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Result & Error Notes

Aamirah Fathima · Physics Practice Paper — Alternating Current · Electromagnetic Waves · Ray Optics · 120 questions · marked +4 / −1 / 0

360out of 480
75.0%percentage
95correct
20wrong
5blank / void
83%accuracy

Section by section

SectionChapterCorrectWrongBlank/voidMarksAccuracy
AAlternating Current3451131/16087%
BElectromagnetic Waves3631141/16092%
CRay Optics2512388/16068%
Total95205360/48083%

Sub-topic map — where the marks went

ChapterSub-topicQ✓✗—Marks
Alternating CurrentAC fundamentals, rms and peak values852118/32
Alternating CurrentAC through R, L and C770028/28
Alternating CurrentSeries LCR circuit and impedance871027/32
Alternating CurrentPower, power factor and wattless current651019/24
Alternating CurrentResonance and sharpness660024/24
Alternating CurrentTransformers and LC oscillations541015/20
Electromagnetic WavesDisplacement current and Maxwell's equations761023/28
Electromagnetic WavesNature and properties of EM waves852118/32
Electromagnetic WavesSpeed, E–B relation and energy990036/36
Electromagnetic WavesThe electromagnetic spectrum990036/36
Electromagnetic WavesSources and uses770028/28
Ray OpticsReflection at plane surfaces743013/28
Ray OpticsRefraction at plane surfaces862022/32
Ray OpticsApparent depth and real depth52305/20
Ray OpticsTotal internal reflection942314/36
Ray OpticsRefraction through a prism752018/28
Ray OpticsDispersion by a prism440016/16

Difficulty profile

LevelQuestionsCorrectWrongBlankAccuracy
Easy45385288%
Moderate503810279%
Hard25195179%

What this paper says

Alternating Current and Electromagnetic Waves are exam-ready. Ray Optics is not. Sections A and B ran at 87% and 92% accuracy and cost 48 marks between them. Section C alone cost 72 — exactly half of everything she lost on a paper that was one-third Ray Optics. There is no general physics problem here; there is one chapter that has not been learnt.

Fix in this order

  1. Total internal reflection — 22 of 36 marks gone, and 3 of the 4 blanks on the whole paper sit here. She did not slow down on this block, she stopped. Rebuild it from two conditions that must both hold: light goes denser → rarer, and i > C. Then C = sin−1(1/n), and three numbers worth knowing cold: n = 1.5 → C ≈ 42°, water n = 4/3 → C ≈ 49°, diamond n = 2.42 → C ≈ 24°. Both her TIR numericals (Q102, Q104) failed on exactly these.
  2. Apparent depth — 15 of 20 marks gone. One relation carries the whole sub-topic: n = real depth / apparent depth, so apparent depth = t/n and the shift = t(1 − 1/n). The consequence she inverted twice: a larger n makes the apparent depth smaller, so the coin appears raised by more, not less.
  3. Plane mirrors — 15 of 28 marks gone. Three facts, nothing else: the image is virtual, erect, laterally inverted and the same size; the minimum mirror height is half the person's height whatever the distance; and for two mirrors at θ, the number of images is 360/θ − 1.
  4. Prism numericals. Both misses (Q113, Q116) were substitution, not concept: n = sin((A + δm)/2) / sin(A/2) at minimum deviation, and δ = (n − 1)A for a thin prism. Write both at the top of the rough sheet before starting any prism block.
  5. Five of the twenty wrongs were tagged Easy — Q82, Q88, Q89, Q101, Q102 are one-line NCERT statements (definition of refractive index, speed in a medium, what a plane-mirror image is). Those 25 marks come back from reading the Ray Optics chapter once, properly. They do not need practice.

On the blanks. Four questions left empty, all four in Ray Optics, three of them consecutive (Q101, Q105, Q106) and ringed in blue. That is the same pattern as Mock 2, where 17 blanks cost 68 marks. A blank scores 0; a guess between two surviving options is worth +4 or −1, which is a bet she should take every time. Ringing the number is the right instinct — but come back and fill it.

