Physics (Theory)

Field of a short dipole on its axis: \( E = \dfrac{2kp}{r^3} \), so force on the charge F = qE ∝ 1/r³.
Doubling r: F' = F/(2)³ = F/8.
Common mistake: using the 1/r² law of a point charge – a dipole field falls as 1/r³.

In a metal, raising temperature increases collisions of free electrons (relaxation time τ decreases), so ρ = m/(ne²τ) increases – this rules out (c).
Over a wide temperature range the rise is not linear: the NCERT plot of resistivity of copper vs T is a curve bending upward, starting small at low temperatures.
The linear relation ρ = ρ₀[1 + α(T − T₀)] is only an approximation over a limited range.
CBSE key: (d).

Open-circuit voltage = emf, E = 8 V.
Terminal voltage V = E − Ir ⇒ 4 = 8 − 4r.
r = 4/4 = 1 Ω.

Current is the same at every cross-section (steady current): I = neAv_d.
n and e are constant, so v_d = I/(neA) ∝ 1/A.
A plot of y ∝ 1/x is a rectangular hyperbola – graph (a).

A diamagnetic substance gets a weak induced magnetisation opposite to the applied field (χ < 0).
So it always moves from a stronger to a weaker field region – away from the magnet – whichever pole is used.
Hence it is repelled by both poles.

Peak emf: e₀ = NBAω.
A = πr² = π(0.08)² = 0.0064π m².
e₀ = 40 × 3.0×10⁻² × 0.0064π × (25/π) = 40 × 0.03 × 0.0064 × 25 = 0.192 V ≈ 0.19 V.
Note how π cancels – that is why ω was given as 25/π.

Motional emf: ε = Blv.
It depends on B, l and v only.
The resistance r decides the current I = ε/r, not the emf.

No charges cross the gap; the changing electric flux there gives Maxwell's displacement current.
\( I_d = \varepsilon_0 \dfrac{d\phi_E}{dt} \).
Check: φ_E = q/ε₀ between plates, so I_d = dq/dt = conduction current in the wires – continuity of current.

When the object is at the centre of curvature (u = −2f), the real image also forms at the centre of curvature (v = −2f).
Object and image coincide, so the distance between them is zero.
Common mistake: answering 4f – that is the minimum object–real-image distance for a convex LENS, not a mirror.

\( r_n = \dfrac{n^2 h^2 \varepsilon_0}{\pi m e^2} = 0.53\,n^2 \) Å.
So r ∝ n².

ΔE = 13.6(1 − 1/25) = 13.6 × 24/25 = 13.056 eV.
λ = hc/ΔE = 1240 eV·nm / 13.056 eV ≈ 95 nm.
(Or 1/λ = R(1 − 1/25) = 1.097×10⁷ × 0.96 ⇒ λ ≈ 9.5×10⁻⁸ m.)

R = R₀A^{1/3} ⇒ volume = (4/3)πR₀³A.
Mass ≈ A·m.
Density = Am / [(4/3)πR₀³A] = 3m/(4πR₀³) – no A left.
So all nuclei have about the same density (~2.3×10¹⁷ kg m⁻³).

A single diode conducts only when forward biased, i.e. during one half of each cycle.
In the other half it is reverse biased and the output is (nearly) zero.
So the load gets one hump every cycle with a gap in between – half-wave rectification, graph (c).
(b) would need a full-wave rectifier (two diodes / bridge).

A trivalent atom (B, Al, In) has only 3 valence electrons, so one bond is incomplete – a hole is created.
The dopant accepts electrons (acceptor), holes become majority carriers ⇒ p-type.
Carrier density rises, so resistance decreases, not increases.

The donor level lies just below the conduction band; for Si it is about 0.01–0.05 eV below E_C.
That is why donors are ionised easily at room temperature (kT ≈ 0.025 eV).
1 eV (≈1.1 eV) is the full band gap of Si – don't confuse the two.

Copper is diamagnetic, not paramagnetic – so R is false.
A diamagnetic bar pushes field lines out (field inside is slightly reduced); lines do not concentrate inside – so A is false.
Both false ⇒ (d).

