Answer
$$\gamma =9327 \frac{\mathrm{N}}{\mathrm{m}^{3}} $$
Work Step by Step
$\begin{array}{l}{\text { The density of the } " \rho^{\prime \prime}=\frac{w}{v}} \\ {\qquad \begin{aligned} &=\frac{(4)[\text { ounces }]\left[0.0283 \frac{K g}{\text { ounces }}\right]}{[1.25 \text { in. }][1.25 \mathrm{in} .][4.65 \mathrm{in} .]\left[(0.0254)^{3} \frac{\mathrm{m}^{3}}{\mathrm{in.}^{3}}\right]} \\ &=\frac{(4)[\text { ounees }]\left[0.0283 \frac{K g}{\text { aunees }}\right]}{[1.25 \mathrm{in} .][4.65 \mathrm{in} .]\left[(0.0254)^{3} \frac{\mathrm{m}^{3}}{\mathrm{in.}^{8}}\right]} \end{aligned}} \\ \quad \quad \quad =950.763 \frac{\mathrm{Kg}}{\mathrm{m}^{3}} \end{array}$
$\begin{aligned} \text { The specific weight } " \gamma^{\prime \prime} &=\rho \cdot g \\ &=950.763 * 9.81 \\ &=9327 \frac{\mathrm{N}}{\mathrm{m}^{3}} \end{aligned}$
