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(1, 1)7λ 5.5× 10t min = =P4n 4×1. 38= 99.6 nm⎛ R ⎞⇒ E = I ⎜ total⎟ (By Ohm's Law)⎝ l ⎠t n mgF 22παI⇒ E =2A3.[D]dqI = ; dt q = it + a;q V = COM(3, 0)it + aV =C∴ V is proportional to timeQ4.[A] Phase difference corresponding to y 1 = –π/2 andthat for y 2 = + π/2∴ Average intensity between y 1 and y 2Fig.π / 212⎛ φ ⎞ ( π + 2)⎛ 13 ⎞= 2 ⎜3 − x=+ ⎟∫x dx0 ∫dxπ ∫I max cos ⎜ ⎟dφ= I max⎝ 2 ⎠2π1⎝2−π / 2⎠1 ⎛ ⎞⎛ 13Hence required ratio =⎞⎜1+ 2 ⎟= 2⎜ 2 3/ 2 1 2 3/ 2x − . (3 − x)⎟2 ⎝ π ⎠⎜ 33 ⎟⎝2M.C.Q. Type questions :01 ⎠5.[A,C]⎡ 1 2 2 4= 2 ⎛3/0 . 2⎞⎤⎛ ⎞a⎢ −⎥ = 2⎜+ ⎟ = 4.⎢⎣3⎜ −2 3⎟dr⎝⎠⎥⎦⎝ 3 3 ⎠rPHYSICSlConsider a cylindrical element of radius r,1.[D]thickness dr. If dR is the resistance of this1element then1′ρ(r)ld/22dR =2πr dr2′3Total resistance of the cylinder is given bya3′1 1 2πC =lR total∫ = dR αl∫r3 dr011 2π⎛ ⎞⇒= ⎜a 4⎟1′R3 C2total αl⎝ 4 ⎠l2′1 2αl3′⇒=4( ∈ A) 2= 2 × 0 × ∈ AR total πa= 4 . 0= 4CddR⇒total 2πα=l ( πa2 ) 2⎡ 2m+ 1⎤λ2.[A] 2t = ⎢2⎥2πα⎣ ⎦ n⇒ R =2AVSince E = (in magnitude)l6.[A,D] Consider only one plate as shown in figure.XtraEdge for IIT-JEE 99APRIL 2010