
(c) the radius of the path traced by the proton.
[Take (q /m) for proton = 1·0 *10⁸ C/kg]
(b) The proton moves perpendicular to magnetic field B = 2.0 mT = 2.0 * 10^-3 T. Magnetic force F = q * v * B = (1.6 * 10^-19) * (1.29 * 10^6) * (2.0 * 10^-3) = 4.13 * 10^-16 N. Acceleration a = F / m = (4.13 * 10^-16) / (1.6 * 10^-27) = 2.58 * 10^11 m/s^2.
(c) Radius r of circular path is given by r = m * v / (q * B) = (1.6 * 10^-27) * (1.29 * 10^6) / (1.6 * 10^-19 * 2.0 * 10^-3) = 6.45 * 10^-3 m or 6.45 mm.
An inductor, a capacitor and a resistor are connected in series with an ac source v = Vₘ sin(ωt). Derive an expression for the average power dissipated in the circuit. Also obtain the expression for the resonant frequency of the circuit.
(b) At resonance, the inductive reactance equals capacitive reactance, so XL = XC, that is ωL = 1 / (ωC). Solving for ω gives the resonant angular frequency, ω0 = 1 / sqrt(LC). The resonant frequency f0 = ω0 / (2π) = 1 / (2π sqrt(LC)). At resonance, the impedance is minimum and equals R, and the circuit current is maximum with voltage and current in phase (φ = 0).
a) The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates. Give two examples to justify this statement.
b) (i) Long distance radio broadcasts use short-wave bands. Why ?
(ii) Optical and radio telescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the Earth. Why ?
Another example is radio waves which have wavelengths more than 0.1 meters and are emitted by large systems like antennas.
(b)(i) Long distance radio broadcasts use short-wave bands because short waves can be reflected by the Earth's ionosphere allowing them to travel long distances around the globe.
(ii) Optical and radio waves can pass through the Earth's atmosphere and so telescopes can be built on the ground. But X-rays are absorbed by the atmosphere, so X-ray astronomy requires satellites orbiting the Earth to detect X-rays from space.
Write the drawbacks of Rutherford's atomic model. How did Bohr remove them? Show that different orbits in Bohr's atom are not equally spaced.
(a) State any two properties of a nucleus.
(b) Why is the density of a nucleus much more than that of an atom ?
(c) Show that the density of the nuclear matter is the same for all nuclei.
What is a Wheatstone bridge ? Obtain the necessary conditions under which the Wheatstone bridge is balanced.
Section E
(i) Derive an expression for potential energy of an electric dipole 𝐩⃗ an external uniform electric field 𝐄⃗ When is the potential energy of the dipole (1) maximum, and (2) minimum?
(ii) An electric dipole consists of point charges - 1.0 PC and + 1.0 pC located at (0, 0) and (3 mm, 4 mm) respectively in x - y plane. An electric field 𝐄⃗ = ((1000V)/m) 𝑖̂ is switched on in the region. Find the torque 𝐫⃗ acting on the dipole.
Consider an electric dipole with charges +q and -q separated by a distance d. The dipole moment p is defined as p = q * d, directed from -q to +q.
When placed in a uniform electric field E, the dipole experiences forces on the charges creating torque and potential energy.
The work done in rotating the dipole from angle 0 to angle theta with the field is stored as potential energy U.
Potential energy U = - p * E * cos(theta), where theta is the angle between dipole moment and electric field.
Maximum and minimum potential energy:
1. U is maximum when cos(theta) = -1, i.e., theta = 180 deg, dipole moment opposite to E.
Maximum potential energy U_max = + p * E.
2. U is minimum when cos(theta) = +1, i.e., theta = 0 deg, dipole moment aligned with E.
Minimum potential energy U_min = - p * E.
(ii) Calculation of torque on the dipole:
Given charges: q = 1.0 pC = 1.0 * 10-12 C,
Positions: negative charge at (0, 0), positive charge at (3 mm, 4 mm) = (3 * 10-3 m, 4 * 10-3 m).
Electric field E = 1000 V/m along x-axis (î).
Dipole vector p = q * displacement vector from -q to +q = q * r;
r = (3 * 10-3) î + (4 * 10-3) ĵ;
Therefore, p = (1.0 * 10-12) * (3 * 10-3 î + 4 * 10-3 ĵ) = 3.0 * 10-15 î + 4.0 * 10-15 ĵ C m.
