Part A: Light - Reflection & Refraction
Q1. Concave Mirror Image
Given: u = -25 cm, f = -15 cm, h = +4.0 cm
Formula: 1/v + 1/u = 1/f
Calc: 1/v = 1/(-15) - 1/(-25) = -1/15 + 1/25 = (-5 + 3)/75 = -2/75
v = -37.5 cm (Screen should be placed here)
m = -v/u = -(-37.5)/(-25) = -1.5
h' = m × h = -1.5 × 4 = -6 cm
Ans: Image is 37.5 cm in front of mirror, Real, Inverted, and Enlarged (6 cm
tall).
Q2. Convex Mirror (Rear View)
Given: R = +3.00 m => f = R/2 = +1.50 m, u = -5.00 m
Calc: 1/v = 1/f - 1/u = 1/1.5 - 1/(-5) = 10/15 + 3/15 = 13/15
v = 15/13 ≈ +1.15 m
m = -v/u = -(1.15)/(-5) = +0.23
Ans: Image is 1.15 m behind mirror, Virtual, Erect, Diminished (0.23 times).
Q3. Magnification Meaning
m = +1 (Plane Mirror): Image is Virtual, Erect (+), and same size (1) as object.
m = -1 (Concave Mirror): Image is Real, Inverted (-), and same size (1) as object
(Object at C).
Q4. Convex Mirror
Given: u = -10 cm, f = +15 cm
Calc: 1/v = 1/15 - 1/(-10) = 2/30 + 3/30 = 5/30 = 1/6
v = +6 cm
Ans: Image is 6 cm behind mirror, Virtual and Erect.
Q5. Concave Mirror (Real Image)
Given: m = -3 (Real/Inverted), u = -10 cm
Formula: m = -v/u => -3 = -v/(-10)
-3 = v/10 => v = -30 cm
Ans: Image is located 30 cm in front of the mirror.
Q6. Concave Mirror (Focal Length)
Given: h = 2 cm, u = -16 cm, h' = -3 cm (Real)
m = h'/h = -3/2 = -1.5
Also m = -v/u => -1.5 = -v/(-16) => v = -24 cm
1/f = 1/(-24) + 1/(-16) = (-2 - 3)/48 = -5/48
f = -9.6 cm
Ans: (i) f = -9.6 cm, (ii) Image at 24 cm in front of mirror.
Q7. Convex Mirror
Given: R = +32 cm => f = +16 cm, u = -20 cm
1/v = 1/16 - 1/(-20) = 5/80 + 4/80 = 9/80
v = 80/9 ≈ +8.89 cm
Ans: f = 16 cm, Image distance = 8.89 cm (behind mirror).
Q8. Convex Mirror (Two Cases)
Case 1: u = -60, m = +1/2. m = -v/u => 1/2 = -v/(-60) => v = 30 cm.
1/f = 1/30 + 1/(-60) = 1/60 => f = +60 cm.
Case 2: m = +1/3, f = +60. m = f/(f-u) (Shortcut formula)
1/3 = 60 / (60 - u) => 60 - u = 180 => u = -120 cm.
Ans: Object should be placed at 120 cm.
Q9. Dentist Mirror
Given: u = -2 cm, R = -6 cm => f = -3 cm
1/v = 1/(-3) - 1/(-2) = -2/6 + 3/6 = 1/6 => v = +6 cm (Virtual)
m = -v/u = -(6)/(-2) = +3
Ans: Magnification is 3.
Q10. Ray Diagram (Between C and F)
Result: Image forms beyond C, Real, Inverted, and Enlarged.
Q11. Speed of Light in Glass
n = c / v => 1.50 = 3×10⁸ / v
v = 3×10⁸ / 1.50 = 2×10⁸ m/s
Ans: 2 × 10⁸ m/s.
Q12. Diamond Refractive Index
Meaning: Light travels 2.42 times slower in diamond than in vacuum.
v = c/n = 3×10⁸ / 2.42 ≈ 1.24 × 10⁸ m/s.
Q13. Refraction Air to Water
Ray bends towards the normal because water is optically denser than air, causing light
to slow down.
