Aieee 2012 Solution (26th May) 5h2c5q
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3i3n4
Overview 26281t
& View Aieee 2012 Solution (26th May) as PDF for free.
More details 6y5l6z
- Words: 2,721
- Pages: 64
Detailed solution given below
Q.1
1) 2) 3) 4) ans: 1 Q.2
1) 2) 3) 4) ans: 4 Q.3
1) 2) 3) 4) ans: 4
Q.4
1) 2) 3) 4) ans:4 Q.5
1) 2) 3) 4) ans:2 Q.6
1) 2) 3) 4) ans:1
Q.7
1) 2) 3) 4) ans:1 Q.8
1) 2) 3) 4) ans:4
Q.9
1) 2) 3) 4) ans:2 NOTE:- actual ans is 0.125m Q.10
1) 2) 3) 4) ans:3
Q.11
1) 2) 3) 4) ans:1 Q.12
1) 2) 3) 4) ans:2 (actual ans is 88.88%)
Q.13
1) 2) 3) 4) ans:3 Q.14
1) 2) 3) 4) ans:1
Q.15
1) 2) 3) 4) ans:3 Q.16
1) 2) 3) 4) ans:4
Q.17
1) 2) 3) 4) ans:4 Q.18
1) 2) 3) 4) ans:3
Q.19
1) 2) 3) 4) ans:2 Q.20
1) 2) 3) 4) ans:3
Q.21
1) 2) 3) 4) ans:4 Q.22
1) 2) 3) 4) ans:3 Q.23
1) 2) 3) 4) ans:2
Q.24
1) 2) 3) 4) ans:4 Q.25
1) 2) 3) 4) ans:4
Q.26
1) 2) 3) 4) ans:3 Q.27
1) 2) 3) 4) ans:2
Q.28
1) 2) 3) 4) ans:2 Q.29
1) 2) 3) 4) ans:3
Q.30
1) 2) 3) 4) ans:2
Q.61
1) 2) 3) 4) ANS:1 Q.62
1) 2) 3) 4) ANS:3
Q.63
1) 2) 3) 4) ANS: 4 Q.64
1) 2) 3) 4) ANS:2 Q.65
1) 2) 3) 4) ANS:4
Q.66
1) 2) 3) 4) ANS:2 Q.67
1) 2) 3) 4) ANS:4 Q.68
1) 2) 3) 4) ANS:3
Q.69
1) 2) 3) 4) ANS:1 Q.70 1) 2) 3) 4) ANS:3 Q.71
1) 2) 3) 4) ANS:2
Q.72
1) 2) 3) 4) ANS:2 Q.73
1) 2) 3) 4) ANS:4 Q.74
1) 2) 3) 4) ANS:1
Q.75
1) 2) 3) 4) ANS:1 Q.76
1) 2) 3) 4) ANS:2 Q.77
1) 2) 3) 4) ANS:1 Q.78 1) 2) 3) 4) ANS:4
Q.79
1) 2) 3) 4) ANS:3 Q.80
1) 2) 3) 4) ANS:2 Q.81
1) 2) 3) 4) ANS:4
Q.82
1) 2) 3) 4) ANS: * (ans is 4) Q.83
1) 2) 3) 4) ANS:3 Q.84
1) 2) 3) 4) ANS:4
Q.85
1) 2) 3) 4) ANS:1 Q.86
1) 2) 3) 4) ANS:1 Q.87
1) 2) 3) 4) ANS:4
Q.88
1) 2) 3) 4) ANS:3 Q.89
1) 2) 3) 4) ANS:4 Q.90
1) 2) 3) 4) ANS:2 By Saurav gupta Electronics & tele comm. Engg. (2nd year) Jadavpur university
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Q.1 Bqv= mv²/R R= mv/Bq
Now, eV= ½ mv² or mv= √2emV
R= √2emV/Bq R∝ √m R’ = √2 R Ans -1
Q.2 q1+q2 = Q ---eq(1) given BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
σ1 =σ2--- given or q1/4πr²=q2/4πR² q1/r²= q2/R²-------- eq(2)
solving eq (1) & (2) we get q1 =Q/(1+r²/R²)
now, V = 1/4πϵ˳ (q1/r+q2/R) = 1/4πϵ˳ q1(1/r+ q1 R/r²) = Q/4πϵ˳ (r+R)/(R²+r²) Ans -4
Q.3 Sinα ≃ tanα ≃α BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) α= d/D Δx=λd/D
d α
=λ/α
D
Ans 4
Q.4 E= [E ͨ cosω ͨt +E ͫcos(ω ͨt) cos(ω ͫ t)]î =[E ͨ cosω ͨt +E ͫ/2 cos(ω ͨ+ ω ͫt) +E ͫ/2 cos(ω ͫ-ω ͨ t)]î Ans 4
Q.5 N M.S.D = (N+1) V.S.D 1 V.S.D = N/(N+1) M.S.D NOW, BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) L.C = 1M.S.D -1V.S.D =[1 – N/(N+1)]M.S.D a/(N+1) Ans 2
Q.6 Case1 Tension /unit length = 100g/l Case2 Tension /unit length= 2mg/l 2m=100 m=50kg Ans -1
Q.7 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) dq/dt ͨ = dq/dtˢ 9KA(100-θ)/(L/2) = KAθ/(L/2) θ=90⁰ Ans -1
Q.8 F = dP/dt =d/dt(mv) = v dm/dt =6 Ans 4
Q.9
s
τ = -mg sinθ R mgsinθ
≃-mgθR Iα=-mgθR BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
mg
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