A wavefront is defined as the locus of points vibrating in the same phase – so A is true.
For a wave from a point source the wavefront is a sphere: all its points are the same distance from the source, the wave reaches them at the same time and they are in phase – R is true and explains A.

A is true: threshold frequency, instantaneous emission and K_max depending only on frequency can be explained only by photons (hν).
R is true: more intensity = more photons = more electrons emitted.
But the current-intensity proportionality alone would also fit the wave picture; the particle nature is proved by the frequency dependence and threshold. So R does not explain A ⇒ (b).

Lorentz force: \( \vec F = q(\vec v\times\vec B) \).
\( \vec v\times\vec B = 3\times10^5\,\hat i\times(0.4\hat i+0.3\hat j) = 3\times10^5\times0.3\,(\hat i\times\hat j) = 0.9\times10^5\,\hat k \) (since î×î = 0).
\( \vec a = \dfrac{q}{m}(\vec v\times\vec B) = 4.8\times10^7\times0.9\times10^5\,\hat k = 4.32\times10^{12}\,\hat k \) m s⁻².
Always state the direction (+z) – the component of B along v gives no force.

Identification: the induced electric field due to a changing magnetic field (Faraday's induction) is easily observed.
Justification: it is produced and detected easily – rotating or moving a coil in a magnetic field, moving a magnet towards a coil, changing the area of a coil – and gives measurable emf/current in everyday devices (generators, transformers).
The magnetic field produced by a changing electric field (displacement current \( I_d=\varepsilon_0\, d\phi_E/dt \)) is extremely small in ordinary situations because ε₀ is tiny (~10⁻¹¹), so it is hard to detect.

Diagram: plane wavefront AB strikes the surface at A at angle i; medium 1 speed v₁, medium 2 speed v₂ (v₂ < v₁ for a denser medium 2).
Let τ be the time for point B to reach C on the surface: BC = v₁τ.
In the same time the wavelet from A spreads into medium 2 as a sphere of radius AE = v₂τ.
The tangent CE from C to this sphere is the refracted wavefront; it makes angle r with the surface.
From triangles ABC and AEC: sin i = BC/AC = v₁τ/AC and sin r = AE/AC = v₂τ/AC.
Dividing: sin i / sin r = v₁/v₂ = constant. With n = c/v this gives n₁ sin i = n₂ sin r – Snell's law.

Any two:
1. A mirror has no chromatic aberration (reflection does not depend on wavelength); a lens objective does.
2. Using a parabolic mirror removes spherical aberration.
3. A large-aperture mirror is much lighter, cheaper and easier to make than an equally large lens, and can be supported over its whole back surface; a big lens sags and is hard to support only at its rim.
4. Large aperture gives high light-gathering power and better resolving power.

Intensity ∝ (amplitude)², so a₁ ∝ √I and a₂ ∝ √(4I) = 2√I.
I_max = (√I₁ + √I₂)² = (√I + 2√I)² = 9I.
I_min = (√I₂ − √I₁)² = (2√I − √I)² = I.
Ratio = 9 : 1.
Common mistake: adding intensities (5I) instead of amplitudes.

Third excited state means n = 4 (ground n=1, first excited n=2, …).
E₄ = −13.6/4² = −0.85 eV.
For a Bohr orbit: K = −E and U = 2E.
K = +0.85 eV; U = 2(−0.85) = −1.70 eV.
Common mistake: taking third excited state as n = 3.

Intrinsic semiconductor:
• pure Si/Ge, no impurity
• electron density = hole density (nₑ = nₕ = nᵢ)
• very low conductivity, which depends only on temperature
Extrinsic semiconductor:
• doped with pentavalent (n-type) or trivalent (p-type) impurity
• nₑ ≠ nₕ (nₑ ≫ nₕ in n-type, nₕ ≫ nₑ in p-type)
• much higher conductivity, controlled by the amount of doping
(Any two clear differences earn the marks.)

Forward bias circuit: battery with potentiometer/rheostat, p-side connected to + terminal, n-side to −; a milliammeter in series and a voltmeter across the diode.
Reverse bias circuit: battery terminals reversed (p to −), a microammeter in series, voltmeter across the diode.
Graph: forward side – current nearly zero up to the threshold (knee) voltage ≈ 0.7 V for Si, then rises steeply (mA scale).
Reverse side – very small, almost constant current (µA) due to minority carriers, until the breakdown voltage, where the reverse current increases suddenly.
Label axes: V (volts) horizontal, I (mA forward / µA reverse) vertical.