Two marks need explaining before the error notes.
Q1 carries two ticks — one on (A) 440 V and one on (D) 220 V, with the question number ringed in blue. (D) is right, but on a real OMR sheet a double mark is read as no answer, so it has been scored 0, not −1. In the hall, one answer per question, and erase properly.
Q50 was attempted on the first printing of this paper, where the EM-wave diagram was drawn in an oblique projection that made the two curves look offset. That figure was corrected in version 2. The physics she wrote (“π/2 out of phase”) follows the misleading picture, so treat this one as a paper defect, not her error — but do re-read the rule: in vacuum E and B oscillate in phase, mutually perpendicular, both perpendicular to the direction of travel.

SECTION A — ALTERNATING CURRENT — 6 lost

AC fundamentals, rms and peak values

Q1Easydouble-marked 0
The peak value of an alternating voltage is 311 V. Its rms value is nearly
  1. (A) 440 V
  2. (B) 311 V
  3. (C) 155 V
  4. (D) 220 V ✓
Why D: 311/√2 = 220 V.V_rms = V₀/√2 = 311/1.414 = 220 V
Q3Moderatewrong −1
The average value of an alternating current over one complete cycle is
  1. (A) zero ✓
  2. (B) equal to its peak value
  3. (C) equal to its rms value
  4. (D) 0.637 times its peak value ✗ her answer
Why A: The positive and negative halves cancel exactly.⟨i⟩ over a full cycle = 0
Q7Hardwrong −1
The form factor of a sinusoidal alternating current, that is the ratio of the rms value to the mean value over a half cycle, is
  1. (A) 1.41 ✗ her answer
  2. (B) 1.11 ✓
  3. (C) 2.00
  4. (D) 0.707
Why B: Form factor = rms ÷ mean over a half cycle.0.707 I₀ / 0.637 I₀ = 1.11

Series LCR circuit and impedance

Q17Moderatewrong −1
Refers back to Q16: In the circuit shown, R = 30 Ω, X_L = 50 Ω and X_C = 10 Ω. The impedance of the circuit is
For that circuit connected to a 200 V source, the current in it is
  1. (A) 2 A
  2. (B) 6.67 A ✗ her answer
  3. (C) 10 A
  4. (D) 4 A ✓
Why D: Ohm's law with the impedance.I = V/Z = 200/50 = 4 A

Power, power factor and wattless current

Q26Moderatewrong −1
Refers back to Q25: For the circuit with R = 30 Ω and Z = 50 Ω, the power factor is
For that circuit with an applied voltage of 200 V and a current of 4 A, the average power consumed is
  1. (A) 400 W
  2. (B) 800 W ✗ her answer
  3. (C) 640 W
  4. (D) 480 W ✓
Why D: Average power uses the power factor.P = V_rms I_rms cos φ = 200 × 4 × 0.6 = 480 W

Transformers and LC oscillations

Q38Moderatewrong −1
A step-down transformer with a turns ratio of 10 : 1 draws a primary current of 1 A. Assuming it is ideal, the secondary current is
  1. (A) 1 A
  2. (B) 0.1 A ✗ her answer
  3. (C) 10 A ✓
  4. (D) 100 A
Why C: An ideal transformer conserves power, so the current rises as the voltage falls.I_s = I_p × N_p/N_s = 1 × 10 = 10 A

SECTION B — ELECTROMAGNETIC WAVES — 4 lost

Displacement current and Maxwell's equations

Q44Hardwrong −1
During the charging of a capacitor, the conduction current in the connecting wire and the displacement current between the plates are
  1. (A) always unequal
  2. (B) equal in magnitude at every instant ✓
  3. (C) in the ratio of the plate area to the plate separation ✗ her answer
  4. (D) equal only at the end of the charging
Why B: Continuity requires the two to be equal at every instant.i_c = i_d = ε₀ dΦ_E/dt