α = -mgRθ/I ω²=mgR/I now, I = mR² + mR² =2mR² ω²=g/2R T = 2π√(2R)/g Putting T=1 & g≃π² R= 1/8 =0.125 Ans 2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
Q.10 υn= (2n-1)v/4L case1 500 = (2n – 1)v/4 x16 ----(1) Case2 500 = (2n’-1)v/4 x 50-----(2) Solving 1 & 2 we get 50n -16n’ = 17 n≃1 n’≃2 now solving for v v = 500 x4 x 50/3 v = 33333.33m/s
now, BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) υ(fundamental ,outside water) = v/λ =v/2l ≃276 Hz υ'(2nd lowest) =2υ =552 Hz Ans -3
Q.11 λav= λα + λβ now, T ½ =1/λ Tav= 1/λα +1/λβ Ans -1
Q.12 Mu = mv + mv’ BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) ½ mu² = ½ mv² + ½ mv’² Solving eq (1) &(2) V = -1/3u V’= 2/3u (ΔE/E )x100= 88.88% Ans 2
Q.13 Sinθ ͨ > 1/μ μg/μw= 1.5/(4/3) = 9/8 Sinθ ͨ>8/9 Ans 3
Q.14 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) K .E= hυ-hυ˳ If υ is doubled K.E is more than doubled Also Photocurrent depends on intensity not on frequency Statement 1- false Statement 2-true Ans -1
Q.15 q(E + v x B)= 0 E=-vxB E=Bxv E = E˳ cos(wt -2π/λy)ẑ Ans -3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
Q.16 I .E³= 13.6 z²/n² = 13.6 x 9 = 122.4 T.E = I.E¹ + I.E² + I.E³ =5.4 +75.6 +122.4 = 203.4 Ans 4
Q.17 P=V²/R R1
P2 Ans -4
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
Q.18 H=I²Rt ΔH= 2ΔI + ΔR +Δt =6% Ans -3
Q.19 v² = u² + 2al V’² = u² + 2a(l/2) =u² + ½ (v²-u²) V’² = (v² + u²)/2 V’ = √(v²+u²)/2 Ans -2
Q.20 Statement 1 –true BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Statement 2- true B near the centre is μ˳ni but decreses as we move towards it’s end and becomes μ˳ni/2 at the extreme end which explains statement 1 Ans -3
Q.21 V = e – iR 120 – 50 x 0.1 = 115 Ans 4
Q.22 Y = A sin(ωt + 2π/3) At t=0 Y = A sin 120⁰ Checking option BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Ans -3
Q.23 Fsinθ
Fκ
Fcosθ mg
Fκ = μ(mg- Fsinθ) Now,
Fcosθ = Fκ F = μW/(cosθ + μsinθ)---(1)
For minimum force BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) dF/dθ = 0 solving we get tanθ = μ or cosθ= 1/√1+μ² or sinθ = μ/√1 +μ²
substituting these values in eq –1 F = μW/√1 +μ² Ans -2
Q.24 As we pull the connector by say x amount there will be growth in current in the circuit which will be opposed by the inductor(due to self inductance) BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Here inductor is analogous to spring(mechanical analog of inductor) Opposing force ∝ -x Hence the connector will execute S.H.M Ans 4
Q.25 P =αv Vf= m Vᵢ W = ∫ p dv = α∫vdv = α/2 (vf²- vᵢ²) = αv² (m²-1)/2 Ans -4
Q.26 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Ф = E.A cos 45⁰ = E πa²/√2 Ans -3
Q.27 L = constant Mvr = constant fMr = (1-f)Mv’r’ fr²ω = (1-f)r’²ω’ ω/ω’ = (1-f)/f x (r’/r)²---(1) now, mr = m’r’ fMr =(1-f)Mr’ r’/r = f/(1-f) putting this in eq ---(1) BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) ω/ω’ = f/(1-f) Ans -2
Q.28 ᶯ(max) = 1- T2/T1 = 1 – 273/373 = 26.80% Hence 30% efficiency is not possible by carnot’s theorem Ans 2
