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Saturday, January 12, 2013

January 2013 VTU Exam Engineering physics Answer key


Our model question paper which published 4 months back carries more than 100 marks question from actual question paper!
Objective questions
1(a) i) B   ii) B   iii) D    iv) A
2(a) i) C   ii) B   iii) A   iv) D
3(a) i) C   ii) A    iii) B    iv) D
4(a) i) C    ii) D   iii) B   iv) C
5(a) i) B   ii) C    iii) C    iv) A
6(a) i) B   ii) D    iii) C    iv) B
7(a) i) B   ii) A    ii) D    iv) C
8(a) i) B   ii) A    iii) C    iv) D
Problems
1(d) m=0.65 MeV/c2 = (0.65×106×1.6×10-19)/(3×108)2 = 1.156×10-30
 λ = h/√2mE = 6.625×10-34/√(2×1.156×10-30×80×1.6×10-19) = 1.218×10-10 m
2(d) Δt = λ2/4πcΔλ = (4000×10-10)2/(4×3.14×3×108×8×10-5×10-10)=5.31×10-9s
3(d) VF = (2EF/m)1/2 = [(2×5.5×1.6×10-19)/(9.1×10-31)]1/ 2 = 1.39×106 m/s
4(d) p = [(k-1)ε0E]/n =  [(1.00014-1) × 8.85×10-12×105]/(6.02×1026/22.4) =4.61×10-36 Coul-m
Α = p/E = 4.61×10-36/105 = 4.61×10-41 Coul-m2/volt
5(d) λ= nhc/p = (5×107×6.625×10-34×3×108)/(10-3×10×10-9) =993.75 nm
6(d) α = 10/L × log (pi/po) = 10/0.6 × log(120mw/90mw) = 2.06 dB/km
7(d) intercepts=3a:4b:∞
Reciprocal = 1/3:1/4:0
Miller indices = 12×(1/3:1/4:0) = (4 3 0)

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