CBSE Class 12 Physics Board Paper 2023 (55/1/1): Solutions

A point charge on the axis of a short dipole, at distance r, feels force F. What is the force when it is moved to 2r?

1. On the axis of a short dipole the field is E = 2kp/r³, so the force on the charge is F = qE ∝ 1/r³.
2. At distance 2r the force becomes F × (r/2r)³ = F × 1/8 = F/8 — option (b).

Answer: The axial field of a short dipole falls as 1/r³, so doubling the distance makes the force F/8 — option (b).

Which R–T graph correctly shows how the resistance of a metallic conductor changes with temperature?

Uses a figure printed in CBSE's paper — what the figure shows: Option graphs (R up, T across): (a) straight rising line starting above zero; (b) straight rising line from the origin; (c) curve falling as T rises; (d) curve from near the origin rising and bending upward (concave up).

1. In a metal, raising the temperature makes the ions vibrate more, so electrons collide more often and the resistance rises; this rules out the falling curve (c).
2. NCERT's graph for copper shows the rise is not exactly linear: the curve bends upward, getting steeper at higher temperature.
3. So the correct graph is the upward-bending curve — option (d).

Answer: Resistance of a metal rises with temperature, and the rise grows faster at higher temperature (a curve bending upward) — option (d).

Note: Over a small temperature range R rises almost linearly, so the straight line (a) can look right, but CBSE's marking scheme accepts only (d).

A cell reads 8 V on open circuit and 4 V while delivering 4 A. Find its internal resistance.

1. Identify the electromotive force E as 8 V and the terminal voltage V as 4 V when current I is 4 A.
2. Calculate the internal resistance using the formula r = (E - V) / I, which gives r = (8 - 4) / 4 = 1 Ω.

Answer: r = (E − V)/I = (8 − 4)/4 = 1 Ω — option (d).

A steady current flows in a metal wire whose cross-section area A grows steadily along its length. Which graph shows the drift speed v_d against A?

Uses a figure printed in CBSE's paper — what the figure shows: Option graphs (v_d up, A across): (a) curve falling steeply then flattening (like 1/A); (b) straight line falling to zero; (c) horizontal line; (d) straight line rising.

1. Use the current formula I = neAv_d, where n and e are constant for the metal.
2. Since the current I is steady, the product Av_d must be constant, meaning v_d is inversely proportional to A.
3. An inverse relationship v_d ∝ 1/A corresponds to a falling hyperbola-like curve, which matches option (a).

Answer: I = neAv_d is the same everywhere, so v_d ∝ 1/A: a falling hyperbola-like curve — option (a).

How does a diamagnetic sample behave when brought close to either pole of a bar magnet?

1. Diamagnetic materials develop an induced magnetic moment opposite to the applied external field.
2. This opposition creates a repulsive force, causing the sample to move from the stronger field region to the weaker field region.
3. Therefore, the sample is pushed away by both the north and south poles of the bar magnet.

Answer: A diamagnetic material moves from stronger to weaker field, so it is pushed away by both poles — option (a).

A 40-turn coil of radius 8.0 cm spins about a vertical diameter at 25/π rad/s in a horizontal uniform field of 3.0 × 10⁻² T. Find the peak induced emf.

1. Identify the formula for peak induced emf in a rotating coil: ε₀ = NBAω.
2. Substitute the given values: N = 40, B = 3.0 × 10⁻² T, A = π(0.08)² m², and ω = 25/π rad/s.
3. Calculate the area: A = π × 0.0064 m².
4. Compute the product: ε₀ = 40 × 3.0 × 10⁻² × π × 0.0064 × (25/π).
5. Simplify the expression by canceling π: ε₀ = 40 × 3.0 × 10⁻² × 0.0064 × 25.
6. Calculate the final value: ε₀ = 0.192 V, which is approximately 0.19 V.

Answer: ε₀ = NBAω = 40 × 3.0 × 10⁻² × π(0.08)² × 25/π = 0.192 V ≈ 0.19 V — option (c).

Arm PQ (resistance r) slides with velocity v on a rectangular frame PSRQ of negligible resistance in a uniform field. The size of the induced emf is independent of which quantity: B, v, r or the length of PQ?

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, a uniform field points into the page; PQ (length l) slides on the two long rails of the frame towards SR with velocity v; the frame's left side is SR.

1. The induced motional emf is given by the formula ε = Blv, where B is the magnetic field, l is the length of PQ, and v is the velocity.
2. Since the expression for emf does not contain the resistance r, the induced emf is independent of r.

Answer: Motional emf is ε = Blv, which has no r in it, so it does not depend on the resistance r — option (c).

While a capacitor charges, what expression gives the current that exists in the gap between its plates?

1. According to Maxwell's modification of Ampere's law, a changing electric field in the gap between capacitor plates produces a displacement current.
2. The expression for this displacement current is i_d = ε₀ dΦ_E/dt, which corresponds to option (c).

Answer: It is the displacement current i_d = ε₀ dΦ_E/dt — option (c).

For a concave mirror of focal length f, what is the least possible gap between an object and its real image?

1. For a concave mirror, an object at the centre of curvature (u = −2f) gives a real, inverted image also at 2f (v = −2f).
2. Object and image are then at the same place, so the least possible distance between them is zero — option (a).

Answer: With the object at the centre of curvature (2f), the real image forms at the same place, so the least gap is zero — option (a).