Q.29 Vol(initially) = vol(finally) n 4/3 πr³ = 4/3 π R³ n = R³/r³ BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) now,
1/2mv²=Δs = n 4πr²T - 4πR²T Solving we get V² = 8πTR²/m [R/r -1] Put m = ρ x 4/3 πR³ V = 6T/ρ[1/r -1/R]⁰·⁵ Ans- 3
Q.30 Self explanatory Ans -2
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.61 Lt x-> 2ˉ x/[x] ≤ limx-> 2ˉ f(x) ≤ lim x->2ˉ √6-x 2 ≤ limx->2ˉ f(x) ≤ 2 Limx->2ˉ f(x) = 2 Statement 1---true f(2)=1 ----given Limx->2ˉ f(x) ≠ f(2) Hence f(x) is not continuous at x=2 Statement 2- ----false Ans 1
Q.62 n(f) = ⁴C1 x 2 x 4 = 32 n(s) = 5! =120 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
p = 32/120 = 4/15 Ans 3
Q.63 g(x) = ax³/3 + bx² + cx g'(x) = ax² + bx + c now, g(1)=0, g(x)= 0 and g(x) is obviously continuous by rolle’s theorem at some point between (0,1) g’(x)=0 hence ax² +bx +c = 0 has it’s root between (0,1) BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 4
Q.64
b²+c²
ab
ac
ab
c²+a²
bc
ac
bc
a²+b²
-a² ab ac = ab -b² bc ac bc -c²
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
-1 1 = a²b²c² 1 1
1
-1 1 1
-1 = 4a²b²c² Ans 2
Q.65 2secθ = sec(θ+ф) + sec(θ-ф)
Cosθ =
2cos(θ+ф)cos(θ-ф) cos(θ+ф) +cos(θ-ф)
= (cos²θ –sin²θ) / cosθcosф Cos²θ(cosф-1) = -sin²ф Cos²θ = 2 cos²ф K= ±√2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 4
Q.66 M = L + (N/2 – c)x h/f 25 = 20 + (50-45) x 10/f F =10 Ans 2
Q.67 X² + sec²x -1 F(x) = ∫ F(x) =
dx
1+x² Sec²x (1+x²) -1 1+x²
∫Sec²x dx - ∫ dx/1+x² BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
dx
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
f(x)= tanx – tanˉ¹x + c f(0) = 0
C=0
f(1) = tan1 – tanˉ¹1 = tan1 – π/4 Ans 4
Q.68 d.r of normal to plane 1 ---(1,0,0) d.r of normal to plane 2 ---(2,-5a,3) d.r of normal to plane 3 ---(3b,1,-3) let d.r of common line (p,q,r) now, line is perpendicular to plane1 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
P=0-----(1)
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
line is perpendicular to plane 2
2p -5aq + 3r = 0----(2)
line is perpendicular to plane 3
3pb + q – 3r = 0----(3) Solving these eq we get a =1/5 Ans 3
Q.69 lim α-0
=
f(1-α) –f(1) α³ + 3α
f’(1-α) 3α² + 3
= f’(1)/3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
= -53/3 Ans 1
Q.70 (x + 3/2)² + y² = (3/2)²---(1) Y = mx+1---(2) Solving eq (1) &(2) Y²(1 +m²) - y(3/m + 2/m²) +1/m²-3/m = 0 Since two points are equidistant from and opp sides of x-axis Sum of the roots of this eqᶯ must be 0 i.e 3/m + 2/m² = 0 3m + 2= 0 Ans 3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.71
W= -a+ib Z=a+ib
θ
π-θ
w = -a – ib -w =a +ib = z Statement 1—true Statement 2---false As |z|=|w| only implies that they are equidistant from centre of argand diagram BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 2
Q.72 Px² + qx + r = 0 αβ = r/p let α= a+ib then β=a-ib αβ = a² + b² |α| +|β| = √a²+b² + √a²+b² =2√a²+b² =2√r/p
Since r>p |α|+|β|>2 Ans 2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.73 (0,1) (a,2a)
