Chapter 1 - Problems - Page 32: 1.37

Let us assume the following: The length of bath tub be $l=1.5 \mathrm{ft}$ The width of the bath tub be $b=0.6 \mathrm{ft}$ The height of the bath tub be $h=0.3 \mathrm{ft}$ Thus calculate the volume of the bath tub. $$ \begin{aligned} V &=1.5 \mathrm{ft} \times 0.6 \mathrm{ft} \times 0.3 \mathrm{ft} \\ &=0.27 \mathrm{ft}^{3} \end{aligned} $$ Now the weight of the bath tub can expressed as follows: $$ \begin{array}{l} W=m g \\ \quad=(\rho V) g \\ \quad=V(\rho g) \\ =V \gamma_{H g} \ldots \ldots(1) \end{array} $$ Here, density of mercury is $\rho$, volume of the bath tub is $V$, specific weigh of mercury is $\gamma_{H g}$ and $W$ is the weight of the bath tub. Substitute $0.27 \mathrm{ft}^{3}$ for $V$ and $847 \mathrm{lb} / \mathrm{ft}^{3}$ for $\gamma_{H g}$ in equation (1). $$ \begin{aligned} W &=V \gamma_{H g} \\ &=0.27 \times 847 \\ &=228.7 \mathrm{lb} \end{aligned} $$ Therefore, the amount of mercury required to fill the bath tub is $228.7 \mathrm{lb}$.

Let us assume the following: The length of bath tub be $l=1.5 \mathrm{ft}$ The width of the bath tub be $b=0.6 \mathrm{ft}$ The height of the bath tub be $h=0.3 \mathrm{ft}$ Thus calculate the volume of the bath tub. $$ \begin{aligned} V &=1.5 \mathrm{ft} \times 0.6 \mathrm{ft} \times 0.3 \mathrm{ft} \\ &=0.27 \mathrm{ft}^{3} \end{aligned} $$ Now the weight of the bath tub can expressed as follows: $$ \begin{array}{l} W=m g \\ \quad=(\rho V) g \\ \quad=V(\rho g) \\ =V \gamma_{H g} \ldots \ldots(1) \end{array} $$ Here, density of mercury is $\rho$, volume of the bath tub is $V$, specific weigh of mercury is $\gamma_{H g}$ and $W$ is the weight of the bath tub. Substitute $0.27 \mathrm{ft}^{3}$ for $V$ and $847 \mathrm{lb} / \mathrm{ft}^{3}$ for $\gamma_{H g}$ in equation (1). $$ \begin{aligned} W &=V \gamma_{H g} \\ &=0.27 \times 847 \\ &=228.7 \mathrm{lb} \end{aligned} $$ Therefore, the amount of mercury required to fill the bath tub is $228.7 \mathrm{lb}$.