Nature and properties of EM waves

Q50Moderatewrong −1
In the electromagnetic wave shown, the solid curve and the dashed curve represent, respectively,direction of propagationPQSolid curve and dashed curve of one electromagnetic wavethe two fields lie in mutually perpendicular planes
  1. (A) two electric fields of different frequency
  2. (B) the magnetic field and the electric field, differing in phase by π
  3. (C) the electric field and the magnetic field, oscillating in phase ✓
  4. (D) the electric field and the magnetic field, differing in phase by π/2 ✗ her answer
Why C: The two curves rise and fall together, so the fields are in phase.E = E₀ sin(kx−ωt), B = B₀ sin(kx−ωt)
Q51Moderatewrong −1
In an electromagnetic wave travelling in vacuum, E and B are
  1. (A) of equal magnitude in SI units
  2. (B) in the same phase, reaching their maxima together ✓
  3. (C) out of phase by π/2 ✗ her answer
  4. (D) out of phase by π
Why B: Both fields reach their maxima and zeros at the same instants.
Q54Hardleft blank 0
A beam of light carrying energy U falls on a perfectly absorbing surface. The momentum delivered to the surface is
  1. (A) U c
  2. (B) U/c ✓
  3. (C) U/c²
  4. (D) 2U/c
Why B: For complete absorption the whole momentum is delivered.p = U/c

SECTION C — RAY OPTICS — 15 lost

Reflection at plane surfaces

Q82Easywrong −1
The image formed by a plane mirror is
  1. (A) real, inverted and magnified
  2. (B) real, erect and of the same size
  3. (C) virtual, inverted and diminished ✗ her answer
  4. (D) virtual, erect and of the same size as the object ✓
Why D: A plane mirror gives a virtual, erect image of the same size, as far behind as the object is in front.
Q84Moderatewrong −1
Two plane mirrors are inclined at θ = 60° as shown. The number of images of the object formed isθobjectTwo plane mirrors inclined at an angle θ
  1. (A) 6
  2. (B) 5 ✓
  3. (C) 3
  4. (D) infinite ✗ her answer
Why B: With 360/θ an even number, the count is one less.360/60 = 6, images = 6 − 1 = 5
Q85Moderatewrong −1
A man 1.8 m tall needs a plane mirror of minimum height
  1. (A) 1.8 m
  2. (B) 3.6 m ✗ her answer
  3. (C) 1.2 m
  4. (D) 0.9 m ✓
Why D: The mirror need only be half the height of the object.h/2 = 1.8/2 = 0.9 m

Refraction at plane surfaces

Q88Easywrong −1
The refractive index of a medium is defined as the ratio of
  1. (A) the frequency in vacuum to that in the medium
  2. (B) the wavelength in the medium to that in vacuum
  3. (C) the speed of light in vacuum to that in the medium ✓
  4. (D) the speed of light in the medium to that in vacuum ✗ her answer
Why C: Refractive index compares the speed in vacuum with that in the medium.n = c/v
Q89Easywrong −1
The speed of light in a medium of refractive index 1.5 is
  1. (A) 3 × 10⁸ m s⁻¹
  2. (B) 1.5 × 10⁸ m s⁻¹
  3. (C) 2 × 10⁸ m s⁻¹ ✓
  4. (D) 4.5 × 10⁸ m s⁻¹ ✗ her answer
Why C: Divide c by the refractive index.v = c/n = 3×10⁸/1.5 = 2×10⁸ m s⁻¹