(α,β) (Y-1)/x= (2a-1)/a Slope of this line m1= (2a-1/a) Slope of given line m2= 2 Since these lines are perpendicular m1 x m2= - 1 we get a = 2/5 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
Y=2x
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
now (a,2a) are middle points of (1,0) and (α,β) we get α = 4/5 β = 3/5 Ans 4
Q.74 ²C1 x ⁴C2 x 3! + ²C2 x ⁴C1 x 3! = 96 Ans 1
Q.75 n p(s) = 2³ n(s) = 3 no. of one one relation is ⁸P3 = 336 Ans 1 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.76 Area= ∫ x dy =˳∫⁸ y¹ ³ dy = 12 Ans 2
Q.77 P q = ~p v q Basic formula Can check it by truth table Ans 1
Q.78 X² = 8y BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
(Y – 2)= -dx/dy|ˣ ⁴ (x-4) y-2 =4-x x +y = 6 Ans 4
Q.79 |a+b| = √ a²+b²+2abcosθ |c| = √ 9+25+2x3x5cosθ 30cosθ = 15 θ= 60⁰ Ans 3 Q.80 ˳∫⁰·⁹ [x -2[x]] dx [x] = 0 ˳∫⁰·⁹ o dx = 0 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
0≤x≤0.9
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 2
Q.81 I .F = exp( ∫ 2/x dx) =x² Now, y.x² = ∫ x².x² dx + c y = x³/5 + cxˉ² Ans 4
Q.82 bf-ec -ae -ab Aˉ¹ =
-1/bd cd -bd
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
-ad 0 0
0
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
0 0 d A =ͭ
0 b e a c f
now, A¯¹ = A ͭ comparing we get c=0, e=0, f=0, a=1,a=b²,a=d² or b² =1, d² =1 or b = ±1 , d= ± 1 hence there are 4 possible matrices Ans *
Q.83 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
N = 2t² + 3 w = t² -t +2 dw/dt = n dw/dt + w dn/dt = 8t³-6t²+14t-3 dw/dt|at (t=1) = 13 Ans 3
Q.84 √ (x-1)² + y²
=½
√ (x+1)² + y² Solving we get X²+y²-10/3 x+1 = 0 Centre = (5/3 , 0) Hence the circumcentre of the triangle is (5/3 , 0) Ans 4
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.85 Given lines x/2 = y/-1 = z/2 (x-1)/4 = (y-1)/-2 = (z-1)/ 4 These 2 lines are parallel p(1,1,1) line 1
q(2λ,-λ,2λ)
line2
line pq and line(1 or2) are perpendicular hence (2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0 λ = 1/3 q = (2/3,-1/3,2/3) length pq = √2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
statement 1----true Ans 1 Q.86 1 + 4/3 + 10/9 + …… 1 +(1+1/3)+ (1 + 1/9) + ….. n + (1/3 +1/9 + 1/27 +….) n + ½[1- 1/3ᶯˉ¹] Ans 1
Q.87 Foci of ellipse= √a²-b² =√16-b² Foci of hyperbola = √a²+b² =3 √16-b² = 3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
b² =7 Ans 4
Q.88 Tanˉ¹(sin(cosˉ¹√2/3)) Tanˉ¹(1/√3) π/6 Ans 3
Q.89 c is perpendicular to a x b c.(a x b) = 0 1
-2 3
2
3 -1
r
1 2r-1
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
=0
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
r= 0 Ans 4
Q.90 n = m(m-1)/2 n-1 = m(m-1)/2 - 1 ᶯC2 = n(n-1)/2 Solving we get ᶯC2 = 3 ͫ ˉ ¹C4 Ans 2
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
Q.1
1) 2) 3) 4) ans: 1 Q.2
1) 2) 3) 4) ans: 4 Q.3
1) 2) 3) 4) ans: 4
Q.4
1) 2) 3) 4) ans:4 Q.5
1) 2) 3) 4) ans:2 Q.6
1) 2) 3) 4) ans:1
Q.7
1) 2) 3) 4) ans:1 Q.8
1) 2) 3) 4) ans:4
Q.9
1) 2) 3) 4) ans:2 NOTE:- actual ans is 0.125m Q.10
