Q. Statement I: Two cubical blocks of same material and of sides a and 2a, respectively are attached rigidley and symmetrically to each other as shown. The system of two blocks is floating in water in such a way that upper surface of bigger blocks is just submerged in the water . If the system of blocks is displaced slightly in vertical directions, then the amplitude of oscillation on either side of equilibrium position would be different. Statement II: The force constant on two sides of equilibrium position in the above described situation is different.

Options

A. Statement I is true statement II is true, Statement II is a correct explanation for Statement I.
B. Statement I is true statement II is true, Statement II is NOT a correct explanation for Statement I.
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true

Answer

A

Explanation

Let us assume that density of material of cubes be `rho_0` and density of liquid be `rho`. Then from equilibrium condition `(8a^2+a^2)rho_0g=(8a^2)rhog`
`implies9rho_0=8rho`
When the block is displaced down by x, the restoring force is `F_1=(8a^3+a^3x)rhog-9a_0^3=rhoa^2gxx x` When the block is displaced by x above the mean position, the restoring force is
`F_2=9a^3rho_0g-(8a^3-4a^2x)rhog=4a^2rhogx`
As force constant on two sides of equilibrium position are not same, amplitudes of oscillations on the two sides are different.