Q14. Snell's Law
n₂₁ = sin i / sin r = sin 60° / sin 45°
= (√3/2) / (1/√2) = √3/√2 = √(1.5) ≈ 1.22
Ans: 1.22
Q15. Relative Refractive Index
n_glass_water = n_g / n_w = (3/2) / (4/3) = 9/8 = 1.125
Ans: 1.125
Q16. Apparent Depth
Light rays from the coin travel from water (denser) to air (rarer), bending away from the normal. When
traced back, they appear to originate from a point higher than the actual coin.
Q17. Speed of Light
v = c/n = 3×10⁸ / 1.33 ≈ 2.26 × 10⁸ m/s.
Q18. Glass Slab
Proof relies on Snell's law at both interfaces. Since medium 1 (air) is same at entry and exit, and slab
sides are parallel, angle of emergence (e) equals angle of incidence (i).
Q19. Concave Lens
Given: f = -15 cm, v = -10 cm (Virtual image always)
1/v - 1/u = 1/f => 1/(-10) - 1/u = 1/(-15)
-1/u = -1/15 + 1/10 = (-2+3)/30 = 1/30 => u = -30 cm
m = v/u = (-10)/(-30) = +1/3 (+0.33)
Ans: Object at 30 cm; Magnification 0.33.
Q20. Convex Lens
Given: h = 2 cm, f = +10 cm, u = -15 cm
1/v = 1/10 + 1/(-15) = 3/30 - 2/30 = 1/30 => v = +30 cm
m = v/u = 30/(-15) = -2. h' = m×h = -4 cm.
Ans: Real, Inverted, Enlarged (4 cm tall) at 30 cm.
Q21. Power of Lens
f = -2 m (Concave). P = 1/f (in meters) = 1/(-2) = -0.5 D.
Q22. Convex Lens (Unit Magnification)
Real, Inverted, Same size => Object is at 2F.
Given v = +50 cm. Thus 2f = 50 => f = 25 cm = 0.25 m.
Object distance u = -50 cm.
P = 1/0.25 = +4.0 D.
Q23. Combination of Lenses
P_net = P1 + P2 = +3.5 - 2.5 = +1.0 D.
f_net = 1/P = 1/1 = 1 m = 100 cm.
Ans: Power +1.0 D, Focal Length 100 cm.
Q24. Lens Identification
Object at 50cm, Virtual image at 10cm. Since |v| < |u| for virtual image, it must be a Concave
Lens.
u = -50, v = -10. 1/f = 1/(-10) - 1/(-50) = -5/50 + 1/50 = -4/50.
f = -12.5 cm.
Q25. Corrective Lens
P = +1.5 D. f = 1/1.5 = 10/15 = 0.67 m = 66.7 cm.
Since Power is positive, it is Converging (Convex).
Q26. Convex Lens (Magnification)
Real image (on screen) => m = -3. v = +80 cm.
m = v/u => -3 = 80/u => u = -80/3 ≈ -26.67 cm.
Nature: Real, Inverted, Enlarged.
Q27. Ray Diagram (F1 to 2F1)
Object between F1 and 2F1 forms a Real, Inverted, Enlarged image beyond 2F2.
Q28. Concave Lens Formula
u = -30, f = -15.
1/v = 1/(-15) + 1/(-30) = -2/30 - 1/30 = -3/30 = -1/10.
v = -10 cm.
Q29. Convex Lens
f = +20, v = +40 (Real).
1/u = 1/v - 1/f = 1/40 - 1/20 = -1/40 => u = -40 cm.
m = v/u = 40/(-40) = -1.
Q30. Lens ID
Forms image on screen => Real Image => Convex Lens.
u = -45, v = +90.
1/f = 1/90 - 1/(-45) = 1/90 + 2/90 = 3/90 = 1/30.
f = +30 cm.
Q31. Lens Covered
Yes, it will produce a complete image, but the intensity (brightness) will be reduced to half because
less light rays can pass through.
Q32. Object at Focus (Concave Lens)
For a concave lens, object at focus: Image forms at f/2 on the same side, Virtual, Erect, Diminished.
Q33. Refractive Index Speed
v = c/n. Speed is inversely proportional to refractive index.
(i) Maximum Speed: Media A (n=1.33, lowest n).
(ii) Minimum Speed: Media D (n=1.65, highest n).
