IIT-JEE-Physics-Paper 2 -2008-Solution

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)

| Next >>

IIT-JEE-Physics-Paper 2 -2008-Solution

Important Links

Related Links

IIT JEE Resources