In Bohr's hydrogen model, the radius of orbit number n varies as which power of n?

1. From the quantization of angular momentum, the orbital velocity is v = n h / (2 π m r).
2. Equating the centripetal force to the electrostatic force gives r = n² h² ε₀ / (π m e²), so r ∝ n².

Answer: r_n = n²h²ε₀/(πme²), so r_n ∝ n² — option (c).

A ground-state hydrogen atom takes in one photon and goes to the n = 5 level. Find the photon's wavelength.

1. Calculate the energy difference between the n=5 and n=1 levels using the formula E = 13.6(1 - 1/n²) eV, giving E = 13.6(1 - 1/25) = 13.06 eV.
2. Convert the energy to joules and use the relation λ = hc/E to find the wavelength, resulting in λ ≈ 95 nm.

Answer: E = 13.6(1 − 1/25) = 13.06 eV; λ = hc/E = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(13.06 × 1.6 × 10⁻¹⁹) ≈ 95 nm — option (d).

How does the mass density of a nucleus depend on its mass number A?

1. The radius of a nucleus is given by R = R₀A^(1/3), where R₀ is a constant.
2. The volume of the nucleus is V = (4/3)πR³, which simplifies to V ∝ A.
3. The mass of the nucleus is proportional to the mass number A (M ∝ A).
4. Density ρ = M/V, so ρ ∝ A/A, meaning the density is constant and independent of A.

Answer: R = R₀A^(1/3), so volume ∝ A and mass ∝ A; the density does not depend on A — option (d).

An ac supply drives a p-n diode and a load resistor joined in series. Which graph shows the voltage across the load?

Uses a figure printed in CBSE's paper — what the figure shows: Option graphs (V against t): (a) full sine wave; (b) continuous positive humps with no gaps (full-wave); (c) positive humps separated by flat zero gaps (half-wave); (d) steady horizontal line.

1. A p-n diode allows current to flow only when it is forward biased, which occurs during the positive half-cycle of the AC supply.
2. During the negative half-cycle, the diode is reverse biased and blocks current, resulting in zero voltage across the load resistor.
3. Therefore, the output voltage consists of positive half-sine waves separated by flat zero intervals, matching graph (c).

Answer: The diode conducts only in alternate half cycles, so the output is positive half-loops with gaps between them (half-wave rectified) — option (c).

What happens when a pure semiconductor gets a little trivalent impurity added?

1. Trivalent atoms are acceptors that create holes in the semiconductor lattice.
2. The presence of these holes makes it a p-type semiconductor.

Answer: Trivalent atoms are acceptors that create holes, so it becomes a p-type semiconductor — option (b).

In the band diagram of n-type silicon, roughly how far below the conduction band edge E_C is the donor level E_D?

1. In n-type silicon each donor atom's fifth electron is very loosely bound, so the donor level E_D lies just below the bottom of the conduction band E_C.
2. The gap E_C − E_D is tiny compared with the band gap (about 1.1 eV for Si); NCERT's band diagram shows it of the order of 0.01 eV — option (d).

Answer: The donor level lies very close to the conduction band, about 0.01 eV below it — option (d).

A: Field lines crowd into a copper bar kept in an external magnetic field. R: Copper is paramagnetic.

1. Reason: copper is diamagnetic, not paramagnetic, so R is false.
2. Assertion: a diamagnetic bar expels field lines, so they are pushed out of the copper bar, not concentrated inside it; A is also false — option (d).

Answer: Copper is diamagnetic, so field lines are pushed out of the bar, not crowded in; both A and R are false — option (d).

A: Any two points on one wavefront have zero phase difference. R: Points of a wavefront are equally far from the source, so they vibrate in step.

1. A wavefront is defined as the surface on which all points oscillate in the same phase, so any two points on it have zero phase difference; A is true.
2. The disturbance from the source takes equal time to reach points at equal distance, so they vibrate in step; R is true and explains A — option (a).

Answer: Both are true and R explains A: a wavefront is a surface of constant phase — option (a).

A: The photoelectric effect shows light behaves as particles. R: Above the threshold frequency, photocurrent is proportional to the light's intensity.

1. Statement A is true because the photoelectric effect demonstrates that light interacts with matter as discrete packets of energy, or photons.
2. Statement R is true because increasing the intensity of light above the threshold frequency increases the number of emitted photoelectrons, making the photocurrent proportional to intensity.
3. However, the particle nature of light is specifically proven by the existence of a threshold frequency and the instantaneous emission of electrons, not by the relationship between current and intensity.
4. Therefore, both statements are true, but R is not the correct explanation for A, which corresponds to option (b).

Answer: Both statements are true, but the particle nature comes from threshold frequency and instant emission, not from current ∝ intensity — option (b).

An alpha particle enters a region with velocity (3.0 × 10⁵ m/s) î where B = (0.4 T) î + (0.3 T) ĵ. With q/m = 4.8 × 10⁷ C/kg, find its acceleration.

1. Use the formula for acceleration of a charged particle in a magnetic field: a = (q/m)(v × B).
2. Calculate the cross product v × B: (3.0 × 10⁵ î) × (0.4 î + 0.3 ĵ) = 0.9 × 10⁵ k̂, since î × î = 0 and î × ĵ = k̂.
3. Substitute the values into the acceleration formula: a = (4.8 × 10⁷) × (0.9 × 10⁵) k̂ = 4.32 × 10¹² k̂ m/s².