Q34. Normal Incidence
If angle of incidence i = 0°, then angle of reflection r = 0°. The ray retraces its path.
Q35. Convex Lens (Virtual Double Size)
m = +2 (Virtual/Erect), f = +10.
m = v/u => v = 2u.
1/f = 1/v - 1/u => 1/10 = 1/2u - 1/u = (1-2)/2u = -1/2u.
2u = -10 => u = -5 cm.
Ans: Object at 5 cm.
Q36. Security Mirror (Convex)
R = +5m, f = +2.5m, u = -20m.
1/v = 1/2.5 - 1/(-20) = 10/25 + 1/20 = 2/5 + 1/20 = 8/20 + 1/20 = 9/20.
v = 20/9 ≈ 2.22 m.
m = -v/u = -(2.22)/(-20) = +0.11.
Ans: Virtual, Erect, Diminished image at 2.22 m.
Q37. Concave Mirror (Virtual Image)
A concave mirror forms a virtual, erect, and enlarged image when the object is placed between
the Pole (P) and the Principal Focus (F).
Q38. Convex Mirror
u = -15, R = +60 => f = +30.
1/v = 1/30 + 1/15 = 1/30 + 2/30 = 3/30 = 1/10 => v = +10 cm.
m = -v/u = -10/(-15) = +2/3 = 0.67.
Q39. Lens Power
P = -4.0 D. f = 1/P = -0.25 m = -25 cm.
Negative power indicates a Concave Lens.
u = -20 cm, f = -25 cm.
1/v = 1/f + 1/u = -1/25 - 1/20 = (-4-5)/100 = -9/100.
v = -100/9 ≈ -11.1 cm (Virtual image).
Q40. Magnifying Glass
Magnifying glass is a Convex Lens.
f = +5 cm, u = -3 cm.
1/v = 1/5 - 1/3 = 3/15 - 5/15 = -2/15 => v = -7.5 cm.
m = v/u = (-7.5)/(-3) = +2.5.
Ans: Magnification is 2.5.
Part B: Electricity
Q41. Charge Calculation
I = 0.5 A, t = 10 min = 600 s.
Q = I × t = 0.5 × 600 = 300 C.
Q42. Work Done
Q = 2 C, V = 12 V.
W = V × Q = 12 × 2 = 24 J.
Q43. Quantization
Q = ne => n = Q/e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons.
Q44. Resistance
V = 6 V, I = 0.25 A.
R = V/I = 6 / 0.25 = 24 Ω.
Q45. Resistivity & Length
d = 0.5 mm = 5×10⁻⁴ m. r = 2.5×10⁻⁴ m. ρ = 1.6×10⁻⁸ Ωm. R = 10 Ω.
A = πr² ≈ 3.14 × (2.5×10⁻⁴)² ≈ 1.96 × 10⁻⁷ m².
L = (R × A) / ρ = (10 × 1.96×10⁻⁷) / 1.6×10⁻⁸ ≈ 122.7 m.
If diameter doubles, Area becomes 4 times. R ∝ 1/A, so R becomes 1/4th.
Q46. Ohm's Law
R = V1/I1 = 60/4 = 15 Ω.
I2 = V2/R = 120/15 = 8 A.
Q47. R Proportionality
R = ρL/A = 4 Ω.
New R' = ρ(L/2) / (2A) = (1/4) (ρL/A) = 1/4 × 4 = 1 Ω.
Q50. Parallel Circuit
R1=5, R2=10, R3=30. V=6V.
(a) I1 = 6/5 = 1.2 A; I2 = 6/10 = 0.6 A; I3 = 6/30 = 0.2 A.
(b) Total I = 1.2 + 0.6 + 0.2 = 2.0 A.
(c) 1/Req = 1/5 + 1/10 + 1/30 = (6+3+1)/30 = 10/30 = 1/3 => Req = 3 Ω.
Q51. Current Calculation (Parallel Diagram)
Given: R1=5Ω, R2=10Ω, R3=30Ω in parallel. V=12V.
I1 = V/R1 = 12/5 = 2.4 A.
I2 = V/R2 = 12/10 = 1.2 A.
I3 = V/R3 = 12/30 = 0.4 A.
Total Current I = 2.4 + 1.2 + 0.4 = 4.0 A.
Q52. Parallel Appliances
1/Rp = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500.
