A spring is attached to the bottom of an swimming pool, with the axis of the spring oriented vertically. An 8.00 kg block of wood `(rho=840 kh//m^(3))` is fixed to the top of the spring and compressed it. Then the pool is filled with water, completely covering the block. the spring is now observed to be stretched twice as much as it had been conpressed. Determine the percentage of the block's total volume that is hollow. Ignore any air in the hollow space.

Answer

A, D

Explanation

Initiallly , `kx=mg` …(i)
In the second case,

`F=mg+2kx`
or, `kx=(F-mg)/(2)` …(ii)
From Eqs. (i) and (ii) we have
`mg=(F-mg)/(2)`
or, `F=3mg`
Let `V` be the total volume then,
`Vrho_(w)(g)=3mg`
`:. V=(3mg)/(rho_(w)g)=((3)(8))/(10^(30))m^(3)`
`=0.024m^(3)`
Volume of wood `=("mass")/("density")`
`=(8)/(840)=0.0095`
`:.` Volume of cavity`=0.024-0.0095`
`=0.0145`
`:.` percentage volume of cavity
`=(0.0145)/(0.024)xx100`
`=60.41%`