Answer: a = (q/m)(v × B); v × B = 3.0 × 10⁵ î × (0.4 î + 0.3 ĵ) = 0.9 × 10⁵ k̂ (since î × î = 0, î × ĵ = k̂). So a = 4.8 × 10⁷ × 0.9 × 10⁵ = 4.32 × 10¹² k̂ m/s², along +z.

Compare a magnetic field produced by a changing electric field with an electric field produced by a changing magnetic field. Which is easier to observe, and why?

1. A changing magnetic field induces an electric field (emf) in a conductor, which is easily observed by moving a magnet near a coil or rotating a coil in a magnetic field.
2. This induced electric field drives a measurable current, making the effect of a changing magnetic field straightforward to detect.
3. A changing electric field produces a magnetic field via displacement current, but this effect is extremely difficult to observe.
4. The magnetic field from a changing electric field is tiny because it involves the factor ε₀μ₀, which equals 1/c² and is a very small value.

Answer: The electric field (emf) induced by a changing magnetic field is easier to observe: it is produced easily by moving a magnet near a coil, rotating a coil in a field or changing its shape, and it drives a measurable current. The magnetic field from a changing electric field (displacement current) is tiny, as it carries the small factor ε₀μ₀ = 1/c².

With Huygens' construction, draw how a plane wavefront refracts at a flat boundary between two media, and use it to prove the law of refraction.

1. Plane wavefront AB meets the boundary at A, making angle i with it; medium 1 has speed v₁ and medium 2 has speed v₂.
2. In time τ the point B moves BC = v₁τ to reach the boundary at C, while the secondary wavelet from A spreads into medium 2 as a sphere of radius AE = v₂τ.
3. The tangent from C to this sphere, CE, is the refracted wavefront; it makes angle r with the boundary.
4. In right triangle ABC (right angle at B), angle BAC = i, so sin i = BC/AC = v₁τ/AC. In right triangle AEC (right angle at E), angle ACE = r, so sin r = AE/AC = v₂τ/AC.
5. Dividing: sin i/sin r = v₁/v₂. Since v₁ = c/n₁ and v₂ = c/n₂, this equals n₂/n₁ = n₂₁, which is Snell's law of refraction.

Answer: Incident wavefront AB meets the boundary at A. In time τ the end B travels BC = v₁τ in medium 1, while a secondary wavelet from A grows to radius AE = v₂τ in medium 2; the tangent CE is the refracted wavefront. sin i = BC/AC = v₁τ/AC and sin r = AE/AC = v₂τ/AC, so sin i/sin r = v₁/v₂ = n₂/n₁ = n₂₁ (Snell's law).

Give two reasons, with justification, why astronomers prefer a reflecting telescope to a refracting one.

1. Reason 1: Reflecting telescopes use mirrors instead of lenses, so they do not suffer from chromatic aberration.
2. Reason 2: The mirrors can be made with a parabolic shape, which eliminates spherical aberration and focuses all parallel rays to a single point.
3. Reason 3: Mirrors are lighter than lenses and can be supported from the entire back, allowing the construction of very large apertures.
4. Justification: A larger aperture increases the resolving power of the telescope and collects more light, resulting in a brighter image.

Answer: Any two: (1) no chromatic aberration, since a mirror does not refract light; (2) a parabolic mirror removes spherical aberration; (3) a mirror is lighter and can be supported over its whole back, so very large apertures are possible; (4) large aperture gives higher resolving power and a brighter image.

Coherent single-wavelength beams of intensity I and 4I overlap. Find I_max : I_min of the combined beam.

1. The ratio of amplitudes is equal to the square root of the ratio of intensities, so A1 : A2 = √I : √4I = 1 : 2.
2. The maximum intensity is proportional to (A1 + A2)², which gives (1 + 2)² = 9.
3. The minimum intensity is proportional to (A1 - A2)², which gives (1 - 2)² = 1.
4. Therefore, the ratio I_max : I_min is 9 : 1.

Answer: Amplitudes are in ratio √4 : √1 = 2 : 1, so I_max/I_min = (2 + 1)²/(2 − 1)² = 9 : 1.

Hydrogen's ground-state energy is −13.6 eV. Find the electron's potential energy and kinetic energy in the third excited state.

1. Identify the principal quantum number for the third excited state as n = 4.
2. Calculate the total energy using the formula Eₙ = -13.6/n² eV, which gives E₄ = -13.6/16 = -0.85 eV.
3. Determine the potential energy using the relation PE = 2Eₙ, resulting in PE = 2 × (-0.85) = -1.7 eV.
4. Determine the kinetic energy using the relation KE = -Eₙ, resulting in KE = -(-0.85) = 0.85 eV.

Answer: Third excited state is n = 4: E₄ = −13.6/16 = −0.85 eV. Potential energy = 2E₄ = −1.7 eV; kinetic energy = −E₄ = 0.85 eV.

State two differences between intrinsic and extrinsic semiconductors.

1. Intrinsic semiconductors are pure elements like silicon, where the number of electrons equals the number of holes (nₑ = n_h) and conductivity is low at room temperature.
2. Extrinsic semiconductors are doped with impurity atoms, resulting in unequal electron and hole concentrations (nₑ ≠ n_h) and much higher, controllable conductivity.

Answer: Intrinsic: pure, with nₑ = n_h and low conductivity at room temperature. Extrinsic: doped with impurity atoms, with nₑ ≠ n_h and much higher conductivity that can be controlled by doping.