Because of the concentration difference, holes diffuse from p to n and electrons from n to p – the diffusion current.
Each electron leaving the n-side leaves a positive donor ion; each hole leaving the p-side leaves a negative acceptor ion. These fixed ions form the depletion layer.
The ions set up an electric field from n to p, i.e. a potential barrier.
This field pushes minority carriers across the junction (electrons p→n, holes n→p) – the drift current, opposite to diffusion current.
At first diffusion is large and drift small; as the layer grows, the field and drift increase until drift current = diffusion current. Then there is no net current and the barrier potential stays fixed.

(a) Every charge is at the same distance R from C. Potential is a scalar, so it simply adds:
V = 12 × (−q)/(4πε₀R) = −12q/(4πε₀R) = −3q/(πε₀R).
Field is a vector: charges are placed symmetrically, so the field of each charge is cancelled by the diametrically opposite one ⇒ E = 0.
(b) Each charge is still at distance R from C, and potential does not depend on direction or spacing:
V = −12q/(4πε₀R) (unchanged).
Key idea: potential (scalar) depends only on distance; field (vector) depends on the arrangement.

Difference: resistance is the opposition of a resistor alone, does not depend on frequency and V, I are in phase; impedance is the total effective opposition of an ac circuit (R together with reactances X_L, X_C), it depends on frequency and in general V and I differ in phase.
Phasor diagram: take current i = i_m sin(ωt + φ) as reference (common to all elements in series).
V_R = i_mR along the current phasor; V_L = i_mX_L leading it by 90°; V_C = i_mX_C lagging it by 90°.
V_L and V_C are opposite, so their resultant is (V_L − V_C), perpendicular to V_R.
By Pythagoras: \( v_m^2 = V_R^2 + (V_L - V_C)^2 = i_m^2[R^2 + (X_L - X_C)^2] \).
So \( i_m = \dfrac{v_m}{\sqrt{R^2+(X_L-X_C)^2}} \) and \( Z = \sqrt{R^2+(X_L-X_C)^2} \), with tan φ = (X_C − X_L)/R.

Current amplitude i_m = v_m/Z with \( Z = \sqrt{R^2+(X_L-X_C)^2} \).
i_m is maximum when Z is minimum, i.e. when X_L = X_C.
ωL = 1/(ωC) ⇒ \( \omega_0 = \dfrac{1}{\sqrt{LC}} \), \( \nu_0 = \dfrac{1}{2\pi\sqrt{LC}} \). Then Z = R and i_max = v_m/R.
Resonant frequency depends only on inductance L and capacitance C (not on R).
Graph: i_m vs ω rises, reaches a sharp peak at ω₀, then falls; a smaller R gives a taller, sharper peak.

Field inside the solenoid near its centre: B = μ₀nI.
Flux through the coil (same cross-section πr²): Φ = NBA = μ₀nNπr² I.
(a) e = −dΦ/dt = −μ₀nNπr² dI/dt = −μ₀nNπr² I₀ω cos ωt.
(b) Compare with e = −M dI/dt (or Φ = MI): M = μ₀nNπr².
Note M depends only on geometry, not on the current.

Explanation: light is a stream of photons of energy hν. An electron absorbs one whole photon. If hν exceeds the work function φ₀, the electron escapes; the extra energy appears as kinetic energy: K_max = hν − φ₀. If hν < φ₀, no emission at any intensity (threshold frequency ν₀ = φ₀/h).
(a) Photocurrent vs collector potential: for each intensity the current rises and saturates for positive potentials; all curves fall to zero at the same negative stopping potential −V₀; higher intensity → higher saturation current.
(b) Photocurrent vs intensity (fixed ν > ν₀): a straight line through the origin (current ∝ intensity).

Draw horizontal levels: 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).
Mark downward arrows:
• Lyman series – transitions ending on n = 1 → ultraviolet.
• Balmer series – transitions ending on n = 2 → visible.
• Paschen (ending n = 3), Brackett (n = 4), Pfund (n = 5) → infrared.
The levels get closer together as n increases.

