Answer
(a) $49.1Mpa$
(b) $-24.6\%$
(c) $1330Kg/m^3$
Work Step by Step
(a) can find the required pressures as
$p=p_{\circ}+\rho gh$
We plug in the known values to obtain:
$p=1\times 10^5+1000\times 9.8\times 5000=49.1MPa$
(b) We know that
$p=-Bf$
This can be rearranged as:
$f=-\frac{p}{B}$
We plug in the known values to obtain:
$f=-\frac{49\times 1\times 10^7}{0.2\times 10^{10}}$
$\implies f=-24.6\%$
(c) We know that
$\rho_n=\frac{\rho}{1+f}$
We plug in the known values to obtain:
$\rho_n=\frac{1}{1-0.246}(1000)$
$\implies \rho_n=1330Kg/m^3$