Sketch the circuits used to study a p-n diode's V–I behaviour in forward and in reverse bias, and plot the V–I curve of a silicon diode.

1. Draw the forward bias circuit with a battery, rheostat in potential divider configuration, and the diode connected with its p-side to the positive terminal.
2. Include a voltmeter connected in parallel across the diode and a milliammeter connected in series to measure the small forward current.
3. Draw the reverse bias circuit by reversing the diode orientation relative to the battery and replacing the milliammeter with a microammeter to measure the tiny reverse saturation current.
4. Plot the V-I curve showing that in forward bias the current remains near zero until the threshold voltage of 0.7 V, after which it rises sharply.
5. Indicate on the curve that in reverse bias a nearly constant microampere current flows until the breakdown voltage, where the current increases steeply.

Answer: Forward bias: battery, rheostat (potential divider), diode with p side to +, voltmeter across the diode and milliammeter in series. Reverse bias: same but diode reversed and a microammeter. Curve: in forward bias current stays near zero until about 0.7 V (threshold) and then rises sharply; in reverse bias only a tiny, nearly constant current (μA) flows until breakdown, where it rises steeply.

Explain in brief the roles of diffusion current and drift current in building up the barrier potential of a p-n junction.

1. Due to the concentration difference, majority carriers (holes and electrons) diffuse across the junction, creating a diffusion current.
2. This diffusion leaves behind immobile ions in the depletion layer, which generates an electric field that drives a drift current opposite to the diffusion current.
3. The depletion layer grows until the drift current equals the diffusion current, establishing equilibrium where the potential difference across the layer is the barrier potential.

Answer: Holes and electrons diffuse across the junction because of the concentration difference (diffusion current), leaving behind immobile ions that form a depletion layer. The field of these ions drives a drift current opposite to diffusion. The layer grows until the two currents are equal; the potential difference across the layer then is the barrier potential.

(a) Twelve equal negative charges (each −q) sit equally spaced on a circle of radius R. Taking V = 0 at infinity, find V and E at the centre C. (b) The same charges are instead bunched unevenly along a 120° arc (radius R); find V at C.

Uses a figure printed in CBSE's paper — what the figure shows: Fig. (i): 12 dots evenly spaced round a full circle of radius R, centre C. Fig. (ii): the 12 dots bunched unevenly along an arc of radius R that subtends 120° at C.

1. Part (a): The electric potential at the centre C is the sum of potentials due to all 12 charges, each at distance R.
2. Since potential is a scalar, V = 12 × (−q)/(4πε₀R) = −12q/(4πε₀R) = −3q/(πε₀R).
3. The electric field vectors from opposite charges are equal in magnitude and opposite in direction, so they cancel in pairs.
4. By symmetry, the net electric field at the centre is E = 0.
5. Part (b): Electric potential is a scalar quantity that depends only on the distance of the charge from the point, not on its angular position.
6. Since all 12 charges remain at distance R from C, the total potential is unchanged: V = −12q/(4πε₀R).

Answer: (a) Every charge is at distance R, so V = 12 × (−q)/(4πε₀R) = −12q/(4πε₀R) = −3q/(πε₀R); by symmetry the fields cancel in pairs, so E = 0. (b) Potential is a scalar and depends only on distance, so V is still −12q/(4πε₀R).

Note: CBSE's marking scheme prints V without the minus sign; since every charge is negative, the potential at C is negative.

How is impedance different from resistance? Using a phasor diagram, find the impedance of a series LCR circuit fed by v = v_m sin ωt.

1. Resistance opposes both dc and ac currents and is independent of frequency, whereas impedance is the total opposition to ac from R, L, and C combined and depends on frequency.
2. In the phasor diagram, the voltage across the resistor V_R = i_m R is in phase with the current i, while V_L = i_m X_L leads by 90° and V_C = i_m X_C lags by 90°.
3. The net voltage phasor is the vector sum of V_R and the difference of the reactive voltages, so v_m² = V_R² + (V_C − V_L)².
4. Substituting the expressions for voltages gives v_m² = (i_m R)² + (i_m X_C − i_m X_L)² = i_m² [R² + (X_C − X_L)²].
5. Comparing this with v_m = i_m Z, the impedance is Z = √(R² + (X_C − X_L)²).

Answer: Resistance opposes both dc and ac and does not depend on frequency; impedance is the total opposition to ac from R, L and C together and depends on frequency. Phasors: V_R = i_mR along i, V_L = i_mX_L leading by 90°, V_C = i_mX_C lagging by 90°. So v_m² = V_R² + (V_C − V_L)² = i_m²[R² + (X_C − X_L)²], giving i_m = v_m/Z with Z = √(R² + (X_C − X_L)²).

For a series LCR circuit on v = v_m sin ωt with variable ω, find the resonance condition, list what the resonant frequency depends on, and sketch current against frequency.

1. Write the impedance of the series LCR circuit as Z = √(R² + (X_L − X_C)²), where X_L = ωL and X_C = 1/(ωC).
2. State that the current is maximum when the impedance Z is minimum, which occurs when the reactance difference is zero, i.e., X_L = X_C.
3. Substitute the expressions for reactance to get ωL = 1/(ωC), and solve for the resonance condition to find ω₀ = 1/√(LC).
4. Note that the resonant frequency ω₀ depends only on the inductance L and capacitance C, and is independent of the resistance R.
5. Describe the graph of current versus frequency: it shows a sharp peak at ω₀, with the current rising to this maximum and falling on both sides; a smaller value of R results in a taller and sharper peak.

