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,T_1/2 = 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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