2nd Year Physics - 20 Most Important Numericals

 

12th Physics Numericals

PHYSICS 12th Federal Board
(20 NUMERICALS) 

Following 20 Numerical are Guess for Physics Numerical in Federal Board. Sample Questions have been provided with relevant topics and being a student you can get detailed answer from any AI Platform (like ChatGPT, Deepseek, Gemini or Meta AI), just copy the question and past it there. So, Important thing is to develop your understanding regarding SLO based exams compromised which types of Questions.

To watch Details Explanation of each topic: Click Here

Don't Forget to Note down the procedure to attempt any Numerical in Question Paper, It matters more than what you expect!

1. A 280 J of work is done in carrying a charge of 2 C from a place where the potential is -12 V to another place where the potential is V. Calculate the value of V?
Step 1: Use the relation W = q(V - Vi)
Step 2: Rearranging gives V = W/q + Vi
Step 3: Substitute values: V = 280 / 2 + (-12) = 140 - 12
Final Answer: V = 128 V

2. Two-point charges of 8 μC and -4 μC are separated by 10 m. At what point on the line joining them is the electric potential zero?
Step 1: Let distance from +8μC charge to point be x.
Step 2: Set potentials equal: 8/x = 4/(10 - x)
Step 3: Solve: 8(10 - x) = 4x => 80 - 8x = 4x => 12x = 80 => x = 6.67 m
Final Answer: 6.67 m from +8 μC charge

3. A 6 μF capacitor is charged to 120 V and connected to an uncharged 4 μF capacitor. Calculate the new P.D.
Step 1: Find total charge: Q = C × V = 6μF × 120V = 720 μC
Step 2: Total capacitance = 6 + 4 = 10 μF
Step 3: V = Q / C = 720 / 10 = 72 V
Final Answer: 72 V

4. A carbon electrode has resistance 0.125 Ω at 20°C. α = -0.0005/°C. Find resistance at 85°C.
Step 1: ΔT = 85 - 20 = 65°C
Step 2: R = R0(1 + αΔT) = 0.125(1 - 0.0005 × 65)
Step 3: R = 0.125(0.9675) = 0.1209 Ω
Final Answer: 0.1209 Ω

5. A 10 W resistor has resistance 120 Ω. Find the current.
Step 1: Use P = I²R ⇒ I = √(P/R)
Step 2: I = √(10 / 120) = √0.0833 = 0.289 A
Final Answer: 0.289 A

6. A 50 Ω resistor has 100 V across it for 1 hour. Calculate (a) Power (b) Energy.
Step 1: P = V² / R = 100² / 50 = 200 W
Step 2: E = P × t = 200 × 3600 = 720,000 J
Final Answer: (a) 200 W, (b) 720,000 J

7. Find distance from wire with 10 A current where B = 5×10⁻⁴ T.
Step 1: B = μ₀I / (2πr) ⇒ r = μ₀I / (2πB)
Step 2: r = (4π×10⁻⁷×10) / (2π×5×10⁻⁴) = 4×10⁻³ m
Final Answer: 4 mm

8. An 8 MeV proton enters 2.5 T field perpendicularly. Find force and radius.
Step 1: Convert E to joules: 8 MeV = 1.28×10⁻¹² J
Step 2: v = √(2E/m) = √(2×1.28×10⁻¹² / 1.67×10⁻²⁷)
Step 3: F = evB = 1.6×10⁻¹⁹ × v × 2.5 = 1.57×10⁻¹¹ N
Step 4: r = mv / (eB) = 0.163 m
Final Answer: (a) 1.57×10⁻¹¹ N, (b) 0.163 m

9. Coil with 250 turns, 6 cm × 4 cm, max torque 0.20 Nm in 0.25 T. Find current.
Step 1: A = 0.06 × 0.04 = 0.0024 m²
Step 2: τ = NIBA ⇒ I = τ / (NBA) = 0.20 / (250 × 0.25 × 0.0024)
Final Answer: 1.33 A

10. Two wires 10 cm apart carry 8 A opposite directions. Find B halfway.
Step 1: B = μ₀I / (2πr), r = 0.05 m (halfway)
Step 2: B_total = 2 × (μ₀I / 2πr) = μ₀I / πr = 6.4×10⁻⁵ T
Final Answer: 6.4×10⁻⁵ T

11. EMF = 200 V, ΔI = 5 A, Δt = 0.1 s. Find L.
Step 1: EMF = L × (ΔI / Δt)
Step 2: L = 200 × 0.1 / 5 = 4 H
Final Answer: 4 H

12. Mutual inductance 1.5 H, ΔI = 20 A in 0.5 s. Find flux linkage.
Step 1: ΔΦ = M × (ΔI / Δt) = 1.5 × (20 / 0.5) = 60 Wb
Final Answer: 60 Wb

13. Voltage: V = 240 sin(1.25×10⁴ t – 30°), C = 0.01 μF. Find current.
Step 1: I = C × dV/dt = ωCV₀ = 1.25×10⁴ × 0.01×10⁻⁶ × 240
Step 2: I = 0.03 A = 30 mA (peak)
Final Answer: 30 mA

14. L = 100 μH, I = 10 mA, V = 6.3 V. Find frequency.
Step 1: V = 2πfLI ⇒ f = V / (2πLI)
Step 2: f = 6.3 / (2π × 100×10⁻⁶ × 0.01) = 1.0027 MHz
Final Answer: 1.0027 MHz

15. IB = 100 μA, β = 100. Find IC, IE, IC/IE.
Step 1: IC = β × IB = 0.01 A
Step 2: IE = IC + IB = 0.0101 A
Step 3: IC/IE = 0.01 / 0.0101 = 0.9901
Final Answer: IC = 0.01 A, IE = 0.0101 A, IC/IE = 0.9901

16. Length appears 1/3rd. Find speed.
Step 1: L = L₀√(1 - v²/c²) ⇒ 1/3 = √(1 - v²/c²)
Step 2: Square both sides: 1/9 = 1 - v²/c² ⇒ v²/c² = 8/9 ⇒ v = c√(8/9)
Step 3: v = 3×10⁸ × √(8/9) = 2.83×10⁸ m/s
Final Answer: 2.83×10⁸ m/s

17. 50 keV X-ray scattered 90°. Find energy after scattering.
Step 1: Δλ = h/(mec)(1 - cosθ) = 0.00243 nm
Step 2: E' = hc / (λ + Δλ) ⇒ E' ≈ 45.56 keV
Final Answer: 45.56 keV

18. Mass of 14N = 13.999234 u. Find binding energy.
Step 1: Mass of parts = 7p + 7n = 7(1.007276) + 7(1.008665) = 14.1115 u
Step 2: Mass defect = 14.1115 - 13.999234 = 0.1123 u
Step 3: BE = Δm × 931.5 = 104.66 MeV
Final Answer: 104.66 MeV

19. Half-life of Ra = 1.6×10³ years. Find decay constant.
Step 1: λ = ln(2)/T₁/₂ = 0.693 / (1.6×10³ × 365 × 24 × 3600)
Step 2: λ = 1.37×10⁻¹¹ s⁻¹
Final Answer: 1.37×10⁻¹¹ s⁻¹

20. Th → Pa β-decay. Mass Th = 234.0436 u, Pa = 234.0428 u. Find energy released.
Step 1: Δm = 234.0436 - 234.0428 = 0.0008 u
Step 2: E = Δm × 931.5 = 0.0008 × 931.5 = 0.745 MeV
Final Answer: 0.745 MeV

Like, Comment & Share!