Torque τ = p × E;
E = 1000 î V/m;
Cross product:
τ = (3.0 * 10-15 î + 4.0 * 10-15 ĵ) × (1000 î) = 3.0 * 10-15 î × 1000 î + 4.0 * 10-15 ĵ × 1000 î = 0 + 4.0 * 10-15 * 1000 (ĵ × î).
Since ĵ × î = - k̂,
τ = - 4.0 * 10-12 k̂ Nm.
The torque magnitude is 4.0 * 10-12 Nm, directed along negative z-axis.
Final answers:
(i) U = - p E cos(theta); maximum at theta = 180 deg, minimum at theta = 0 deg.
(ii) Torque τ = - 4.0 * 10-12 k̂ Nm.
(i) A resistor and a capacitor are connected in series to an ac source v = vₘ sin t. Derive an expression for the impedance of the circuit.
(ii) When does an inductor act as a conductor in a circuit ? Give reason for it.
(iii) An electric lamp is designed to operate at 110 V dc and 11 A current. If the lamp is operated on 220 V, 50 Hz ac source with a coil in series, then find the inductance of the coil.
The impedance Z of the circuit is the vector sum of resistance R and capacitive reactance X_C. Since voltage across capacitor lags current by 90 degrees and voltage across resistor is in phase with current,
Impedance Z = sqrt(R2 + X_C2), but since capacitive reactance causes phase lag, it is Z = sqrt(R2 + (-X_C)2) = sqrt(R2 + X_C2). The phase angle phi = - tan-1(X_C / R).
Hence, Z = sqrt(R2 + (1 / (omega C))2).
(ii) An inductor acts as a conductor only when there is a changing current in the circuit, that is, with an AC source. This is because inductors oppose changes in current due to self induced emf, but allow steady current to pass without opposition in DC case as inductive reactance X_L = omega L = 0 at zero frequency.
Thus, in AC, inductor presents inductive reactance and allows current to flow depending on frequency.
(iii) Given, lamp rated voltage V = 110 V DC and rated current I = 11 A. Resistance of lamp R = V / I = 110 / 11 = 10 ohms.
Lamp is operated on 220 V, 50 Hz AC source with coil of inductance L in series.
Total voltage V = 220 V, current remains 11 A.
Total impedance Z = V / I = 220 / 11 = 20 ohms.
Since lamp resistance is R = 10 ohms, coil impedance Z_L = sqrt(Z2 - R2) = sqrt(202 - 102) = sqrt(400 - 100) = sqrt(300) = 17.32 ohms.
Inductive reactance X_L = omega L = 2 pi f L = 17.32.
Therefore, L = 17.32 / (2 pi * 50) = 0.055 henry or 55 mH.
Hence, inductance of coil is 55 millihenry.
(i) A ray of light passes through a triangular prism. Show graphically, how the angle of deviation varies with the angle of incidence ? Hence define the angle of minimum deviation.
(ii) A ray of light is incident normally on a refracting face of a prism of prism angle A and suffers a deviation of angle δ. Prove that the refractive index n of the material of the prism is given by n=sin (A+δ)/ sin A.
(iii) The refractive index of the material of a prism is √2. If the refracting angle of the prism is 60 degree find the
1.Angle of minimum deviation, and
2.Angle of incidence.
Angle of minimum deviation: It is the smallest angle through which light is deviated by the prism when passing through it.
(ii) Given a ray incident normally on one face of a prism of angle A, let the deviation suffered be δ.
In this case, angle of incidence i = 0 and angle of refraction r = 0 at the first face.
Using geometry inside the prism, the angle of refraction at the second face = A.
The ray emerges with angle of incidence i' at the second face, which equals the angle of refraction inside from prism air interface.
The deviation δ = i' - A (since the ray is incident normally on first face)
By Snell's law at second face: n = sin i' / sin r' = sin i' / sin A
Since i' = A + δ, we get n = sin (A + δ) / sin A.
(iii) Given n = sqrt(2), A = 60 degrees.