Answer: Z = √(R² + (X_L − X_C)²) is least, and current greatest, when X_L = X_C, i.e. ω₀L = 1/(ω₀C), so ω₀ = 1/√(LC). The resonant frequency depends only on L and C (not on R). Graph: current rises to a sharp peak at ω₀ and falls on both sides; smaller R gives a taller, sharper peak.

A long solenoid (radius r, n turns per unit length) carries I = I₀ sin ωt. A coil of N turns is wound closely round its middle. Find (a) the emf induced in the coil and (b) the mutual inductance.

1. The magnetic field inside a long solenoid is given by B = μ₀nI.
2. The magnetic flux linked with the coil of N turns is Φ = N × B × Area = N(μ₀nI)(πr²).
3. According to Faraday's law, the induced emf is e = −dΦ/dt = −μ₀nNπr² (dI/dt).
4. Substituting I = I₀ sin ωt, we get dI/dt = I₀ω cos ωt, so e = −μ₀nNπr²I₀ω cos ωt.
5. Comparing this with the standard formula e = −M(dI/dt), the mutual inductance is M = μ₀nNπr².

Answer: (a) Inside, B = μ₀nI, so the coil's flux is Φ = Nπr²μ₀nI and e = −dΦ/dt = −μ₀nNπr² dI/dt = −μ₀nNπr²I₀ω cos ωt. (b) Comparing with e = −M dI/dt gives M = μ₀nNπr².

Briefly explain electron emission from a metal using Einstein's photoelectric equation. Plot photocurrent against (a) collector potential for different intensities and (b) intensity of the light.

1. According to Einstein's photoelectric equation, one electron absorbs one photon of energy hν; if hν exceeds the work function φ₀, the electron escapes with maximum kinetic energy K_max = hν − φ₀.
2. No emission occurs below the threshold frequency regardless of intensity, as the photon energy is insufficient to overcome the work function.
3. For part (a), plot photocurrent versus collector potential: the current rises with potential and levels off at a saturation value, where higher intensity yields a higher saturation current but all curves intersect the axis at the same stopping potential.
4. For part (b), plot photocurrent versus intensity: the graph is a straight line passing through the origin, indicating that photocurrent is directly proportional to the intensity of the incident light.

Answer: One electron absorbs one photon of energy hν; if hν exceeds the work function φ₀ it escapes with K_max = hν − φ₀ (no emission below threshold, whatever the intensity). (a) Current rises with collector potential and levels off at a saturation value; higher intensity gives a higher saturation current, but all curves meet the axis at the same stopping potential. (b) Photocurrent is a straight line through the origin, proportional to intensity.

Draw hydrogen's energy level diagram and mark the transitions of the series in the ultraviolet, visible and infrared regions.

1. Draw horizontal lines for the levels n = 1, 2, 3, 4, 5 … ∞, with n = 1 at the bottom and ∞ at the top.
2. Label their energies: n = 1 (−13.6 eV), n = 2 (−3.4 eV), n = 3 (−1.51 eV), n = 4 (−0.85 eV), n = 5 (−0.54 eV), n = ∞ (0 eV).
3. Arrows from n = 2, 3, 4 … down to n = 1 form the Lyman series, in the ultraviolet.
4. Arrows from n = 3, 4, 5 … down to n = 2 form the Balmer series, in the visible region.
5. Arrows ending on n = 3 (from n ≥ 4) form the Paschen series, ending on n = 4 (from n ≥ 5) the Brackett series, and ending on n = 5 (from n ≥ 6) the Pfund series; all three are in the infrared.

Answer: Levels: n = 1 (−13.6 eV), 2 (−3.4 eV), 3 (−1.51 eV), 4 (−0.85 eV), 5 (−0.54 eV) … ∞ (0 eV). Transitions ending on n = 1 form the Lyman series (ultraviolet); ending on n = 2, the Balmer series (visible); ending on n = 3, 4, 5, the Paschen, Brackett and Pfund series (infrared).

Sketch binding energy per nucleon against mass number, give two features of the curve, and say why light nuclei tend to fuse.

1. Draw a graph with mass number A on the x-axis and binding energy per nucleon (BE/A) in MeV on the y-axis.
2. Sketch a curve that rises steeply from A = 2, peaks at about 8.75 MeV near A = 56, and falls slowly to about 7.6 MeV at A = 238.
3. State two features: the curve is nearly flat for 30 < A < 170, and it has a maximum near iron (A ≈ 56).
4. Explain that light nuclei fuse because the heavier product has a larger binding energy per nucleon, making it more tightly bound and releasing energy.

Answer: The curve rises steeply for light nuclei, peaks at about 8.75 MeV near A = 56 and falls slowly to about 7.6 MeV at A = 238. Features (any two): nearly flat for 30 < A < 170; maximum near iron (A ≈ 56); lower for very light (A < 30) and very heavy (A > 170) nuclei. Light nuclei fuse because the heavier product has a larger binding energy per nucleon, so it is more tightly bound and energy is released.

(i) State Coulomb's law and give its vector form for two charges. (ii) Explain why Gauss's law relies on the 1/r² dependence in Coulomb's law. (iii) Charge q at A(0, 0) and charge 2q at B(a, a): find the force of A on B in terms of î and ĵ.

1. Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
2. The vector form of the force exerted by charge q1 on charge q2 is F₂₁ = (1/4πε₀)(q₁q₂/r²) r̂₂₁, where r̂₂₁ is the unit vector from q1 to q2.
3. Gauss's law relies on the 1/r² dependence because the electric field E is proportional to 1/r² while the surface area of a sphere is proportional to r².
4. Since E ∝ 1/r² and Area ∝ r², their product (flux) is independent of the radius r, ensuring the flux depends only on the enclosed charge.
5. For part (iii), the position vector from A(0,0) to B(a,a) is r_AB = aî + aĵ, and its magnitude is |r| = √(a² + a²) = √2 a.
6. The force magnitude is F = (1/4πε₀)(q × 2q / (√2 a)²) = (1/4πε₀)(2q² / 2a²) = q² / (4πε₀a²).
7. The force vector is F = F × (r_AB / |r|) = [q² / (4πε₀a²)] × [(aî + aĵ) / (√2 a)] = q²(î + ĵ) / (4√2 πε₀a²).

Answer: (i) The force between two point charges is proportional to the product of the charges, inversely proportional to the square of their separation, and acts along the line joining them: F₂₁ = (1/4πε₀)(q₁q₂/r²) r̂₂₁. (ii) Flux through a sphere = E × area; E ∝ 1/r² and area ∝ r², so the flux is independent of r and depends only on the charge enclosed — true only for an inverse-square law. (iii) r_AB = aî + aĵ, |r| = √2 a, so F = (1/4πε₀)(2q²/2a²)(î + ĵ)/√2 = q²(î + ĵ)/(4√2 πε₀a²), repulsive, along AB.

(i) For a dipole of +q and −q, 2a apart, derive E at an equatorial point. (ii) For a far point on this plane, what happens to the field if its distance is halved? (iii) Charges +q at A, −q at B, −q at C, +q at D sit at the corners of a square ABCD (side √2 m), making two identical dipoles on its diagonals. Find the size and direction of the net field at the centre O.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, square ABCD has A top-left (+q), B top-right (−q), C bottom-right (−q), D bottom-left (+q); diagonals AC and BD meet at centre O; side √2 m.

1. For the dipole, the distance from each charge to the equatorial point is R = √(r² + a²), so the field magnitude from each charge is E₁ = kq / (r² + a²).
2. The perpendicular components of the fields cancel, and the axial components add, giving E = 2E₁ cos θ where cos θ = a / √(r² + a²).
3. Substituting these values yields E = 2kqa / (r² + a²)^(3/2), which is opposite to the dipole moment p = 2qa.
4. For a far point where r >> a, the field scales as E ∝ 1/r³, so halving the distance r makes the field 8 times larger.
5. In the square of side √2 m, the distance from each corner to the center O is 1 m, so the field from each dipole at O is 2kq.
6. The field from dipole AC is directed towards C, and the field from dipole DB is directed towards B; these two vectors are perpendicular.
7. The net field is the vector sum E = √[(2kq)² + (2kq)²] = 2√2 kq ≈ 25.5 × 10⁹ q N/C, directed parallel to side DC.

Answer: (i) Each charge gives q/[4πε₀(r² + a²)]; the components perpendicular to the axis cancel and those along the axis add, so E = −p/[4πε₀(r² + a²)^(3/2)], opposite to p (p = 2qa). (ii) For r ≫ a, E ∝ 1/r³, so halving r makes the field 8 times larger. (iii) The half-diagonal is 1 m. A and C together give 2kq towards C; D and B together give 2kq towards B; these are perpendicular, so E = 2√2 kq = 2√2 q/(4πε₀) ≈ 25.5 × 10⁹ q N/C (q in coulomb), directed parallel to DC (from D towards C).

(i) State the Biot–Savart law and use it to find B at the centre of a circular loop of radius a carrying current I; sketch the field lines of a current loop showing their direction. (ii) An electron circles a nucleus at 10⁷ m/s in an orbit of radius 10⁻¹⁰ m. Find the equivalent current.

1. Biot–Savart law: a current element I dl produces at a point at distance r the field dB = (μ₀/4π) I dl × r / r³, of size (μ₀/4π) I dl sin θ / r².
2. At the centre of a loop of radius a every element is at distance a and dl is perpendicular to r (θ = 90°), so each dB = (μ₀/4π) I dl / a², all pointing along the axis.
3. Adding over the whole length 2πa: B = (μ₀/4π)(I/a²)(2πa) = μ₀I/(2a), along the axis of the loop.
4. Field lines: closed loops circle each side of the wire; near the centre they become a straight line along the axis, with direction given by the right-hand thumb rule (curl fingers along the current, thumb gives B).
5. For the electron, one revolution takes T = 2πr/v, so the equivalent current is I = e/T = ev/(2πr).
6. I = (1.6 × 10⁻¹⁹ × 10⁷)/(2π × 10⁻¹⁰) = (1.6 × 10⁻¹²)/(6.28 × 10⁻¹⁰) ≈ 2.55 × 10⁻³ A = 2.55 mA.

Answer: (i) dB = (μ₀/4π) I dl × r / r³, magnitude (μ₀/4π) I dl sin θ / r². At the centre every element is at distance a with θ = 90°, so B = (μ₀/4π)(I/a²)(2πa) = μ₀I/(2a), along the axis; field lines circle each side of the wire and pass straight through the loop's centre (direction by the right-hand rule). (ii) I = e/T = ev/(2πr) = 1.6 × 10⁻¹⁹ × 10⁷/(2π × 10⁻¹⁰) ≈ 2.55 mA.

