Answer
(a) $3.09\times 10^{-4}m^3$
(b) water level drops
(c) $2.2\times 10^{-4}m^3$
(d) $1.4\times 10^{-2}m$
Work Step by Step
(a) We know that
$V_{wood}=\frac{T}{(\rho_w-\rho_{wood})g}$
We plug in the known values to obtain:
$V_{wood}=\frac{0.89N}{(1000Kg/m^3-706Kg/m^3)(9.81m/s^2)}=3.09\times 10^{-4}m^3$
(b) We know that the buoyant force increases when the wood is immersed in the water; then more water is displaced from the flask. Thus, the water level in the flask will drop when the block rises to the surface by a volume equal to that of the wood above the surface.
(c) We know that
$V_{sub}=\frac{\rho_{wood}V_{wood}}{\rho_w}$
We plug in the known values to obtain:
$V_{sub}=\frac{706Kg/m^3(3.09\times 10^{-4}m^3)}{1000Kg/m^3}$
$V_{sub}=2.2\times 10^{-4}m^3$
(d) We know that
$V_w=V_{wood}-V_{sub}$
$\implies V_w=3.09\times 10^{-4}m^3-2.2\times 10^{-4}m^3$
$\implies V_w=8.9\times 10^{-5}m^3$
Now the change in water level after the string breaks is given as
$\Delta h=\frac{V_w}{A}$
We plug in the known values to obtain:
$\Delta h=\frac{8.9\times 10^{-5}m^3}{62\times 10^{-4}m^2}=1.4\times 10^{-2}m$