1. Using the formula for refractive index at minimum deviation:
n = sin[(A + δ_min) / 2] / sin (A / 2)
=> sqrt(2) = sin[(60 + δ_min)/2] / sin 30
sin 30 = 0.5
So, sin[(60 + δ_min)/2] = sqrt(2) * 0.5 = sqrt(2)/2 = 0.707
Therefore, (60 + δ_min)/2 = 45 degrees
=> 60 + δ_min = 90 degrees
=> δ_min = 30 degrees.
2. For minimum deviation, angle of incidence i = angle of emergence e and inside the prism angle of refraction r = A/2 = 30 degrees.
Using Snell's law, sin i = n sin r = sqrt(2) * sin 30 = sqrt(2) * 0.5 = 0.707
Therefore, i = 45 degrees.
Thus, angle of minimum deviation is 30 degrees and angle of incidence is 45 degrees.
i) An electric dipole (dipole moment 𝐩⃗ =𝐩⃗ hat hat i ) consisting of charges q and q, separated by distance 2a, is placed along the x-axis, with its centre at the origin. Show that the potential V, due to this dipole, at a point x, (x >>a) is equal to 1/4πε₀. 𝐩⃗. 𝑖̂ /x²
(ii) Two isolated metallic spheres S₁ and S₂ of radii 1 cm and 3 cm respectively are charged such that both have the same charge density (2/π ×10 ⁻⁹) C / m ². They are placed far away from each other and connected by a thin wire. Calculate the new charge on sphere S₁.
V = (1 / 4πε₀) [q / (x - a) - q / (x + a)]
Using binomial expansion for large x, 1 / (x ± a) ≈ 1 / x ∓ a / x²,
So, V = (1 / 4πε₀) [q (1 / x + a / x²) - q (1 / x - a / x²)] = (1 / 4πε₀) (2aq / x²) = (1 / 4πε₀) (p / x²)
This shows the potential V at point x due to the dipole is (1 / 4πε₀) p / x² along the x-axis.
(ii) Given: Radius of sphere S₁, r₁ = 1 cm = 0.01 m, radius of sphere S₂, r₂ = 3 cm = 0.03 m, charge density σ = 2 / π * 10-9 C / m².
Charge on sphere S = surface charge density * surface area = σ * 4πr².
So, initial charges:
Q₁ = σ * 4π r₁² = (2 / π * 10-9) * 4π * (0.01)2 = 2 * 4 * 10-9 * (0.01)2 = 8 * 10-9 * 10-4 = 8 * 10-13 C
Q₂ = σ * 4π r₂² = similarly 2 / π * 10-9 * 4π * (0.03)2 = 2 * 4 * 10-9 * 0.0009 = 8 * 10-9 * 0.0009 = 7.2 * 10-12 C
Since spheres are connected by a thin wire, they share charge until their potentials are equal:
Potential on sphere = Q / (4πε₀ r).
Let new charges be Q₁' and Q₂'. Since charge is conserved,
Q₁' + Q₂' = Q₁ + Q₂ = 8 * 10-13 + 7.2 * 10-12 = 8 * 10-12 C (approx).
Equal potentials: Q₁' / r₁ = Q₂' / r₂ => Q₂' = Q₁' * r₂ / r₁ = Q₁' * 3.
So, Q₁' + 3 Q₁' = 4 Q₁' = 8 * 10-12 => Q₁' = 2 * 10-12 C.
Therefore, new charge on sphere S₁ is 2 * 10-12 C.
(i) Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer.
(ii) A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy ? Explain.
(iii) A step-up transformer has 200 and 3000 turns in its primary and secondary coils respectively. The input voltage given to the primary coil is 90 V. Calculate :
The output voltage across the secondary coil
The current in the primary coil if the current in the secondary coil is 2·0 A.
(i) State Huygens' principle. A plane wave is incident at an angle i on a reflecting surface. Construct the corresponding reflected wavefront. Using this diagram, prove that the angle of reflection is equal to the angle of incidence.
(ii) What are the coherent sources of light ? Can two independent sodium lamps act like coherent sources ? Explain.
(iii) A beam of light consisting of a known wavelength 520 nm and an unknown wavelength λ, used in Young's double slit experiment produces two interference patterns such that the fourth bright fringe of unknown wavelength coincides with the fifth bright fringe of known wavelength. Find the value of λ.
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