1) 2) 3) 4) ans:3
Q.11
1) 2) 3) 4) ans:1 Q.12
1) 2) 3) 4) ans:2 (actual ans is 88.88%)
Q.13
1) 2) 3) 4) ans:3 Q.14
1) 2) 3) 4) ans:1
Q.15
1) 2) 3) 4) ans:3 Q.16
1) 2) 3) 4) ans:4
Q.17
1) 2) 3) 4) ans:4 Q.18
1) 2) 3) 4) ans:3
Q.19
1) 2) 3) 4) ans:2 Q.20
1) 2) 3) 4) ans:3
Q.21
1) 2) 3) 4) ans:4 Q.22
1) 2) 3) 4) ans:3 Q.23
1) 2) 3) 4) ans:2
Q.24
1) 2) 3) 4) ans:4 Q.25
1) 2) 3) 4) ans:4
Q.26
1) 2) 3) 4) ans:3 Q.27
1) 2) 3) 4) ans:2
Q.28
1) 2) 3) 4) ans:2 Q.29
1) 2) 3) 4) ans:3
Q.30
1) 2) 3) 4) ans:2
Q.61
1) 2) 3) 4) ANS:1 Q.62
1) 2) 3) 4) ANS:3
Q.63
1) 2) 3) 4) ANS: 4 Q.64
1) 2) 3) 4) ANS:2 Q.65
1) 2) 3) 4) ANS:4
Q.66
1) 2) 3) 4) ANS:2 Q.67
1) 2) 3) 4) ANS:4 Q.68
1) 2) 3) 4) ANS:3
Q.69
1) 2) 3) 4) ANS:1 Q.70 1) 2) 3) 4) ANS:3 Q.71
1) 2) 3) 4) ANS:2
Q.72
1) 2) 3) 4) ANS:2 Q.73
1) 2) 3) 4) ANS:4 Q.74
1) 2) 3) 4) ANS:1
Q.75
1) 2) 3) 4) ANS:1 Q.76
1) 2) 3) 4) ANS:2 Q.77
1) 2) 3) 4) ANS:1 Q.78 1) 2) 3) 4) ANS:4
Q.79
1) 2) 3) 4) ANS:3 Q.80
1) 2) 3) 4) ANS:2 Q.81
1) 2) 3) 4) ANS:4
Q.82
1) 2) 3) 4) ANS: * (ans is 4) Q.83
1) 2) 3) 4) ANS:3 Q.84
1) 2) 3) 4) ANS:4
Q.85
1) 2) 3) 4) ANS:1 Q.86
1) 2) 3) 4) ANS:1 Q.87
1) 2) 3) 4) ANS:4
Q.88
1) 2) 3) 4) ANS:3 Q.89
1) 2) 3) 4) ANS:4 Q.90
1) 2) 3) 4) ANS:2 By Saurav gupta Electronics & tele comm. Engg. (2nd year) Jadavpur university
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Q.1 Bqv= mv²/R R= mv/Bq
Now, eV= ½ mv² or mv= √2emV
R= √2emV/Bq R∝ √m R’ = √2 R Ans -1
Q.2 q1+q2 = Q ---eq(1) given BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
σ1 =σ2--- given or q1/4πr²=q2/4πR² q1/r²= q2/R²-------- eq(2)
solving eq (1) & (2) we get q1 =Q/(1+r²/R²)
now, V = 1/4πϵ˳ (q1/r+q2/R) = 1/4πϵ˳ q1(1/r+ q1 R/r²) = Q/4πϵ˳ (r+R)/(R²+r²) Ans -4
Q.3 Sinα ≃ tanα ≃α BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) α= d/D Δx=λd/D
d α
=λ/α
D
Ans 4
Q.4 E= [E ͨ cosω ͨt +E ͫcos(ω ͨt) cos(ω ͫ t)]î =[E ͨ cosω ͨt +E ͫ/2 cos(ω ͨ+ ω ͫt) +E ͫ/2 cos(ω ͫ-ω ͨ t)]î Ans 4
Q.5 N M.S.D = (N+1) V.S.D 1 V.S.D = N/(N+1) M.S.D NOW, BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) L.C = 1M.S.D -1V.S.D =[1 – N/(N+1)]M.S.D a/(N+1) Ans 2
Q.6 Case1 Tension /unit length = 100g/l Case2 Tension /unit length= 2mg/l 2m=100 m=50kg Ans -1
Q.7 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) dq/dt ͨ = dq/dtˢ 9KA(100-θ)/(L/2) = KAθ/(L/2) θ=90⁰ Ans -1
Q.8 F = dP/dt =d/dt(mv) = v dm/dt =6 Ans 4
Q.9
s
τ = -mg sinθ R mgsinθ
≃-mgθR Iα=-mgθR BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
mg
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
α = -mgRθ/I ω²=mgR/I now, I = mR² + mR² =2mR² ω²=g/2R T = 2π√(2R)/g Putting T=1 & g≃π² R= 1/8 =0.125 Ans 2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
Q.10 υn= (2n-1)v/4L case1 500 = (2n – 1)v/4 x16 ----(1) Case2 500 = (2n’-1)v/4 x 50-----(2) Solving 1 & 2 we get 50n -16n’ = 17 n≃1 n’≃2 now solving for v v = 500 x4 x 50/3 v = 33333.33m/s
now, BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) υ(fundamental ,outside water) = v/λ =v/2l ≃276 Hz υ'(2nd lowest) =2υ =552 Hz Ans -3