Apparent depth and real depth

Q96Easywrong −1
In the figure shown, the object O in water is seen from above at the position I. The apparent depth isO (real)I (seen)airwaterAn object seen from above the surface
  1. (A) greater than the real depth
  2. (B) less than the real depth ✓
  3. (C) zero
  4. (D) equal to the real depth ✗ her answer
Why B: Refraction at the surface makes the object appear nearer than it is.apparent < real
Q99Moderatewrong −1
The relation between the real depth and the apparent depth for near-normal viewing is
  1. (A) n = apparent depth / real depth
  2. (B) n = real depth × apparent depth
  3. (C) n = real depth − apparent depth ✗ her answer
  4. (D) n = real depth / apparent depth ✓
Why D: For near-normal viewing the refractive index is the ratio of the two depths.n = real depth / apparent depth
Q100Hardwrong −1
A coin at the bottom of a beaker appears raised. If the water is replaced by a liquid of greater refractive index, the coin appears
  1. (A) at its true position
  2. (B) lowered
  3. (C) raised by a larger amount ✓
  4. (D) raised by a smaller amount ✗ her answer
Why C: A larger n gives a smaller apparent depth, so the coin seems raised further.shift = t(1 − 1/n) increases with n

Total internal reflection

Q101Easyleft blank 0
Total internal reflection can take place only when light travels from
  1. (A) a denser medium to a rarer medium, with the angle of incidence greater than the critical angle ✓
  2. (B) a denser medium to a rarer medium, with the angle of incidence less than the critical angle
  3. (C) a rarer medium to a denser medium, with the angle of incidence greater than the critical angle
  4. (D) either medium to the other, at any angle
Why A: Both conditions are needed: denser to rarer, and beyond the critical angle.i > C with n₁ > n₂
Q102Easywrong −1
The critical angle for a medium of refractive index 1.5 with respect to air is about
  1. (A) 42° ✓
  2. (B) 48° ✗ her answer
  3. (C) 30°
  4. (D) 60°
Why A: Take the inverse sine of the reciprocal of n.C = sin⁻¹(1/1.5) = 41.8° ≈ 42°
Q104Moderatewrong −1
The critical angle for water of refractive index 4/3 is about
  1. (A) 42° ✗ her answer
  2. (B) 60°
  3. (C) 30°
  4. (D) 49° ✓
Why D: Same relation for water.C = sin⁻¹(3/4) = 48.6° ≈ 49°
Q105Moderateleft blank 0
Three rays strike the same boundary from the denser side, as shown. Total internal reflection occurs atPQRrarerdenserThree rays meeting the same boundary from the denser side
  1. (A) P only
  2. (B) R only ✓
  3. (C) P and Q
  4. (D) Q only
Why B: Only the ray incident beyond the critical angle is turned back into the denser medium.i_R > C
Q106Moderateleft blank 0
Refers back to Q105: Three rays strike the same boundary from the denser side, as shown. Total internal reflection occurs at
In the same figure, the ray at Q grazes along the boundary. The angle of incidence at Q is thereforePQRrarerdenserThree rays meeting the same boundary from the denser side
  1. (A) exactly equal to the critical angle ✓
  2. (B) equal to 90°
  3. (C) less than the critical angle
  4. (D) greater than the critical angle
Why A: A refracted ray grazing the surface means the angle of refraction is 90°, which defines the critical angle.sin C = n₂/n₁, r = 90°

Refraction through a prism

Q113Hardwrong −1
A prism of refracting angle 60° is made of a material of refractive index √2. Its angle of minimum deviation is
  1. (A) 60°
  2. (B) 15° ✗ her answer
  3. (C) 45°
  4. (D) 30° ✓
Why D: Prism formula at minimum deviation.n = sin((A+δ_m)/2)/sin(A/2); √2 = sin((60+δ_m)/2)/sin 30° ⇒ δ_m = 30°
Q116Hardwrong −1
A thin prism of refracting angle 4° is made of glass of refractive index 1.5. The deviation it produces is
  1. (A) 1° ✗ her answer
  2. (B) 2° ✓
  3. (C) 4°
  4. (D) 6°
Why B: Apply the thin-prism formula.δ = (1.5 − 1) × 4° = 2°