Answer
(b) $2.20\times10^{10}\;N.m$
(c) $11.64\;m$
Work Step by Step
$(b)$ The horizontal force acting on a horizontal strip at depth $h$ having a vertical thickness $dh$ and width $W$ is given by
$dF=\rho ghwdh$
The corresponding torque about O is
$d\tau=\rho ghwdh(D-h)$
Therefore, the net torque due to that force about a horizontal line through O parallel to the (long) width of the dam is given by
$\tau=\int_0^D \rho ghwdh(D-h)$
or, $\tau=\rho gw\int_0^D h(D-h)dh$
or, $\tau=\rho gw[\frac{Dh^2}{2}-\frac{h^3}{3}]_0^D$
or, $\tau=\rho gw[\frac{D^3}{2}-\frac{D^3}{3}]$
Substituting the given values
$\tau=1000\times9.81\times314\times[\frac{35^3}{2}-\frac{35^3}{3}]\;N.m$
or, $\boxed{\tau=2.20\times10^{10}\;N.m}$
$(c)$ Let $r$ be the moment arm of the torque
Therefore, $\tau=rF$
or, $r=\frac{\tau}{F}$
Substituting the given values
$r=\frac{2.20\times10^{10}}{1.89\times10^{9}}\;m$
or, $\boxed{r=11.64\;m}$