Q.11 λav= λα + λβ now, T ½ =1/λ Tav= 1/λα +1/λβ Ans -1
Q.12 Mu = mv + mv’ BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) ½ mu² = ½ mv² + ½ mv’² Solving eq (1) &(2) V = -1/3u V’= 2/3u (ΔE/E )x100= 88.88% Ans 2
Q.13 Sinθ ͨ > 1/μ μg/μw= 1.5/(4/3) = 9/8 Sinθ ͨ>8/9 Ans 3
Q.14 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) K .E= hυ-hυ˳ If υ is doubled K.E is more than doubled Also Photocurrent depends on intensity not on frequency Statement 1- false Statement 2-true Ans -1
Q.15 q(E + v x B)= 0 E=-vxB E=Bxv E = E˳ cos(wt -2π/λy)ẑ Ans -3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
Q.16 I .E³= 13.6 z²/n² = 13.6 x 9 = 122.4 T.E = I.E¹ + I.E² + I.E³ =5.4 +75.6 +122.4 = 203.4 Ans 4
Q.17 P=V²/R R1
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012)
Q.18 H=I²Rt ΔH= 2ΔI + ΔR +Δt =6% Ans -3
Q.19 v² = u² + 2al V’² = u² + 2a(l/2) =u² + ½ (v²-u²) V’² = (v² + u²)/2 V’ = √(v²+u²)/2 Ans -2
Q.20 Statement 1 –true BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Statement 2- true B near the centre is μ˳ni but decreses as we move towards it’s end and becomes μ˳ni/2 at the extreme end which explains statement 1 Ans -3
Q.21 V = e – iR 120 – 50 x 0.1 = 115 Ans 4
Q.22 Y = A sin(ωt + 2π/3) At t=0 Y = A sin 120⁰ Checking option BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Ans -3
Q.23 Fsinθ
Fκ
Fcosθ mg
Fκ = μ(mg- Fsinθ) Now,
Fcosθ = Fκ F = μW/(cosθ + μsinθ)---(1)
For minimum force BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) dF/dθ = 0 solving we get tanθ = μ or cosθ= 1/√1+μ² or sinθ = μ/√1 +μ²
substituting these values in eq –1 F = μW/√1 +μ² Ans -2
Q.24 As we pull the connector by say x amount there will be growth in current in the circuit which will be opposed by the inductor(due to self inductance) BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Here inductor is analogous to spring(mechanical analog of inductor) Opposing force ∝ -x Hence the connector will execute S.H.M Ans 4
Q.25 P =αv Vf= m Vᵢ W = ∫ p dv = α∫vdv = α/2 (vf²- vᵢ²) = αv² (m²-1)/2 Ans -4
Q.26 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) Ф = E.A cos 45⁰ = E πa²/√2 Ans -3
Q.27 L = constant Mvr = constant fMr = (1-f)Mv’r’ fr²ω = (1-f)r’²ω’ ω/ω’ = (1-f)/f x (r’/r)²---(1) now, mr = m’r’ fMr =(1-f)Mr’ r’/r = f/(1-f) putting this in eq ---(1) BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) ω/ω’ = f/(1-f) Ans -2
Q.28 ᶯ(max) = 1- T2/T1 = 1 – 273/373 = 26.80% Hence 30% efficiency is not possible by carnot’s theorem Ans 2
Q.29 Vol(initially) = vol(finally) n 4/3 πr³ = 4/3 π R³ n = R³/r³ BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE PHYSICS SOLUTION (EXAM-26TH MAY 2012) now,
1/2mv²=Δs = n 4πr²T - 4πR²T Solving we get V² = 8πTR²/m [R/r -1] Put m = ρ x 4/3 πR³ V = 6T/ρ[1/r -1/R]⁰·⁵ Ans- 3
Q.30 Self explanatory Ans -2
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.61 Lt x-> 2ˉ x/[x] ≤ limx-> 2ˉ f(x) ≤ lim x->2ˉ √6-x 2 ≤ limx->2ˉ f(x) ≤ 2 Limx->2ˉ f(x) = 2 Statement 1---true f(2)=1 ----given Limx->2ˉ f(x) ≠ f(2) Hence f(x) is not continuous at x=2 Statement 2- ----false Ans 1
