Solution - Career Point

7.[B,D]8.[B,C]9.[C]The field on both sides of plate is shown in figure.Applying Ampere's Circuital Law to the contourC, we getL2BL = µ 0 (jL)µ jB = 02BSuperimposing, the field due to two plates we getat P both fields cancel each other and at Q, theyadded to give B 0 = µ 0 jv < g(t)⇒ v < 2g {due to Lenz's Law}Also, s < 21 gt2⇒ s < 2g {due to Lenz's Law}R 0 RR 3SR 2R 1C1CVR 2R3⇒ R 0 =R + R2⎛⇒ τ = C(R 1 + R 0 ) = C ⎜R⎝⇒ q = q 0 (1 – e –t/τ )⇒ q = CV 0 (1 – e –t/τ )CVR 2⇒ q = (1 – e –t/τ )R + R233V 0BCVR 2and V 0 =R + R123R 2R3+R + RIf it was AY → 2ZReactant : R = 60 × 8.5 = 510 MeVProduct : P = 2 × 30 × 5 = 300 MeV∆E = – 210 MeVENDOTHERMICIf it was BW → X + ZR = 120 × 7.5 = 900 MeVP = 90 × 8 + 30 × 5 = 870 MeV∆E = – 30 MeVENDOTHERMICIf it was CW → 2Y23⎞⎟⎠R = 120 × 7.5 = 900 MeVP = 2 × 60 × 8.5 = 1020 MeV∆E = 120 MeVEXOTHERMICIf it was DX → Y + ZR = 90 × 8.0 = 720 MeVP = 60 × 8.5 + 30 × 5.0 = 660 MeV∆E = – 60 MeVENDOTHERMICColumn Matching :10. [A] → p ; [B] → q, r, s; [C] → p,s ; [D] → q,rWhen cohesive forces are greater then adhesiveforces shape of meniscus is concave from liquidside and pressure is greater in concave side due tosurface tension.11. [A] → r ; [B] → p, s; [C] → q ; [D] → s⎛ dV ⎞ dT(A) ⎜ ⎟ = ⇒⎝ V ⎠PTkT(B) λ =2πd2 ρ1 ⎡dV⎤dT⎢V⎥⎣ ⎦P1=T(C) Ideal gas law are valid at all temperatures.(D) conceptualNumerical Response type questions :12. [1]13. [5]Mv12α v' 12v 1 α30ºN–vO230º→In the figure v 12 = velocity of ball w.r.t. wedge→before coolision, and v' 12 = velocity of ball w.r.t.wedge after collision, which must be in vergicallyupward direction as shown.→ →In elastic collision v 12 and v' 12 will make equalangle (say α) with the normal to the plane.We can show that α = 30º∴ ∠MON = 30ºv 11Now = tan 30º = ≈ 1v32hABhCXtraEdge for IIT-JEE 100APRIL 2010