(i) Derive the force on a straight current-carrying conductor in a magnetic field, name the rule that gives its direction, and state when the force is greatest and least. (ii) Long parallel wires A and B, 2.5 cm apart in air, have opposite currents, 5.0 A in A and 2.5 A in B. Find the force of A on 10 cm of B.

1. Take a conductor of length l, cross-section A, with n free carriers per unit volume, each of charge q moving with drift velocity v_d; it holds nlA carriers.
2. Each carrier feels the force q(v_d × B), so the total force is F = nlAq(v_d × B). Since I = nAqv_d, this is F = I(l × B), with size F = IlB sin θ.
3. Its direction is given by Fleming's left-hand rule. The force is greatest, IlB, when the conductor is perpendicular to B (θ = 90°), and zero when it is parallel to B (θ = 0° or 180°).
4. For the parallel wires, force on length l of B: F = μ₀I₁I₂l/(2πd) = 2 × 10⁻⁷ × 5.0 × 2.5 × 0.10/0.025.
5. F = 2 × 10⁻⁷ × 1.25/0.025 = 2 × 10⁻⁷ × 50 = 1 × 10⁻⁵ N.
6. The currents are opposite, so the force is repulsive: B is pushed away from A.

Answer: (i) A rod of length l and area A has nlA carriers, each feeling q(v_d × B); total F = nlAq(v_d × B) = I(l × B), |F| = IlB sin θ. Direction by Fleming's left-hand rule. Maximum (IlB) when θ = 90°; zero when θ = 0° or 180°. (ii) F = μ₀I₁I₂l/(2πd) = 2 × 10⁻⁷ × 5.0 × 2.5 × 0.10/0.025 = 1 × 10⁻⁵ N, repulsive (opposite currents).

(i) Give two differences between interference and diffraction patterns, and two factors that set the fringe width in Young's experiment. (ii) In Young's experiment the slit gap is 100 times the wavelength. Find (1) the angle in radians between the central and the next maximum and (2) their separation on a screen 50 cm away.

1. Interference fringes are equally spaced and of equal brightness, while diffraction has a central maximum twice as wide as others with decreasing brightness.
2. Interference arises from two coherent sources, whereas diffraction results from wavelets across a continuous aperture.
3. The fringe width is given by β = λD/d, so it depends on the wavelength of light and the distance of the screen from the slits.
4. For the first maximum, the path difference equals the wavelength, so sin θ = λ/d.
5. Given d = 100λ, sin θ = 1/100 = 0.01, and for small angles θ ≈ 0.01 rad.
6. The separation on the screen is y = Dθ = 50 cm × 0.01 = 0.5 cm.
7. Converting to millimeters, the separation is 5 mm.

Answer: (i) Interference has many equally spaced fringes of equal brightness; diffraction has a central maximum twice as wide as the others, with brightness falling for higher orders. Interference comes from two slits; diffraction from a continuous set of wavelets across one slit. Fringe width β = λD/d depends on wavelength λ, screen distance D and slit separation d (any two). (ii) (1) θ = λ/d = 1/100 = 0.01 rad. (2) Separation = Dθ = 50 cm × 0.01 = 0.5 cm = 5 mm.

(i) A spherical surface of radius R divides media of indices n₁ and n₂; an object at distance u in n₁ gives an image at distance v in n₂. Derive the u–v relation. (ii) A glass sphere (radius 6.0 cm, μ = 1.5) has an air bubble 3.0 cm from its centre. Find where the bubble appears when viewed from air through the nearer surface.

Uses a figure printed in CBSE's paper — what the figure shows: In the figure, the viewer E is outside on the left; the bubble O lies on the line EC, 3 cm from the centre C on E's side, so it is 3 cm inside the near surface.

1. For small aperture, angles are i = MN/OM + MN/MC and r = MN/MC − MN/MI.
2. Applying Snell's law n₁i = n₂r gives n₁/OM + n₂/MI = (n₂ − n₁)/MC.
3. Using sign convention OM = −u, MI = v, MC = R, the relation is n₂/v − n₁/u = (n₂ − n₁)/R.
4. For the bubble, refraction is glass to air: n₁ = 1.5, n₂ = 1, u = −(6 − 3) = −3 cm, R = −6 cm.
5. Substitute values: 1/v = (1 − 1.5)/(−6) + 1.5/(−3) = 1/12 − 1/2.
6. Solving gives v = −2.4 cm, so the bubble appears 2.4 cm inside the sphere from the near surface.
7. The distance from the centre is 6.0 − 2.4 = 3.6 cm.

Answer: (i) For small aperture, with N on the surface, i = MN/OM + MN/MC and r = MN/MC − MN/MI; Snell's law n₁i = n₂r gives n₁/OM + n₂/MI = (n₂ − n₁)/MC; with OM = −u, MI = v, MC = R: n₂/v − n₁/u = (n₂ − n₁)/R. (ii) Refraction glass → air: n₁ = 1.5, n₂ = 1, u = −(6 − 3) = −3 cm, R = −6 cm. 1/v = (1 − 1.5)/(−6) + 1.5/(−3) = 1/12 − 1/2, so v = −2.4 cm: the bubble appears 2.4 cm inside the sphere from the near surface (3.6 cm from the centre).

