Solution - Career Point

SOLUTION FOR MOCK TESTPAPER - II BIT-SAT1.[B]2.[C]PHYSICSIn case of damped vibrations, amplitudedecreases exponentially with time∴ A = A 0 e –bt Aor = e–btororHere1 = e b×2 and2A ´ = (e –2b ) 3 =A 0k e= k1A 0A ´A 0⎛ 1 ⎞⎜ ⎟⎠⎝ 2= e–b×b3= 811 1 1 = + + + ....... ∞k 21k 4k⎛11 1 ⎞⎜ + + + ...... ∞⎟⎝12 4 ⎠1 ⎛ 1 ⎞ 2= ⎜ ⎟ =k ⎝1−1/2 ⎠ k3.[ B] KE max = 21 kA2= 21 Mω 2 A 28 × 10 –3 = 21 × 1 × ω 2 × (0.1) 2i.e. k e = 2kω = 4 rad/secy = A sin(ωt + φ) or y = 0.1 sin (4t + π/4)4. [B] Here → → →→L = m ( r × v)= m v y(– k )(Where y is the vertical distance of particle fromx axis)Here m,v and y all are fixed so → L . remainsconstant.5. [B] Using, weight of floating body = weight ofliquid displaced.⎛ V ⎞ Vwe get V ρ g = ⎜ ⎟ (13.6) g – (0.8g)⎝ 2 ⎠ 2(buoyant forces of mercury and oil act inopposite direction)13.6− 0.8Then, ρ = = 6.426. [C] v´ =v + vv − v9 340 + v´i.e. = 8 340 − v´i.e. v´ = 20 ms –10sv We get νν´ =v − v0v − v∴ (v 0 = v s = v´)I2I27. [A] I log = dB; i.e. 4 = 10 log−94I1(10×10 × 10 )Then I 2 = 2.5 × 10 –4 Wm –28. [A] Here work done = pdv and area under the curvegiven work done∴ 10 + W CA = 5 or W CA = –5 J9. [C] For equilibriumF = qE = mg1/ 34or neE = πr 3 ⎛ 3neE ⎞ρg or r =3 ⎜4 g⎟⎝ πρ ⎠1.5 + 1.5 −1.510.. [A] i == 0.5 A1+1+1As the current has to from A to C to B,for kirchhoff's law,V A = 0.5 × 1 + 1.5 = 1V (Q v = E – ir)V B = 0.5 × 1 + 1.5 = 1VV C = 0.5 × 1 – (–1.5) = 2V11. [D] R = R 1 + (273 – T) α ...(i)or 2R = R 0 [1 + T´α]...(ii)1 1(273- T) αDeciding (i) by (ii) = 2 1+T´ αor 1 + T´α = 2 + (273 – T)2α(273- T)2α + 1or T´ =α12. [A] Current, I =2ER + r 1 + r 22Er1P.O. across cell 1 = Ir 1 =R + r1+ r2For zero p.d. the fall of potential should beequal to in emf.2Er1E =i.e. R = r 1 – r 2R + r + r12sXtraEdge for IIT-JEE 110APRIL 2010