Rp = 500/16 = 31.25 Ω. (Resistance of Iron)
Current I = V/R = 220 / 31.25 = 7.04 A.
Q53. Combinations
(a) 4 Ω: Connect 3Ω and 6Ω in parallel (gives 2Ω), then in series with 2Ω. (2 + 2 = 4).
(b) 1 Ω: Connect all 2Ω, 3Ω, 6Ω in parallel. (1/2 + 1/3 + 1/6 = 1).
Q54. Series Current
Total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω.
I = V/R = 9 / 13.4 = 0.67 A.
In series, current is same through all resistors. Ans: 0.67 A.
Q55. Equivalent Resistance
Diagram Analysis: Two 4Ω resistors are in parallel, followed by one 4Ω resistor in
series.
Step 1: Parallel combination of 4Ω and 4Ω.
Rp = (4×4)/(4+4) = 16/8 = 2 Ω.
Step 2: Series combination of Rp and 4Ω.
Req = 2 Ω + 4 Ω = 6 Ω.
Ans: 6 Ω
Q56. Potential Difference
Diagram Analysis: 2Ω and 4Ω resistors in series with 6V battery. Voltmeter across
4Ω.
Total Resistance Req = 2 + 4 = 6 Ω.
Total Current I = V/Req = 6/6 = 1 A.
Potential Difference across 4Ω (V4) = I × R = 1 A × 4 Ω = 4 V.
Ans: 4 V
Q57. Parallel Lamps
P1 = 100W, P2 = 60W, V = 220V.
Total Power P = 160 W.
Total Current I = P/V = 160/220 = 8/11 ≈ 0.73 A.
Q58. Electric Iron Power
Max rate: P=840W, V=220V. I = 840/220 = 3.82 A. R = 220/3.82 = 57.6 Ω.
Min rate: P=360W, V=220V. I = 360/220 = 1.64 A. R = 220/1.64 = 134.15 Ω.
Q59. Heat Production
H = 100 J, t = 1s, R = 4Ω.
H = I²Rt => 100 = I² × 4 × 1 => I² = 25 => I = 5 A.
V = IR = 5 × 4 = 20 V.
Q60. Heat Generated
Q = 96000 C, V = 50 V.
H = V × Q = 50 × 96000 = 4,800,000 J = 4.8 × 10⁶ J.
Q61. Power of Bulb
V = 220 V, I = 0.50 A.
P = V × I = 220 × 0.50 = 110 W.
Q62. Cost of Energy
P = 400 W = 0.4 kW.
Time per month = 8 hours × 30 days = 240 hours.
Energy = 0.4 kW × 240 h = 96 kWh.
Cost = 96 × 3.00 = Rs 288.
Q63. Bulb Count
P = 10W, V = 220V. I_bulb = P/V = 10/220 = 1/22 A.
Max I = 5 A. Let n bulbs be connected.
n × (1/22) = 5 => n = 5 × 22 = 110 bulbs.
Q64. Hot Plate
R = 24 Ω each. V = 220 V.
(i) Separately: I = 220/24 ≈ 9.17 A.
(ii) Series: R_s = 48 Ω. I = 220/48 ≈ 4.58 A.
(iii) Parallel: R_p = 12 Ω. I = 220/12 ≈ 18.33 A.
Q65. Energy Comparison
(i) TV: P = 250 W = 0.25 kW, t = 1 h. E = 0.25 kWh.
(ii) Toaster: P = 1200 W = 1.2 kW, t = 10/60 = 1/6 h. E = 1.2/6 = 0.20 kWh.
Ans: TV uses more energy.
Q66. Heat Rate
R = 8 Ω, I = 15 A.
Rate of heat (Power) P = I²R = 15² × 8 = 225 × 8 = 1800 W (J/s).
Q67. Heat Ratio
Series: Rs = 2R. Power_s = V²/2R.
Parallel: Rp = R/2. Power_p = V²/(R/2) = 2V²/R.
Ratio Hs/Hp = Ps/Pp = (V²/2R) / (2V²/R) = 1/4.
Ans: 1:4
Q68. Fuse Wire (Advanced)
Heat produced per sec H ∝ I²R = I²(ρL/πr²).
Heat lost per sec from surface ∝ Surface Area = 2πrL.