Q.62 n(f) = ⁴C1 x 2 x 4 = 32 n(s) = 5! =120 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
p = 32/120 = 4/15 Ans 3
Q.63 g(x) = ax³/3 + bx² + cx g'(x) = ax² + bx + c now, g(1)=0, g(x)= 0 and g(x) is obviously continuous by rolle’s theorem at some point between (0,1) g’(x)=0 hence ax² +bx +c = 0 has it’s root between (0,1) BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 4
Q.64
b²+c²
ab
ac
ab
c²+a²
bc
ac
bc
a²+b²
-a² ab ac = ab -b² bc ac bc -c²
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
-1 1 = a²b²c² 1 1
1
-1 1 1
-1 = 4a²b²c² Ans 2
Q.65 2secθ = sec(θ+ф) + sec(θ-ф)
Cosθ =
2cos(θ+ф)cos(θ-ф) cos(θ+ф) +cos(θ-ф)
= (cos²θ –sin²θ) / cosθcosф Cos²θ(cosф-1) = -sin²ф Cos²θ = 2 cos²ф K= ±√2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 4
Q.66 M = L + (N/2 – c)x h/f 25 = 20 + (50-45) x 10/f F =10 Ans 2
Q.67 X² + sec²x -1 F(x) = ∫ F(x) =
dx
1+x² Sec²x (1+x²) -1 1+x²
∫Sec²x dx - ∫ dx/1+x² BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
dx
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
f(x)= tanx – tanˉ¹x + c f(0) = 0
C=0
f(1) = tan1 – tanˉ¹1 = tan1 – π/4 Ans 4
Q.68 d.r of normal to plane 1 ---(1,0,0) d.r of normal to plane 2 ---(2,-5a,3) d.r of normal to plane 3 ---(3b,1,-3) let d.r of common line (p,q,r) now, line is perpendicular to plane1 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
P=0-----(1)
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
line is perpendicular to plane 2
2p -5aq + 3r = 0----(2)
line is perpendicular to plane 3
3pb + q – 3r = 0----(3) Solving these eq we get a =1/5 Ans 3
Q.69 lim α-0
=
f(1-α) –f(1) α³ + 3α
f’(1-α) 3α² + 3
= f’(1)/3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
= -53/3 Ans 1
Q.70 (x + 3/2)² + y² = (3/2)²---(1) Y = mx+1---(2) Solving eq (1) &(2) Y²(1 +m²) - y(3/m + 2/m²) +1/m²-3/m = 0 Since two points are equidistant from and opp sides of x-axis Sum of the roots of this eqᶯ must be 0 i.e 3/m + 2/m² = 0 3m + 2= 0 Ans 3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.71
W= -a+ib Z=a+ib
θ
π-θ
w = -a – ib -w =a +ib = z Statement 1—true Statement 2---false As |z|=|w| only implies that they are equidistant from centre of argand diagram BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 2
Q.72 Px² + qx + r = 0 αβ = r/p let α= a+ib then β=a-ib αβ = a² + b² |α| +|β| = √a²+b² + √a²+b² =2√a²+b² =2√r/p
Since r>p |α|+|β|>2 Ans 2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.73 (0,1) (a,2a)
(α,β) (Y-1)/x= (2a-1)/a Slope of this line m1= (2a-1/a) Slope of given line m2= 2 Since these lines are perpendicular m1 x m2= - 1 we get a = 2/5 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
Y=2x
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
now (a,2a) are middle points of (1,0) and (α,β) we get α = 4/5 β = 3/5 Ans 4
Q.74 ²C1 x ⁴C2 x 3! + ²C2 x ⁴C1 x 3! = 96 Ans 1
