Solution - Career Point

SOLUTION FOR MOCK TESTPAPER - II AIEEEPHYSICS12.[B]13.[B]1.[C]2.[B] v ∝ n ∝ T because λ = constantN + 4 324 18= =N + 1 289 1717N + 68 = 18N + 1850 = N3.[B]4.[A]5.[B] u = 0a = constantv 0 = u + at = at⎛ v + v0⎞ a n tn' = ⎜ ⎟ n = n +⎝ v ⎠ v6.[C] LC = 1 MSD – 1VSD⎛ 29 ⎞ 1LC = ⎜1 − ⎟ MSD = MSD⎝ 30 ⎠ 30720 MSD = 360°360°1 MSD = = 1/2° 7201 1° 1° 60 'So LC = × = = = 1'30 2 60 607.[C]8.[D]Ticker timer is a better device than a stop watch.9.[C]M MM =1 + 2 10 .30+12.62=22= 11.46 g10.[B]11.[B]∆ Y 2∆ D ∆ l= +Y D l∆ YY⎛ 0.01⎞⎛ 0.05 ⎞= 2 ⎜ ⎟ + ⎜ ⎟⎝ 0.4 ⎠ ⎝ 0.8 ⎠= 2× 0.025 + 0.0625∆ Y= 0.05 + 0.0625 = 0.1125Y∆Y = 2× 10 11 × 0. 1125 = 0.225 × 10 11So (2 ± 0.2) × 10 11 N/m 2fdIt is only possible if object and image coincide.14.[D] I = I 0 cos 2 φ215.[C]FI 0= I 0 cos 2 φ2 2φ 1cos = 2 22πφ = 90° = ∆xλπ 2 = ∆x2 λπλ y d λ∆x = ⇒ = 4 D 4−7λ D 5×10 × 1y = =4d−44×1×10y = 1.25 mmf= 45 × 10–3→ τ =→ p ×→E = q (2→ a ) ×→EHere 2 → a = (2 – 1) î + (– 1–0) ĵ + (5 – 4) kˆ= î – ĵ + kˆ→E = 0.20 î V/cm = 20 î V/m∴16.[D]→ τ = (4 × 10 –6 ) [( î – ĵ kˆ ) × 20 î]( kˆ + ĵ)Magnitude of torque τ = 8CBAc+ σ– σ+ σba+ = 8 × 10 –52 × 10 –5 N-mXtraEdge for IIT-JEE 102APRIL 2010