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IIT-JEE-Physics-Paper 2 -2008-Solution
1. Correct Choice: (C) 2. λ1/λ2 =A1/N1 .N2/2 =5/(2N2 ).N2/10; λ1/λ2 =5/20,T1/2 = 1/λ Correct Choice: (A) 3. The slope at point P is negative, and vP = –v (slope) Correct Choice: (A) 4. By conservation of mechanical energy 1/2 kx2 = 1/2 (4k) y2 => y/x=1/2 Correct Choice: (C) 5. By conservation of mechanical energy 1/2 mv2 = 1/2 m(v/2)2 + mgL(1-cos θ ) and v2 = 5gL cos θ = -7/8 => 3π/4 < θ < π Correct choice: (D) 6. Since radius of curvature of sub-hemispherical surface is more than that of hemispherical surface and ΔP µ 1/R Correct choice: (B) 7. f = 3v/4l = (3×340)/(4×0.75) = 340 Hz |f-n|=4 n = 344 Hz or 336 Hz Since on increasing tension, f increases and number of beats decreases, => n = 344 Hz Correct choice: (A)
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