Q.75 n p(s) = 2³ n(s) = 3 no. of one one relation is ⁸P3 = 336 Ans 1 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.76 Area= ∫ x dy =˳∫⁸ y¹ ³ dy = 12 Ans 2
Q.77 P q = ~p v q Basic formula Can check it by truth table Ans 1
Q.78 X² = 8y BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
(Y – 2)= -dx/dy|ˣ ⁴ (x-4) y-2 =4-x x +y = 6 Ans 4
Q.79 |a+b| = √ a²+b²+2abcosθ |c| = √ 9+25+2x3x5cosθ 30cosθ = 15 θ= 60⁰ Ans 3 Q.80 ˳∫⁰·⁹ [x -2[x]] dx [x] = 0 ˳∫⁰·⁹ o dx = 0 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
0≤x≤0.9
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Ans 2
Q.81 I .F = exp( ∫ 2/x dx) =x² Now, y.x² = ∫ x².x² dx + c y = x³/5 + cxˉ² Ans 4
Q.82 bf-ec -ae -ab Aˉ¹ =
-1/bd cd -bd
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
-ad 0 0
0
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
0 0 d A =ͭ
0 b e a c f
now, A¯¹ = A ͭ comparing we get c=0, e=0, f=0, a=1,a=b²,a=d² or b² =1, d² =1 or b = ±1 , d= ± 1 hence there are 4 possible matrices Ans *
Q.83 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
N = 2t² + 3 w = t² -t +2 dw/dt = n dw/dt + w dn/dt = 8t³-6t²+14t-3 dw/dt|at (t=1) = 13 Ans 3
Q.84 √ (x-1)² + y²
=½
√ (x+1)² + y² Solving we get X²+y²-10/3 x+1 = 0 Centre = (5/3 , 0) Hence the circumcentre of the triangle is (5/3 , 0) Ans 4
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
Q.85 Given lines x/2 = y/-1 = z/2 (x-1)/4 = (y-1)/-2 = (z-1)/ 4 These 2 lines are parallel p(1,1,1) line 1
q(2λ,-λ,2λ)
line2
line pq and line(1 or2) are perpendicular hence (2λ-1)2 + (-λ-1)-1 +(2λ-1)2 = 0 λ = 1/3 q = (2/3,-1/3,2/3) length pq = √2 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
statement 1----true Ans 1 Q.86 1 + 4/3 + 10/9 + …… 1 +(1+1/3)+ (1 + 1/9) + ….. n + (1/3 +1/9 + 1/27 +….) n + ½[1- 1/3ᶯˉ¹] Ans 1
Q.87 Foci of ellipse= √a²-b² =√16-b² Foci of hyperbola = √a²+b² =3 √16-b² = 3 BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
b² =7 Ans 4
Q.88 Tanˉ¹(sin(cosˉ¹√2/3)) Tanˉ¹(1/√3) π/6 Ans 3
Q.89 c is perpendicular to a x b c.(a x b) = 0 1
-2 3
2
3 -1
r
1 2r-1
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
=0
AIEEE 2012 ONLINE DETAILED SOLUTION (DATED 26TH MAY)
r= 0 Ans 4
Q.90 n = m(m-1)/2 n-1 = m(m-1)/2 - 1 ᶯC2 = n(n-1)/2 Solving we get ᶯC2 = 3 ͫ ˉ ¹C4 Ans 2
BY SAURAV GUPTA ELECTRONICS & TELECOMM. ENGG.(2ND YEAR) JADAVPUR UNIVERSITY
Related Documents 3h463d
Aieee 2012 Solution (26th May) 5h2c5q
December 2019 48
Aieee 2012 Solution (19th May) 3n1v6x
October 2019 85
Aieee 2012 Online Solution (7th May) 2h5t2
October 2019 62
Aieee 2012 Question Paper (19th May) 3h492v
December 2019 56
Aieee 2012 Question Paper And Solution 2m2x6j
February 2022 0
Aieee 2010 Solution k4m36
November 2019 43More Documents from "saurav gupta" 6k5y5a
Aieee 2012 Online Solution (7th May) 2h5t2
October 2019 62
Aipmt 2010 Final Exam Solution By Aakash 6t1d53
November 2019 46
Aieee 2012 Solution (19th May) 3n1v6x
October 2019 85
Iit Jee Screening Chemistry 2005 Solution 54l1a
December 2019 50
Bi Tools Comparison 4f3727
July 2021 0