At equilibrium: I²ρL/πr² = k(2πrL) => I² ∝ r³.
(I2/I1)² = (r2/r1)³ => (10/5)² = (r2/r1)³ => 4 = (ratio)³.
Ratio = ∛4 ≈ 1.58. Ans: Radius must be larger.
Q69. Total Resistance & Current
Diagram Analysis: 10Ω resistor in series with a parallel block of (20Ω || 20Ω). V =
24V.
Step 1: Parallel Block Rp = (20×20)/(20+20) = 400/40 = 10 Ω.
Step 2: Total Resistance R_total = 10 (series) + 10 (parallel) = 20 Ω.
Step 3: Total Current I = V/R_total = 24/20 = 1.2 A.
Ans: Resistance = 20 Ω, Current = 1.2 A.
Q70. Circular Wire Resistance
Total R = 20 Ω. Ends of diameter divide it into two semicircles of 10 Ω each.
These two 10 Ω resistors are in parallel.
Req = (10 × 10) / (10 + 10) = 100/20 = 5 Ω.
Q71. Potential Difference (AB & AC)
Given: R1=5Ω, R2=10Ω (Between A & B), R3=15Ω (Between B & C). Series Current I=1A.
(i) Across AB (Voltage across R2): V_AB = I × R2 = 1 × 10 = 10 V.
(ii) Across AC (Voltage across R2 + R3): R_AC = 10 + 15 = 25 Ω. V_AC = I × R_AC = 1 × 25 = 25 V.
Ans: V_AB = 10 V, V_AC = 25 V.
Q72. Meter Reading
Diagram Analysis: 2Ω and 3Ω resistors are in series. Voltmeter is across 3Ω. Ammeter is
in series. V_source = 6V.
Step 1: Total Resistance Req = 2 + 3 = 5 Ω.
Step 2: Current (Ammeter Reading) I = V/Req = 6/5 = 1.2 A.
Step 3: Voltmeter Reading (across 3Ω) V3 = I × R = 1.2 × 3 = 3.6 V.
Ans: Ammeter = 1.2 A, Voltmeter = 3.6 V.
Q73. Torch Bulb
V = 2.5V, I = 750mA = 0.75A.
(a) P = VI = 2.5 × 0.75 = 1.875 W.
(b) R = V/I = 2.5 / 0.75 = 3.33 Ω.
(c) E = P × t = 1.875W × 4h = 7.5 Wh = 0.0075 kWh.
Q74. Energy Consumption
Bulbs 1: 2 × 60W × 4h = 480 Wh.
Bulbs 2: 3 × 100W × 5h = 1500 Wh.
Daily Total = 1980 Wh = 1.98 kWh.
Monthly (30 days) = 1.98 × 30 = 59.4 kWh (units).
Q75. Series Derivation
In series, current I is constant. Total V = V1 + V2 + V3.
By Ohm's Law: IR = IR1 + IR2 + IR3.
Dividing by I: R_eq = R1 + R2 + R3.
Q76. Wire Stretching/Volume
Volume remains concept: V = A × l = A' × 2l => A' = A/2.
Ratio A'/A = 1/2.
Q77. Electric Kettle
P = 1000 W, t = 5 min = 300 s.
H = P × t = 1000 × 300 = 300,000 J = 300 kJ.
Q78. Filament Heating
H ∝ R (for series current).
Resistance of filament is very high compared to connecting wires.
Therefore, H_filament >> H_wires.
Q79. Cut Wire Resistance
Wire R cut into 5 parts => each part is R/5.
Connected in parallel: 1/R' = 5 / (R/5) = 25/R => R' = R/25.
Ratio R/R' = R / (R/25) = 25.
Ans: 25:1
Q80. Total Resistance & Current
Diagram Analysis: R3=4Ω is in series with a parallel pair (R1=3Ω, R2=6Ω). V=12V.
(i) Total Resistance:
Rp (for 3Ω || 6Ω) = (3×6)/(3+6) = 18/9 = 2 Ω.
Total R = R3 + Rp = 4 + 2 = 6 Ω.
(ii) Current (Ammeter A):
I = V/Total R = 12 / 6 = 2 A.
Ans: Resistance = 6 Ω, Current = 2 A.