Graph: BE/A (MeV) vs A – rises sharply for light nuclei (with peaks at He-4, C-12, O-16), flattens for 30 < A < 170, maximum ≈ 8.75 MeV near A = 56 (Fe), then slowly falls to ≈ 7.6 MeV at A = 238.
Features (any two):
1. BE/A is nearly constant for middle mass numbers (30 < A < 170) – shows saturation/short range of nuclear force.
2. Maximum ≈ 8.75 MeV at A ≈ 56.
3. BE/A is lower for both light (A < 30) and heavy (A > 170) nuclei.
Fusion: when two light nuclei join, the product nucleus has a higher binding energy per nucleon, so it is more tightly bound; the difference in binding energy is released. Hence light nuclei gain stability by fusion.

(i) The force between two point charges is directly proportional to the product of the charges, inversely proportional to the square of the distance between them, and acts along the line joining them.
Vector form: \( \vec F_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{21}^2}\hat r_{21} \), force on q₂ due to q₁, \( \hat r_{21} \) pointing from q₁ to q₂.
(ii) To get Gauss's law, flux through a sphere around a point charge is E × 4πr² = (q/4πε₀r²) × 4πr². The 1/r² of Coulomb's field cancels the r² of the area, so flux = q/ε₀ is independent of r and of the surface shape. If the force were not exactly inverse-square, the flux would depend on r and Gauss's law would fail.
(iii) \( \vec r_{AB} = a\hat i + a\hat j \), |r| = √2 a, unit vector = (î + ĵ)/√2.
\( \vec F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q(2q)}{(\sqrt2 a)^2}\cdot\dfrac{\hat i+\hat j}{\sqrt2} = \dfrac{q^2}{4\sqrt2\,\pi\varepsilon_0 a^2}(\hat i+\hat j) \).
Magnitude = q²/(4πε₀a²), directed away from A (repulsion, both positive).

(i) Point P at distance r from the centre on the perpendicular bisector. Each charge is at distance √(r²+a²).
|E₊| = |E₋| = kq/(r²+a²).
Components perpendicular to the dipole axis cancel; components along the axis add (both pointing from + to −):
E = 2 × kq/(r²+a²) × a/√(r²+a²) = k(2qa)/(r²+a²)^{3/2}.
With p = 2qa: \( \vec E = -\dfrac{k\vec p}{(r^2+a^2)^{3/2}} \) (opposite to p). For r ≫ a, E = kp/r³.
(ii) E ∝ 1/r³ ⇒ r → r/2 gives E → 8E.
(iii) Half-diagonal = (√2 × √2)/2 = 1 m, so each charge is 1 m from O.
Dipole A–C: field of +q at A points away from A (towards C); field of −q at C points towards C. Sum = 2kq along OC.
Dipole D–B: similarly 2kq along OB.
OB and OC are perpendicular, so E_net = √[(2kq)² + (2kq)²] = 2√2 kq.
Direction: bisector of OB and OC, i.e. parallel to DC (pointing from D towards C / towards side BC).
E_net = 2√2q/(4πε₀) N/C.

(i) Biot–Savart law: \( d\vec B = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\vec r}{r^3} \), magnitude dB = (μ₀/4π) I dl sin θ / r².
At the centre of a loop every element dl is at distance R and perpendicular to the radius (θ = 90°), and all dB point the same way (along the axis).
B = ∮ (μ₀/4π) I dl / R² = (μ₀I/4πR²) × 2πR = μ₀I/(2R).
Field lines: closed loops encircling the wire, passing straight through the centre along the axis; direction given by right-hand rule (anticlockwise current as seen → field towards the viewer).
(ii) I = e/T = ev/(2πr).
I = (1.6×10⁻¹⁹ × 10⁷)/(2π × 10⁻¹⁰) = (1.6×10⁻¹²)/(6.28×10⁻¹⁰) ≈ 2.55×10⁻³ A.