In the circuit described, which points have equal potential?

Uses a figure printed in CBSE's paper — what the figure shows: Circuit: left branch a–h has cell E = 6.0 V with internal r = 1.0 Ω; top wire a–b–c; middle branch b–i–R₁ (10 Ω)–j–g; right branch c–R₂ (5.0 Ω)–d–e–R₃ (5.0 Ω)–f; bottom wire h–g–f. Points i, d, e, j are on plain wire next to the resistors.

1. Points connected by plain wire segments without any resistance or cell have the same potential.
2. Thus, a, b, c and i are at equal potential; d and e are at equal potential; and f, g, h and j are at equal potential.

Answer: Points joined only by plain wire: a, b, c (and i); d and e; f, g, h and j.

In the same circuit (6.0 V cell with 1.0 Ω; 10 Ω in branch bg; 5.0 Ω + 5.0 Ω in branch cf), find the current in arm bg.

Uses a figure printed in CBSE's paper — what the figure shows: Same circuit as 34(a): cell 6.0 V, r = 1.0 Ω; bg has R₁ = 10 Ω; cf has R₂ = 5.0 Ω and R₃ = 5.0 Ω in series; bg and cf are in parallel across the cell.

1. Calculate the resistance of branch cf: 5.0 Ω + 5.0 Ω = 10 Ω.
2. Find the equivalent resistance of the parallel branches (10 Ω and 10 Ω): (10 × 10)/(10 + 10) = 5 Ω.
3. Determine the total current from the cell: I = 6.0 V / (5 Ω + 1.0 Ω) = 1 A.
4. Calculate the current in arm bg: Since the parallel branches have equal resistance, the current splits equally, so I_bg = 1 A / 2 = 0.5 A.

Answer: The two branches (10 Ω and 5 + 5 = 10 Ω) in parallel give 5 Ω; total current = 6/(5 + 1) = 1 A, which splits equally, so the current in bg is 0.5 A.

In the same circuit, find the voltage across R₃ = 5.0 Ω.

Uses a figure printed in CBSE's paper — what the figure shows: Same circuit as 34(a); R₃ = 5.0 Ω is in branch cf in series with R₂ = 5.0 Ω.

1. The current in branch cf is found by subtracting the current in the other branch from the total current: I = 1 − 0.5 = 0.5 A.
2. The voltage across R₃ is calculated using Ohm's law: V = 0.5 × 5.0 = 2.5 V.

Answer: Current in branch cf = 1 − 0.5 = 0.5 A, so V = 0.5 × 5.0 = 2.5 V.

In the same circuit, find the power dissipated in R₂ = 5.0 Ω.

Uses a figure printed in CBSE's paper — what the figure shows: Same circuit as 34(a); R₂ = 5.0 Ω is in branch cf in series with R₃ = 5.0 Ω.

1. The current flowing through R₂ is given as 0.5 A.
2. Power dissipated is calculated using P = I²R, so P = (0.5)² × 5.0 = 1.25 W.

Answer: Current in R₂ is 0.5 A, so P = I²R = 0.5² × 5.0 = 1.25 W.

Strontium titanate has refractive index 2.41 and critical angle 24.5°. A ray falls normally on one face of an equilateral prism made of it. Trace the ray through the prism.

1. Since the ray falls normally on the first face, it enters the prism undeviated and travels straight to the second face.
2. In an equilateral prism, the angle of incidence at the second face is 60°.
3. As 60° is greater than the critical angle of 24.5°, the ray undergoes total internal reflection at the second face.
4. The reflected ray strikes the third face normally and emerges undeviated.

Answer: The ray enters the first face undeviated and meets the second face at 60°. Since 60° > 24.5°, it is totally internally reflected, then strikes the third face along the normal and leaves undeviated.

Find the speed of light inside the strontium titanate prism (n = 2.41).

1. The speed of light in a medium is given by the formula v = c/n, where c is the speed of light in vacuum and n is the refractive index.
2. Substituting the given values, v = (3 × 10⁸ m/s) / 2.41 ≈ 1.24 × 10⁸ m/s.

Answer: v = c/n = 3 × 10⁸/2.41 ≈ 1.24 × 10⁸ m/s.

Briefly explain two uses of total internal reflection.

1. Use 1: Optical fibres use total internal reflection to trap light within the core, allowing it to be transmitted over long distances with minimal loss.
2. Use 2: The sparkle of a diamond is due to total internal reflection, where its small critical angle causes light to reflect multiple times inside before exiting.

Answer: Any two: optical fibres — light is trapped inside the core by repeated total internal reflection and carried long distances; sparkle of diamond — its small critical angle makes light reflect many times inside before leaving; mirage — light from the sky is totally reflected near hot ground, appearing as water.

Define total internal reflection and give the two conditions needed for it.

1. Total internal reflection is the phenomenon where light traveling from a denser to a rarer medium is completely reflected back into the denser medium at the boundary.
2. The first condition is that the light must travel from an optically denser medium to an optically rarer medium.
3. The second condition is that the angle of incidence must be greater than the critical angle for that specific pair of media.

Answer: When light going from a denser to a rarer medium is reflected wholly back into the denser medium at the boundary, it is total internal reflection. Conditions: (1) light must travel from the optically denser to the rarer medium; (2) the angle of incidence must exceed the critical angle for that pair of media.