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PhysicsClass 12CBSE

Bullet fires with velocity u; after 24 cm velocity = u/3. Then comes to rest at end of block. Find total length.

Step-1: Find retardation (v² = u² − 2as):(u/3)² = u² − 2a(0.24)u²/9 = u² − 0.48a0.48a = u²(1 − 1/9) = 8u²/9 ⇒ a = 50u²/27Step-2: Additional distance to stop from v = u/3:0 = (u/3)² − 2a·s₂s₂ = (u²/9) / (2 × 50u²/27) = (u²/9) × (27/100u²) = 0.03 m = 3 cmStep-3: Total length:L = 24 cm + 3 cm = 27 cmAnswer: Total length of the block = 27 cm

GeneralClass 12CBSE

Electric Field at Centre of Ring with Non-Uniform Charge Distribution. Four quadrants carry linear charge densities: +2λ, −2λ, +λ, −λ. Find electric field at centre.

For a quarter-circle arc of linear charge density λ and radius R, the field at the centre has magnitude:E_quadrant = (√2 λ) / (4πε₀R)Direction: along the bisector of the quadrant, pointing away from arc (for +λ) or toward arc (for −λ).Resolving all four quadrant fields into x and y components:Ex = [−2 − 2 + 1 + 1] × (λK/√2) = −2λK/√2Ey = [−2 + 2 + 1 − 1] × (λK/√2) = 0Magnitude:|E| = 2λK/√2 = λ / (2πε₀R)Answer: E = λ / (2πε₀R), directed along the negative x-axis

PhysicsClass 12CBSE

Magnetic Moment of a Current-Carrying Wire Bent into a CircleWire length = 2 m, current = 3.14 A (≈ π A). Find magnetic moment M.

Step-1: Radius of circle:2πr = 2 ⇒ r = 1/π mStep-2: Area:A = πr² = π × (1/π)² = 1/π m²Step 3 — Magnetic moment:M = I × A = π × (1/π) = 1 A·m²Answer: M = 1 A·m²

GeneralClass 12CBSE

P°(hexane) = 408 Torr, P°(heptane) = 141 Torr. x(hexane) = 0.300. Find Y₆ and Y₇.

Step-1: Partial pressures:P(hexane) = 0.300 × 408 = 122.4 TorrP(heptane) = 0.700 × 141 = 98.7 TorrStep-2: Total pressure:P_total = 122.4 + 98.7 = 221.1 TorrStep-3: Vapour mole fractions:Y₆ (hexane) = 122.4 / 221.1 ≈ 0.554Y₇ (heptane) = 98.7 / 221.1 ≈ 0.446Answer: Y₆ ≈ 0.554, Y₇ ≈ 0.446

GeneralClass 12CBSE

2N₂O₅(g) → 4NO₂(g) + O₂(g). Initial P = 50 mmHg; P at 30 min = 87.5 mmHg. Find P at 60 min.

Step-1: Let N₂O₅ decrease by 2p. Total pressure = 50 + 3p:50 + 3p = 87.5 ⇒ p = 12.5 mmHgP(N₂O₅) at 30 min = 50 − 2(12.5) = 25 mmHgStep-2: Rate constant (first order):k = (1/30) × ln(50/25) = ln(2)/30 = 0.02310 min⁻¹Step-3: P(N₂O₅) at 60 min (= two half-lives = 25% of initial):P(N₂O₅) at 60 min = 50/4 = 12.5 mmHgStep-4: Total pressure at 60 min:Decrease = 50 − 12.5 = 37.5 = 2p ⇒ p = 18.75P_total = 50 + 3(18.75) = 106.25 mmHgAnswer: Total pressure after 60 minutes = 106.25 mm Hg

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