(i) Rod of length l, area A, carrier density n, charge q, drift velocity v_d. Number of carriers = nAl.
Force on each: q(v_d × B). Total: \( \vec F = nAl\,q(\vec v_d\times\vec B) \).
Since I = nqAv_d, \( \vec F = I\vec l\times\vec B \), magnitude F = IlB sin θ.
Direction: Fleming's left-hand rule – forefinger along B, middle finger along current, thumb gives force.
(1) Maximum F = IlB when θ = 90° (conductor ⟂ field).
(2) Minimum F = 0 when θ = 0° or 180° (conductor ∥ field).
(ii) Force per length: F/l = μ₀I₁I₂/(2πd).
F = (2×10⁻⁷ × 5.0 × 2.5 × 0.10)/(2.5×10⁻²) = (2.5×10⁻⁷)/(2.5×10⁻²) = 1.0×10⁻⁵ N.
Opposite currents ⇒ the force is repulsive.

(i)(1) Any two:
• Interference: equally spaced bright and dark fringes; diffraction: central maximum twice as wide as the others.
• Interference: all bright fringes of (nearly) equal intensity; diffraction: intensity of maxima falls rapidly away from the centre.
• Interference comes from superposition of waves from two coherent sources; diffraction from waves from different points of the same wavefront (single slit).
(2) Fringe width β = λD/d – depends on wavelength λ, slit-to-screen distance D and slit separation d (any two).
(ii)(1) First maximum: d sin θ = λ ⇒ θ ≈ λ/d = 1/100 = 0.01 rad.
(2) y = Dθ = 0.50 m × 0.01 = 5×10⁻³ m = 5 mm.

(i) Take a convex surface, pole M, centre C, object O on the axis, ray ON hitting at N, image I. Small-angle approximation.
Angles: i = ∠NOM + ∠NCM ≈ MN/OM + MN/MC; r = ∠NCM − ∠NIM ≈ MN/MC − MN/MI.
Snell (small angles): n₁i = n₂r ⇒ n₁/OM + n₂/MI = (n₂ − n₁)/MC.
Sign convention: OM = −u, MI = +v, MC = +R ⇒ \( \dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R} \).
(ii) Light goes from the bubble (inside glass, n₁ = 1.5) out to air (n₂ = 1) through the surface nearest E.
Distance of bubble from that surface = 6 − 3 = 3 cm ⇒ u = −3 cm. Centre C is on the glass side ⇒ R = −6 cm.
1/v − 1.5/(−3) = (1 − 1.5)/(−6)
1/v + 0.5 = 1/12 ⇒ 1/v = −5/12 ⇒ v = −2.4 cm.
The image is virtual, 2.4 cm from the surface inside the sphere (i.e. 3.6 cm from C). So the bubble appears closer to the viewing surface (2.4 cm instead of its real 3 cm).
Common mistake: taking R = +6 cm – the centre lies on the incoming-light (glass) side, so R is negative.

(a) Points joined by plain wire (no resistance) are at the same potential: a, b, c (and i); d and e; h, g, f (and j).
(b) R₂ + R₃ = 10 Ω, in parallel with R₁ = 10 Ω ⇒ R_ext = 5 Ω.
Total current I = E/(R_ext + r) = 6/(5 + 1) = 1 A.
Kirchhoff check: loop abgha: 10I₂ + 1·I = 6; loop acfha: 10(I − I₂) + I = 6 ⇒ I = 1 A, I₂ = 0.5 A.
Current in bg = 0.5 A.
(c) Current in branch c–f = 1 − 0.5 = 0.5 A ⇒ V(R₃) = 0.5 × 5 = 2.5 V.

(a) At AB the ray is normal ⇒ no bending.
Geometry of an equilateral prism: the ray then meets face AC at an angle of incidence of 60°.
60° > critical angle 24.5° ⇒ total internal reflection at AC.
The reflected ray travels perpendicular to BC, so it leaves through BC without deviation.
(b) v = c/n = 3×10⁸/2.41 ≈ 1.24×10⁸ m/s.
(c) Any two:
• Optical fibres – light repeatedly totally reflected at the core–cladding boundary travels long distances with little loss (communication, endoscopy).
• Sparkling of diamond – small critical angle (≈24°) causes multiple TIR inside a well-cut diamond.
• Mirage; totally reflecting prisms in periscopes/binoculars.